당신은 주제를 찾고 있습니까 “**angular momentum quantization – Angular Momentum Quantization in Bohr Model (Quantum Mechanics – De Broglie’s Wave-particle Duality)**“? 다음 카테고리의 웹사이트 ppa.diaochoangduong.vn 에서 귀하의 모든 질문에 답변해 드립니다: https://ppa.diaochoangduong.vn/blog/. 바로 아래에서 답을 찾을 수 있습니다. 작성자 Elucyda 이(가) 작성한 기사에는 조회수 12,619회 및 좋아요 546개 개의 좋아요가 있습니다.

Table of Contents

## angular momentum quantization 주제에 대한 동영상 보기

여기에서 이 주제에 대한 비디오를 시청하십시오. 주의 깊게 살펴보고 읽고 있는 내용에 대한 피드백을 제공하세요!

## d여기에서 Angular Momentum Quantization in Bohr Model (Quantum Mechanics – De Broglie’s Wave-particle Duality) – angular momentum quantization 주제에 대한 세부정보를 참조하세요

Link to Quantum Playlist:

https://www.youtube.com/playlist?list=PLl0eQOWl7mnWPTQF7lgLWZmb5obvOowVw

Wave-particle duality and the De Broglie wavelength are used to show why angular momentum is quantized in the Bohr Model.

#Quantum

#Bohr

#AngularMomentum

#KonstantinLakic

## angular momentum quantization 주제에 대한 자세한 내용은 여기를 참조하세요.

### Quantized Angular Momentum

Quantized Angular Momentum … This general form applies to orbital angular momentum, spin angular momentum, and the total angular momentum for an atomic system.

Source: hyperphysics.phy-astr.gsu.edu

Date Published: 12/21/2022

View: 1283

### Quantization of Electronic Angular Momentum

Quantization of Electronic Angular Momentum. Rutherford proposed that electrons orbit about the nucleus of an atom. One problem with this model is that, …

Source: vergil.chemistry.gatech.edu

Date Published: 10/27/2021

View: 9022

### Angular momentum operator – Wikipedia

In quantum mechanics, the angular momentum operator is one of several related operators analogous to ical angular momentum. The angular momentum …

Source: en.wikipedia.org

Date Published: 2/17/2022

View: 9268

### Space-Quantization of Angular Momentum

The definite magnitude and direction of one component of angular momentum is known as “space quantization”. Restriction of to integer values was exploited in …

Source: demonstrations.wolfram.com

Date Published: 5/1/2021

View: 8189

### Angular Momentum Quantization – Galileo

This means that if we measure the angle between the total angular momentum and the z-axis, there can only be 2l+1 possible answers, the total angular momentum …

Source: galileo.phys.virginia.edu

Date Published: 5/19/2021

View: 7699

### Angular Momentum Of Electron – De Broglie’s Explanation, FAQs

According to Bohr’s atomic model, the angular momentum of electrons orbiting around the nucleus is quantized. He further added that electrons move only in …

Source: byjus.com

Date Published: 11/4/2022

View: 8485

### QUANTIZATION OF ANGULAR MOMENTUM – De Anza College

Thus, for all potential energies where U=U(r) the angular momentum will be quantized and given by the above equations. SPACE QUANTIZATION. The physical …

Source: www.deanza.edu

Date Published: 1/19/2022

View: 1838

### Why is angular momentum quantized? – Quora

quantization means “ available in fixed amount” so · quantization of angular momentum means that the angular momentum is present in fixed amounts. · according to …

Source: www.quora.com

Date Published: 8/30/2021

View: 261

### Is all angular momentum quantized? – Physics Stack Exchange

Angular momentum is definitely quantized in elementary particles and electrons in atoms. Molecules also have characteristic rotation spectra.

Source: physics.stackexchange.com

Date Published: 5/5/2022

View: 1594

## 주제와 관련된 이미지 angular momentum quantization

주제와 관련된 더 많은 사진을 참조하십시오 **Angular Momentum Quantization in Bohr Model (Quantum Mechanics – De Broglie’s Wave-particle Duality)**. 댓글에서 더 많은 관련 이미지를 보거나 필요한 경우 더 많은 관련 기사를 볼 수 있습니다.

## 주제에 대한 기사 평가 angular momentum quantization

- Author: Elucyda
- Views: 조회수 12,619회
- Likes: 좋아요 546개
- Date Published: 2020. 11. 26.
- Video Url link: https://www.youtube.com/watch?v=7V6j4sZ7w_E

## Quantized Angular Momentum

Quantized Angular Momentum

In the process of solving the Schrodinger equation for the hydrogen atom, it is found that the orbital angular momentum is quantized according to the relationship:

It is a characteristic of angular momenta in quantum mechanics that the magnitude of the angular momentum in terms of the orbital quantum number is of the form

and that the z-component of the angular momentum in terms of the magnetic quantum number takes the form

## Is all angular momentum quantized?

$\begingroup$

I’m going to disagree with the other answers: I think that the angular momentum of macroscopic “classical” objects is not quantized.

Consider an automobile tire spinning in a wheel well. On the tire is a device that triggers whenever a certain point on the wheel crosses a certain point on the well, adding one to an internal counter if it passes it clockwise and subtracting one if it crosses it counterclockwise. (Alternately, you could dispense with the wheel well and say that the device tracks its own position by inertial navigation.) The state of this system can be described by a value $θ\in\mathbb R$, where the current value of the counter is $\lfloor θ/2π\rfloor$ and the angle of the wheel is $θ\text{ (mod }2π\text{)}$. In the absence of external forces, the Hamiltonian of the system is essentially that of a free particle in $\mathbb R$, and the spectrum of angular momenta is continuous just like the free particle’s momentum spectrum.

That’s a 2+1 dimensional system. In 3+1 dimensions, there’s the Dirac belt trick to worry about. Does it matter? I don’t think so. There’s no reason to limit the device to holding a single integer, or to being reversible. It could simply store the entire history of its orientation readings internally, or broadcast them by radio, indelibly recording them in the universal wave function. That’s a very noncompact state space, and it’s an accurate enough model of bodies like the earth.

The angular momentum operator on this monstrosity obviously violates the assumptions of any proof of the quantization of angular momentum, but that’s no reason not to call it angular momentum. We do call it angular momentum, and it’s what the question was about.

In response to comments I’ll try to clarify my answer.

These are quantum systems, but the earth system is “classical” in the sense of being a quantum system with emergent classical behavior.

The reason that high temperature systems behave classically is that they constantly leak which-path information into the environment. If you do a double-slit experiment with the earth, it will emit different patterns of light going through one slit than through the other. You can literally see which slit it goes through, but even if you don’t look, the which-path information is there in the patterns of light, or in patterns of heat if the light is absorbed by the walls of the lab, and that’s all that’s necessary to make the final states orthogonal and destroy the interference pattern.

It’s sometimes said that you can’t see an interference pattern in the earth double-slit experiment simply because its de Broglie wavelength is so small. That would be correct for a supermassive stable particle that doesn’t radiate, but it’s wrong for the earth. For the earth there’s no interference pattern at all, for the same reason there’s no interference pattern when there’s a detector at one of the slits. Earth’s thermal radiation is the “detector”.

The case of rotation is similar. Different rotations are different paths through the state space (it’s the state space, not physical 3D space, that the wave function is defined on and which matters here). If you consider two different paths ending in the same physical orientation (analogous to the same position on the screen in the double-slit case), these paths will interfere if no information about which path was taken is recorded anywhere. In the case of the earth, this means they’ll interfere if there’s no way for anyone to tell whether the earth rotated around its axis or not. If there’s any record of it – if any animals remember the day-night cycle, or don’t but could in principle, or if aliens see it rotate through a telescope, or don’t but could in principle – then there’s no interference.

The proof that angular momentum is quantized depends on the compactness of the space of orientations. This is fine if the space of orientations is the phase space, i.e., if the system is memoryless. If it has a memory, rotating the system through $2π$ or $4π$ doesn’t leave it in the same state as not rotating it.

The tire example in the second paragraph may have been a mistake since it seems to have only caused confusion. But it’s a perfectly good quantum system in the abstract, and its state space is $\mathbb R$, not $S^1$.

## Angular momentum operator

Quantum mechanical operator related to rotational symmetry

In quantum mechanics, the angular momentum operator is one of several related operators analogous to classical angular momentum. The angular momentum operator plays a central role in the theory of atomic and molecular physics and other quantum problems involving rotational symmetry. Such an operator is applied to a mathematical representation of the physical state of a system and yields an angular momentum value if the state has a definite value for it. In both classical and quantum mechanical systems, angular momentum (together with linear momentum and energy) is one of the three fundamental properties of motion.[1]

There are several angular momentum operators: total angular momentum (usually denoted J), orbital angular momentum (usually denoted L), and spin angular momentum (spin for short, usually denoted S). The term angular momentum operator can (confusingly) refer to either the total or the orbital angular momentum. Total angular momentum is always conserved, see Noether’s theorem.

Overview [ edit ]

J (purple), orbital L (blue), and spin S (green). The cones arise due to “Vector cones” of total angular momentum(purple), orbital(blue), and spin(green). The cones arise due to quantum uncertainty between measuring angular momentum components ( see below ).

In quantum mechanics, angular momentum can refer to one of three different, but related things.

Orbital angular momentum [ edit ]

The classical definition of angular momentum is L = r × p {\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} } . The quantum-mechanical counterparts of these objects share the same relationship:

L = r × p {\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }

L = ( L x , L y , L z ) {\displaystyle \mathbf {L} =\left(L_{x},L_{y},L_{z}\right)}

x

y

z

whereis the quantum position operator is the quantum momentum operator , × is cross product , andis the(just likeand) is a(a vector whose components are operators), i.e.whereare three different quantum-mechanical operators.

In the special case of a single particle with no electric charge and no spin, the orbital angular momentum operator can be written in the position basis as:

L = − i ℏ ( r × ∇ ) {\displaystyle \mathbf {L} =-i\hbar (\mathbf {r} \times

abla )}

∇

Spin angular momentum [ edit ]

whereis the vector differential operator, del

There is another type of angular momentum, called spin angular momentum (more often shortened to spin), represented by the spin operator S = ( S x , S y , S z ) {\displaystyle \mathbf {S} =\left(S_{x},S_{y},S_{z}\right)} . Spin is often depicted as a particle literally spinning around an axis, but this is only a metaphor: spin is an intrinsic property of a particle, unrelated to any sort of (yet experimentally observable) motion in space. All elementary particles have a characteristic spin, which is usually nonzero. For example, electrons always have “spin 1/2” while photons always have “spin 1” (details below).

Total angular momentum [ edit ]

Finally, there is total angular momentum J = ( J x , J y , J z ) {\displaystyle \mathbf {J} =\left(J_{x},J_{y},J_{z}\right)} , which combines both the spin and orbital angular momentum of a particle or system:

J = L + S . {\displaystyle \mathbf {J} =\mathbf {L} +\mathbf {S} .}

Conservation of angular momentum states that J for a closed system, or J for the whole universe, is conserved. However, L and S are not generally conserved. For example, the spin–orbit interaction allows angular momentum to transfer back and forth between L and S, with the total J remaining constant.

Commutation relations [ edit ]

Commutation relations between components [ edit ]

The orbital angular momentum operator is a vector operator, meaning it can be written in terms of its vector components L = ( L x , L y , L z ) {\displaystyle \mathbf {L} =\left(L_{x},L_{y},L_{z}\right)} . The components have the following commutation relations with each other:[2] [ L x , L y ] = i ℏ L z , [ L y , L z ] = i ℏ L x , [ L z , L x ] = i ℏ L y , {\displaystyle \left[L_{x},L_{y}\right]=i\hbar L_{z},\;\;\left[L_{y},L_{z}\right]=i\hbar L_{x},\;\;\left[L_{z},L_{x}\right]=i\hbar L_{y},}

where [ , ] denotes the commutator

[ X , Y ] ≡ X Y − Y X . {\displaystyle [X,Y]\equiv XY-YX.}This can be written generally as

[ L l , L m ] = i ℏ ∑ n = 1 3 ε l m n L n , {\displaystyle \left[L_{l},L_{m}\right]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}L_{n},}ε lmn

whereare the component indices (1 for, 2 for, 3 for), anddenotes the Levi-Civita symbol

A compact expression as one vector equation is also possible:[3]

L × L = i ℏ L {\displaystyle \mathbf {L} \times \mathbf {L} =i\hbar \mathbf {L} }

The commutation relations can be proved as a direct consequence of the canonical commutation relations [ x l , p m ] = i ℏ δ l m {\displaystyle [x_{l},p_{m}]=i\hbar \delta _{lm}} , where δ lm is the Kronecker delta.

There is an analogous relationship in classical physics:[4]

{ L i , L j } = ε i j k L k {\displaystyle \left\{L_{i},L_{j}\right\}=\varepsilon _{ijk}L_{k}}

n

{ , } {\displaystyle \{,\}}

whereis a component of theangular momentum operator, andis the Poisson bracket

The same commutation relations apply for the other angular momentum operators (spin and total angular momentum):[5] [ S l , S m ] = i ℏ ∑ n = 1 3 ε l m n S n , [ J l , J m ] = i ℏ ∑ n = 1 3 ε l m n J n . {\displaystyle \left[S_{l},S_{m}\right]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}S_{n},\quad \left[J_{l},J_{m}\right]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}J_{n}.}

These can be assumed to hold in analogy with L. Alternatively, they can be derived as discussed below.

These commutation relations mean that L has the mathematical structure of a Lie algebra, and the ε lmn are its structure constants. In this case, the Lie algebra is SU(2) or SO(3) in physics notation ( su ( 2 ) {\displaystyle \operatorname {su} (2)} or so ( 3 ) {\displaystyle \operatorname {so} (3)} respectively in mathematics notation), i.e. Lie algebra associated with rotations in three dimensions. The same is true of J and S. The reason is discussed below. These commutation relations are relevant for measurement and uncertainty, as discussed further below.

In molecules the total angular momentum F is the sum of the rovibronic (orbital) angular momentum N, the electron spin angular momentum S, and the nuclear spin angular momentum I. For electronic singlet states the rovibronic angular momentum is denoted J rather than N. As explained by Van Vleck,[6] the components of the molecular rovibronic angular momentum referred to molecule-fixed axes have different commutation relations from those given above which are for the components about space-fixed axes.

Commutation relations involving vector magnitude [ edit ]

Like any vector, the square of a magnitude can be defined for the orbital angular momentum operator,

L 2 ≡ L x 2 + L y 2 + L z 2 . {\displaystyle L^{2}\equiv L_{x}^{2}+L_{y}^{2}+L_{z}^{2}.}

L 2 {\displaystyle L^{2}} is another quantum operator. It commutes with the components of L {\displaystyle \mathbf {L} } ,

[ L 2 , L x ] = [ L 2 , L y ] = [ L 2 , L z ] = 0. {\displaystyle \left[L^{2},L_{x}\right]=\left[L^{2},L_{y}\right]=\left[L^{2},L_{z}\right]=0.}One way to prove that these operators commute is to start from the [L ℓ , L m ] commutation relations in the previous section:

Proof of [L2, L x ] = 0, starting from the [L ℓ , L m ] commutation relations[7] [ L 2 , L x ] = [ L x 2 , L x ] + [ L y 2 , L x ] + [ L z 2 , L x ] = L y [ L y , L x ] + [ L y , L x ] L y + L z [ L z , L x ] + [ L z , L x ] L z = L y ( − i ℏ L z ) + ( − i ℏ L z ) L y + L z ( i ℏ L y ) + ( i ℏ L y ) L z = 0 {\displaystyle {\begin{aligned}\left[L^{2},L_{x}\right]&=\left[L_{x}^{2},L_{x}\right]+\left[L_{y}^{2},L_{x}\right]+\left[L_{z}^{2},L_{x}\right]\\&=L_{y}\left[L_{y},L_{x}\right]+\left[L_{y},L_{x}\right]L_{y}+L_{z}\left[L_{z},L_{x}\right]+\left[L_{z},L_{x}\right]L_{z}\\&=L_{y}\left(-i\hbar L_{z}\right)+\left(-i\hbar L_{z}\right)L_{y}+L_{z}\left(i\hbar L_{y}\right)+\left(i\hbar L_{y}\right)L_{z}\\&=0\end{aligned}}}

Mathematically, L 2 {\displaystyle L^{2}} is a Casimir invariant of the Lie algebra SO(3) spanned by L {\displaystyle \mathbf {L} } .

