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Commutator of a Cross Product

  1. Nov 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Given that [tex]\vec{V} [/tex] and [tex] \vec{W}[/tex] are vector operators, show that [tex] \vec{V}\times \vec{W}[/tex] is also a vector operator.

    2. The attempt at a solution
    The only way I know how to do this is by showing that the commutator with the angular momentum vector operator ([itex] \vec{J}[/itex]) is zero. Namely that [itex] [\vec{V}\times \vec{W} , \vec{J}] = 0[/itex]. I want to start the problem by expressing the commutator as I would usually do by writing [A,B] = AB - BA, but I don't know exactly which type of multiplication to use here. Intuition tells me the dot product, but I want to be sure.
     
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  3. Nov 29, 2014 #2

    mfb

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    Dot product is right.

    Edit: Or not, see below.
     
    Last edited: Nov 29, 2014
  4. Nov 29, 2014 #3

    stevendaryl

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    No, it's not the dot-product. The usual meaning of [itex][A, \vec{J}][/itex] is that the result is a composite object (a tensor) with three components:

    [itex][A, J_x][/itex], [itex][A, J_y][/itex], [itex][A, J_z][/itex]

    If [itex]A[/itex] is itself a vector, then you get 9 components:

    [itex][\vec{A}, \vec{J}] = T[/itex]

    where [itex]T_{ij} = [A_i, J_j][/itex] and where [itex]i[/itex] and [itex]j[/itex] are either [itex]x[/itex], [itex]y[/itex], or [itex]z[/itex].

    As to your claim that for a vector operator, [itex][\vec{A}, \vec{J}] = 0[/itex], you should try an example with the momentum operator, [itex]\vec{p}[/itex]. Try [itex][p_x, J_y][/itex].
     
  5. Nov 29, 2014 #4

    stevendaryl

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  6. Nov 29, 2014 #5
    Thank you. That is a mistake above. I meant that the only way I know of to show that it is a vector operator, is to show that [itex]
    [\vec{V}\times \vec{W} , \vec{J}] \neq 0[/itex]. I was able to show that this is the case. This means it is not a scalar operator, but I am not sure if this is sufficient to show that it is a vector operator. It does mean that the rotation generator operator U won't commute with V x W.

    Edit: I believe now that the condition to show that it is a vector operator, is to show that: [tex]
    [(\vec{V}\times \vec{W})_{i}, \vec{J}_{j}] = i \hbar \epsilon_{ijk} (\vec{V}\times \vec{W})_{k}
    [/tex]
     
    Last edited: Nov 29, 2014
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