4
$\begingroup$

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

$\endgroup$

3 Answers 3

7
$\begingroup$

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}

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – Eric Angle
    Jan 13, 2014 at 22:20
  • $\begingroup$ @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. $\endgroup$ Jan 13, 2014 at 23:00
  • $\begingroup$ 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. $\endgroup$
    – Eric Angle
    Jan 14, 2014 at 15:04
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 14, 2014 at 20:12
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 14, 2014 at 20:25
4
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

  1. $[A,BC] = B[A,C] +[A,B]C$
  2. $A_iA_i = A^2$, 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!

$\endgroup$
1
  • 1
    $\begingroup$ It seems you’re working too hard. Since $X^2$ is a scalar it should be quicker to show that $[L_i,X^2]=0$ which the result on $L^2$ follows immediately. $\endgroup$ Sep 29, 2023 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.