As above, there is an analogous relationship in classical physics:

{ L 2 , L x } = { L 2 , L y } = { L 2 , L z } = 0 {\displaystyle \left\{L^{2},L_{x}\right\}=\left\{L^{2},L_{y}\right\}=\left\{L^{2},L_{z}\right\}=0}

L i {\displaystyle L_{i}}

{ , } {\displaystyle \{,\}}

whereis a component of theangular momentum operator, andis the Poisson bracket

Returning to the quantum case, the same commutation relations apply to the other angular momentum operators (spin and total angular momentum), as well,

[ S 2 , S i ] = 0 , [ J 2 , J i ] = 0. {\displaystyle {\begin{aligned}\left[S^{2},S_{i}\right]&=0,\\\left[J^{2},J_{i}\right]&=0.\end{aligned}}}Uncertainty principle [ edit ]

In general, in quantum mechanics, when two observable operators do not commute, they are called complementary observables. Two complementary observables cannot be measured simultaneously; instead they satisfy an uncertainty principle. The more accurately one observable is known, the less accurately the other one can be known. Just as there is an uncertainty principle relating position and momentum, there are uncertainty principles for angular momentum.

The Robertson–Schrödinger relation gives the following uncertainty principle:

σ L x σ L y ≥ ℏ 2 | ⟨ L z ⟩ | . {\displaystyle \sigma _{L_{x}}\sigma _{L_{y}}\geq {\frac {\hbar }{2}}\left|\langle L_{z}\rangle \right|.}

σ X {\displaystyle \sigma _{X}}

⟨ X ⟩ {\displaystyle \langle X\rangle }

whereis the standard deviation in the measured values ofanddenotes the expectation value of. This inequality is also true ifare rearranged, or ifis replaced byor

Therefore, two orthogonal components of angular momentum (for example L x and L y ) are complementary and cannot be simultaneously known or measured, except in special cases such as L x = L y = L z = 0 {\displaystyle L_{x}=L_{y}=L_{z}=0} .

It is, however, possible to simultaneously measure or specify L2 and any one component of L; for example, L2 and L z . This is often useful, and the values are characterized by the azimuthal quantum number (l) and the magnetic quantum number (m). In this case the quantum state of the system is a simultaneous eigenstate of the operators L2 and L z , but not of L x or L y . The eigenvalues are related to l and m, as shown in the table below.

Quantization [ edit ]

In quantum mechanics, angular momentum is quantized – that is, it cannot vary continuously, but only in “quantum leaps” between certain allowed values. For any system, the following restrictions on measurement results apply, where ℏ {\displaystyle \hbar } is reduced Planck constant:[9]

If you measure… …the result can be… Notes L 2 {\displaystyle L^{2}} ℏ 2 ℓ ( ℓ + 1 ) {\displaystyle \hbar ^{2}\ell (\ell +1)} where ℓ = 0 , 1 , 2 , … {\displaystyle \ell =0,1,2,\ldots } ℓ {\displaystyle \ell } azimuthal quantum number or orbital quantum number. L z {\displaystyle L_{z}} ℏ m ℓ {\displaystyle \hbar m_{\ell }} where m ℓ = − ℓ , ( − ℓ + 1 ) , … , ( ℓ − 1 ) , ℓ {\displaystyle m_{\ell }=-\ell ,(-\ell +1),\ldots ,(\ell -1),\ell } m ℓ {\displaystyle m_{\ell }} magnetic quantum number. This same quantization rule holds for any component of L {\displaystyle \mathbf {L} } ; e.g., L x o r L y {\displaystyle L_{x}\,or\,L_{y}} . This rule is sometimes called spatial quantization.[10] S 2 {\displaystyle S^{2}} ℏ 2 s ( s + 1 ) {\displaystyle \hbar ^{2}s(s+1)} where s = 0 , 1 2 , 1 , 3 2 , … {\displaystyle s=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\ldots } s is called spin quantum number or just spin. For example, a spin-1⁄2 particle is a particle where s = 1⁄2. S z {\displaystyle S_{z}} ℏ m s {\displaystyle \hbar m_{s}} where m s = − s , ( − s + 1 ) , … , ( s − 1 ) , s {\displaystyle m_{s}=-s,(-s+1),\ldots ,(s-1),s} m s {\displaystyle m_{s}} spin projection quantum number. This same quantization rule holds for any component of S {\displaystyle \mathbf {S} } ; e.g., S x o r S y {\displaystyle S_{x}\,or\,S_{y}} . J 2 {\displaystyle J^{2}} ℏ 2 j ( j + 1 ) {\displaystyle \hbar ^{2}j(j+1)} where j = 0 , 1 2 , 1 , 3 2 , … {\displaystyle j=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\ldots } j is sometimes called total angular momentum quantum number. J z {\displaystyle J_{z}} ℏ m j {\displaystyle \hbar m_{j}} where m j = − j , ( − j + 1 ) , … , ( j − 1 ) , j {\displaystyle m_{j}=-j,(-j+1),\ldots ,(j-1),j} m j {\displaystyle m_{j}} total angular momentum projection quantum number. This same quantization rule holds for any component of J {\displaystyle \mathbf {J} } ; e.g., J x o r J y {\displaystyle J_{x}\,or\,J_{y}} .

cannot have a non-integer number of wavelengths like 8.3. In quantum mechanics, angular momentum is quantized for a similar reason. In this standing wave on a circular string, the circle is broken into exactly 8 wavelengths . A standing wave like this can have 0, 1, 2, or any integer number of wavelengths around the circle, but ithave a non-integer number of wavelengths like 8.3. In quantum mechanics, angular momentum is quantized for a similar reason.

Derivation using ladder operators [ edit ]

A common way to derive the quantization rules above is the method of ladder operators.[11] The ladder operators for the total angular momentum J = ( J x , J y , J z ) {\displaystyle \mathbf {J} =\left(J_{x},J_{y},J_{z}\right)} are defined as:

J + ≡ J x + i J y , J − ≡ J x − i J y {\displaystyle {\begin{aligned}J_{+}&\equiv J_{x}+iJ_{y},\\J_{-}&\equiv J_{x}-iJ_{y}\end{aligned}}}

Suppose | ψ ⟩ {\displaystyle |\psi \rangle } is a simultaneous eigenstate of J 2 {\displaystyle J^{2}} and J z {\displaystyle J_{z}} (i.e., a state with a definite value for J 2 {\displaystyle J^{2}} and a definite value for J z {\displaystyle J_{z}} ). Then using the commutation relations for the components of J {\displaystyle \mathbf {J} } , one can prove that each of the states J + | ψ ⟩ {\displaystyle J_{+}|\psi \rangle } and J − | ψ ⟩ {\displaystyle J_{-}|\psi \rangle } is either zero or a simultaneous eigenstate of J 2 {\displaystyle J^{2}} and J z {\displaystyle J_{z}} , with the same value as | ψ ⟩ {\displaystyle |\psi \rangle } for J 2 {\displaystyle J^{2}} but with values for J z {\displaystyle J_{z}} that are increased or decreased by ℏ {\displaystyle \hbar } respectively. The result is zero when the use of a ladder operator would otherwise result in a state with a value for J z {\displaystyle J_{z}} that is outside the allowable range. Using the ladder operators in this way, the possible values and quantum numbers for J 2 {\displaystyle J^{2}} and J z {\displaystyle J_{z}} can be found.

Derivation of the possible values and quantum numbers for J z {\displaystyle J_{z}} J 2 {\displaystyle J^{2}} [12] Let ψ ( J 2 ′ J z ′ ) {\displaystyle \psi ({J^{2}}’J_{z}’)} be a state function for the system with eigenvalue J 2 ′ {\displaystyle {J^{2}}’} for J 2 {\displaystyle J^{2}} and eigenvalue J z ′ {\displaystyle J_{z}’} for J z {\displaystyle J_{z}} .[note 1] From J 2 = J x 2 + J y 2 + J z 2 {\displaystyle J^{2}=J_{x}^{2}+J_{y}^{2}+J_{z}^{2}} is obtained, J x 2 + J y 2 = J 2 − J z 2 . {\displaystyle J_{x}^{2}+J_{y}^{2}=J^{2}-J_{z}^{2}.} ψ ( J 2 ′ J z ′ ) {\displaystyle \psi ({J^{2}}’J_{z}’)} ( J x 2 + J y 2 ) ψ ( J 2 ′ J z ′ ) = ( J 2 ′ − J z ′ 2 ) ψ ( J 2 ′ J z ′ ) . {\displaystyle (J_{x}^{2}+J_{y}^{2})\;\psi ({J^{2}}’J_{z}’)=({J^{2}}’-J_{z}’^{2})\;\psi ({J^{2}}’J_{z}’).} J x {\displaystyle J_{x}} J y {\displaystyle J_{y}} J 2 ′ − J z ′ 2 {\displaystyle {J^{2}}’-J_{z}’^{2}} | J z ′ | ≤ J 2 ′ {\textstyle |J_{z}’|\leq {\sqrt {{J^{2}}’}}} J z ′ {\displaystyle J_{z}’} Applying both sides of the above equation toSinceandare real observables,is not negative and. Thushas an upper and lower bound. Two of the commutation relations for the components of J {\displaystyle \mathbf {J} } are, [ J y , J z ] = i ℏ J x , [ J z , J x ] = i ℏ J y . {\displaystyle [J_{y},J_{z}]=i\hbar J_{x},\;\;[J_{z},J_{x}]=i\hbar J_{y}.} ± {\displaystyle \pm } J z ( J x ± i J y ) = ( J x ± i J y ) ( J z ± ℏ ) , {\displaystyle J_{z}(J_{x}\pm iJ_{y})=(J_{x}\pm iJ_{y})(J_{z}\pm \hbar ),} + {\displaystyle +} − {\displaystyle -} ψ ( J 2 ′ J z ′ ) {\displaystyle \psi ({J^{2}}’J_{z}’)} J z ( J x ± i J y ) ψ ( J 2 ′ J z ′ ) = ( J x ± i J y ) ( J z ± ℏ ) ψ ( J 2 ′ J z ′ ) = ( J z ′ ± ℏ ) ( J x ± i J y ) ψ ( J 2 ′ J z ′ ) . {\displaystyle {\begin{aligned}J_{z}(J_{x}\pm iJ_{y})\;\psi ({J^{2}}’J_{z}’)&=(J_{x}\pm iJ_{y})(J_{z}\pm \hbar )\;\psi ({J^{2}}’J_{z}’)\\&=(J_{z}’\pm \hbar )(J_{x}\pm iJ_{y})\;\psi ({J^{2}}’J_{z}’)\;.\\\end{aligned}}} ( J x ± i J y ) ψ ( J 2 ′ J z ′ ) {\displaystyle (J_{x}\pm iJ_{y})\;\psi ({J^{2}}’J_{z}’)} J z {\displaystyle J_{z}} J z ′ ± ℏ {\displaystyle {J_{z}}’\pm \hbar } ψ ( J 2 ′ J z ′ ± ℏ ) = ( J x ± i J y ) ψ ( J 2 ′ J z ′ ) . {\displaystyle \psi ({J^{2}}’J_{z}’\pm \hbar )=(J_{x}\pm iJ_{y})\;\psi ({J^{2}}’J_{z}’).} J z {\displaystyle J_{z}} J x ± i J y {\displaystyle J_{x}\pm iJ_{y}} ≤ J 2 ′ {\displaystyle \leq {\sqrt {{J^{2}}’}}} J z {\displaystyle J_{z}} J z 0 {\displaystyle J_{z}^{0}} J z 1 {\displaystyle J_{z}^{1}} ( J x − i J y ) ψ ( J 2 ′ J z 0 ) = 0 {\displaystyle (J_{x}-iJ_{y})\;\psi ({J^{2}}’J_{z}^{0})=0} ( J x + i J y ) ψ ( J 2 ′ J z 1 ) = 0 , {\displaystyle (J_{x}+iJ_{y})\;\psi ({J^{2}}’J_{z}^{1})=0,} J z {\displaystyle J_{z}} < J z 0 {\displaystyle

J z 1 {\displaystyle >J_{z}^{1}} ( J x + i J y ) {\displaystyle (J_{x}+iJ_{y})} ( J x − i J y ) {\displaystyle (J_{x}-iJ_{y})} J x 2 + J y 2 = J 2 − J z 2 {\displaystyle J_{x}^{2}+J_{y}^{2}=J^{2}-J_{z}^{2}} J 2 ′ − ( J z 0 ) 2 + ℏ J z 0 = 0 {\displaystyle {J^{2}}’-(J_{z}^{0})^{2}+\hbar J_{z}^{0}=0} J 2 ′ − ( J z 1 ) 2 − ℏ J z 1 = 0. {\displaystyle {J^{2}}’-(J_{z}^{1})^{2}-\hbar J_{z}^{1}=0.} ( J z 1 + J z 0 ) ( J z 0 − J z 1 − ℏ ) = 0. {\displaystyle (J_{z}^{1}+J_{z}^{0})(J_{z}^{0}-J_{z}^{1}-\hbar )=0.} J z 1 ≥ J z 0 {\displaystyle J_{z}^{1}\geq J_{z}^{0}} J z 0 = − J z 1 {\displaystyle J_{z}^{0}=-J_{z}^{1}} They can be combined to obtain two equations, which are written together usingsigns in the following,where one of the equations uses thesigns and the other uses thesigns. Applying both sides of the above toThe above shows thatare two eigenfunctions ofwith respective eigenvalues, unless one of the functions is zero, in which case it is not an eigenfunction. For the functions that are not zero,Further eigenfunctions ofand corresponding eigenvalues can be found by repeatedly applyingas long as the magnitude of the resulting eigenvalue is. Since the eigenvalues ofare bounded, letbe the lowest eigenvalue andbe the highest. Thenandsince there are no states where the eigenvalue ofisor. By applyingto the first equation,to the second, and using, it can be shown thatandSubtracting the first equation from the second and rearranging,Since, the second factor is negative. Then the first factor must be zero and thus The difference J z 1 − J z 0 {\displaystyle J_{z}^{1}-J_{z}^{0}} comes from successive application of J x − i J y {\displaystyle J_{x}-iJ_{y}} or J x + i J y {\displaystyle J_{x}+iJ_{y}} which lower or raise the eigenvalue of J z {\displaystyle J_{z}} by ℏ {\displaystyle \hbar } so that, J z 1 − J z 0 = 0 , ℏ , 2 ℏ , … {\displaystyle J_{z}^{1}-J_{z}^{0}=0,\hbar ,2\hbar ,\dots } J z 1 − J z 0 = 2 j ℏ , {\displaystyle J_{z}^{1}-J_{z}^{0}=2j\hbar ,} j = 0 , 1 2 , 1 , 3 2 , … . {\displaystyle j=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\dots \;.} J z 0 = − J z 1 {\displaystyle J_{z}^{0}=-J_{z}^{1}} J z 0 = − j ℏ {\displaystyle J_{z}^{0}=-j\hbar } J z 1 = j ℏ , {\displaystyle J_{z}^{1}=j\hbar ,} J z {\displaystyle J_{z}} J z ′ = − j ℏ , − j ℏ + ℏ , − j ℏ + 2 ℏ , … , j ℏ . {\displaystyle J_{z}’=-j\hbar ,-j\hbar +\hbar ,-j\hbar +2\hbar ,\dots ,j\hbar .} J z ′ {\displaystyle J_{z}’} m j {\displaystyle m_{j}\;} J z 0 = − j ℏ {\displaystyle J_{z}^{0}=-j\hbar } J 2 ′ − ( J z 0 ) 2 + ℏ J z 0 = 0 {\displaystyle {J^{2}}’-(J_{z}^{0})^{2}+\hbar J_{z}^{0}=0} J z ′ = m j ℏ m j = − j , − j + 1 , − j + 2 , … , j J 2 ′ = j ( j + 1 ) ℏ 2 j = 0 , 1 2 , 1 , 3 2 , … . {\displaystyle {\begin{aligned}J_{z}’&=m_{j}\hbar &m_{j}&=-j,-j+1,-j+2,\dots ,j\\{J^{2}}’&=j(j+1)\hbar ^{2}&j&=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\dots \;.\end{aligned}}} LetwhereThen usingand the above,andand the allowable eigenvalues ofareExpressingin terms of a quantum number, and substitutingintofrom above, Since S {\displaystyle \mathbf {S} } and L {\displaystyle \mathbf {L} } have the same commutation relations as J {\displaystyle \mathbf {J} } , the same ladder analysis can be applied to them, except that for L {\displaystyle \mathbf {L} } there is a further restriction on the quantum numbers that they must be integers.

