Angular Momentum Spectrum from Commutators

Derivation

Angular Momentum Spectrum from Commutators

D-215 Home PU-301 Threads symmetry · matter Depends on commutator-algebra-canonical, Spectral Theorem for Hermitian Observables

Statement

Assuming only the angular-momentum algebra \(\left[\hat{J}_i,\hat{J}_j\right]=i\hbar\,\varepsilon_{ijk}\hat{J}_k\) and self-adjointness of the components, the simultaneous eigenvalues of the commuting pair \(\left(\hat{J}^2,\hat{J}_z\right)\) are \(\hat{J}^2\to\hbar^2 j(j+1)\) and \(\hat{J}_z\to\hbar m\), where \(j\in\{0,\tfrac12,1,\tfrac32,\dots\}\) and \(m\) runs in unit steps from \(-j\) to \(+j\); no property of orbital motion, coordinates, or wavefunctions is used.

Why it matters

Every quantum theory of rotation — atomic fine structure, nuclear spin, the classification of elementary particles by spin — rests on this single algebraic result. It shows that the discreteness and half-integer spectrum of angular momentum are dictated by the Lie algebra of \(\mathrm{SO}(3)\) (more precisely \(\mathrm{SU}(2)\)) alone, not by any specific dynamics or potential.

The method — pick a maximal commuting set, build ladder operators from the remaining generators, and let normalisability truncate the ladder — is the archetype for representation-theoretic spectra throughout physics, reused verbatim for the harmonic oscillator, the hydrogen \(\mathrm{SO}(4)\) symmetry, and \(\mathrm{SU}(3)\) flavour.

Assumptions

The three components close under the \(\mathrm{SO}(3)\) bracket \(\left[\hat{J}_i,\hat{J}_j\right]=i\hbar\,\varepsilon_{ijk}\hat{J}_k\).Without closure there is no \(\hat{J}^2\) that commutes with every component, so no common eigenbasis exists and the ladder construction never starts.

Each \(\hat{J}_i\) is self-adjoint, so \(\hat{J}^2=\sum_i\hat{J}_i^2\) is self-adjoint and positive.If the components are not Hermitian the eigenvalues can be complex and \(\langle\hat{J}^2\rangle\ge 0\) fails, destroying the bound that truncates the ladder.

The representation acts on a Hilbert space where \(\hat{J}^2\) has at least one normalisable eigenvector (states have finite norm).On a space admitting infinite-norm or non-terminating states the top and bottom rungs need not exist, and the quantisation \(j=n/2\) is lost — this is the rigour hinge of the whole argument.

Derivation

\[ \left[\hat{J}^2,\hat{J}_i\right]=\sum_k\left[\hat{J}_k^2,\hat{J}_i\right]=\sum_k\left(\hat{J}_k\left[\hat{J}_k,\hat{J}_i\right]+\left[\hat{J}_k,\hat{J}_i\right]\hat{J}_k\right)=0 \]

Using the Leibniz rule for commutators and the total antisymmetry of \(\varepsilon_{kij}\), the sum cancels pairwise; \(\hat{J}^2\) is a Casimir. A

\[ \hat{J}^2\lvert\lambda,\mu\rangle=\hbar^2\lambda\,\lvert\lambda,\mu\rangle,\qquad \hat{J}_z\lvert\lambda,\mu\rangle=\hbar\mu\,\lvert\lambda,\mu\rangle \]

Since \(\left[\hat{J}^2,\hat{J}_z\right]=0\) and both are self-adjoint, the spectral theorem provides a common orthonormal eigenbasis; \(\hbar\) is inserted so \(\lambda,\mu\) are dimensionless. A

\[ \hat{J}_\pm\equiv\hat{J}_x\pm i\hat{J}_y,\qquad \hat{J}_\pm^\dagger=\hat{J}_\mp \]

