Angular Momentum Commutator relation

[andre220]

Homework Statement

Calculate the commutator $[\hat{L}_i, (\mathbf{rp})^2]$

Homework Equations

$\hat{\vec{L}} = \sum\limits_{a=1}^N \vec{r}_a \times \hat{\vec{p}}$
$[r_i,p_k] = i\hbar\delta_{ik}$

The Attempt at a Solution

Okay so here is what I have so far:

$$
\begin{eqnarray}
[\hat{L}_i, (\mathbf{rp})^2] & = &\left [\epsilon_{ikl}r_k p_l, \sum\limits_j r_j^2 p_j^2\right] \\
& = & \epsilon_{ikl}\left(r_k \left[p_l, \sum\limits_j r_j^2 p_j^2\right] + \left[r_k, \sum\limits_j r_j^2 p_j^2\right]p_l\right)
\end{eqnarray}
$$

My next step is I think would be to expand the summation terms, but I am not quite sure if that would the correct step.

 

[BvU]

I don't follow your first relevant equation. What is a ? All I ever learned was ${\bf L} = {\bf r} \times {\bf p}$.

Although your last step is ingenious, it only seems to make the expression longer. Wouldn't it be easier to make use of the basic commutation relations, such as $[L_x, y] =[yp_z-zp_y,y]=-z[p_y,y]=i \hbar z$ ?

 

[andre220]

From Landau Liftshitz pg 82 I have that $-i\hbar \mathbf{r}\times \vec{\nabla} = \mathbf{r}\times\hat{\mathbf{p}}$, my mistake on that first equation I was looking at something else in my notes and confused it. But regardless my thinking with the last step was that I could get it in terms of $p^2[p,r^2] +[p,p^2]r^2$ or more explicitly
$$ \epsilon_{ikl}\left(r_k\sum\limits_j p_j^2\left[p_l,r_j^2\right]+\sum\limits_j\left[r_k,p_j^2\right]r_j^2 p_l\right),$$
but you are correct that it does make it longer. Where are you saying that the expression you wrote should be applied?

 

[BvU]

I was just giving one example. There's also $[L_x, p_y]$, $[L_x, x]$, $[L_x, p_x]$ (latter two are 0).

One more thing: $(rp)^2 = \sum\limits_j r_j p_j r_j p_j$ ; what do you use to be able to write $(rp)^2 = \sum\limits_j r_j r_j p_j p_j$ ? because I seem to remember that $p_j r_j = r_j p_j - [r_j, p_j]$ ...

 

[andre220]

Ahh yes you are right that will change things, also I suppose I could just drop the sum altogether since it is implied. That may make it easier to split it up. Ill take a look at it. Thanks for your input.

 

[andre220]

Okay so I retried it and I have a result, though I am not sure that it is correct here is what I do:

$$
\begin{eqnarray}
[\hat{L}_i, (\mathbf{rp})^2] & = & \left[\epsilon_{ikl} r_k p_l, r_j p_j r_j p_j\right]\\
& = & \epsilon_{ikl}\left(r_k\left[p_l, r_j p_j r_j p_j\right] + \left[ r_k, r_j p_j r_j p_j\right]p_l\right)\\
& = & \epsilon_{ikl}\left(r_k\left[p_l, r_j p_j \right] r_j p_j+ r_j p_j \left[ r_k, r_j p_j \right]p_l\right)\\
& = & \epsilon_{ikl}\left(r_k\left(-i\hbar p_j\delta_{jl}\right)r_j p_j + r_j p_j\left(i\hbar r_j\delta_{kj}\right) p_l \right) \\
& = & \epsilon_{ikl}\left(-i\hbar r_k p_l r_j p_j + i\hbar r_k p_l r_j p_j\right) \\
& = & \boxed{ 0 }
\end{eqnarray}
$$

Any flaws in that logic?

 

[BvU]

Just a detail: you left out the summation

Well, it is strange that it should change anything: $[r_j, p_j]$ is a number, so that should commute with $L_i$...meaning $\left [\epsilon_{ikl}r_k p_l, \sum\limits_j r_j^2 p_j^2\right] \$ should come out 0 too, isn't it ?

 

[nrqed]

Okay so I retried it and I have a result, though I am not sure that it is correct here is what I do:

$$
\begin{eqnarray}
[\hat{L}_i, (\mathbf{rp})^2] & = & \left[\epsilon_{ikl} r_k p_l, r_j p_j r_j p_j\right]\\
& = & \epsilon_{ikl}\left(r_k\left[p_l, r_j p_j r_j p_j\right] + \left[ r_k, r_j p_j r_j p_j\right]p_l\right)\\
& = & \epsilon_{ikl}\left(r_k\left[p_l, r_j p_j \right] r_j p_j+ r_j p_j \left[ r_k, r_j p_j \right]p_l\right)\\
& = & \epsilon_{ikl}\left(r_k\left(-i\hbar p_j\delta_{jl}\right)r_j p_j + r_j p_j\left(i\hbar r_j\delta_{kj}\right) p_l \right) \\
& = & \epsilon_{ikl}\left(-i\hbar r_k p_l r_j p_j + i\hbar r_k p_l r_j p_j\right) \\
& = & \boxed{ 0 }
\end{eqnarray}
$$

Any flaws in that logic?

One has to wrote $(\vec{r} \cdot \vec{p})^2$ as $r_j p_j r_a p_a$ i.e. one must use two different sets of indices.
Also, on your third line you left out a couple of terms since, as you know, [A,BC] = B[A,C] + [A,B] C.