Traditional derivation of the restriction to integer quantum numbers for L z {\displaystyle L_{z}} L 2 {\displaystyle L^{2}} [13] In the Schroedinger representation, the z component of the orbital angular momentum operator can be expressed in spherical coordinates as,[14] L z = − i ℏ ∂ ∂ ϕ . {\displaystyle L_{z}=-i\hbar {\frac {\partial }{\partial \phi }}.} L z {\displaystyle L_{z}} ψ {\displaystyle \psi } L z ′ {\displaystyle L_{z}’} − i ℏ ∂ ∂ ϕ ψ = L z ′ ψ . {\displaystyle -i\hbar {\frac {\partial }{\partial \phi }}\psi =L_{z}’\psi .} ψ {\displaystyle \psi } ψ = A e i L z ′ ϕ / ℏ , {\displaystyle \psi =Ae^{iL_{z}’\phi /\hbar },} A {\displaystyle A} ϕ {\displaystyle \phi } ψ {\displaystyle \psi } 2 π {\displaystyle 2\pi } ϕ {\displaystyle \phi } A e i L z ′ ( ϕ + 2 π ) / ℏ = A e i L z ′ ϕ / ℏ , e i L z ′ 2 π / ℏ = 1. {\displaystyle {\begin{aligned}Ae^{iL_{z}'(\phi +2\pi )/\hbar }&=Ae^{iL_{z}’\phi /\hbar },\\e^{iL_{z}’2\pi /\hbar }&=1.\end{aligned}}} L z ′ {\displaystyle L_{z}’} L z ′ = m l ℏ , {\displaystyle L_{z}’=m_{l}\hbar \;,} m l {\displaystyle m_{l}} [15] From the above and the relation m ℓ = − ℓ , ( − ℓ + 1 ) , … , ( ℓ − 1 ) , ℓ {\displaystyle m_{\ell }=-\ell ,(-\ell +1),\ldots ,(\ell -1),\ell \ \ } ℓ {\displaystyle \ell } m ℓ {\displaystyle m_{\ell }} ℓ {\displaystyle \ell } L {\displaystyle \mathbf {L} } J {\displaystyle \mathbf {J} } S {\displaystyle \mathbf {S} } [16] Forand eigenfunction with eigenvalueSolving forwhereis independent of. Sinceis required to be single valued, and addingtoresults in a coordinate for the same point in space,Solving for the eigenvaluewhereis an integer.From the above and the relation, it follows thatis also an integer. This shows that the quantum numbersandfor the orbital angular momentumare restricted to integers, unlike the quantum numbers for the total angular momentumand spin, which can have half-integer values. An alternative derivation which does not assume single-valued wave functions follows and another argument using Lie groups is below.

Alternative derivation of the restriction to integer quantum numbers for L z {\displaystyle L_{z}} L 2 {\displaystyle L^{2}} A key part of the traditional derivation above is that the wave function must be single-valued. This is now recognised by many as not being completely correct: a wave function ψ {\displaystyle \psi } is not observable and only the probability density ψ ∗ ψ {\displaystyle \psi ^{*}\psi } is required to be single-valued. The possible double-valued half-integer wave functions have a single-valued probability density.[17] This was recognised by Pauli in 1939 (cited by Japaridze et al[18]) … there is no a priori convincing argument stating that the wave functions which describe some physical states must be single valued functions. For physical quantities, which are expressed by squares of wave functions, to be single valued it is quite sufficient that after moving around a closed contour these functions gain a factor exp(iα) Double-valued wave functions have been found, such as | 1 2 , 1 2 ⟩ = e i ϕ / 2 sin 1 2 θ {\displaystyle \left|{\tfrac {1}{2}},{\tfrac {1}{2}}\right\rangle =e^{i\phi /2}\sin ^{\frac {1}{2}}\theta } and | 1 2 , − 1 2 ⟩ = e − i ϕ / 2 sin 1 2 θ {\displaystyle \left|{\tfrac {1}{2}},-{\tfrac {1}{2}}\right\rangle =e^{-i\phi /2}\sin ^{\frac {1}{2}}\theta } .[19][20] These do not behave well under the ladder operators, but have been found to be useful in describing rigid quantum particles[21] Ballentine[22] gives an argument based solely on the operator formalism and which does not rely on the wave function being single-valued. The azimuthal angular momentum is defined as L z = Q x P y − Q y P x {\displaystyle L_{z}=Q_{x}P_{y}-Q_{y}P_{x}} q 1 = Q x + P y 2 q 2 = Q x − P y 2 p 1 = P x − Q y 2 p 2 = P x + Q y 2 {\displaystyle {\begin{aligned}q_{1}&={\frac {Q_{x}+P_{y}}{\sqrt {2}}}&q_{2}&={\frac {Q_{x}-P_{y}}{\sqrt {2}}}\\p_{1}&={\frac {P_{x}-Q_{y}}{\sqrt {2}}}&p_{2}&={\frac {P_{x}+Q_{y}}{\sqrt {2}}}\end{aligned}}} L z = 1 2 ( p 1 2 + q 1 2 ) − 1 2 ( p 2 2 + q 2 2 ) {\displaystyle L_{z}={\frac {1}{2}}\left(p_{1}^{2}+q_{1}^{2}\right)-{\frac {1}{2}}\left(p_{2}^{2}+q_{2}^{2}\right)} H 1 = 1 2 ( p 1 2 + q 1 2 ) H 2 = 1 2 ( p 2 2 + q 2 2 ) {\displaystyle {\begin{aligned}H_{1}&={\frac {1}{2}}\left(p_{1}^{2}+q_{1}^{2}\right)\\H_{2}&={\frac {1}{2}}\left(p_{2}^{2}+q_{2}^{2}\right)\end{aligned}}} L 2 {\displaystyle L^{2}} L z {\displaystyle L_{z}} H 1 {\displaystyle H_{1}} H 2 {\displaystyle H_{2}} Define new operators(Dimensional correctness may be maintained by inserting factors of mass and unit angular frequency numerically equal to one.) ThenBut the two terms on the right are just the Hamiltonians for the quantum harmonic oscillator with unit mass and angular frequencyandandall commute. For commuting Hermitian operators a complete set of basis vectors can be chosen that are eigenvectors for all four operators. (The argument by Glorioso[23] can easily be generalised to any number of commuting operators.) For any of these eigenvectors ψ {\displaystyle \psi } with L 2 ψ = ℏ ℓ ( ℓ + 1 ) ψ L z ψ = ℏ m ψ H 1 ψ = ℏ ( n 1 + 1 2 ) ψ H 2 ψ = ℏ ( n 2 + 1 2 ) ψ {\displaystyle {\begin{aligned}L^{2}\psi &=\hbar \ell (\ell +1)\psi \\L_{z}\psi &=\hbar m\psi \\H_{1}\psi &=\hbar \left(n_{1}+{\tfrac {1}{2}}\right)\psi \\H_{2}\psi &=\hbar \left(n_{2}+{\tfrac {1}{2}}\right)\psi \end{aligned}}} n 1 , n 2 {\displaystyle n_{1},n_{2}} m = ( n 1 + 1 2 ) − ( n 2 + 1 2 ) = n 1 − n 2 {\displaystyle m=\left(n_{1}+{\tfrac {1}{2}}\right)-\left(n_{2}+{\tfrac {1}{2}}\right)=n_{1}-n_{2}} m {\displaystyle m} ℓ {\displaystyle \ell } for some integers, we findAs a difference of two integers,must be an integer, from whichis also integral. A more complex version of this argument using the ladder operators of the quantum harmonic oscillator has been given by Buchdahl.[24]

Visual interpretation [ edit ]

Illustration of the vector model of orbital angular momentum.

Since the angular momenta are quantum operators, they cannot be drawn as vectors like in classical mechanics. Nevertheless, it is common to depict them heuristically in this way. Depicted on the right is a set of states with quantum numbers ℓ = 2 {\displaystyle \ell =2} , and m ℓ = − 2 , − 1 , 0 , 1 , 2 {\displaystyle m_{\ell }=-2,-1,0,1,2} for the five cones from bottom to top. Since | L | = L 2 = ℏ 6 {\displaystyle |L|={\sqrt {L^{2}}}=\hbar {\sqrt {6}}} , the vectors are all shown with length ℏ 6 {\displaystyle \hbar {\sqrt {6}}} . The rings represent the fact that L z {\displaystyle L_{z}} is known with certainty, but L x {\displaystyle L_{x}} and L y {\displaystyle L_{y}} are unknown; therefore every classical vector with the appropriate length and z-component is drawn, forming a cone. The expected value of the angular momentum for a given ensemble of systems in the quantum state characterized by ℓ {\displaystyle \ell } and m ℓ {\displaystyle m_{\ell }} could be somewhere on this cone while it cannot be defined for a single system (since the components of L {\displaystyle L} do not commute with each other).

Quantization in macroscopic systems [ edit ]

The quantization rules are widely thought to be true even for macroscopic systems, like the angular momentum L of a spinning tire. However they have no observable effect so this has not been tested. For example, if L z / ℏ {\displaystyle L_{z}/\hbar } is roughly 100000000, it makes essentially no difference whether the precise value is an integer like 100000000 or 100000001, or a non-integer like 100000000.2—the discrete steps are currently too small to measure.

Angular momentum as the generator of rotations [ edit ]

The most general and fundamental definition of angular momentum is as the generator of rotations.[5] More specifically, let R ( n ^ , ϕ ) {\displaystyle R({\hat {n}},\phi )} be a rotation operator, which rotates any quantum state about axis n ^ {\displaystyle {\hat {n}}} by angle ϕ {\displaystyle \phi } . As ϕ → 0 {\displaystyle \phi \rightarrow 0} , the operator R ( n ^ , ϕ ) {\displaystyle R({\hat {n}},\phi )} approaches the identity operator, because a rotation of 0° maps all states to themselves. Then the angular momentum operator J n ^ {\displaystyle J_{\hat {n}}} about axis n ^ {\displaystyle {\hat {n}}} is defined as:[5]

J n ^ ≡ i ℏ lim ϕ → 0 R ( n ^ , ϕ ) − 1 ϕ = i ℏ ∂ R ( n ^ , ϕ ) ∂ ϕ | ϕ = 0 {\displaystyle J_{\hat {n}}\equiv i\hbar \lim _{\phi \rightarrow 0}{\frac {R\left({\hat {n}},\phi \right)-1}{\phi }}=\left.i\hbar {\frac {\partial R\left({\hat {n}},\phi \right)}{\partial \phi }}\right|_{\phi =0}}

where 1 is the identity operator. Also notice that R is an additive morphism : R ( n ^ , ϕ 1 + ϕ 2 ) = R ( n ^ , ϕ 1 ) R ( n ^ , ϕ 2 ) {\displaystyle R\left({\hat {n}},\phi _{1}+\phi _{2}\right)=R\left({\hat {n}},\phi _{1}\right)R\left({\hat {n}},\phi _{2}\right)} ; as a consequence[5]

R ( n ^ , ϕ ) = exp ( − i ϕ J n ^ ℏ ) {\displaystyle R\left({\hat {n}},\phi \right)=\exp \left(-{\frac {i\phi J_{\hat {n}}}{\hbar }}\right)}

where exp is matrix exponential

In simpler terms, the total angular momentum operator characterizes how a quantum system is changed when it is rotated. The relationship between angular momentum operators and rotation operators is the same as the relationship between Lie algebras and Lie groups in mathematics, as discussed further below.

The operator R, related to J, rotates the entire system. The operator R spatial , related to L, rotates the particle positions without altering their internal spin states. The operator R internal , related to S, rotates the particles’ internal spin states without changing their positions. The different types of rotation operators . The top box shows two particles, with spin states indicated schematically by the arrows.

Just as J is the generator for rotation operators, L and S are generators for modified partial rotation operators. The operator

R spatial ( n ^ , ϕ ) = exp ( − i ϕ L n ^ ℏ ) , {\displaystyle R_{\text{spatial}}\left({\hat {n}},\phi \right)=\exp \left(-{\frac {i\phi L_{\hat {n}}}{\hbar }}\right),}

R internal ( n ^ , ϕ ) = exp ( − i ϕ S n ^ ℏ ) , {\displaystyle R_{\text{internal}}\left({\hat {n}},\phi \right)=\exp \left(-{\frac {i\phi S_{\hat {n}}}{\hbar }}\right),}

R ( n ^ , ϕ ) = R internal ( n ^ , ϕ ) R spatial ( n ^ , ϕ ) {\displaystyle R\left({\hat {n}},\phi \right)=R_{\text{internal}}\left({\hat {n}},\phi \right)R_{\text{spatial}}\left({\hat {n}},\phi \right)}

rotates the position (in space) of all particles and fields, without rotating the internal (spin) state of any particle. Likewise, the operatorrotates the internal (spin) state of all particles, without moving any particles or fields in space. The relationcomes from:

i.e. if the positions are rotated, and then the internal states are rotated, then altogether the complete system has been rotated.

SU(2), SO(3), and 360° rotations [ edit ]

Although one might expect R ( n ^ , 360 ∘ ) = 1 {\displaystyle R\left({\hat {n}},360^{\circ }\right)=1} (a rotation of 360° is the identity operator), this is not assumed in quantum mechanics, and it turns out it is often not true: When the total angular momentum quantum number is a half-integer (1/2, 3/2, etc.), R ( n ^ , 360 ∘ ) = − 1 {\displaystyle R\left({\hat {n}},360^{\circ }\right)=-1} , and when it is an integer, R ( n ^ , 360 ∘ ) = + 1 {\displaystyle R\left({\hat {n}},360^{\circ }\right)=+1} .[5] Mathematically, the structure of rotations in the universe is not SO(3), the group of three-dimensional rotations in classical mechanics. Instead, it is SU(2), which is identical to SO(3) for small rotations, but where a 360° rotation is mathematically distinguished from a rotation of 0°. (A rotation of 720° is, however, the same as a rotation of 0°.)[5]

On the other hand, R spatial ( n ^ , 360 ∘ ) = + 1 {\displaystyle R_{\text{spatial}}\left({\hat {n}},360^{\circ }\right)=+1} in all circumstances, because a 360° rotation of a spatial configuration is the same as no rotation at all. (This is different from a 360° rotation of the internal (spin) state of the particle, which might or might not be the same as no rotation at all.) In other words, the R spatial {\displaystyle R_{\text{spatial}}} operators carry the structure of SO(3), while R {\displaystyle R} and R internal {\displaystyle R_{\text{internal}}} carry the structure of SU(2).

From the equation + 1 = R spatial ( z ^ , 360 ∘ ) = exp ( − 2 π i L z / ℏ ) {\displaystyle +1=R_{\text{spatial}}\left({\hat {z}},360^{\circ }\right)=\exp \left(-2\pi iL_{z}/\hbar \right)} , one picks an eigenstate L z | ψ ⟩ = m ℏ | ψ ⟩ {\displaystyle L_{z}|\psi \rangle =m\hbar |\psi \rangle } and draws

e − 2 π i m = 1 {\displaystyle e^{-2\pi im}=1}

Connection to representation theory [ edit ]

which is to say that the orbital angular momentum quantum numbers can only be integers, not half-integers.

