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

Question

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 $[L^2, P^2] = 0$?

Answer 1

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}

Answer 2

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: $$[L^2, X^2]=\sum_{i=1}^3[L^2_i, X^2]=\sum_{i=1}^3 L_i[L_i, X^2] +\sum_{i=1}^3[L_i, X^2]L_i =0$$

Answer 3

I use the following facts to derive the commutation relations asked in question:-

1. $[X_i,L_j] = i\hbar\ \epsilon_{ijk}X_k$
2. $[P_i,L_j] = i\hbar\ \epsilon_{ijk}P_k$
3. $[X_i,X_j] = 0 = [P_i,P_j]$

where $\epsilon_{ijk}$ is the Levi-Civita symbol defined as

$\epsilon_{ijk} = \begin{cases} 1& \text{even permutations of (1,2,3)} \\ -1& \text {odd permutations of (1,2,3)} \\ 0& \text{any index repeated} \end{cases}$

4. $[A,BC] = B[A,C] +[A,B]C$
5. $A_iA_i = A^2$, that is we will implement Einstein's Summation convention.

With these, we get the following:-

$[X^2,L^2] = [X_iX_i,L_jL_j] $

$ = X_iL_j\ [X_i,L_j] + X_i\ [X_i,L_j]\ L_j + L_j\ [X_i,L_j]\ X_i + [X_i,L_j]\ L_jX_i$

$= i\hbar\ (\epsilon_{ijk}X_iL_jX_k + \epsilon_{ijk}X_iX_kL_j + \epsilon_{ijk}L_jX_kX_i + \epsilon_{ijk}X_kL_jX_i)$

The $i$ and $k$ indices in the last term are dummy labels which are summed over. So, interchanging $i \leftrightarrow k$ in the last term, the first and last term above becomes

$i\hbar\ (\epsilon_{ijk}X_iL_jX_k + \epsilon_{kji}X_iL_jX_k) $

$= i\hbar\ (\epsilon_{ijk}X_iL_jX_k - \epsilon_{ijk}X_iL_jX_k) $

$= 0$

For the remaining terms, we use :-

$X_kL_j = i\hbar\ \epsilon_{kjm}X_m + L_jX_k $

With this, the second and third term above becomes

$i\hbar\ (i\hbar\ \epsilon_{ijk}\epsilon_{kjm} X_iX_m + \epsilon_{ijk}X_iL_jX_k +\epsilon_{ijk}X_kL_jX_i - i\hbar\ \epsilon_{ijk}\epsilon_{kjm}X_mX_i)$

The second and third term adds to $0$ just like above, while the first and last term also adds to $0$ because $X_i$ and $X_m$ commute with each other.

This proves $[X^2,L^2] = 0$

One follows exactly the same line of proof as above with $X_i$ replaced by $P_i$ to obtain $[P^2,L^2] = 0$

Cheers!