Define ladder operators as the complex combinations that diagonalise the adjoint action of \(\hat{J}_z\); Hermiticity of \(\hat{J}_x,\hat{J}_y\) gives the stated conjugation. A

\[ \left[\hat{J}_z,\hat{J}_\pm\right]=\left[\hat{J}_z,\hat{J}_x\right]\pm i\left[\hat{J}_z,\hat{J}_y\right]=i\hbar\hat{J}_y\pm i(-i\hbar\hat{J}_x)=\pm\hbar\hat{J}_\pm \]

Direct substitution of the algebra; the eigen-equation \(\left[\hat{J}_z,\hat{J}_\pm\right]=\pm\hbar\hat{J}_\pm\) is what makes \(\hat{J}_\pm\) a raising/lowering operator. A

\[ \hat{J}_z\left(\hat{J}_\pm\lvert\lambda,\mu\rangle\right)=\left(\hat{J}_\pm\hat{J}_z\pm\hbar\hat{J}_\pm\right)\lvert\lambda,\mu\rangle=\hbar(\mu\pm1)\left(\hat{J}_\pm\lvert\lambda,\mu\rangle\right) \]

Rearranging step 4 as \(\hat{J}_z\hat{J}_\pm=\hat{J}_\pm\hat{J}_z\pm\hbar\hat{J}_\pm\) and acting on the eigenstate: \(\hat{J}_\pm\) shifts \(\mu\) by \(\pm1\). Since \(\left[\hat{J}^2,\hat{J}_\pm\right]=0\), \(\lambda\) is unchanged. A

\[ \hat{J}_\mp\hat{J}_\pm=\hat{J}_x^2+\hat{J}_y^2\pm i\left[\hat{J}_x,\hat{J}_y\right]=\hat{J}^2-\hat{J}_z^2\mp\hbar\hat{J}_z \]

Expanding the product and using \(\left[\hat{J}_x,\hat{J}_y\right]=i\hbar\hat{J}_z\); this identity converts operator norms into eigenvalues of the commuting pair. B

\[ \bigl\lVert\hat{J}_\pm\lvert\lambda,\mu\rangle\bigr\rVert^2=\langle\lambda,\mu\rvert\hat{J}_\mp\hat{J}_\pm\lvert\lambda,\mu\rangle=\hbar^2\left(\lambda-\mu^2\mp\mu\right)\ge0 \]

Using \(\hat{J}_\pm^\dagger=\hat{J}_\mp\) (step 3) the left side is a squared norm, hence non-negative; substituting step 6 gives the inequality \(\lambda\ge\mu^2\pm\mu\). BC

\[ \lambda\ge\mu(\mu+1)\ \text{and}\ \lambda\ge\mu(\mu-1)\ \Rightarrow\ -\tfrac{1}{2}\!\left(1+\sqrt{1+4\lambda}\right)\le\mu\le\tfrac{1}{2}\!\left(-1+\sqrt{1+4\lambda}\right) \]

For fixed \(\lambda\ge0\), the two quadratic inequalities bound \(\mu\) both above and below; the ladder cannot run forever. C

\[ \hat{J}_+\lvert\lambda,\mu_{\max}\rangle=0\ \Rightarrow\ \lambda=\mu_{\max}(\mu_{\max}+1);\qquad \hat{J}_-\lvert\lambda,\mu_{\min}\rangle=0\ \Rightarrow\ \lambda=\mu_{\min}(\mu_{\min}-1) \]

Because \(\mu\) is bounded, there must exist a top state annihilated by \(\hat{J}_+\) and a bottom state annihilated by \(\hat{J}_-\); setting the norm in step 7 to zero fixes \(\lambda\) in terms of each extreme. C

\[ \mu_{\max}(\mu_{\max}+1)=\mu_{\min}(\mu_{\min}-1)\ \Rightarrow\ \mu_{\min}=-\mu_{\max} \]

Equating the two expressions for \(\lambda\) and discarding the unphysical root \(\mu_{\min}=\mu_{\max}+1\) (which would put the bottom above the top). Write \(j\equiv\mu_{\max}\). C

\[ \mu_{\max}-\mu_{\min}=2j=N\in\{0,1,2,\dots\}\ \Rightarrow\ j=\frac{N}{2},\qquad \lambda=j(j+1) \]

