Angular momentum operator

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

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

where r is the quantum position operator, p is the quantum momentum operator, × is cross product, and L is the orbital angular momentum operator. L (just like p and r) is a vector operator (a vector whose components are operators), i.e. ${\displaystyle \mathbf {L} =\left(L_{x},L_{y},L_{z}\right)}$ where L_x, L_y, L_z are 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:

$${\displaystyle \mathbf {L} =-i\hbar (\mathbf {r} \times \nabla )}$$

where ∇ is 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 ${\displaystyle \mathbf {S} =\left(S_{x},S_{y},S_{z}\right)}$. Spin is often depicted as a particle literally spinning around an axis, but this is only a metaphor: the closest classical analog is based on wave circulation.[2] All elementary particles have a characteristic spin (scalar bosons have zero spin). For example, electrons always have "spin 1/2" while photons always have "spin 1" (details below).

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

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

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.

The orbital angular momentum operator is a vector operator, meaning it can be written in terms of its vector components ${\displaystyle \mathbf {L} =\left(L_{x},L_{y},L_{z}\right)}$. The components have the following commutation relations with each other:[3]

$${\displaystyle \left[L_{x},L_{y}\right]=i\hbar L_{z},\;\;\left[L_{y},L_{z}\right]=i\hbar L_{x},\;\;\left[L_{z},L_{x}\right]=i\hbar L_{y},}$$

where [ , ] denotes the commutator

$${\displaystyle [X,Y]\equiv XY-YX.}$$

This can be written generally as

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

where l, m, n are the component indices (1 for x, 2 for y, 3 for z), and ε_lmn denotes the Levi-Civita symbol.

A compact expression as one vector equation is also possible:[4]

$${\displaystyle \mathbf {L} \times \mathbf {L} =i\hbar \mathbf {L} }$$

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

There is an analogous relationship in classical physics:[5]

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

where L_n is a component of the classical angular momentum operator, and ${\displaystyle \{,\}}$ is the Poisson bracket.

The same commutation relations apply for the other angular momentum operators (spin and total angular momentum):[6]

$${\displaystyle \left[S_{l},S_{m}\right]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}S_{n},\quad \left[J_{l},J_{m}\right]=i\hbar \sum _{n=1}^{3}\varepsilon _{lmn}J_{n}.}$$

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 (${\displaystyle \operatorname {su} (2)}$ or ${\displaystyle \operatorname {so} (3)}$ 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,[7] 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.

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

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

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

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

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[8]

$${\displaystyle {\begin{aligned}\left[L^{2},L_{x}\right]&=\left[L_{x}^{2},L_{x}\right]+\left[L_{y}^{2},L_{x}\right]+\left[L_{z}^{2},L_{x}\right]\\&=L_{y}\left[L_{y},L_{x}\right]+\left[L_{y},L_{x}\right]L_{y}+L_{z}\left[L_{z},L_{x}\right]+\left[L_{z},L_{x}\right]L_{z}\\&=L_{y}\left(-i\hbar L_{z}\right)+\left(-i\hbar L_{z}\right)L_{y}+L_{z}\left(i\hbar L_{y}\right)+\left(i\hbar L_{y}\right)L_{z}\\&=0\end{aligned}}}$$

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

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

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

where ${\displaystyle L_{i}}$ is a component of the classical angular momentum operator, and ${\displaystyle \{,\}}$ is the Poisson bracket.[9]

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

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

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:

$${\displaystyle \sigma _{L_{x}}\sigma _{L_{y}}\geq {\frac {\hbar }{2}}\left|\langle L_{z}\rangle \right|.}$$

where ${\displaystyle \sigma _{X}}$ is the standard deviation in the measured values of X and ${\displaystyle \langle X\rangle }$ denotes the expectation value of X. This inequality is also true if x, y, z are rearranged, or if L is replaced by J or S.

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 ${\displaystyle L_{x}=L_{y}=L_{z}=0}$.

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.

A common way to derive the quantization rules above is the method of ladder operators.[12] The ladder operators for the total angular momentum ${\displaystyle \mathbf {J} =\left(J_{x},J_{y},J_{z}\right)}$ are defined as:

$${\displaystyle {\begin{aligned}J_{+}&\equiv J_{x}+iJ_{y},\\J_{-}&\equiv J_{x}-iJ_{y}\end{aligned}}}$$

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

Derivation of the possible values and quantum numbers for ${\displaystyle J_{z}}$ and ${\displaystyle J^{2}}$.[13]

Let ${\displaystyle \psi ({J^{2}}'J_{z}')}$ be a state function for the system with eigenvalue ${\displaystyle {J^{2}}'}$ for ${\displaystyle J^{2}}$ and eigenvalue ${\displaystyle J_{z}'}$ for ${\displaystyle J_{z}}$.[note 1]

