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Commutation relations of angular momentum with position, momentum.

  1. Dec 5, 2011 #1

    B L

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    1. The problem statement, all variables and given/known data
    Using the position space representation, prove that:
    [itex] \left[L_i, x_j\right] = i\hbar\epsilon_{ijk}x_k [/itex].
    Similarly for [itex] \left[L_i, p_j\right] [/itex].

    2. Relevant equations
    Presumably, [itex] L_i = \epsilon_{ijk}x_jp_k [/itex].
    [itex] \left[x_i, p_j\right] = i\hbar\delta_{ij} [/itex].

    3. The attempt at a solution
    [itex] \left[L_i, x_j\right] = \epsilon_{ijk}\left[x_jp_k, x_j\right]
    = \epsilon_{ijk}\left(x_jp_kx_j - x_jx_jp_k\right)
    = \epsilon_{ijk}x_j\left(p_kx_j - x_jp_k\right)
    = \epsilon_{ijk}x_j\left[p_k, x_j\right]
    = -i\hbar\epsilon_{ijk}x_j\delta_{jk} [/itex]
    which is where I become confused - it seems to me that the right hand side is always zero (if the Kronecker delta is nonzero, the Levi-Civita symbol is zero, and vice-versa).

    Any help is much appreciated.
     
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  3. Dec 5, 2011 #2

    dextercioby

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    You should respect the laws of tensor algebra. Don't use the same index 3 times in a tensor product.

    So [itex] [L_i,x_j]=\epsilon_{ikl}[x_k p_l, x_j] = ... [/itex]
     
  4. Dec 5, 2011 #3

    dextercioby

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    I can't edit my post ??? Ok...

    It's not true that epsilon times (in a tensor way) delta =0. This happens iff there's a double contracted tensor product among them.
     
  5. Dec 5, 2011 #4

    B L

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    That's the thing: when I asked my TA, he said that we are not to assume the summation convention. That's the source of my confusion - I have no idea what this question even means without the summation convention.

    (I'm literally the only person in the course who knows what a tensor is).
     
  6. Dec 5, 2011 #5

    dextercioby

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    You need the summation convention and know a little about cartesian vectors/tensors, if you're approaching the identity/equality from the all-components-at-the-same-time perspective
     
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