Applying \(\hat{J}_+\) an integer number \(N\) of times steps from bottom to top in unit increments (step 5), so the total span \(2j\) is a non-negative integer; hence \(j\) is integer or half-integer. C

Result

\[ \hat{J}^2\lvert j,m\rangle=\hbar^2\,j(j+1)\,\lvert j,m\rangle,\qquad \hat{J}_z\lvert j,m\rangle=\hbar\,m\,\lvert j,m\rangle \]\[ j\in\Bigl\{0,\tfrac12,1,\tfrac32,\dots\Bigr\},\qquad m\in\{-j,-j+1,\dots,j-1,j\} \]\[ \hat{J}_\pm\lvert j,m\rangle=\hbar\sqrt{j(j+1)-m(m\pm1)}\;\lvert j,m\pm1\rangle \]

Reading. The magnitude of angular momentum is quantised as \(\lvert\vec{J}\rvert=\hbar\sqrt{j(j+1)}\), and its projection on any chosen axis takes the \(2j+1\) equally spaced values \(\hbar m\). The strict inequality \(j(j+1)>j^2\) means the vector can never point exactly along \(z\) — a purely algebraic statement of the uncertainty between components. The half-integer branch, forbidden for orbital motion, is what the algebra permits and what spin realises.

Units check. \(\hat{J}_i\) carries the dimension of action, \(\mathrm{J\,s}=\mathrm{kg\,m^2\,s^{-1}}\), the same as \(\hbar\); \(j(j+1)\) and \(m\) are dimensionless, so \(\hbar^2 j(j+1)\) has units of (action)\(^2\), matching \(\hat{J}^2\), and \(\hbar m\) has units of action, matching \(\hat{J}_z\). The matrix element coefficient \(\hbar\sqrt{\cdots}\) carries one power of action, as \(\hat{J}_\pm\) must.

Limiting cases

\(j=0\): a single non-degenerate state \(\lvert0,0\rangle\), rotationally invariant, annihilated by all \(\hat{J}_i\) — the scalar (spin-0) representation.
\(j=\tfrac12\): two states \(m=\pm\tfrac12\); \(\hat{J}_i=\tfrac{\hbar}{2}\sigma_i\) reproduces the Pauli algebra, the fundamental representation of \(\mathrm{SU}(2)\).
Large \(j\): \(\sqrt{j(j+1)}\to j+\tfrac12\approx j\), and the \(2j+1\) closely spaced projections approach a continuous classical cone — the correspondence limit.
Integer \(j\): coincides with the orbital spectrum \(\ell=0,1,2,\dots\) obtained from single-valued spherical harmonics.

Breaks when

The state space contains no normalisable extreme rung — e.g. a non-unitary or infinite-dimensional representation of \(\mathrm{sl}(2,\mathbb{C})\) — then step 7's non-negativity bound is unavailable, the ladder does not terminate, and \(j\) is not quantised (continuous or complex "spin" appears).
The rotation generators fail to close on \(\mathrm{SO}(3)\), as when a magnetic monopole or non-abelian gauge field adds a term so that \(\left[\hat{J}_i,\hat{J}_j\right]=i\hbar\varepsilon_{ijk}(\hat{J}_k-\text{extra})\); the Casimir is modified and the spectrum shifts (monopole harmonics start at \(j=\lvert q\rvert\), not \(0\)).
Relativistic settings where boosts mix with rotations: the full Lorentz algebra \(\mathrm{so}(3,1)\) is non-compact, its finite-dimensional reps are non-unitary, and the compact-group truncation argument no longer forces real half-integer \(j\) for the boost sector.