From ${\displaystyle J^{2}=J_{x}^{2}+J_{y}^{2}+J_{z}^{2}}$ is obtained,

$${\displaystyle J_{x}^{2}+J_{y}^{2}=J^{2}-J_{z}^{2}.}$$

Applying both sides of the above equation to ${\displaystyle \psi ({J^{2}}'J_{z}')}$,

$${\displaystyle (J_{x}^{2}+J_{y}^{2})\;\psi ({J^{2}}'J_{z}')=({J^{2}}'-J_{z}'^{2})\;\psi ({J^{2}}'J_{z}').}$$

Since ${\displaystyle J_{x}}$ and ${\displaystyle J_{y}}$ are real observables, ${\displaystyle {J^{2}}'-J_{z}'^{2}}$ is not negative and ${\textstyle |J_{z}'|\leq {\sqrt {{J^{2}}'}}}$. Thus ${\displaystyle J_{z}'}$ has an upper and lower bound.

Two of the commutation relations for the components of ${\displaystyle \mathbf {J} }$ are,

$${\displaystyle [J_{y},J_{z}]=i\hbar J_{x},\;\;[J_{z},J_{x}]=i\hbar J_{y}.}$$

They can be combined to obtain two equations, which are written together using ${\displaystyle \pm }$ signs in the following,

$${\displaystyle J_{z}(J_{x}\pm iJ_{y})=(J_{x}\pm iJ_{y})(J_{z}\pm \hbar ),}$$

where one of the equations uses the ${\displaystyle +}$ signs and the other uses the ${\displaystyle -}$ signs. Applying both sides of the above to ${\displaystyle \psi ({J^{2}}'J_{z}')}$,

$${\displaystyle {\begin{aligned}J_{z}(J_{x}\pm iJ_{y})\;\psi ({J^{2}}'J_{z}')&=(J_{x}\pm iJ_{y})(J_{z}\pm \hbar )\;\psi ({J^{2}}'J_{z}')\\&=(J_{z}'\pm \hbar )(J_{x}\pm iJ_{y})\;\psi ({J^{2}}'J_{z}')\;.\\\end{aligned}}}$$

The above shows that ${\displaystyle (J_{x}\pm iJ_{y})\;\psi ({J^{2}}'J_{z}')}$ are two eigenfunctions of ${\displaystyle J_{z}}$ with respective eigenvalues ${\displaystyle {J_{z}}'\pm \hbar }$ , unless one of the functions is zero, in which case it is not an eigenfunction. For the functions that are not zero,

$${\displaystyle \psi ({J^{2}}'J_{z}'\pm \hbar )=(J_{x}\pm iJ_{y})\;\psi ({J^{2}}'J_{z}').}$$

Further eigenfunctions of ${\displaystyle J_{z}}$ and corresponding eigenvalues can be found by repeatedly applying ${\displaystyle J_{x}\pm iJ_{y}}$ as long as the magnitude of the resulting eigenvalue is ${\displaystyle \leq {\sqrt {{J^{2}}'}}}$. Since the eigenvalues of ${\displaystyle J_{z}}$ are bounded, let ${\displaystyle J_{z}^{0}}$ be the lowest eigenvalue and ${\displaystyle J_{z}^{1}}$ be the highest. Then

$${\displaystyle (J_{x}-iJ_{y})\;\psi ({J^{2}}'J_{z}^{0})=0}$$

and

$${\displaystyle (J_{x}+iJ_{y})\;\psi ({J^{2}}'J_{z}^{1})=0,}$$

since there are no states where the eigenvalue of ${\displaystyle J_{z}}$ is ${\displaystyle <J_{z}^{0}}$ or ${\displaystyle >J_{z}^{1}}$. By applying ${\displaystyle (J_{x}+iJ_{y})}$ to the first equation, ${\displaystyle (J_{x}-iJ_{y})}$ to the second, and using ${\displaystyle J_{x}^{2}+J_{y}^{2}=J^{2}-J_{z}^{2}}$, it can be shown that

$${\displaystyle {J^{2}}'-(J_{z}^{0})^{2}+\hbar J_{z}^{0}=0}$$

and

$${\displaystyle {J^{2}}'-(J_{z}^{1})^{2}-\hbar J_{z}^{1}=0.}$$

Subtracting the first equation from the second and rearranging,

$${\displaystyle (J_{z}^{1}+J_{z}^{0})(J_{z}^{0}-J_{z}^{1}-\hbar )=0.}$$

Since ${\displaystyle J_{z}^{1}\geq J_{z}^{0}}$, the second factor is negative. Then the first factor must be zero and thus ${\displaystyle J_{z}^{0}=-J_{z}^{1}}$.

