# Commutator of $L^2$ and $X^2$, $P^2$

In our quantum mechanics script, it states that $[L^2, X^2] = 0$ and $[L^2, P^2] = 0$, therefore for the following Hamiltonan

$$H = \frac{P^2}{2m} + V(X^2)$$

it is that $[H, L^2] = 0$ therefore $H$ and $L^2$ have the same eigenvectors, and then it continues to calculate orbitals of the hydrogen atom.

My question is: Is there a nice and simple proof of $[L^2, X^2] = 0$ and $[L^2, P^2] = 0$?

Yes there is.

I use the summation convention throughout; repeated indices are summed over 1,2,3.

1. Begin with the canonical commutation relations (CCRs) \begin{align} [X_i, P_j] = i\hbar\delta_{ij} I, \qquad [X_i, X_j] = 0, \qquad [P_i, P_j] = 0. \end{align}
2. Define the components of orbital angular momentum as follows: \begin{align} L_i = \epsilon_{ijk}X_jP_k \end{align}
3. Prove your desired identities by applying the definition of the angular momentum components and by repeatedly using the CCRs. I'll leave the details to you; it's a good exercise to get comfortable with using the CCRs to prove stuff.

It actually turns out to prove useful to first prove the following identities which encode the fact that the $X_i$ are components of a "vector operator" and so are the $P_i$. \begin{align} [X_i, L_j] =i\hbar\epsilon_{ijk}X_k \end{align}

• I don't want to derail this question, but I've never quite understood the last part of your answer about how an identity like $\left[X_i, L_j\right] = i \hbar \epsilon_{i j k} X_k$ means $X_k$ is a vector operator. Any good references or intuitive explanations for that? – Eric Angle Jan 13 '14 at 22:20
• @EricAngle Yes definitely. I'm on a phone though at the moment. I'll try to write an addendum. Feel free to remind me if I forget. – joshphysics Jan 13 '14 at 23:00
• I just went through the derivation of that identity, starting from (1) $U^{\dagger}\left(R\right) {\hat V}_i U\left(R\right) = R_{ij} {\hat V}_j$ with $U\left(R\right) = 1 - i \delta\theta \hat{\bf n} \cdot {\hat{\bf L}} / \hbar$ and $R_{i j} = \delta_{ij} + \delta\theta n_k \epsilon_{k j i}$. So I just need to think a bit more about where (1) comes from, and why this means $\hat{V}$ is a vector operator. I guess from the usual definition of a vector as something that transforms like a vector under rotations. – Eric Angle Jan 14 '14 at 15:04
• @EricAngle Yes that's essentially it. The idea behind (1) is that $U(R)$ represents how rotations act on states, and the similarity transformation $U(R)^\dagger O U(R)$ represents how rotations act on operators $O$. As a result, (1) says that the three tuple of operators $(\hat V_1, \hat V_2, \hat V_3)$ is such that when each "component" is rotated, it mixes with the others in exactly the same way that the components of a three-vector do in three, Euclidean dimensions. Ultimately, it's really a definition, but that's the motivation behind it. – joshphysics Jan 14 '14 at 20:12
• @EricAngle By the way, in case you're interested in math, this is all related to the representation theory of Lie Groups and Lie Algebras; a really cool subject. In particular, then we talk about scalar, vector, and tensor operators, we need to have a representation of a Lie group or Lie algebra in hand. In this particular example, the mapping $R\to U(R)$ is a unitary representation of $\mathrm{SO}(3)$, the 3d rotation group, acting on the Hilbert space of the theory, and we can speak of scalar, vector, tensor operators with respect to this representation. – joshphysics Jan 14 '14 at 20:25

Another possibility. I directly deal with $X^2$, but the same reasoning can obviously be used for $P^2$, $V(X^2)$ and $H$ (that is a linear combination of scalars) in place of $X^2$ below.

Consider a unitary representation of a rotation $U_\theta = e^{-i\theta L_k}$ along the ${\bf e}_k$ axis. As $X^2$ is a scalar under rotations: $$U_\theta X^2 U_\theta^\dagger =X^2$$ Therefore, taking the derivative at $\theta=0$: $$[L_i,X^2]=0\:,\quad i=1,2,3$$ and so: $$[L^2, X^2]=\sum_{i=1}^3[L^2_i, X^2]=\sum_{i=1}^3 L_i[L_i, X^2] +\sum_{i=1}^3[L_i, X^2]L_i =0$$