Starting with a certain quantum state | ψ 0 ⟩ {\displaystyle |\psi _{0}\rangle } , consider the set of states R ( n ^ , ϕ ) | ψ 0 ⟩ {\displaystyle R\left({\hat {n}},\phi \right)\left|\psi _{0}\right\rangle } for all possible n ^ {\displaystyle {\hat {n}}} and ϕ {\displaystyle \phi } , i.e. the set of states that come about from rotating the starting state in every possible way. The linear span of that set is a vector space, and therefore the manner in which the rotation operators map one state onto another is a representation of the group of rotation operators.

When rotation operators act on quantum states, it forms a representation of the Lie group SU(2) (for R and R internal ), or SO(3) (for R spatial ).

From the relation between J and rotation operators,

When angular momentum operators act on quantum states, it forms a representation of the Lie algebra s u ( 2 ) {\displaystyle {\mathfrak {su}}(2)} s o ( 3 ) {\displaystyle {\mathfrak {so}}(3)}

(The Lie algebras of SU(2) and SO(3) are identical.)

The ladder operator derivation above is a method for classifying the representations of the Lie algebra SU(2).

Connection to commutation relations [ edit ]

Classical rotations do not commute with each other: For example, rotating 1° about the x-axis then 1° about the y-axis gives a slightly different overall rotation than rotating 1° about the y-axis then 1° about the x-axis. By carefully analyzing this noncommutativity, the commutation relations of the angular momentum operators can be derived.[5]

(This same calculational procedure is one way to answer the mathematical question “What is the Lie algebra of the Lie groups SO(3) or SU(2)?”)

Conservation of angular momentum [ edit ]

The Hamiltonian H represents the energy and dynamics of the system. In a spherically symmetric situation, the Hamiltonian is invariant under rotations:

R H R − 1 = H {\displaystyle RHR^{-1}=H}

[ H , R ] = 0 {\displaystyle [H,R]=0} [ H , J ] = 0 {\displaystyle [H,\mathbf {J} ]=\mathbf {0} }whereis a rotation operator . As a consequence,, and thendue to the relationship betweenand. By the Ehrenfest theorem , it follows thatis conserved.

To summarize, if H is rotationally-invariant (spherically symmetric), then total angular momentum J is conserved. This is an example of Noether’s theorem.

If H is just the Hamiltonian for one particle, the total angular momentum of that one particle is conserved when the particle is in a central potential (i.e., when the potential energy function depends only on | r | {\displaystyle \left|\mathbf {r} \right|} ). Alternatively, H may be the Hamiltonian of all particles and fields in the universe, and then H is always rotationally-invariant, as the fundamental laws of physics of the universe are the same regardless of orientation. This is the basis for saying conservation of angular momentum is a general principle of physics.

For a particle without spin, J = L, so orbital angular momentum is conserved in the same circumstances. When the spin is nonzero, the spin–orbit interaction allows angular momentum to transfer from L to S or back. Therefore, L is not, on its own, conserved.

Angular momentum coupling [ edit ]

Often, two or more sorts of angular momentum interact with each other, so that angular momentum can transfer from one to the other. For example, in spin–orbit coupling, angular momentum can transfer between L and S, but only the total J = L + S is conserved. In another example, in an atom with two electrons, each has its own angular momentum J 1 and J 2 , but only the total J = J 1 + J 2 is conserved.

In these situations, it is often useful to know the relationship between, on the one hand, states where ( J 1 ) z , ( J 1 ) 2 , ( J 2 ) z , ( J 2 ) 2 {\displaystyle \left(J_{1}\right)_{z},\left(J_{1}\right)^{2},\left(J_{2}\right)_{z},\left(J_{2}\right)^{2}} all have definite values, and on the other hand, states where ( J 1 ) 2 , ( J 2 ) 2 , J 2 , J z {\displaystyle \left(J_{1}\right)^{2},\left(J_{2}\right)^{2},J^{2},J_{z}} all have definite values, as the latter four are usually conserved (constants of motion). The procedure to go back and forth between these bases is to use Clebsch–Gordan coefficients.

One important result in this field is that a relationship between the quantum numbers for ( J 1 ) 2 , ( J 2 ) 2 , J 2 {\displaystyle \left(J_{1}\right)^{2},\left(J_{2}\right)^{2},J^{2}} :

j ∈ { | j 1 − j 2 | , ( | j 1 − j 2 | + 1 ) , … , ( j 1 + j 2 ) } . {\displaystyle j\in \left\{\left|j_{1}-j_{2}\right|,\left(\left|j_{1}-j_{2}\right|+1\right),\ldots ,\left(j_{1}+j_{2}\right)\right\}.}

For an atom or molecule with J = L + S, the term symbol gives the quantum numbers associated with the operators L 2 , S 2 , J 2 {\displaystyle L^{2},S^{2},J^{2}} .

Orbital angular momentum in spherical coordinates [ edit ]

Angular momentum operators usually occur when solving a problem with spherical symmetry in spherical coordinates. The angular momentum in the spatial representation is[25][26]

L = i ℏ ( θ ^ sin ( θ ) ∂ ∂ ϕ − ϕ ^ ∂ ∂ θ ) = i ℏ ( x ^ ( sin ( ϕ ) ∂ ∂ θ + cot ( θ ) cos ( ϕ ) ∂ ∂ ϕ ) + y ^ ( − cos ( ϕ ) ∂ ∂ θ + cot ( θ ) sin ( ϕ ) ∂ ∂ ϕ ) − z ^ ∂ ∂ ϕ ) L + = ℏ e i ϕ ( ∂ ∂ θ + i cot ( θ ) ∂ ∂ ϕ ) , L − = ℏ e − i ϕ ( − ∂ ∂ θ + i cot ( θ ) ∂ ∂ ϕ ) , L 2 = − ℏ 2 ( 1 sin ( θ ) ∂ ∂ θ ( sin ( θ ) ∂ ∂ θ ) + 1 sin 2 ( θ ) ∂ 2 ∂ ϕ 2 ) , L z = − i ℏ ∂ ∂ ϕ . {\displaystyle {\begin{aligned}\mathbf {L} &=i\hbar \left({\frac {\hat {\boldsymbol {\theta }}}{\sin(\theta )}}{\frac {\partial }{\partial \phi }}-{\hat {\boldsymbol {\phi }}}{\frac {\partial }{\partial \theta }}\right)\\&=i\hbar \left({\hat {\mathbf {x} }}\left(\sin(\phi ){\frac {\partial }{\partial \theta }}+\cot(\theta )\cos(\phi ){\frac {\partial }{\partial \phi }}\right)+{\hat {\mathbf {y} }}\left(-\cos(\phi ){\frac {\partial }{\partial \theta }}+\cot(\theta )\sin(\phi ){\frac {\partial }{\partial \phi }}\right)-{\hat {\mathbf {z} }}{\frac {\partial }{\partial \phi }}\right)\\L_{+}&=\hbar e^{i\phi }\left({\frac {\partial }{\partial \theta }}+i\cot(\theta ){\frac {\partial }{\partial \phi }}\right),\\L_{-}&=\hbar e^{-i\phi }\left(-{\frac {\partial }{\partial \theta }}+i\cot(\theta ){\frac {\partial }{\partial \phi }}\right),\\L^{2}&=-\hbar ^{2}\left({\frac {1}{\sin(\theta )}}{\frac {\partial }{\partial \theta }}\left(\sin(\theta ){\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}(\theta )}}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right),\\L_{z}&=-i\hbar {\frac {\partial }{\partial \phi }}.\end{aligned}}}

In spherical coordinates the angular part of the Laplace operator can be expressed by the angular momentum. This leads to the relation

Δ = 1 r 2 ∂ ∂ r ( r 2 ∂ ∂ r ) − L 2 ℏ 2 r 2 . {\displaystyle \Delta ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}\,{\frac {\partial }{\partial r}}\right)-{\frac {L^{2}}{\hbar ^{2}r^{2}}}.}

When solving to find eigenstates of the operator L 2 {\displaystyle L^{2}} , we obtain the following

L 2 | l , m ⟩ = ℏ 2 l ( l + 1 ) | l , m ⟩ L z | l , m ⟩ = ℏ m | l , m ⟩ {\displaystyle {\begin{aligned}L^{2}|l,m\rangle &=\hbar ^{2}l(l+1)|l,m\rangle \\L_{z}|l,m\rangle &=\hbar m|l,m\rangle \end{aligned}}}

⟨ θ , ϕ | l , m ⟩ = Y l , m ( θ , ϕ ) {\displaystyle \left\langle \theta ,\phi |l,m\right\rangle =Y_{l,m}(\theta ,\phi )}

See also [ edit ]

Notes [ edit ]

^ Γ {\displaystyle \Gamma } J 2 {\displaystyle J^{2}} J z {\displaystyle J_{z}} Γ {\displaystyle \Gamma } J x {\displaystyle J_{x}} J y {\displaystyle J_{y}} [12] The present derivation is simplified by not including the set Γ {\displaystyle \Gamma } γ {\displaystyle \gamma } In the derivation of Condon and Shortley that the current derivation is based on, a set of observablesalong withandform a complete set of commuting observables. Additionally they required thatcommutes withandThe present derivation is simplified by not including the setor its corresponding set of eigenvalues

References [ edit ]

Further reading [ edit ]

Quantum Mechanics Demystified , D. McMahon, Mc Graw Hill (USA), 2006, ISBN 0-07-145546 9

, D. McMahon, Mc Graw Hill (USA), 2006, ISBN 0-07-145546 9 Quantum mechanics , E. Zaarur, Y. Peleg, R. Pnini, Schaum’s Easy Outlines Crash Course, Mc Graw Hill (USA), 2006, ISBN 007-145533-7 ISBN 978-007-145533-6

, E. Zaarur, Y. Peleg, R. Pnini, Schaum’s Easy Outlines Crash Course, Mc Graw Hill (USA), 2006, ISBN 007-145533-7 ISBN 978-007-145533-6 Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles (2nd Edition) , R. Eisberg, R. Resnick, John Wiley & Sons, 1985, ISBN 978-0-471-87373-0

, R. Eisberg, R. Resnick, John Wiley & Sons, 1985, ISBN 978-0-471-87373-0 Quantum Mechanics , E. Abers, Pearson Ed., Addison Wesley, Prentice Hall Inc, 2004, ISBN 978-0-13-146100-0

, E. Abers, Pearson Ed., Addison Wesley, Prentice Hall Inc, 2004, ISBN 978-0-13-146100-0 Physics of Atoms and Molecules , B.H. Bransden, C.J.Joachain, Longman, 1983, ISBN 0-582-44401-2

, B.H. Bransden, C.J.Joachain, Longman, 1983, ISBN 0-582-44401-2 Angular Momentum. Understanding Spatial Aspects in Chemistry and Physics, R. N. Zare, Wiley-Interscience, 1991, ISBN 978-0-47-1858928

whereare the spherical harmonics

## Angular momentum operator

Quantum mechanical operator related to rotational symmetry

In quantum mechanics, the angular momentum operator is one of several related operators analogous to classical angular momentum. The angular momentum operator plays a central role in the theory of atomic and molecular physics and other quantum problems involving rotational symmetry. Such an operator is applied to a mathematical representation of the physical state of a system and yields an angular momentum value if the state has a definite value for it. In both classical and quantum mechanical systems, angular momentum (together with linear momentum and energy) is one of the three fundamental properties of motion.[1]

There are several angular momentum operators: total angular momentum (usually denoted J), orbital angular momentum (usually denoted L), and spin angular momentum (spin for short, usually denoted S). The term angular momentum operator can (confusingly) refer to either the total or the orbital angular momentum. Total angular momentum is always conserved, see Noether’s theorem.

Overview [ edit ]

J (purple), orbital L (blue), and spin S (green). The cones arise due to “Vector cones” of total angular momentum(purple), orbital(blue), and spin(green). The cones arise due to quantum uncertainty between measuring angular momentum components ( see below ).

In quantum mechanics, angular momentum can refer to one of three different, but related things.

Orbital angular momentum [ edit ]

The classical definition of angular momentum is L = r × p {\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} } . The quantum-mechanical counterparts of these objects share the same relationship:

L = r × p {\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }

L = ( L x , L y , L z ) {\displaystyle \mathbf {L} =\left(L_{x},L_{y},L_{z}\right)}

x

y

z

whereis the quantum position operator is the quantum momentum operator , × is cross product , andis the(just likeand) is a(a vector whose components are operators), i.e.whereare three different quantum-mechanical operators.

In the special case of a single particle with no electric charge and no spin, the orbital angular momentum operator can be written in the position basis as:

L = − i ℏ ( r × ∇ ) {\displaystyle \mathbf {L} =-i\hbar (\mathbf {r} \times

abla )}

∇

Spin angular momentum [ edit ]

whereis the vector differential operator, del

There is another type of angular momentum, called spin angular momentum (more often shortened to spin), represented by the spin operator S = ( S x , S y , S z ) {\displaystyle \mathbf {S} =\left(S_{x},S_{y},S_{z}\right)} . Spin is often depicted as a particle literally spinning around an axis, but this is only a metaphor: spin is an intrinsic property of a particle, unrelated to any sort of (yet experimentally observable) motion in space. All elementary particles have a characteristic spin, which is usually nonzero. For example, electrons always have “spin 1/2” while photons always have “spin 1” (details below).

Total angular momentum [ edit ]

Finally, there is total angular momentum J = ( J x , J y , J z ) {\displaystyle \mathbf {J} =\left(J_{x},J_{y},J_{z}\right)} , which combines both the spin and orbital angular momentum of a particle or system:

J = L + S . {\displaystyle \mathbf {J} =\mathbf {L} +\mathbf {S} .}

Conservation of angular momentum states that J for a closed system, or J for the whole universe, is conserved. However, L and S are not generally conserved. For example, the spin–orbit interaction allows angular momentum to transfer back and forth between L and S, with the total J remaining constant.

Commutation relations [ edit ]

Commutation relations between components [ edit ]

The orbital angular momentum operator is a vector operator, meaning it can be written in terms of its vector components L = ( L x , L y , L z ) {\displaystyle \mathbf {L} =\left(L_{x},L_{y},L_{z}\right)} . The components have the following commutation relations with each other:[2] [ L x , L y ] = i ℏ L z , [ L y , L z ] = i ℏ L x , [ L z , L x ] = i ℏ L y , {\displaystyle \left[L_{x},L_{y}\right]=i\hbar L_{z},\;\;\left[L_{y},L_{z}\right]=i\hbar L_{x},\;\;\left[L_{z},L_{x}\right]=i\hbar L_{y},}

where [ , ] denotes the commutator

[ X , Y ] ≡ X Y − Y X . {\displaystyle [X,Y]\equiv XY-YX.}This can be written generally as

[ L l , L m ] = i ℏ ∑ n = 1 3 ε l m n L n , {\displaystyle \left[L_{l},L_{m}\right]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}L_{n},}ε lmn

whereare the component indices (1 for, 2 for, 3 for), anddenotes the Levi-Civita symbol

A compact expression as one vector equation is also possible:[3]

L × L = i ℏ L {\displaystyle \mathbf {L} \times \mathbf {L} =i\hbar \mathbf {L} }

The commutation relations can be proved as a direct consequence of the canonical commutation relations [ x l , p m ] = i ℏ δ l m {\displaystyle [x_{l},p_{m}]=i\hbar \delta _{lm}} , where δ lm is the Kronecker delta.

There is an analogous relationship in classical physics:[4]

{ L i , L j } = ε i j k L k {\displaystyle \left\{L_{i},L_{j}\right\}=\varepsilon _{ijk}L_{k}}

n

{ , } {\displaystyle \{,\}}

whereis a component of theangular momentum operator, andis the Poisson bracket

The same commutation relations apply for the other angular momentum operators (spin and total angular momentum):[5] [ S l , S m ] = i ℏ ∑ n = 1 3 ε l m n S n , [ J l , J m ] = i ℏ ∑ n = 1 3 ε l m n J n . {\displaystyle \left[S_{l},S_{m}\right]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}S_{n},\quad \left[J_{l},J_{m}\right]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}J_{n}.}

These can be assumed to hold in analogy with L. Alternatively, they can be derived as discussed below.

