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Angular Momentum Commutator relation

  1. Oct 9, 2014 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. 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.
     
  2. jcsd
  3. Oct 9, 2014 #2

    BvU

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    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## ?
     
  4. Oct 9, 2014 #3
    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?
     
  5. Oct 9, 2014 #4

    BvU

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    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] ## ...
     
  6. Oct 9, 2014 #5
    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.
     
  7. Oct 9, 2014 #6
    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?
     
  8. Oct 10, 2014 #7

    BvU

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    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 ?
     
  9. Oct 10, 2014 #8

    nrqed

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    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.
     
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