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I'm in a quantum mechanics class, and it is given in the book that the operators $\hat{L^{2}}$ and $\hat{H}$ commute for the 3D Harmonic Oscillator, but no definite mathematical proof is given, and I'm having a hard time proving it myself, and conceptualizing why this must be true.

I've been trying to use spherical coordinate to prove it, and I know that in spherical coordinates $$\hat{L^{2}}= \frac{-h^{2}}{4\pi^{2}}(\frac{1}{\sin\theta}\frac{d}{d\theta}(\sin\theta \frac{d}{d\theta})+\frac{1}{\sin^{2}\theta}\frac{d^{2}}{d\phi^{2}})$$

And $$\hat{H}=\frac{-h^{2}}{4\pi^{2}(2m)}\Delta+\frac{1}{2}kr^{2}.$$

I've been trying to prove it using the very basic $[\hat{L^{2}},\hat{H}]f= \hat{L^{2}}\hat{H}f-\hat{H}\hat{L^{2}}f$ method of showing the commutation relationship. My first thought was that by applying $\hat{L^{2}}$ to $\hat{H}$ all the terms that are dependent on r or $\frac{r}{dr}$ would disappear, but if some arbitrary function f had cross terms, this isn't necessarily true, and the algebra got pretty messy after that. Is there a better way to prove this?

Qmechanic
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Mecury-197
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2 Answers2

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You want to show that $[L^2,H]=0$. There are some ways to do this. The easiest and more direct one is to notice that in spherical coordinates

$$H = -\dfrac{\hbar^2}{2m}\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}r + \dfrac{1}{2mr^2}L^2+ V(r), $$

where $V(r)$ is the potential energy you are considering. This is easy to see, you just substitute the laplacian in spherical coordinates and you'll see that the $L^2$ term appears naturally. In that case it's obvious the operators commute. Recall that $L^2$ just affects angular coordinates, for the purpose of $L^2$ anything $r$-related is a constant. In that case we have that

$$[L^2,V(r)]f = L^2V(r)f - V(r)L^2f = V(r)L^2f-V(r)L^2f = 0.$$

The same happens to the other term involving just $r$ in the $H$ operator. The other piece is obvious also because $[L^2,L^2]=0$. In that case $[L^2,H]=0$.

EDIT: The first term commutes with $L^2$ because it just involves operations over $r$. I think you can see it better applying the commutator to a function

$$[L^2,\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}r]f =L^2\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}(rf)-\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}(rL^2f),$$

Now look that since $L^2$ do not act on the $r$-dependence we get $rL^2f = L^2(rf)$. Concerning the derivative: the operator $\partial^2/\partial r^2$ does not act on angular variables. Because of that we can exchange it's order with $L^2$. This gives us

$$[L^2,\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}r]f =L^2\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}(rf)-\dfrac{1}{r}L^2\dfrac{\partial^2}{\partial r^2}(rf),$$

and again because $L^2$ doesn't act on $r$-dependence you can bring the $1/r$ inside. This leaves us with

$$ [L^2,\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}r]f =L^2\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}(rf)-L^2\dfrac{1}{r}\dfrac{\partial^2}{\partial r^2}(rf)=0.$$

Gold
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As an alternative proof, let $\mathcal D (R)$ be the unitary operator associated with a tridimensional rotation $R$ and define a vector operator $\mathbf A$ by $$\mathcal D (R)^\dagger \mathbf A \mathcal D (R)=R\mathbf A,$$ or $$A_i'\equiv\mathcal D (R)^\dagger A_i \mathcal D (R)=R_{ij} A_j.$$ Note that, since $R$ is orthogonal: $$A_i'A_i'=R_{ij}A_jR_{ik}A_k=A_jA_j,$$ that is $$\mathcal D (R)^\dagger \mathbf A ^2 \mathcal D (R)=(D (R)^\dagger \mathbf A \mathcal D (R))\cdot (D (R)^\dagger \mathbf A \mathcal D (R)) = \mathbf A^2$$ (this is obvious).

Since $\mathcal D (R)$ commutes with $\mathbf A ^2$ for every $R$, it follows that the angular momentum itself commutes with $\mathbf A ^2$.

Note now that $\mathbf p = - i\nabla$ is a vector operator. The rotation operator is realized in the $L^2$ Hilbert space of a single spinless particle by $$\mathcal D (R)\psi (\mathbf x )=\psi (R^{-1}\mathbf x ),$$ so this is the ordinary statement that the gradient of a scalar function is a vector. It follows at once that the kinetic energy operator $T=\frac{\mathbf p ^2}{2m}$ commutes with the angular momentum. Since a spherically symmetric potential also commutes with every rotation, hence with the angular momentum, it follows that the full hamiltonian commutes with the angular momentum.

Note: likely you won't find this derivation immediately useful. The point I want to make is that the commutation relations of an operator like $\mathbf J$ which generates some symmetry operation, can be easily read off from the definition of the symmetry. An easier example is that of translations: the definition of the translation operator $\mathcal T(a)^\dagger q\mathcal T(a)=q+a$, together with $T(a)=e^{-ipa}$ immediately gives the canonical commutation relations $[q,p]=i$; just differentiate by $a$.

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