These commutation relations mean that L has the mathematical structure of a Lie algebra, and the ε lmn are its structure constants. In this case, the Lie algebra is SU(2) or SO(3) in physics notation ( su ( 2 ) {\displaystyle \operatorname {su} (2)} or so ( 3 ) {\displaystyle \operatorname {so} (3)} respectively in mathematics notation), i.e. Lie algebra associated with rotations in three dimensions. The same is true of J and S. The reason is discussed below. These commutation relations are relevant for measurement and uncertainty, as discussed further below.

In molecules the total angular momentum F is the sum of the rovibronic (orbital) angular momentum N, the electron spin angular momentum S, and the nuclear spin angular momentum I. For electronic singlet states the rovibronic angular momentum is denoted J rather than N. As explained by Van Vleck,[6] the components of the molecular rovibronic angular momentum referred to molecule-fixed axes have different commutation relations from those given above which are for the components about space-fixed axes.

Commutation relations involving vector magnitude [ edit ]

Like any vector, the square of a magnitude can be defined for the orbital angular momentum operator,

L 2 ≡ L x 2 + L y 2 + L z 2 . {\displaystyle L^{2}\equiv L_{x}^{2}+L_{y}^{2}+L_{z}^{2}.}

L 2 {\displaystyle L^{2}} is another quantum operator. It commutes with the components of L {\displaystyle \mathbf {L} } ,

[ L 2 , L x ] = [ L 2 , L y ] = [ L 2 , L z ] = 0. {\displaystyle \left[L^{2},L_{x}\right]=\left[L^{2},L_{y}\right]=\left[L^{2},L_{z}\right]=0.}One way to prove that these operators commute is to start from the [L ℓ , L m ] commutation relations in the previous section:

Proof of [L2, L x ] = 0, starting from the [L ℓ , L m ] commutation relations[7] [ L 2 , L x ] = [ L x 2 , L x ] + [ L y 2 , L x ] + [ L z 2 , L x ] = L y [ L y , L x ] + [ L y , L x ] L y + L z [ L z , L x ] + [ L z , L x ] L z = L y ( − i ℏ L z ) + ( − i ℏ L z ) L y + L z ( i ℏ L y ) + ( i ℏ L y ) L z = 0 {\displaystyle {\begin{aligned}\left[L^{2},L_{x}\right]&=\left[L_{x}^{2},L_{x}\right]+\left[L_{y}^{2},L_{x}\right]+\left[L_{z}^{2},L_{x}\right]\\&=L_{y}\left[L_{y},L_{x}\right]+\left[L_{y},L_{x}\right]L_{y}+L_{z}\left[L_{z},L_{x}\right]+\left[L_{z},L_{x}\right]L_{z}\\&=L_{y}\left(-i\hbar L_{z}\right)+\left(-i\hbar L_{z}\right)L_{y}+L_{z}\left(i\hbar L_{y}\right)+\left(i\hbar L_{y}\right)L_{z}\\&=0\end{aligned}}}

Mathematically, L 2 {\displaystyle L^{2}} is a Casimir invariant of the Lie algebra SO(3) spanned by L {\displaystyle \mathbf {L} } .

As above, there is an analogous relationship in classical physics:

{ L 2 , L x } = { L 2 , L y } = { L 2 , L z } = 0 {\displaystyle \left\{L^{2},L_{x}\right\}=\left\{L^{2},L_{y}\right\}=\left\{L^{2},L_{z}\right\}=0}

L i {\displaystyle L_{i}}

{ , } {\displaystyle \{,\}}

whereis a component of theangular momentum operator, andis the Poisson bracket

Returning to the quantum case, the same commutation relations apply to the other angular momentum operators (spin and total angular momentum), as well,

[ S 2 , S i ] = 0 , [ J 2 , J i ] = 0. {\displaystyle {\begin{aligned}\left[S^{2},S_{i}\right]&=0,\\\left[J^{2},J_{i}\right]&=0.\end{aligned}}}Uncertainty principle [ edit ]

In general, in quantum mechanics, when two observable operators do not commute, they are called complementary observables. Two complementary observables cannot be measured simultaneously; instead they satisfy an uncertainty principle. The more accurately one observable is known, the less accurately the other one can be known. Just as there is an uncertainty principle relating position and momentum, there are uncertainty principles for angular momentum.

The Robertson–Schrödinger relation gives the following uncertainty principle:

σ L x σ L y ≥ ℏ 2 | ⟨ L z ⟩ | . {\displaystyle \sigma _{L_{x}}\sigma _{L_{y}}\geq {\frac {\hbar }{2}}\left|\langle L_{z}\rangle \right|.}

σ X {\displaystyle \sigma _{X}}

⟨ X ⟩ {\displaystyle \langle X\rangle }

whereis the standard deviation in the measured values ofanddenotes the expectation value of. This inequality is also true ifare rearranged, or ifis replaced byor

Therefore, two orthogonal components of angular momentum (for example L x and L y ) are complementary and cannot be simultaneously known or measured, except in special cases such as L x = L y = L z = 0 {\displaystyle L_{x}=L_{y}=L_{z}=0} .

It is, however, possible to simultaneously measure or specify L2 and any one component of L; for example, L2 and L z . This is often useful, and the values are characterized by the azimuthal quantum number (l) and the magnetic quantum number (m). In this case the quantum state of the system is a simultaneous eigenstate of the operators L2 and L z , but not of L x or L y . The eigenvalues are related to l and m, as shown in the table below.

Quantization [ edit ]

In quantum mechanics, angular momentum is quantized – that is, it cannot vary continuously, but only in “quantum leaps” between certain allowed values. For any system, the following restrictions on measurement results apply, where ℏ {\displaystyle \hbar } is reduced Planck constant:[9]

If you measure… …the result can be… Notes L 2 {\displaystyle L^{2}} ℏ 2 ℓ ( ℓ + 1 ) {\displaystyle \hbar ^{2}\ell (\ell +1)} where ℓ = 0 , 1 , 2 , … {\displaystyle \ell =0,1,2,\ldots } ℓ {\displaystyle \ell } azimuthal quantum number or orbital quantum number. L z {\displaystyle L_{z}} ℏ m ℓ {\displaystyle \hbar m_{\ell }} where m ℓ = − ℓ , ( − ℓ + 1 ) , … , ( ℓ − 1 ) , ℓ {\displaystyle m_{\ell }=-\ell ,(-\ell +1),\ldots ,(\ell -1),\ell } m ℓ {\displaystyle m_{\ell }} magnetic quantum number. This same quantization rule holds for any component of L {\displaystyle \mathbf {L} } ; e.g., L x o r L y {\displaystyle L_{x}\,or\,L_{y}} . This rule is sometimes called spatial quantization.[10] S 2 {\displaystyle S^{2}} ℏ 2 s ( s + 1 ) {\displaystyle \hbar ^{2}s(s+1)} where s = 0 , 1 2 , 1 , 3 2 , … {\displaystyle s=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\ldots } s is called spin quantum number or just spin. For example, a spin-1⁄2 particle is a particle where s = 1⁄2. S z {\displaystyle S_{z}} ℏ m s {\displaystyle \hbar m_{s}} where m s = − s , ( − s + 1 ) , … , ( s − 1 ) , s {\displaystyle m_{s}=-s,(-s+1),\ldots ,(s-1),s} m s {\displaystyle m_{s}} spin projection quantum number. This same quantization rule holds for any component of S {\displaystyle \mathbf {S} } ; e.g., S x o r S y {\displaystyle S_{x}\,or\,S_{y}} . J 2 {\displaystyle J^{2}} ℏ 2 j ( j + 1 ) {\displaystyle \hbar ^{2}j(j+1)} where j = 0 , 1 2 , 1 , 3 2 , … {\displaystyle j=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\ldots } j is sometimes called total angular momentum quantum number. J z {\displaystyle J_{z}} ℏ m j {\displaystyle \hbar m_{j}} where m j = − j , ( − j + 1 ) , … , ( j − 1 ) , j {\displaystyle m_{j}=-j,(-j+1),\ldots ,(j-1),j} m j {\displaystyle m_{j}} total angular momentum projection quantum number. This same quantization rule holds for any component of J {\displaystyle \mathbf {J} } ; e.g., J x o r J y {\displaystyle J_{x}\,or\,J_{y}} .

cannot have a non-integer number of wavelengths like 8.3. In quantum mechanics, angular momentum is quantized for a similar reason. In this standing wave on a circular string, the circle is broken into exactly 8 wavelengths . A standing wave like this can have 0, 1, 2, or any integer number of wavelengths around the circle, but ithave a non-integer number of wavelengths like 8.3. In quantum mechanics, angular momentum is quantized for a similar reason.

Derivation using ladder operators [ edit ]

A common way to derive the quantization rules above is the method of ladder operators.[11] The ladder operators for the total angular momentum J = ( J x , J y , J z ) {\displaystyle \mathbf {J} =\left(J_{x},J_{y},J_{z}\right)} are defined as:

J + ≡ J x + i J y , J − ≡ J x − i J y {\displaystyle {\begin{aligned}J_{+}&\equiv J_{x}+iJ_{y},\\J_{-}&\equiv J_{x}-iJ_{y}\end{aligned}}}

Suppose | ψ ⟩ {\displaystyle |\psi \rangle } is a simultaneous eigenstate of J 2 {\displaystyle J^{2}} and J z {\displaystyle J_{z}} (i.e., a state with a definite value for J 2 {\displaystyle J^{2}} and a definite value for J z {\displaystyle J_{z}} ). Then using the commutation relations for the components of J {\displaystyle \mathbf {J} } , one can prove that each of the states J + | ψ ⟩ {\displaystyle J_{+}|\psi \rangle } and J − | ψ ⟩ {\displaystyle J_{-}|\psi \rangle } is either zero or a simultaneous eigenstate of J 2 {\displaystyle J^{2}} and J z {\displaystyle J_{z}} , with the same value as | ψ ⟩ {\displaystyle |\psi \rangle } for J 2 {\displaystyle J^{2}} but with values for J z {\displaystyle J_{z}} that are increased or decreased by ℏ {\displaystyle \hbar } respectively. The result is zero when the use of a ladder operator would otherwise result in a state with a value for J z {\displaystyle J_{z}} that is outside the allowable range. Using the ladder operators in this way, the possible values and quantum numbers for J 2 {\displaystyle J^{2}} and J z {\displaystyle J_{z}} can be found.

Derivation of the possible values and quantum numbers for J z {\displaystyle J_{z}} J 2 {\displaystyle J^{2}} [12] Let ψ ( J 2 ′ J z ′ ) {\displaystyle \psi ({J^{2}}’J_{z}’)} be a state function for the system with eigenvalue J 2 ′ {\displaystyle {J^{2}}’} for J 2 {\displaystyle J^{2}} and eigenvalue J z ′ {\displaystyle J_{z}’} for J z {\displaystyle J_{z}} .[note 1] From J 2 = J x 2 + J y 2 + J z 2 {\displaystyle J^{2}=J_{x}^{2}+J_{y}^{2}+J_{z}^{2}} is obtained, J x 2 + J y 2 = J 2 − J z 2 . {\displaystyle J_{x}^{2}+J_{y}^{2}=J^{2}-J_{z}^{2}.} ψ ( J 2 ′ J z ′ ) {\displaystyle \psi ({J^{2}}’J_{z}’)} ( J x 2 + J y 2 ) ψ ( J 2 ′ J z ′ ) = ( J 2 ′ − J z ′ 2 ) ψ ( J 2 ′ J z ′ ) . {\displaystyle (J_{x}^{2}+J_{y}^{2})\;\psi ({J^{2}}’J_{z}’)=({J^{2}}’-J_{z}’^{2})\;\psi ({J^{2}}’J_{z}’).} J x {\displaystyle J_{x}} J y {\displaystyle J_{y}} J 2 ′ − J z ′ 2 {\displaystyle {J^{2}}’-J_{z}’^{2}} | J z ′ | ≤ J 2 ′ {\textstyle |J_{z}’|\leq {\sqrt {{J^{2}}’}}} J z ′ {\displaystyle J_{z}’} Applying both sides of the above equation toSinceandare real observables,is not negative and. Thushas an upper and lower bound. Two of the commutation relations for the components of J {\displaystyle \mathbf {J} } are, [ J y , J z ] = i ℏ J x , [ J z , J x ] = i ℏ J y . {\displaystyle [J_{y},J_{z}]=i\hbar J_{x},\;\;[J_{z},J_{x}]=i\hbar J_{y}.} ± {\displaystyle \pm } J z ( J x ± i J y ) = ( J x ± i J y ) ( J z ± ℏ ) , {\displaystyle J_{z}(J_{x}\pm iJ_{y})=(J_{x}\pm iJ_{y})(J_{z}\pm \hbar ),} + {\displaystyle +} − {\displaystyle -} ψ ( J 2 ′ J z ′ ) {\displaystyle \psi ({J^{2}}’J_{z}’)} J z ( J x ± i J y ) ψ ( J 2 ′ J z ′ ) = ( J x ± i J y ) ( J z ± ℏ ) ψ ( J 2 ′ J z ′ ) = ( J z ′ ± ℏ ) ( J x ± i J y ) ψ ( J 2 ′ J z ′ ) . {\displaystyle {\begin{aligned}J_{z}(J_{x}\pm iJ_{y})\;\psi ({J^{2}}’J_{z}’)&=(J_{x}\pm iJ_{y})(J_{z}\pm \hbar )\;\psi ({J^{2}}’J_{z}’)\\&=(J_{z}’\pm \hbar )(J_{x}\pm iJ_{y})\;\psi ({J^{2}}’J_{z}’)\;.\\\end{aligned}}} ( J x ± i J y ) ψ ( J 2 ′ J z ′ ) {\displaystyle (J_{x}\pm iJ_{y})\;\psi ({J^{2}}’J_{z}’)} J z {\displaystyle J_{z}} J z ′ ± ℏ {\displaystyle {J_{z}}’\pm \hbar } ψ ( J 2 ′ J z ′ ± ℏ ) = ( J x ± i J y ) ψ ( J 2 ′ J z ′ ) . {\displaystyle \psi ({J^{2}}’J_{z}’\pm \hbar )=(J_{x}\pm iJ_{y})\;\psi ({J^{2}}’J_{z}’).} J z {\displaystyle J_{z}} J x ± i J y {\displaystyle J_{x}\pm iJ_{y}} ≤ J 2 ′ {\displaystyle \leq {\sqrt {{J^{2}}’}}} J z {\displaystyle J_{z}} J z 0 {\displaystyle J_{z}^{0}} J z 1 {\displaystyle J_{z}^{1}} ( J x − i J y ) ψ ( J 2 ′ J z 0 ) = 0 {\displaystyle (J_{x}-iJ_{y})\;\psi ({J^{2}}’J_{z}^{0})=0} ( J x + i J y ) ψ ( J 2 ′ J z 1 ) = 0 , {\displaystyle (J_{x}+iJ_{y})\;\psi ({J^{2}}’J_{z}^{1})=0,} J z {\displaystyle J_{z}} < J z 0 {\displaystyle

