Eigenvalues of L² and L_z

We now derive the central result of angular-momentum theory: the allowed eigenvalues of $\hat{L}^2$ and $\hat{L}_z$ . The argument uses only the commutation algebra and the fact that the ladder must terminate — no differential equations required.

Setting up the eigenvalue problem

Because $\hat{L}^2$ and $\hat{L}_z$ commute, they share simultaneous eigenstates. Label them $|\lambda, \mu\rangle$ by their eigenvalues,

\hat{L}^2|\lambda,\mu\rangle = \hbar^2\lambda\,|\lambda,\mu\rangle, \qquad \hat{L}_z|\lambda,\mu\rangle = \hbar\mu\,|\lambda,\mu\rangle,

with $\lambda, \mu$ dimensionless. Since $\hat{L}^2 - \hat{L}_z^2 = \hat{L}_x^2 + \hat{L}_y^2$ is a sum of squares of Hermitian operators, its expectation value is non-negative, giving the bound

\lambda \geq \mu^2.

So for fixed $\lambda$ the value of $\mu$ cannot grow without limit — there must be a largest and a smallest $\mu$ .

The ladder must terminate

Let $\mu_{\max}$ be the top rung. Then $\hat{L}_+|\lambda,\mu_{\max}\rangle = 0$ , otherwise it would produce a state with larger $\mu$ . Using $\hat{L}_-\hat{L}_+ = \hat{L}^2 - \hat{L}_z^2 - \hbar \hat{L}_z$ and acting on the top state gives

0 = \hbar^2\big(\lambda - \mu_{\max}^2 - \mu_{\max}\big) \;\Longrightarrow\; \lambda = \mu_{\max}(\mu_{\max}+1).

Similarly, the bottom rung $\mu_{\min}$ satisfies $\hat{L}_+\hat{L}_-|\lambda,\mu_{\min}\rangle = 0$ , which gives $\lambda = \mu_{\min}(\mu_{\min}-1)$ . Equating the two expressions for $\lambda$ yields $\mu_{\min} = -\mu_{\max}$ .

Quantization

Starting at $\mu_{\min} = -\mu_{\max}$ and applying $\hat{L}_+$ repeatedly must land exactly on $\mu_{\max}$ after an integer number $N$ of steps:

\mu_{\max} = \mu_{\min} + N = -\mu_{\max} + N \;\Longrightarrow\; \mu_{\max} = \frac{N}{2}, \quad N = 0, 1, 2, \dots

Defining $\ell \equiv \mu_{\max}$ , this means $\ell$ can take the values $0, \tfrac{1}{2}, 1, \tfrac{3}{2}, 2, \dots$ . The eigenvalues are therefore

\boxed{\;\hat{L}^2|\ell,m\rangle = \hbar^2\,\ell(\ell+1)\,|\ell,m\rangle, \qquad \hat{L}_z|\ell,m\rangle = \hbar\,m\,|\ell,m\rangle\;}

with $m$ running in unit steps from $-\ell$ to $+\ell$ .

Half-integers and orbital motion

The algebra permits both integer and half-integer $\ell$ . Orbital angular momentum, tied to a single-valued spatial wavefunction $e^{im\phi}$ , allows only integer $\ell$ . Half-integer values are realized by intrinsic spin, which has no spatial wavefunction to constrain it — a distinction we explore later in the module.

Try it

For the $d$ -orbital multiplet $\ell = 2$ , compute the eigenvalue of $\hat{L}^2$ in units of $\hbar^2$ (that is, $\ell(\ell+1)$ ) and return it.

Run your code to see the quantum state.