The difference ${\displaystyle J_{z}^{1}-J_{z}^{0}}$ comes from successive application of ${\displaystyle J_{x}-iJ_{y}}$ or ${\displaystyle J_{x}+iJ_{y}}$ which lower or raise the eigenvalue of ${\displaystyle J_{z}}$ by ${\displaystyle \hbar }$ so that,

$${\displaystyle J_{z}^{1}-J_{z}^{0}=0,\hbar ,2\hbar ,\dots }$$

Let

$${\displaystyle J_{z}^{1}-J_{z}^{0}=2j\hbar ,}$$

where

$${\displaystyle j=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\dots \;.}$$

Then using ${\displaystyle J_{z}^{0}=-J_{z}^{1}}$ and the above,

$${\displaystyle J_{z}^{0}=-j\hbar }$$

and

$${\displaystyle J_{z}^{1}=j\hbar ,}$$

and the allowable eigenvalues of ${\displaystyle J_{z}}$ are

$${\displaystyle J_{z}'=-j\hbar ,-j\hbar +\hbar ,-j\hbar +2\hbar ,\dots ,j\hbar .}$$

Expressing ${\displaystyle J_{z}'}$ in terms of a quantum number ${\displaystyle m_{j}\;}$, and substituting ${\displaystyle J_{z}^{0}=-j\hbar }$ into ${\displaystyle {J^{2}}'-(J_{z}^{0})^{2}+\hbar J_{z}^{0}=0}$ from above,

${\displaystyle {\begin{aligned}J_{z}'&=m_{j}\hbar &m_{j}&=-j,-j+1,-j+2,\dots ,j\\{J^{2}}'&=j(j+1)\hbar ^{2}&j&=0,{\tfrac {1}{2}},1,{\tfrac {3}{2}},\dots \;.\end{aligned}}}$

Since ${\displaystyle \mathbf {S} }$ and ${\displaystyle \mathbf {L} }$ have the same commutation relations as ${\displaystyle \mathbf {J} }$, the same ladder analysis can be applied to them, except that for ${\displaystyle \mathbf {L} }$ there is a further restriction on the quantum numbers that they must be integers.

Traditional derivation of the restriction to integer quantum numbers for ${\displaystyle L_{z}}$ and ${\displaystyle L^{2}}$.[14]

In the Schroedinger representation, the z component of the orbital angular momentum operator can be expressed in spherical coordinates as,[15]

$${\displaystyle L_{z}=-i\hbar {\frac {\partial }{\partial \phi }}.}$$

For ${\displaystyle L_{z}}$ and eigenfunction ${\displaystyle \psi }$ with eigenvalue ${\displaystyle L_{z}'}$,

$${\displaystyle -i\hbar {\frac {\partial }{\partial \phi }}\psi =L_{z}'\psi .}$$

Solving for ${\displaystyle \psi }$,

$${\displaystyle \psi =Ae^{iL_{z}'\phi /\hbar },}$$

where ${\displaystyle A}$ is independent of ${\displaystyle \phi }$. Since ${\displaystyle \psi }$ is required to be single valued, and adding ${\displaystyle 2\pi }$ to ${\displaystyle \phi }$ results in a coordinate for the same point in space,

$${\displaystyle {\begin{aligned}Ae^{iL_{z}'(\phi +2\pi )/\hbar }&=Ae^{iL_{z}'\phi /\hbar },\\e^{iL_{z}'2\pi /\hbar }&=1.\end{aligned}}}$$

Solving for the eigenvalue ${\displaystyle L_{z}'}$,

$${\displaystyle L_{z}'=m_{l}\hbar \;,}$$

where ${\displaystyle m_{l}}$ is an integer.[16] From the above and the relation ${\displaystyle m_{\ell }=-\ell ,(-\ell +1),\ldots ,(\ell -1),\ell \ \ }$, it follows that ${\displaystyle \ell }$ is also an integer. This shows that the quantum numbers ${\displaystyle m_{\ell }}$ and ${\displaystyle \ell }$ for the orbital angular momentum ${\displaystyle \mathbf {L} }$ are restricted to integers, unlike the quantum numbers for the total angular momentum ${\displaystyle \mathbf {J} }$ and spin ${\displaystyle \mathbf {S} }$, which can have half-integer values.[17]

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 ${\displaystyle L_{z}}$ and ${\displaystyle L^{2}}$

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 ${\displaystyle \psi }$ is not observable and only the probability density ${\displaystyle \psi ^{*}\psi }$ is required to be single-valued. The possible double-valued half-integer wave functions have a single-valued probability density.[18] This was recognised by Pauli in 1939 (cited by Japaridze et al[19])