J z 1 {\displaystyle >J_{z}^{1}} ( J x + i J y ) {\displaystyle (J_{x}+iJ_{y})} ( J x − i J y ) {\displaystyle (J_{x}-iJ_{y})} J x 2 + J y 2 = J 2 − J z 2 {\displaystyle J_{x}^{2}+J_{y}^{2}=J^{2}-J_{z}^{2}} J 2 ′ − ( J z 0 ) 2 + ℏ J z 0 = 0 {\displaystyle {J^{2}}’-(J_{z}^{0})^{2}+\hbar J_{z}^{0}=0} J 2 ′ − ( J z 1 ) 2 − ℏ J z 1 = 0. {\displaystyle {J^{2}}’-(J_{z}^{1})^{2}-\hbar J_{z}^{1}=0.} ( J z 1 + J z 0 ) ( J z 0 − J z 1 − ℏ ) = 0. {\displaystyle (J_{z}^{1}+J_{z}^{0})(J_{z}^{0}-J_{z}^{1}-\hbar )=0.} J z 1 ≥ J z 0 {\displaystyle J_{z}^{1}\geq J_{z}^{0}} J z 0 = − J z 1 {\displaystyle J_{z}^{0}=-J_{z}^{1}} They can be combined to obtain two equations, which are written together usingsigns in the following,where one of the equations uses thesigns and the other uses thesigns. Applying both sides of the above toThe above shows thatare two eigenfunctions ofwith respective eigenvalues, unless one of the functions is zero, in which case it is not an eigenfunction. For the functions that are not zero,Further eigenfunctions ofand corresponding eigenvalues can be found by repeatedly applyingas long as the magnitude of the resulting eigenvalue is. Since the eigenvalues ofare bounded, letbe the lowest eigenvalue andbe the highest. Thenandsince there are no states where the eigenvalue ofisor. By applyingto the first equation,to the second, and using, it can be shown thatandSubtracting the first equation from the second and rearranging,Since, the second factor is negative. Then the first factor must be zero and thus The difference J z 1 − J z 0 {\displaystyle J_{z}^{1}-J_{z}^{0}} comes from successive application of J x − i J y {\displaystyle J_{x}-iJ_{y}} or J x + i J y {\displaystyle J_{x}+iJ_{y}} which lower or raise the eigenvalue of J z {\displaystyle J_{z}} by ℏ {\displaystyle \hbar } so that, J z 1 − J z 0 = 0 , ℏ , 2 ℏ , … {\displaystyle J_{z}^{1}-J_{z}^{0}=0,\hbar ,2\hbar ,\dots } J z 1 − J z 0 = 2 j ℏ , {\displaystyle J_{z}^{1}-J_{z}^{0}=2j\hbar ,} j = 0 , 1 2 , 1 , 3 2 , … . {\displaystyle j=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\dots \;.} J z 0 = − J z 1 {\displaystyle J_{z}^{0}=-J_{z}^{1}} J z 0 = − j ℏ {\displaystyle J_{z}^{0}=-j\hbar } J z 1 = j ℏ , {\displaystyle J_{z}^{1}=j\hbar ,} J z {\displaystyle J_{z}} J z ′ = − j ℏ , − j ℏ + ℏ , − j ℏ + 2 ℏ , … , j ℏ . {\displaystyle J_{z}’=-j\hbar ,-j\hbar +\hbar ,-j\hbar +2\hbar ,\dots ,j\hbar .} J z ′ {\displaystyle J_{z}’} m j {\displaystyle m_{j}\;} J z 0 = − j ℏ {\displaystyle J_{z}^{0}=-j\hbar } J 2 ′ − ( J z 0 ) 2 + ℏ J z 0 = 0 {\displaystyle {J^{2}}’-(J_{z}^{0})^{2}+\hbar J_{z}^{0}=0} J z ′ = m j ℏ m j = − j , − j + 1 , − j + 2 , … , j J 2 ′ = j ( j + 1 ) ℏ 2 j = 0 , 1 2 , 1 , 3 2 , … . {\displaystyle {\begin{aligned}J_{z}’&=m_{j}\hbar &m_{j}&=-j,-j+1,-j+2,\dots ,j\\{J^{2}}’&=j(j+1)\hbar ^{2}&j&=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\dots \;.\end{aligned}}} LetwhereThen usingand the above,andand the allowable eigenvalues ofareExpressingin terms of a quantum number, and substitutingintofrom above, Since S {\displaystyle \mathbf {S} } and L {\displaystyle \mathbf {L} } have the same commutation relations as J {\displaystyle \mathbf {J} } , the same ladder analysis can be applied to them, except that for L {\displaystyle \mathbf {L} } there is a further restriction on the quantum numbers that they must be integers.

Traditional derivation of the restriction to integer quantum numbers for L z {\displaystyle L_{z}} L 2 {\displaystyle L^{2}} [13] In the Schroedinger representation, the z component of the orbital angular momentum operator can be expressed in spherical coordinates as,[14] L z = − i ℏ ∂ ∂ ϕ . {\displaystyle L_{z}=-i\hbar {\frac {\partial }{\partial \phi }}.} L z {\displaystyle L_{z}} ψ {\displaystyle \psi } L z ′ {\displaystyle L_{z}’} − i ℏ ∂ ∂ ϕ ψ = L z ′ ψ . {\displaystyle -i\hbar {\frac {\partial }{\partial \phi }}\psi =L_{z}’\psi .} ψ {\displaystyle \psi } ψ = A e i L z ′ ϕ / ℏ , {\displaystyle \psi =Ae^{iL_{z}’\phi /\hbar },} A {\displaystyle A} ϕ {\displaystyle \phi } ψ {\displaystyle \psi } 2 π {\displaystyle 2\pi } ϕ {\displaystyle \phi } A e i L z ′ ( ϕ + 2 π ) / ℏ = A e i L z ′ ϕ / ℏ , e i L z ′ 2 π / ℏ = 1. {\displaystyle {\begin{aligned}Ae^{iL_{z}'(\phi +2\pi )/\hbar }&=Ae^{iL_{z}’\phi /\hbar },\\e^{iL_{z}’2\pi /\hbar }&=1.\end{aligned}}} L z ′ {\displaystyle L_{z}’} L z ′ = m l ℏ , {\displaystyle L_{z}’=m_{l}\hbar \;,} m l {\displaystyle m_{l}} [15] From the above and the relation m ℓ = − ℓ , ( − ℓ + 1 ) , … , ( ℓ − 1 ) , ℓ {\displaystyle m_{\ell }=-\ell ,(-\ell +1),\ldots ,(\ell -1),\ell \ \ } ℓ {\displaystyle \ell } m ℓ {\displaystyle m_{\ell }} ℓ {\displaystyle \ell } L {\displaystyle \mathbf {L} } J {\displaystyle \mathbf {J} } S {\displaystyle \mathbf {S} } [16] Forand eigenfunction with eigenvalueSolving forwhereis independent of. Sinceis required to be single valued, and addingtoresults in a coordinate for the same point in space,Solving for the eigenvaluewhereis an integer.From the above and the relation, it follows thatis also an integer. This shows that the quantum numbersandfor the orbital angular momentumare restricted to integers, unlike the quantum numbers for the total angular momentumand spin, which can have half-integer values. An alternative derivation which does not assume single-valued wave functions follows and another argument using Lie groups is below.

Alternative derivation of the restriction to integer quantum numbers for L z {\displaystyle L_{z}} L 2 {\displaystyle L^{2}} A key part of the traditional derivation above is that the wave function must be single-valued. This is now recognised by many as not being completely correct: a wave function ψ {\displaystyle \psi } is not observable and only the probability density ψ ∗ ψ {\displaystyle \psi ^{*}\psi } is required to be single-valued. The possible double-valued half-integer wave functions have a single-valued probability density.[17] This was recognised by Pauli in 1939 (cited by Japaridze et al[18]) … there is no a priori convincing argument stating that the wave functions which describe some physical states must be single valued functions. For physical quantities, which are expressed by squares of wave functions, to be single valued it is quite sufficient that after moving around a closed contour these functions gain a factor exp(iα) Double-valued wave functions have been found, such as | 1 2 , 1 2 ⟩ = e i ϕ / 2 sin 1 2 θ {\displaystyle \left|{\tfrac {1}{2}},{\tfrac {1}{2}}\right\rangle =e^{i\phi /2}\sin ^{\frac {1}{2}}\theta } and | 1 2 , − 1 2 ⟩ = e − i ϕ / 2 sin 1 2 θ {\displaystyle \left|{\tfrac {1}{2}},-{\tfrac {1}{2}}\right\rangle =e^{-i\phi /2}\sin ^{\frac {1}{2}}\theta } .[19][20] These do not behave well under the ladder operators, but have been found to be useful in describing rigid quantum particles[21] Ballentine[22] gives an argument based solely on the operator formalism and which does not rely on the wave function being single-valued. The azimuthal angular momentum is defined as L z = Q x P y − Q y P x {\displaystyle L_{z}=Q_{x}P_{y}-Q_{y}P_{x}} q 1 = Q x + P y 2 q 2 = Q x − P y 2 p 1 = P x − Q y 2 p 2 = P x + Q y 2 {\displaystyle {\begin{aligned}q_{1}&={\frac {Q_{x}+P_{y}}{\sqrt {2}}}&q_{2}&={\frac {Q_{x}-P_{y}}{\sqrt {2}}}\\p_{1}&={\frac {P_{x}-Q_{y}}{\sqrt {2}}}&p_{2}&={\frac {P_{x}+Q_{y}}{\sqrt {2}}}\end{aligned}}} L z = 1 2 ( p 1 2 + q 1 2 ) − 1 2 ( p 2 2 + q 2 2 ) {\displaystyle L_{z}={\frac {1}{2}}\left(p_{1}^{2}+q_{1}^{2}\right)-{\frac {1}{2}}\left(p_{2}^{2}+q_{2}^{2}\right)} H 1 = 1 2 ( p 1 2 + q 1 2 ) H 2 = 1 2 ( p 2 2 + q 2 2 ) {\displaystyle {\begin{aligned}H_{1}&={\frac {1}{2}}\left(p_{1}^{2}+q_{1}^{2}\right)\\H_{2}&={\frac {1}{2}}\left(p_{2}^{2}+q_{2}^{2}\right)\end{aligned}}} L 2 {\displaystyle L^{2}} L z {\displaystyle L_{z}} H 1 {\displaystyle H_{1}} H 2 {\displaystyle H_{2}} Define new operators(Dimensional correctness may be maintained by inserting factors of mass and unit angular frequency numerically equal to one.) ThenBut the two terms on the right are just the Hamiltonians for the quantum harmonic oscillator with unit mass and angular frequencyandandall commute. For commuting Hermitian operators a complete set of basis vectors can be chosen that are eigenvectors for all four operators. (The argument by Glorioso[23] can easily be generalised to any number of commuting operators.) For any of these eigenvectors ψ {\displaystyle \psi } with L 2 ψ = ℏ ℓ ( ℓ + 1 ) ψ L z ψ = ℏ m ψ H 1 ψ = ℏ ( n 1 + 1 2 ) ψ H 2 ψ = ℏ ( n 2 + 1 2 ) ψ {\displaystyle {\begin{aligned}L^{2}\psi &=\hbar \ell (\ell +1)\psi \\L_{z}\psi &=\hbar m\psi \\H_{1}\psi &=\hbar \left(n_{1}+{\tfrac {1}{2}}\right)\psi \\H_{2}\psi &=\hbar \left(n_{2}+{\tfrac {1}{2}}\right)\psi \end{aligned}}} n 1 , n 2 {\displaystyle n_{1},n_{2}} m = ( n 1 + 1 2 ) − ( n 2 + 1 2 ) = n 1 − n 2 {\displaystyle m=\left(n_{1}+{\tfrac {1}{2}}\right)-\left(n_{2}+{\tfrac {1}{2}}\right)=n_{1}-n_{2}} m {\displaystyle m} ℓ {\displaystyle \ell } for some integers, we findAs a difference of two integers,must be an integer, from whichis also integral. A more complex version of this argument using the ladder operators of the quantum harmonic oscillator has been given by Buchdahl.[24]

Visual interpretation [ edit ]

Illustration of the vector model of orbital angular momentum.

Since the angular momenta are quantum operators, they cannot be drawn as vectors like in classical mechanics. Nevertheless, it is common to depict them heuristically in this way. Depicted on the right is a set of states with quantum numbers ℓ = 2 {\displaystyle \ell =2} , and m ℓ = − 2 , − 1 , 0 , 1 , 2 {\displaystyle m_{\ell }=-2,-1,0,1,2} for the five cones from bottom to top. Since | L | = L 2 = ℏ 6 {\displaystyle |L|={\sqrt {L^{2}}}=\hbar {\sqrt {6}}} , the vectors are all shown with length ℏ 6 {\displaystyle \hbar {\sqrt {6}}} . The rings represent the fact that L z {\displaystyle L_{z}} is known with certainty, but L x {\displaystyle L_{x}} and L y {\displaystyle L_{y}} are unknown; therefore every classical vector with the appropriate length and z-component is drawn, forming a cone. The expected value of the angular momentum for a given ensemble of systems in the quantum state characterized by ℓ {\displaystyle \ell } and m ℓ {\displaystyle m_{\ell }} could be somewhere on this cone while it cannot be defined for a single system (since the components of L {\displaystyle L} do not commute with each other).

Quantization in macroscopic systems [ edit ]

The quantization rules are widely thought to be true even for macroscopic systems, like the angular momentum L of a spinning tire. However they have no observable effect so this has not been tested. For example, if L z / ℏ {\displaystyle L_{z}/\hbar } is roughly 100000000, it makes essentially no difference whether the precise value is an integer like 100000000 or 100000001, or a non-integer like 100000000.2—the discrete steps are currently too small to measure.

Angular momentum as the generator of rotations [ edit ]

The most general and fundamental definition of angular momentum is as the generator of rotations.[5] More specifically, let R ( n ^ , ϕ ) {\displaystyle R({\hat {n}},\phi )} be a rotation operator, which rotates any quantum state about axis n ^ {\displaystyle {\hat {n}}} by angle ϕ {\displaystyle \phi } . As ϕ → 0 {\displaystyle \phi \rightarrow 0} , the operator R ( n ^ , ϕ ) {\displaystyle R({\hat {n}},\phi )} approaches the identity operator, because a rotation of 0° maps all states to themselves. Then the angular momentum operator J n ^ {\displaystyle J_{\hat {n}}} about axis n ^ {\displaystyle {\hat {n}}} is defined as:[5]

J n ^ ≡ i ℏ lim ϕ → 0 R ( n ^ , ϕ ) − 1 ϕ = i ℏ ∂ R ( n ^ , ϕ ) ∂ ϕ | ϕ = 0 {\displaystyle J_{\hat {n}}\equiv i\hbar \lim _{\phi \rightarrow 0}{\frac {R\left({\hat {n}},\phi \right)-1}{\phi }}=\left.i\hbar {\frac {\partial R\left({\hat {n}},\phi \right)}{\partial \phi }}\right|_{\phi =0}}

where 1 is the identity operator. Also notice that R is an additive morphism : R ( n ^ , ϕ 1 + ϕ 2 ) = R ( n ^ , ϕ 1 ) R ( n ^ , ϕ 2 ) {\displaystyle R\left({\hat {n}},\phi _{1}+\phi _{2}\right)=R\left({\hat {n}},\phi _{1}\right)R\left({\hat {n}},\phi _{2}\right)} ; as a consequence[5]

R ( n ^ , ϕ ) = exp ( − i ϕ J n ^ ℏ ) {\displaystyle R\left({\hat {n}},\phi \right)=\exp \left(-{\frac {i\phi J_{\hat {n}}}{\hbar }}\right)}

where exp is matrix exponential

In simpler terms, the total angular momentum operator characterizes how a quantum system is changed when it is rotated. The relationship between angular momentum operators and rotation operators is the same as the relationship between Lie algebras and Lie groups in mathematics, as discussed further below.

The operator R, related to J, rotates the entire system. The operator R spatial , related to L, rotates the particle positions without altering their internal spin states. The operator R internal , related to S, rotates the particles’ internal spin states without changing their positions. The different types of rotation operators . The top box shows two particles, with spin states indicated schematically by the arrows.

Just as J is the generator for rotation operators, L and S are generators for modified partial rotation operators. The operator

R spatial ( n ^ , ϕ ) = exp ( − i ϕ L n ^ ℏ ) , {\displaystyle R_{\text{spatial}}\left({\hat {n}},\phi \right)=\exp \left(-{\frac {i\phi L_{\hat {n}}}{\hbar }}\right),}

R internal ( n ^ , ϕ ) = exp ( − i ϕ S n ^ ℏ ) , {\displaystyle R_{\text{internal}}\left({\hat {n}},\phi \right)=\exp \left(-{\frac {i\phi S_{\hat {n}}}{\hbar }}\right),}

R ( n ^ , ϕ ) = R internal ( n ^ , ϕ ) R spatial ( n ^ , ϕ ) {\displaystyle R\left({\hat {n}},\phi \right)=R_{\text{internal}}\left({\hat {n}},\phi \right)R_{\text{spatial}}\left({\hat {n}},\phi \right)}

rotates the position (in space) of all particles and fields, without rotating the internal (spin) state of any particle. Likewise, the operatorrotates the internal (spin) state of all particles, without moving any particles or fields in space. The relationcomes from:

i.e. if the positions are rotated, and then the internal states are rotated, then altogether the complete system has been rotated.