Failure modes

Writing \(\hat{J}^2=\hbar^2 j^2\) instead of \(\hbar^2 j(j+1)\) — forgetting the \(+j\) from the \(\mp\hbar\hat{J}_z\) term in step 6.
Concluding \(j\) must be an integer by importing the orbital single-valuedness argument; the pure algebra permits half-integers, and discarding them is an error.
Taking \(\hat{J}_\pm\) to be Hermitian; they are not (\(\hat{J}_\pm^\dagger=\hat{J}_\mp\)), which breaks the norm calculation in step 7.
Assuming \(\hat{J}_+\lvert j,j\rangle\) is nonzero and normalising it — the top state is annihilated, and dividing by its zero norm gives nonsense.
Keeping the spurious root \(\mu_{\min}=\mu_{\max}+1\) in step 10, which places the ladder's floor above its ceiling.
Confusing the label \(m\) (eigenvalue of \(\hat{J}_z\)) with a mass or with the magnetic quantum number of a specific \(\ell\); here it is defined solely by the algebra.

Discussion

The result is a statement about a Lie algebra, not about space. Nowhere did we invoke coordinates, a wavefunction, or a Hamiltonian; the only inputs were the structure constants \(\varepsilon_{ijk}\) and Hermiticity. This is why the same numbers govern the intrinsic spin of an electron, which has no orbital wavefunction analogue. The double-valued (half-integer) representations are single-valued reps of the covering group \(\mathrm{SU}(2)\), and their existence is the algebraic root of the spin-statistics distinction and of the sign a fermion picks up under a \(2\pi\) rotation.

The dimension count \(2j+1\) is the dimension of the irreducible representation labelled by \(j\); the Casimir eigenvalue \(j(j+1)\) is what labels the irrep invariantly. Adding two angular momenta corresponds to decomposing the tensor product \(\mathbf{j_1}\otimes\mathbf{j_2}=\bigoplus_{J=\lvert j_1-j_2\rvert}^{j_1+j_2}\mathbf{J}\), the Clebsch-Gordan series — again fixed entirely by the algebra derived here.

Physically, the impossibility of a state with \(\vec{J}\) exactly along \(z\) (since \(m_{\max}=j<\sqrt{j(j+1)}\)) is the geometric face of the commutator \(\left[\hat{J}_x,\hat{J}_y\right]\ne0\): the transverse components cannot simultaneously vanish. In the vector model the angular momentum precesses on a cone of half-angle \(\cos\theta=m/\sqrt{j(j+1)}\), which only in the \(j\to\infty\) limit closes onto the axis.

The deeper structure is that \(\hat{J}^2\) generates the centre of the universal enveloping algebra \(\mathcal{U}(\mathrm{su}(2))\): it is the unique (up to scaling) quadratic Casimir, and Schur's lemma guarantees it acts as a scalar on each irrep. The ladder argument is precisely the highest-weight construction of representation theory — \(\lvert j,j\rangle\) is the highest-weight vector, \(\hat{J}_-\) generates the weight string, and the finite-dimensionality of the compact group \(\mathrm{SU}(2)\) is what forces the string to terminate. Generalised to any semisimple Lie algebra, this becomes the theorem of the highest weight, with \(2j\in\mathbb{Z}_{\ge0}\) replaced by dominant integral weights.

Common misconceptions. Quantisation here is not imposed by a boundary condition on a wavefunction; it emerges from requiring finite-norm states in a representation of a compact group. And angular momentum is not "an integer number of \(\hbar\)": the magnitude is \(\hbar\sqrt{j(j+1)}\), an irrational multiple of \(\hbar\) for every \(j>0\).

Worked examples

Spin-\(\tfrac12\): construct \(\hat{J}_+\) matrix element and verify \(\hat{J}^2\).