... 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 ${\displaystyle \left|{\tfrac {1}{2}},{\tfrac {1}{2}}\right\rangle =e^{i\phi /2}\sin ^{\frac {1}{2}}\theta }$ and ${\displaystyle \left|{\tfrac {1}{2}},-{\tfrac {1}{2}}\right\rangle =e^{-i\phi /2}\sin ^{\frac {1}{2}}\theta }$.[20][21] These do not behave well under the ladder operators, but have been found to be useful in describing rigid quantum particles[22]

Ballentine[23] 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

$${\displaystyle L_{z}=Q_{x}P_{y}-Q_{y}P_{x}}$$

Define new operators

$${\displaystyle {\begin{aligned}q_{1}&={\frac {Q_{x}+P_{y}}{\sqrt {2}}}&q_{2}&={\frac {Q_{x}-P_{y}}{\sqrt {2}}}\\p_{1}&={\frac {P_{x}-Q_{y}}{\sqrt {2}}}&p_{2}&={\frac {P_{x}+Q_{y}}{\sqrt {2}}}\end{aligned}}}$$

(Dimensional correctness may be maintained by inserting factors of mass and unit angular frequency numerically equal to one.) Then

$${\displaystyle L_{z}={\tfrac {1}{2}}\left(p_{1}^{2}+q_{1}^{2}\right)-{\tfrac {1}{2}}\left(p_{2}^{2}+q_{2}^{2}\right)}$$

But the two terms on the right are just the Hamiltonians for the quantum harmonic oscillator with unit mass and angular frequency

$${\displaystyle {\begin{aligned}H_{1}&={\tfrac {1}{2}}\left(p_{1}^{2}+q_{1}^{2}\right)\\H_{2}&={\tfrac {1}{2}}\left(p_{2}^{2}+q_{2}^{2}\right)\end{aligned}}}$$

and ${\displaystyle L^{2}}$, ${\displaystyle L_{z}}$, ${\displaystyle H_{1}}$ and ${\displaystyle H_{2}}$ all 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[24] can easily be generalised to any number of commuting operators.)

For any of these eigenvectors ${\displaystyle \psi }$ with

$${\displaystyle {\begin{aligned}L^{2}\psi &=\hbar \ell (\ell +1)\psi \\L_{z}\psi &=\hbar m\psi \\H_{1}\psi &=\hbar \left(n_{1}+{\tfrac {1}{2}}\right)\psi \\H_{2}\psi &=\hbar \left(n_{2}+{\tfrac {1}{2}}\right)\psi \end{aligned}}}$$

for some integers ${\displaystyle n_{1},n_{2}}$, we find

$${\displaystyle m=\left(n_{1}+{\tfrac {1}{2}}\right)-\left(n_{2}+{\tfrac {1}{2}}\right)=n_{1}-n_{2}}$$

As a difference of two integers, ${\displaystyle m}$ must be an integer, from which ${\displaystyle \ell }$ is also integral.

A more complex version of this argument using the ladder operators of the quantum harmonic oscillator has been given by Buchdahl.[25]

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 ${\displaystyle \ell =2}$, and ${\displaystyle m_{\ell }=-2,-1,0,1,2}$ for the five cones from bottom to top. Since ${\displaystyle |L|={\sqrt {L^{2}}}=\hbar {\sqrt {6}}}$, the vectors are all shown with length ${\displaystyle \hbar {\sqrt {6}}}$. The rings represent the fact that ${\displaystyle L_{z}}$ is known with certainty, but ${\displaystyle L_{x}}$ and ${\displaystyle L_{y}}$ 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 ${\displaystyle \ell }$ and ${\displaystyle m_{\ell }}$ could be somewhere on this cone while it cannot be defined for a single system (since the components of ${\displaystyle L}$ do not commute with each other).

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 ${\displaystyle L_{z}/\hbar }$ 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.

Although one might expect ${\displaystyle R\left({\hat {n}},360^{\circ }\right)=1}$ (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.), ${\displaystyle R\left({\hat {n}},360^{\circ }\right)=-1}$, and when it is an integer, ${\displaystyle R\left({\hat {n}},360^{\circ }\right)=+1}$.[6] 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°.)[6]

On the other hand, ${\displaystyle R_{\text{spatial}}\left({\hat {n}},360^{\circ }\right)=+1}$ 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 ${\displaystyle R_{\text{spatial}}}$ operators carry the structure of SO(3), while ${\displaystyle R}$ and ${\displaystyle R_{\text{internal}}}$ carry the structure of SU(2).

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

$${\displaystyle e^{-2\pi im}=1}$$

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

Starting with a certain quantum state ${\displaystyle |\psi _{0}\rangle }$, consider the set of states ${\displaystyle R\left({\hat {n}},\phi \right)\left|\psi _{0}\right\rangle }$ for all possible ${\displaystyle {\hat {n}}}$ and ${\displaystyle \phi }$, 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.

From the relation between J and rotation operators,

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

(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).

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.[6]

(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)?")