SU(2), SO(3), and 360° rotations [ edit ]

Although one might expect R ( n ^ , 360 ∘ ) = 1 {\displaystyle R\left({\hat {n}},360^{\circ }\right)=1} (a rotation of 360° is the identity operator), this is not assumed in quantum mechanics, and it turns out it is often not true: When the total angular momentum quantum number is a half-integer (1/2, 3/2, etc.), R ( n ^ , 360 ∘ ) = − 1 {\displaystyle R\left({\hat {n}},360^{\circ }\right)=-1} , and when it is an integer, R ( n ^ , 360 ∘ ) = + 1 {\displaystyle R\left({\hat {n}},360^{\circ }\right)=+1} .[5] Mathematically, the structure of rotations in the universe is not SO(3), the group of three-dimensional rotations in classical mechanics. Instead, it is SU(2), which is identical to SO(3) for small rotations, but where a 360° rotation is mathematically distinguished from a rotation of 0°. (A rotation of 720° is, however, the same as a rotation of 0°.)[5]

On the other hand, R spatial ( n ^ , 360 ∘ ) = + 1 {\displaystyle R_{\text{spatial}}\left({\hat {n}},360^{\circ }\right)=+1} in all circumstances, because a 360° rotation of a spatial configuration is the same as no rotation at all. (This is different from a 360° rotation of the internal (spin) state of the particle, which might or might not be the same as no rotation at all.) In other words, the R spatial {\displaystyle R_{\text{spatial}}} operators carry the structure of SO(3), while R {\displaystyle R} and R internal {\displaystyle R_{\text{internal}}} carry the structure of SU(2).

From the equation + 1 = R spatial ( z ^ , 360 ∘ ) = exp ( − 2 π i L z / ℏ ) {\displaystyle +1=R_{\text{spatial}}\left({\hat {z}},360^{\circ }\right)=\exp \left(-2\pi iL_{z}/\hbar \right)} , one picks an eigenstate L z | ψ ⟩ = m ℏ | ψ ⟩ {\displaystyle L_{z}|\psi \rangle =m\hbar |\psi \rangle } and draws

e − 2 π i m = 1 {\displaystyle e^{-2\pi im}=1}

Connection to representation theory [ edit ]

which is to say that the orbital angular momentum quantum numbers can only be integers, not half-integers.

Starting with a certain quantum state | ψ 0 ⟩ {\displaystyle |\psi _{0}\rangle } , consider the set of states R ( n ^ , ϕ ) | ψ 0 ⟩ {\displaystyle R\left({\hat {n}},\phi \right)\left|\psi _{0}\right\rangle } for all possible n ^ {\displaystyle {\hat {n}}} and ϕ {\displaystyle \phi } , i.e. the set of states that come about from rotating the starting state in every possible way. The linear span of that set is a vector space, and therefore the manner in which the rotation operators map one state onto another is a representation of the group of rotation operators.

When rotation operators act on quantum states, it forms a representation of the Lie group SU(2) (for R and R internal ), or SO(3) (for R spatial ).

From the relation between J and rotation operators,

When angular momentum operators act on quantum states, it forms a representation of the Lie algebra s u ( 2 ) {\displaystyle {\mathfrak {su}}(2)} s o ( 3 ) {\displaystyle {\mathfrak {so}}(3)}

(The Lie algebras of SU(2) and SO(3) are identical.)

The ladder operator derivation above is a method for classifying the representations of the Lie algebra SU(2).

Connection to commutation relations [ edit ]

Classical rotations do not commute with each other: For example, rotating 1° about the x-axis then 1° about the y-axis gives a slightly different overall rotation than rotating 1° about the y-axis then 1° about the x-axis. By carefully analyzing this noncommutativity, the commutation relations of the angular momentum operators can be derived.[5]

(This same calculational procedure is one way to answer the mathematical question “What is the Lie algebra of the Lie groups SO(3) or SU(2)?”)

Conservation of angular momentum [ edit ]

The Hamiltonian H represents the energy and dynamics of the system. In a spherically symmetric situation, the Hamiltonian is invariant under rotations:

R H R − 1 = H {\displaystyle RHR^{-1}=H}

[ H , R ] = 0 {\displaystyle [H,R]=0} [ H , J ] = 0 {\displaystyle [H,\mathbf {J} ]=\mathbf {0} }whereis a rotation operator . As a consequence,, and thendue to the relationship betweenand. By the Ehrenfest theorem , it follows thatis conserved.

To summarize, if H is rotationally-invariant (spherically symmetric), then total angular momentum J is conserved. This is an example of Noether’s theorem.

If H is just the Hamiltonian for one particle, the total angular momentum of that one particle is conserved when the particle is in a central potential (i.e., when the potential energy function depends only on | r | {\displaystyle \left|\mathbf {r} \right|} ). Alternatively, H may be the Hamiltonian of all particles and fields in the universe, and then H is always rotationally-invariant, as the fundamental laws of physics of the universe are the same regardless of orientation. This is the basis for saying conservation of angular momentum is a general principle of physics.

For a particle without spin, J = L, so orbital angular momentum is conserved in the same circumstances. When the spin is nonzero, the spin–orbit interaction allows angular momentum to transfer from L to S or back. Therefore, L is not, on its own, conserved.

Angular momentum coupling [ edit ]

Often, two or more sorts of angular momentum interact with each other, so that angular momentum can transfer from one to the other. For example, in spin–orbit coupling, angular momentum can transfer between L and S, but only the total J = L + S is conserved. In another example, in an atom with two electrons, each has its own angular momentum J 1 and J 2 , but only the total J = J 1 + J 2 is conserved.

In these situations, it is often useful to know the relationship between, on the one hand, states where ( J 1 ) z , ( J 1 ) 2 , ( J 2 ) z , ( J 2 ) 2 {\displaystyle \left(J_{1}\right)_{z},\left(J_{1}\right)^{2},\left(J_{2}\right)_{z},\left(J_{2}\right)^{2}} all have definite values, and on the other hand, states where ( J 1 ) 2 , ( J 2 ) 2 , J 2 , J z {\displaystyle \left(J_{1}\right)^{2},\left(J_{2}\right)^{2},J^{2},J_{z}} all have definite values, as the latter four are usually conserved (constants of motion). The procedure to go back and forth between these bases is to use Clebsch–Gordan coefficients.

One important result in this field is that a relationship between the quantum numbers for ( J 1 ) 2 , ( J 2 ) 2 , J 2 {\displaystyle \left(J_{1}\right)^{2},\left(J_{2}\right)^{2},J^{2}} :

j ∈ { | j 1 − j 2 | , ( | j 1 − j 2 | + 1 ) , … , ( j 1 + j 2 ) } . {\displaystyle j\in \left\{\left|j_{1}-j_{2}\right|,\left(\left|j_{1}-j_{2}\right|+1\right),\ldots ,\left(j_{1}+j_{2}\right)\right\}.}

For an atom or molecule with J = L + S, the term symbol gives the quantum numbers associated with the operators L 2 , S 2 , J 2 {\displaystyle L^{2},S^{2},J^{2}} .

Orbital angular momentum in spherical coordinates [ edit ]

Angular momentum operators usually occur when solving a problem with spherical symmetry in spherical coordinates. The angular momentum in the spatial representation is[25][26]

L = i ℏ ( θ ^ sin ( θ ) ∂ ∂ ϕ − ϕ ^ ∂ ∂ θ ) = i ℏ ( x ^ ( sin ( ϕ ) ∂ ∂ θ + cot ( θ ) cos ( ϕ ) ∂ ∂ ϕ ) + y ^ ( − cos ( ϕ ) ∂ ∂ θ + cot ( θ ) sin ( ϕ ) ∂ ∂ ϕ ) − z ^ ∂ ∂ ϕ ) L + = ℏ e i ϕ ( ∂ ∂ θ + i cot ( θ ) ∂ ∂ ϕ ) , L − = ℏ e − i ϕ ( − ∂ ∂ θ + i cot ( θ ) ∂ ∂ ϕ ) , L 2 = − ℏ 2 ( 1 sin ( θ ) ∂ ∂ θ ( sin ( θ ) ∂ ∂ θ ) + 1 sin 2 ( θ ) ∂ 2 ∂ ϕ 2 ) , L z = − i ℏ ∂ ∂ ϕ . {\displaystyle {\begin{aligned}\mathbf {L} &=i\hbar \left({\frac {\hat {\boldsymbol {\theta }}}{\sin(\theta )}}{\frac {\partial }{\partial \phi }}-{\hat {\boldsymbol {\phi }}}{\frac {\partial }{\partial \theta }}\right)\\&=i\hbar \left({\hat {\mathbf {x} }}\left(\sin(\phi ){\frac {\partial }{\partial \theta }}+\cot(\theta )\cos(\phi ){\frac {\partial }{\partial \phi }}\right)+{\hat {\mathbf {y} }}\left(-\cos(\phi ){\frac {\partial }{\partial \theta }}+\cot(\theta )\sin(\phi ){\frac {\partial }{\partial \phi }}\right)-{\hat {\mathbf {z} }}{\frac {\partial }{\partial \phi }}\right)\\L_{+}&=\hbar e^{i\phi }\left({\frac {\partial }{\partial \theta }}+i\cot(\theta ){\frac {\partial }{\partial \phi }}\right),\\L_{-}&=\hbar e^{-i\phi }\left(-{\frac {\partial }{\partial \theta }}+i\cot(\theta ){\frac {\partial }{\partial \phi }}\right),\\L^{2}&=-\hbar ^{2}\left({\frac {1}{\sin(\theta )}}{\frac {\partial }{\partial \theta }}\left(\sin(\theta ){\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}(\theta )}}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right),\\L_{z}&=-i\hbar {\frac {\partial }{\partial \phi }}.\end{aligned}}}

In spherical coordinates the angular part of the Laplace operator can be expressed by the angular momentum. This leads to the relation

Δ = 1 r 2 ∂ ∂ r ( r 2 ∂ ∂ r ) − L 2 ℏ 2 r 2 . {\displaystyle \Delta ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}\,{\frac {\partial }{\partial r}}\right)-{\frac {L^{2}}{\hbar ^{2}r^{2}}}.}

When solving to find eigenstates of the operator L 2 {\displaystyle L^{2}} , we obtain the following

L 2 | l , m ⟩ = ℏ 2 l ( l + 1 ) | l , m ⟩ L z | l , m ⟩ = ℏ m | l , m ⟩ {\displaystyle {\begin{aligned}L^{2}|l,m\rangle &=\hbar ^{2}l(l+1)|l,m\rangle \\L_{z}|l,m\rangle &=\hbar m|l,m\rangle \end{aligned}}}

⟨ θ , ϕ | l , m ⟩ = Y l , m ( θ , ϕ ) {\displaystyle \left\langle \theta ,\phi |l,m\right\rangle =Y_{l,m}(\theta ,\phi )}

See also [ edit ]

Notes [ edit ]

^ Γ {\displaystyle \Gamma } J 2 {\displaystyle J^{2}} J z {\displaystyle J_{z}} Γ {\displaystyle \Gamma } J x {\displaystyle J_{x}} J y {\displaystyle J_{y}} [12] The present derivation is simplified by not including the set Γ {\displaystyle \Gamma } γ {\displaystyle \gamma } In the derivation of Condon and Shortley that the current derivation is based on, a set of observablesalong withandform a complete set of commuting observables. Additionally they required thatcommutes withandThe present derivation is simplified by not including the setor its corresponding set of eigenvalues

References [ edit ]

Further reading [ edit ]

Quantum Mechanics Demystified , D. McMahon, Mc Graw Hill (USA), 2006, ISBN 0-07-145546 9

, D. McMahon, Mc Graw Hill (USA), 2006, ISBN 0-07-145546 9 Quantum mechanics , E. Zaarur, Y. Peleg, R. Pnini, Schaum’s Easy Outlines Crash Course, Mc Graw Hill (USA), 2006, ISBN 007-145533-7 ISBN 978-007-145533-6

, E. Zaarur, Y. Peleg, R. Pnini, Schaum’s Easy Outlines Crash Course, Mc Graw Hill (USA), 2006, ISBN 007-145533-7 ISBN 978-007-145533-6 Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles (2nd Edition) , R. Eisberg, R. Resnick, John Wiley & Sons, 1985, ISBN 978-0-471-87373-0

, R. Eisberg, R. Resnick, John Wiley & Sons, 1985, ISBN 978-0-471-87373-0 Quantum Mechanics , E. Abers, Pearson Ed., Addison Wesley, Prentice Hall Inc, 2004, ISBN 978-0-13-146100-0

, E. Abers, Pearson Ed., Addison Wesley, Prentice Hall Inc, 2004, ISBN 978-0-13-146100-0 Physics of Atoms and Molecules , B.H. Bransden, C.J.Joachain, Longman, 1983, ISBN 0-582-44401-2

, B.H. Bransden, C.J.Joachain, Longman, 1983, ISBN 0-582-44401-2 Angular Momentum. Understanding Spatial Aspects in Chemistry and Physics, R. N. Zare, Wiley-Interscience, 1991, ISBN 978-0-47-1858928

whereare the spherical harmonics

## Space-Quantization of Angular Momentum

The commutation relations for angular momentum in quantum mechanics are given by , , with cyclic permutations. From these, the allowed values of quantized angular momentum can be derived, namely, and , with , . Customarily, the component is singled out, with the other two components retaining indefinite or fluctuating values (except when . The definite magnitude and direction of one component of angular momentum is known as “space quantization”. Restriction of to integer values was exploited in Bohr’s model of the hydrogen atom. When spin is involved, and can also take half-integer values.

The vector model of angular momentum pictures the total angular momentum vector as precessing about its constant component. This is also consistent with the fluctuating values of and . The fact that the quantized value of equals , rather than , can be rationalized by the fact that the average value of the sum of the squares of the three components is given by . [less]

## Angular Momentum Quantization

Angular Momentum Quantization: Physical Manifestations and Chemical Consequences

Michael Fowler, University of Virginia

The Stern-Gerlach Experiment

We’ve established that for the hydrogen atom, the angular momentum of the electron’s orbital motion has values l ( l + 1 ) ℏ , where l = 0 , 1 , 2 , … , and the component of angular momentum in the z -direction is m ℏ , where m takes integer values − l , − l + 1 , − l + 2 , … , + l . This means that if we measure the angle between the total angular momentum and the z -axis, there can only be 2 l + 1 possible answers, the total angular momentum cannot point in an arbitrary direction relative to the z -axis, odd though this conclusion seems. This is sometimes called “space quantization”.

Is there any way we can actually see some effect of this directional quantization? The answer is yes—because the electron moving around its orbit is a tiny loop of electric current, and, therefore, an electromagnet. So, if we switch on a magnetic field in the direction of the z -axis, the energy of the atom will depend on the degree of alignment of its magnetic moment with the external applied magnetic field. The magnetic field from a small current loop is like that from a small bar magnet aligned along the axis of the loop.

The simplest way to see how the potential energy of the little magnet depends on which way it’s pointing relative to the field is to take a little bar with an N pole of strength + p at one end, an S pole of strength − p at the other end. Think of a compass needle, of length d , say. The magnetic moment is defined as pole strength multiplied by distance between the poles, μ = p d , ,and is considered to be a vector pointing along the axis of the magnet, from S to N. The potential energy of this little magnet in an external field H is − μ → ⋅ H → , lowest when the magnet is fully aligned with the field. It is easy to check this: counting as zero potential energy the magnet at right angles with the field, the work needed to point it at an angle θ is 2 ⋅ p ⋅ ( d / 2 ) ⋅ cos θ .