Take \(j=\tfrac12\), states \(m=+\tfrac12,-\tfrac12\). Compute the raising coefficient and the Casimir eigenvalue numerically. A

\[ \hat{J}_+\lvert\tfrac12,-\tfrac12\rangle=\hbar\sqrt{j(j+1)-m(m+1)}\;\lvert\tfrac12,+\tfrac12\rangle \]

Substitute the ladder formula with symbols before numbers. A

\[ =\hbar\sqrt{\tfrac12\!\cdot\!\tfrac32-\left(-\tfrac12\right)\!\left(\tfrac12\right)}\;\lvert\tfrac12,+\tfrac12\rangle=\hbar\sqrt{\tfrac34+\tfrac14}\;\lvert\tfrac12,+\tfrac12\rangle=\hbar\,\lvert\tfrac12,+\tfrac12\rangle \]

Insert numbers: \(j(j+1)=\tfrac34\), \(m(m+1)=-\tfrac14\); the coefficient is \(\hbar\), matching \(\hat{J}_+=\hbar\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)\). A

\[ \hat{J}^2=\hbar^2 j(j+1)=\hbar^2\cdot\tfrac34=\tfrac34\hbar^2 \]

The Casimir on both states; with \(\hbar=1.055\times10^{-34}\,\mathrm{J\,s}\), \(\hat{J}^2=8.35\times10^{-69}\,\mathrm{J^2\,s^2}\). A

\[ \hat{J}_+\lvert\tfrac12,-\tfrac12\rangle=\hbar\,\lvert\tfrac12,+\tfrac12\rangle,\qquad \lvert\vec J\rvert=\tfrac{\sqrt3}{2}\hbar=9.14\times10^{-35}\,\mathrm{J\,s} \]

Reading. The single raising step reaches the top rung with coefficient exactly \(\hbar\), and the spin magnitude \(\tfrac{\sqrt3}{2}\hbar\) exceeds its maximal projection \(\tfrac12\hbar\), as required.

Units check. \(\hbar\) has units \(\mathrm{J\,s}\); \(\hat{J}^2=\tfrac34\hbar^2\) has \(\mathrm{J^2 s^2}\). Consistent.

Spin-1 (\(j=1\)): projection cone half-angle and top-rung annihilation.

Take \(j=1\), states \(m=-1,0,+1\). Find the smallest angle between \(\vec J\) and the \(z\)-axis, and confirm \(\hat J_+\) annihilates the top. B

\[ \cos\theta_{\min}=\frac{m_{\max}}{\sqrt{j(j+1)}}=\frac{j}{\sqrt{j(j+1)}} \]

The tightest cone uses the largest projection \(m=j\); symbols first. B

\[ \cos\theta_{\min}=\frac{1}{\sqrt{1\cdot2}}=\frac{1}{\sqrt2}=0.7071\ \Rightarrow\ \theta_{\min}=45.0^\circ \]

Insert \(j=1\): magnitude \(\sqrt2\,\hbar\), projection \(1\,\hbar\). B

\[ \hat J_+\lvert1,1\rangle=\hbar\sqrt{j(j+1)-m(m+1)}\;\lvert1,2\rangle=\hbar\sqrt{2-1\cdot2}\;\lvert1,2\rangle=0 \]

The radicand vanishes at \(m=j\), confirming the ladder terminates — no \(m=2\) state exists. B

\[ \theta_{\min}=45.0^\circ,\qquad \hat J_+\lvert1,1\rangle=0,\qquad \lvert\vec J\rvert=\sqrt2\,\hbar=1.49\times10^{-34}\,\mathrm{J\,s} \]

Reading. Even in its most-aligned state the spin-1 vector sits \(45^\circ\) off the axis; the algebra self-consistently caps the ladder at \(m=1\).

Units check. \(\theta\) is dimensionless (an angle); \(\lvert\vec J\rvert=\sqrt2\,\hbar\) carries units \(\mathrm{J\,s}\), matching action. Consistent.

