The Total Angular Momentum L²

The individual components $L_x, L_y, L_z$ cannot be known simultaneously. But there is one more natural observable: the total angular momentum squared, and remarkably it can be measured together with any single component.

Definition

The total angular momentum operator is

It is Hermitian (a sum of squares of Hermitian operators) and positive, so its eigenvalues are real and non-negative. Physically $\hat{L}^2$ measures the magnitude of the angular-momentum vector, independent of its orientation.

L² commutes with every component

The key theorem of this lesson is

To prove $[\hat{L}^2, \hat{L}_z] = 0$, expand using $\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$. The $\hat{L}_z^2$ term obviously commutes with $\hat{L}_z$. For the others, use the identity $[A^2, B] = A[A,B] + [A,B]A$ together with the fundamental relations $[\hat{L}_x, \hat{L}_z] = -i\hbar\hat{L}_y$ and $[\hat{L}_y, \hat{L}_z] = i\hbar\hat{L}_x$:

These two terms are equal and opposite, so they cancel exactly, leaving $[\hat{L}^2, \hat{L}_z] = 0$. The cyclic symmetry gives the same result for $\hat{L}_x$ and $\hat{L}_y$.

Why this matters

A complete description of an angular-momentum state needs a complete set of commuting observables. Since the three components mutually fail to commute, we cannot use all of them. Instead we use the largest commuting set: $\{\hat{L}^2, \hat{L}_z\}$. This is exactly why atomic states are labeled by $\ell$ (from $\hat{L}^2$) and $m$ (from $\hat{L}_z$) and never by all three components at once.

A concrete check with spin-1/2

For spin-1/2 with $\hbar = 1$, the operator $S^2 = S_x^2 + S_y^2 + S_z^2$ evaluates to $\tfrac{3}{4}\,\mathbb{1}$, a multiple of the identity. Any operator commutes with the identity, so $[S^2, S_z] = 0$ trivially. In the exercise you will compute this commutator matrix and confirm it is the zero matrix.

Try it

Build $S^2$ for spin-1/2 and compute $[S^2, S_z]$. Return the sum of the absolute values of its entries; it should be $0$, demonstrating that $L^2$ commutes with $L_z$.

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