The magnetic moment of a current I going in a circle around an area A is just I A . The electron has charge e , and speed v , so goes around v / 2 π r times per second. In other words, if you stand at one point in the orbit, the total charge passing you per second is e v / 2 π r = I . Hence the magnetic moment, usually denoted μ L = I A = π r 2 ⋅ ( e v / 2 π r ) = e r v / 2. The angular momentum is L = r m v so

μ L = ( e 2 m ) L .

Thus if the electron is in an l = 1 orbit, the current will generate a magnetic moment ( e / 2 m ) ℏ , which is 9.3 × 10-24 joules per tesla, or 5.8 × 10-5 eV per tesla. Note that this means in a one tesla field an atomic energy level will move ~10-4 eV, an easily detectable shift in spectral lines will result.

But there is a more direct way to see how the atoms are oriented, the Stern Gerlach apparatus (1922). In this experiment, a beam of atoms is sent into a nonuniform magnetic field. This means the north and south poles of a small bar magnet would feel different strength forces, so there would be a net force on a small magnet, and hence on an atom. Furthermore, the direction of this force would depend on the orientation of the dipole.

Suppose the nonuniform field is pointing upwards, and is stronger at the top. Then a small bar magnet oriented vertically with the north pole on top will be pushed upwards, because the north pole will be experiencing the stronger force. If the south pole is on top, the magnet will be pushed downwards. If the magnet is horizontal, there will be no net force (assuming magnetic field strength varies only negligibly in the horizontal direction).

Imagine, then, a stream of atoms with magnetic moments entering a region of magnetic field as described. Each atom will feel a vertical force depending on the orientation of its magnetic moment. If with no magnetic field present the stream of atoms formed a dot on a screen after passing through the apparatus, on switching on the field one would expect the dot to be stretched into a vertical line, if one assumed equal likelihood of all orientations of the magnetic moment. However, the quantum theory predicts that this is not the case—we have argued that for l = 1 , say, there are only three allowed orientations of the magnet (atom) relative to the field. Therefore, we would predict that three dots (or, more realistically, blobs) would appear on the screen, not a continuous line.

In fact, when the experiment was carried out, there was a very surprising result. Perhaps the most dramatic form of the new result came later, in 1927, when ground state l = 0 hydrogen atoms were used (Phipps and Taylor, Phys Rev 29, 309). Such atoms have no orbital angular momentum, and therefore no orbital current, and were not expected to show magnetic effects. Yet on going through the Stern-Gerlach apparatus, the beam of hydrogen atoms split into two! This was difficult to interpret, because the least allowed angular momentum, l = 1 , would give three blobs, and l = 0 would give only one. You might expect a mixture to give one strong blob and two weak ones, but two equal blobs didn’t seem possible, theoretically. Stern and Gerlach had themselves seen two blobs with silver atoms in 1922. We mention the hydrogen case first because it was by far the best understood atom (and still is!) so the need for new physics was clearest.

The solution to the problem was suggested by two graduate students, Goudsmit and Uhlenbeck. They suggested that the electron itself had a spin. That is to say, the electron both orbited the proton and spun on its own axis, just as the earth orbits the sun once a year and also spins on its own axis once a day. If the electron spin is assumed to be ℏ / 2 , and we assume as before that the z -component can only change by whole units of ℏ , then there are only two allowed values of the z -component, ± ℏ / 2. Of course, this is a hand waving argument — the reason the z -component only changed by integers was that the wave function had to fit a whole number of wavelengths on going around the z -axis. But our wave function for spin one-half , if it is of the same form as those for angular momentum, must have a term e i φ / 2 , and so is multiplied by − 1 on rotating through 2 π ! (In fact, that the z -component can only change by whole units of ℏ follows from very general properties of angular momentum.)

Further difficulties arose when people tried to construct models of how a spinning electron would have its own magnetic moment. It’s not too difficult to see how this might occur — if the electron is a charged sphere, or has charge on its surface, then its rotation implies that this charge is going around in circles, little current loops, and so will give a magnetic field. The problem was, it was known that the electron was a very small object. It turned out that the equatorial speed of the electron in this kind of model would have to be greater than the speed of light for the magnetic moment to be of the observed strength.

These difficulties in understanding the electron spin and magnetic moment were far from trivial, and in fact were not resolved until around 1930, by Dirac, who gave a fully relativistic treatment of the problem, which, remarkably, predicted the magnetic moment correctly and at the same time treated the electron as a point particle. There is no simple picture presenting this in classical or semiclassical terms, but Dirac’s work is the basis of our modern understanding of particle physics. It is unfortunately beyond the scope of this course.

The bottom line, as far as we are concerned, is that assuming the electron has spin one-half and hence two possible spin orientations with respect to a given axis explains the observed Stern-Gerlach results, and also, more importantly, helps us construct the periodic table, as we shall see below.

Building the Periodic Table

The atomic number (usually denoted Z) of an element denotes its place in the periodic table, so H has Z = 1; He, Z = 2; Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, F = 9, Ne = 10, and so on. This number is equal to the number of protons in the nucleus, and also equal to the number of electrons orbiting around the nucleus, to preserve electrical neutrality.

To try to understand how the electrons orbit the nucleus, we need to make some simplifying assumptions. We are not going to be able to solve Schrödinger’s equation for even two electrons exactly, if we include their repulsion of each other. However, the presence of the other electrons is clearly important — their repulsion to some extent counteracts the attraction the nucleus has for a given electron. For an electron imagined to be in some outer orbit, the electrons in closer to the nucleus orbits lower the effective nuclear charge. Thinking now about the force felt by one electron, a simple approximation is to imagine all the other electrons as changing the electrical attraction the one electron feels from the nucleus to a shielded attraction, so that the further it is away from the nucleus, the weaker an attractive charge it sees.

We then make the naïve assumption that all the electrons see the same potential, this shielded Coulomb potential, so we have Z electrons all in the same potential well, but we assume they are independent particles, in the sense that they do not repel each other, except to the extent already taken into account by changing to a shielded potential. So the question is, what are the possible wave functions of Z independent electrons in this well? The crucial point is that although they do not interact with each other, they are identical, so the wave function must be antisymmetrized, as we discussed for the two particle case earlier. This means that the electrons must be in different bound states in the well—the Pauli exclusion principle. But what do the bound state wave functions look like in this potential? Since the shielded Coulomb potential is still spherically symmetric, all our arguments about the θ , φ behavior of the ordinary Coulomb potential apply equally to the shielded case, in particular the angular momentum has values l ( l + 1 ) ℏ , where l = 0 , 1 , 2 , … , and the component of angular momentum in the z -direction is m ℏ , , where m takes integer values − l , − l + 1 , − l + 2 , … , + l . Furthermore, each electron has spin ℏ / 2 , and there are two allowed values of the spin z -component, ± ℏ / 2. The radial wave functions R ( r ) are clearly somewhat different from those in the pure Coulomb case. The main difference is that states of different angular momentum which were degenerate in the Coulomb case are no longer the same energy in the shielded Coulomb case. If you examine wave functions corresponding to the same energy but different values of l , you will see that the higher the l value, the smaller the wave function is near the nucleus. This means the higher l wave functions do not feel the powerful unshielded potential near the nucleus, and so are not as strongly bound as the lower l functions.

Notation

A standard notation is used by atomic physicists to describe these states. The different angular momenta are denoted by letters, s for l = 0 , p for l = 1 , d for l = 2 , f for l = 3 , g for l = 4 and then on alphabetically. The Principal Quantum Number n , such that for the hydrogen atom E = − 1 / n 2 in Rydberg units, is given as a number, so the lowest hydrogen atom state is written 1s. The two n = 2 orbital states are 2s and 2p, then come 3s, 3p and 3d and so on. From the discussion immediately above, 2s and 2p have the same energy in the hydrogen atom, but for the shielded potential used to approximate for the presence of other electrons in bigger atoms 2s would be more tightly bound, and so at a lower energy, than 2p.

Filling an Atom with Electrons

Let us now consider taking a bare nucleus, charge Z, and adding Z electrons to it one by one. From the Pauli Exclusion Principle, each electron must be in a different state. But remember that having a different spin counts as different (you could tell them apart) so we can put two electrons, with opposite spins, into each orbital state. Thus He has two electrons in the 1s state. Li must have two electrons in 1s, and one electron in 2s. This suggests a picture of one electron outside of a “closed shell” of two 1s electrons. The next occurrence of a similar picture is Na, having Z = 11, which is chemically very similar to Li. This means that 10 electrons fill closed shells. We can understand this because 2 go into 1s, 2 go into 2s and 6 fill 2p. But notice by saying it takes 6 electrons to fill 3p, we are saying there are three distinct l = 1 orbitals. In other words, the chemical properties of the elements support and confirm the hypothesis of “space quantization” — that there are three and only three distinct l = 1 angular wave functions, those given by m = 1 , 0 , − 1.

Atoms interact chemically by sharing or partially transferring electrons. It’s easier to transfer an electron that is loosely bound, and easier to accept one if there’s a “hole” in a shell. Not surprisingly, atoms with filled shells only, like He and Ne, are chemically unreactive. The valency, roughly speaking, is the number of electrons available for transfer (so Li and Na have valency 1) or available sites for reception of electrons — fluorine has an outer shell with one vacancy, so a valency of 1. To some extent, valency can vary depending on the strength of attraction of other atoms in the chemical environment.

Filling a Box with Electrons

When many Li atoms are put together to form a solid, it is found that the loosely attached outer electrons leave their original atoms and wander freely throughout the metal. Their wave functions are well represented by standing plane waves in a box (let’s take a cube of metal, of side L ). Each such plane wave state in the box can be represented by three numbers n x , n y , n z representing the number of nodes of the standing wave in the x , y , z directions respectively. Extending slightly our analysis of an electron in a two-dimensional box, the energy of such a state will be E = ℏ 2 2 m π 2 L 2 ( n x 2 + n y 2 + n z 2 ) . Thus if we imagine pouring electrons into an empty lattice of Li atoms each with one electron missing(not a physically realistic procedure!) two electrons (opposite spins) will go into each state, first ( 0 , 0 , 0 ) then ( 1 , 0 , 0 ) or equally ( 0 , 1 , 0 ) etc., and from the form of the energy we can see that in ( n x , n y , n z ) space, the electrons will fill up all the positive integer points within a sphere up to some maximum energy determined by how many electrons we put in. Notice that since the n ’s are all positive integers, the filled space is only the one-eighth of the sphere’s volume corresponding to n x ≥ 0 , n y ≥ 0 , n z ≥ 0 for the sphere centered at the origin. (The fraction having some quantum numbers zero is vanishingly small for macroscopic bodies, so the one-eighth is accurate).

## Angular Momentum Of Electron

In this article, we will be learning about the angular momentum of electrons in detail. We will also be learning about De Broglie’s Explanation of the Quantization of Angular Momentum of Electrons.

What is Angular Momentum of Electron?

Bohr’s atomic model laid down various postulates for the arrangement of electrons in different orbits around the nucleus. According to Bohr’s atomic model, the angular momentum of electrons orbiting around the nucleus is quantized. He further added that electrons move only in those orbits where the angular momentum of an electron is an integral multiple of h/2. This postulate regarding the quantisation of angular momentum of an electron was later explained by Louis de Broglie. According to him, a moving electron in its circular orbit behaves like a particle-wave.

The angular momentum of an electron by Bohr is given by mvr or nh/2π (where v is the velocity, n is the orbit in which the electron is revolving, m is mass of the electron, and r is the radius of the nth orbit).

Among the various proposed models over the years, the Quantum Mechanical Model seems to best fit all properties. Watch the video to learn more about the features of the quantum mechanical model.

## Is all angular momentum quantized?

$\begingroup$

I’m going to disagree with the other answers: I think that the angular momentum of macroscopic “classical” objects is not quantized.

Consider an automobile tire spinning in a wheel well. On the tire is a device that triggers whenever a certain point on the wheel crosses a certain point on the well, adding one to an internal counter if it passes it clockwise and subtracting one if it crosses it counterclockwise. (Alternately, you could dispense with the wheel well and say that the device tracks its own position by inertial navigation.) The state of this system can be described by a value $θ\in\mathbb R$, where the current value of the counter is $\lfloor θ/2π\rfloor$ and the angle of the wheel is $θ\text{ (mod }2π\text{)}$. In the absence of external forces, the Hamiltonian of the system is essentially that of a free particle in $\mathbb R$, and the spectrum of angular momenta is continuous just like the free particle’s momentum spectrum.

That’s a 2+1 dimensional system. In 3+1 dimensions, there’s the Dirac belt trick to worry about. Does it matter? I don’t think so. There’s no reason to limit the device to holding a single integer, or to being reversible. It could simply store the entire history of its orientation readings internally, or broadcast them by radio, indelibly recording them in the universal wave function. That’s a very noncompact state space, and it’s an accurate enough model of bodies like the earth.

The angular momentum operator on this monstrosity obviously violates the assumptions of any proof of the quantization of angular momentum, but that’s no reason not to call it angular momentum. We do call it angular momentum, and it’s what the question was about.

In response to comments I’ll try to clarify my answer.

These are quantum systems, but the earth system is “classical” in the sense of being a quantum system with emergent classical behavior.

The reason that high temperature systems behave classically is that they constantly leak which-path information into the environment. If you do a double-slit experiment with the earth, it will emit different patterns of light going through one slit than through the other. You can literally see which slit it goes through, but even if you don’t look, the which-path information is there in the patterns of light, or in patterns of heat if the light is absorbed by the walls of the lab, and that’s all that’s necessary to make the final states orthogonal and destroy the interference pattern.

It’s sometimes said that you can’t see an interference pattern in the earth double-slit experiment simply because its de Broglie wavelength is so small. That would be correct for a supermassive stable particle that doesn’t radiate, but it’s wrong for the earth. For the earth there’s no interference pattern at all, for the same reason there’s no interference pattern when there’s a detector at one of the slits. Earth’s thermal radiation is the “detector”.

The case of rotation is similar. Different rotations are different paths through the state space (it’s the state space, not physical 3D space, that the wave function is defined on and which matters here). If you consider two different paths ending in the same physical orientation (analogous to the same position on the screen in the double-slit case), these paths will interfere if no information about which path was taken is recorded anywhere. In the case of the earth, this means they’ll interfere if there’s no way for anyone to tell whether the earth rotated around its axis or not. If there’s any record of it – if any animals remember the day-night cycle, or don’t but could in principle, or if aliens see it rotate through a telescope, or don’t but could in principle – then there’s no interference.

The proof that angular momentum is quantized depends on the compactness of the space of orientations. This is fine if the space of orientations is the phase space, i.e., if the system is memoryless. If it has a memory, rotating the system through $2π$ or $4π$ doesn’t leave it in the same state as not rotating it.

The tire example in the second paragraph may have been a mistake since it seems to have only caused confusion. But it’s a perfectly good quantum system in the abstract, and its state space is $\mathbb R$, not $S^1$.

## 키워드에 대한 정보 angular momentum quantization

다음은 Bing에서 **angular momentum quantization** 주제에 대한 검색 결과입니다. 필요한 경우 더 읽을 수 있습니다.

이 기사는 인터넷의 다양한 출처에서 편집되었습니다. 이 기사가 유용했기를 바랍니다. 이 기사가 유용하다고 생각되면 공유하십시오. 매우 감사합니다!

## 사람들이 주제에 대해 자주 검색하는 키워드 Angular Momentum Quantization in Bohr Model (Quantum Mechanics – De Broglie’s Wave-particle Duality)

- 동영상
- 공유
- 카메라폰
- 동영상폰
- 무료
- 올리기

Angular #Momentum #Quantization #in #Bohr #Model #(Quantum #Mechanics #- #De #Broglie’s #Wave-particle #Duality)

YouTube에서 angular momentum quantization 주제의 다른 동영상 보기

주제에 대한 기사를 시청해 주셔서 감사합니다 **Angular Momentum Quantization in Bohr Model (Quantum Mechanics – De Broglie’s Wave-particle Duality) | angular momentum quantization**, 이 기사가 유용하다고 생각되면 공유하십시오, 매우 감사합니다.