Problems

For \(j=\tfrac32\), list all allowed \(m\) and evaluate \(\hat J_-\lvert\tfrac32,\tfrac12\rangle\).
Solution
Allowed \(m=-\tfrac32,-\tfrac12,+\tfrac12,+\tfrac32\) (four states, \(2j+1=4\)). The lowering coefficient is \(\hbar\sqrt{j(j+1)-m(m-1)}\) with \(j=\tfrac32,\ m=\tfrac12\): \(j(j+1)=\tfrac{15}{4}\), \(m(m-1)=\tfrac12\cdot(-\tfrac12)=-\tfrac14\), so the radicand is \(\tfrac{15}{4}+\tfrac14=4\). Hence \(\hat J_-\lvert\tfrac32,\tfrac12\rangle=2\hbar\,\lvert\tfrac32,-\tfrac12\rangle\).
Show that \(\hat J_+\hat J_-+\hat J_-\hat J_+=2(\hat J^2-\hat J_z^2)\) and use it to find \(\langle\hat J_x^2\rangle\) in the state \(\lvert j,m\rangle\).
Solution
From step 6, \(\hat J_-\hat J_+=\hat J^2-\hat J_z^2-\hbar\hat J_z\) and \(\hat J_+\hat J_-=\hat J^2-\hat J_z^2+\hbar\hat J_z\); adding gives \(2(\hat J^2-\hat J_z^2)\). Also \(\hat J_+\hat J_-+\hat J_-\hat J_+=2(\hat J_x^2+\hat J_y^2)\). By symmetry \(\langle\hat J_x^2\rangle=\langle\hat J_y^2\rangle=\tfrac12\langle\hat J_x^2+\hat J_y^2\rangle=\tfrac12\langle\hat J^2-\hat J_z^2\rangle=\tfrac{\hbar^2}{2}\left[j(j+1)-m^2\right]\).
A system has \(\hat J^2\) eigenvalue \(12\hbar^2\). Find \(j\), the number of \(m\)-states, and \(\lvert\vec J\rvert\).
Solution
Solve \(j(j+1)=12\Rightarrow j^2+j-12=0\Rightarrow j=\tfrac{-1+\sqrt{49}}{2}=3\). Number of states \(2j+1=7\). Magnitude \(\lvert\vec J\rvert=\hbar\sqrt{12}=2\sqrt3\,\hbar\approx3.46\hbar=3.65\times10^{-34}\,\mathrm{J\,s}\).
Prove the two solutions of \(\mu_{\max}(\mu_{\max}+1)=\mu_{\min}(\mu_{\min}-1)\) are \(\mu_{\min}=-\mu_{\max}\) and \(\mu_{\min}=\mu_{\max}+1\), and explain why the second is rejected.
Solution
Write \(a=\mu_{\max},b=\mu_{\min}\): \(a^2+a=b^2-b\Rightarrow a^2-b^2+a+b=0\Rightarrow(a+b)(a-b)+(a+b)=0\Rightarrow(a+b)(a-b+1)=0\). Thus \(b=-a\) or \(b=a+1\). Since by construction \(\mu_{\min}\le\mu_{\max}\), i.e. \(b\le a\), the root \(b=a+1>a\) is impossible; only \(b=-a\) survives, giving the symmetric ladder from \(-j\) to \(+j\).
An electron (\(s=\tfrac12\)) sits in a \(0.50\,\mathrm{T}\) field along \(z\); the interaction is \(\hat H=-\gamma\hbar\,\hat J_z/\hbar\cdot B\) with \(\gamma=1.76\times10^{11}\,\mathrm{s^{-1}T^{-1}}\) (electron gyromagnetic ratio). Compute the energy splitting between \(m=\pm\tfrac12\) and the Larmor frequency.
Solution
The two levels are \(E_m=-\gamma\hbar B\,m\) with \(m=\pm\tfrac12\). Splitting \(\Delta E=E_{-1/2}-E_{+1/2}=\gamma\hbar B\,[\tfrac12-(-\tfrac12)]=\gamma\hbar B\). Numerically \(\Delta E=(1.76\times10^{11})(1.055\times10^{-34})(0.50)=9.28\times10^{-24}\,\mathrm{J}=5.79\times10^{-5}\,\mathrm{eV}\). Larmor frequency \(f=\Delta E/h=\gamma B/2\pi=(1.76\times10^{11})(0.50)/(2\pi)=1.40\times10^{10}\,\mathrm{Hz}\approx14.0\,\mathrm{GHz}\). The two-fold splitting is a direct manifestation of the \(2j+1=2\) spectrum derived here.