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?

  • $\begingroup$ Use the identity for commutators $[AB, C] = A[B,C] + [A,C]B$, then find $[L_i, H]$. This is easier in Cartesian coordinates, unless you can come up with a clever argument as to why you only need to look at $[L_z, H]$ (there is such an argument). $\endgroup$ – Robin Ekman Feb 22 '16 at 2:06
  • $\begingroup$ The correct argument to use, however, is this. 1. $L_i$ is the generator of rotations. 2. $x^2$ and $p^2$ are scalars, i.e., invariant under rotations. 3. Therefore $[L_i, x^2] = [L_i, p^2] = 0$. $\endgroup$ – Robin Ekman Feb 22 '16 at 2:26

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

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

  • $\begingroup$ Thank you for the comprehensive explanation, and I can see why the the second two terms of $\hat{H}$commute with $\hat{L^{2}}$ but I still can't see why the first one does. Can you explain that in more detail? I don't see why that therm is entirely r- dependent if the function has cross terms with r and an angular component. $\endgroup$ – Mecury-197 Feb 22 '16 at 3:03
  • $\begingroup$ I've added one edit. I believe the trickiest part would be to see that $L^2$ and the second partial derivative with respect to $r$ commute. You can see this expanding the $L^2$ operator in spherical coordinates and performing the calculation. You'll see you can exchange the order of operations. Perhaps this approach is not quite obvious. Of course, if you already know that $L$ is the generator of rotations, the approach suggested by @RobinEkman in comment is the more direct one. $\endgroup$ – user1620696 Feb 22 '16 at 3:13
  • $\begingroup$ thank you so much! It was the exchanging the order part that I was getting stuck on before, and you've clarified that really well. $\endgroup$ – Mecury-197 Feb 22 '16 at 4:12

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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