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Spin-1 operator identity

  1. Jan 22, 2010 #1
    I have a simple technical problem. I'm following a paper [Shore, G. Ann Phys. 137, 262-305 (1981)], and I am unable to show a very simple identity for the non-abelian fluctuation operator (eq 4.37):

    [tex]D_\mu\left[-D^2\delta_{\mu\nu}+D_\mu D_\nu-2F_{\mu\nu}\right]\,\phi=-(D_\mu F_{\mu\nu})\,\phi[/tex] , (typo fixed)​

    where [itex]\phi[/itex] is a test function and [itex](F_{\mu\nu})^{ab}\equiv gf^{abc}F_{\mu\nu}^{c}=[D_\mu,\,D_\nu][/itex], and hence [itex]D_\mu F_{\mu\nu}=D^2D_\nu-D_\mu D_\nu D_\mu[/itex] (color indices suppressed). So far, I have worked on the LHS, and I'm almost there:

    [tex]\text{LHS}=(-D_\nu D^2+D^2 D_\nu-2D_\mu F_{\mu\nu})\phi[/itex]
    [tex]\phantom{LHS}=(-\underline{D_\mu D_\nu D_\mu}-[D_\nu,\,D_\mu]D_\mu+\underline{D^2D_\nu}-2D_\mu F_{\mu\nu})\phi[/tex]
    combine underlined terms using identity stated above
    [tex]=(-[D_\nu,\,D_\mu]D_\mu+D_\mu F_{\mu\nu}-2D_\mu F_{\mu\nu})\phi[/tex]
    then first term is [itex]-[D_\nu,\,D_\mu]D_\mu=+F_{\mu\nu}D_\mu[/itex], and 2nd and 3rd terms add
    [tex]=(F_{\mu\nu}D_\mu-D_\mu F_{\mu\nu})\phi[/tex]
    Finally, use product rule in 2nd term: [itex]D_\mu(fg)=(D_\mu f)g+f\partial_\mu g[/itex].
    [tex]=F_{\mu\nu}(\partial+A)_\mu\phi-(D_\mu F_{\mu\nu})\phi-F_{\mu\nu}\,\partial_\mu\phi[/tex]
    to get
    [tex]=F_{\mu\nu} A_\mu \phi-(D_\mu F_{\mu\nu})\phi[/tex].

    This is almost equal to RHS, except for that stupid [itex]A_\mu[/itex] term. How the hell do I get rid of it?!?
    Last edited: Jan 23, 2010
  2. jcsd
  3. Jan 23, 2010 #2
    \text{LHS}=(-D_\nu D^2+D^2 D_\nu-2D_\mu F_{\mu\nu})\phi

    how did you get this from
    D_\mu\left[-D^2\delta_{\mu\nu}+D_\mu D_\nu-2D_\mu F_{\mu\nu}\right]


    be careful with that last term in the square bracket, you have in total three mu index... also be carefull with upper and lower index.
  4. Jan 23, 2010 #3
    There's a typo in my post. The identity should read:

    [tex] D_\mu\left[-D^2\delta_{\mu\nu}+D_\mu D_\nu-2F_{\mu\nu}\right]\,\phi=-(D_\mu F_{\mu\nu})\,\phi [/tex]

    otherwise, the dimensions (and indices) don't work.
    Also, I'm in Euclidean spacetime, where I don't need to worry about upper and lower indices.
    I'm still stuck.
    Last edited: Jan 23, 2010
  5. Jan 23, 2010 #4


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    Clearly, this is wrong, do you know why? you have a tensor on the LHS (which should be just [itex]F_{\mu\nu}[/itex]) and vector on the RHS (which is [itex]\D_{\mu}F_{\mu\nu}[/itex]).


    Last edited: Jan 23, 2010
  6. Jan 23, 2010 #5
    Whoops! Another typo. The identity should have read
    [tex]D_\mu F_{\mu\nu}=D^2D_\nu-D_\mu D_\nu D_\mu[/tex]. I'm fixing this in my original post.
    I'm still stuck.
  7. Jan 23, 2010 #6


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    If your test function takes values in the lie algebra of the gauge group,i.e., matrix-valued function;[itex]\Phi = \phi^{a}T^{a}[/itex], then

    [tex]D_{\mu}\Phi = \partial_{\mu}\Phi + [A_{\mu},\Phi][/tex]

    If it is a c-number function, then

    [tex]D_{\mu}\Phi = \partial_{\mu}\Phi[/tex]

    In both cases, the covariant derivative is distributive;

    D_{\mu}(F_{\mu\nu}\Phi) = (D_{\mu}F_{\mu\nu})\Phi + F_{\mu\nu}D_{\mu}\Phi

    So, your LHS is equal to [itex] -(D_{\mu}F_{\mu\nu})\Phi[/itex]


  8. Jan 24, 2010 #7
    This is very helpful, but I don't quite understand. Naïvely, I would expect the covariant derivative not to be distributive because the vector potential, [itex]A_\mu[/itex] is not an object that behaves like the derivative.

    If my test function were a column vector, shouldn't I have
    [tex]=(\partial_\mu F_{\mu\nu})\phi+F_{\mu\nu}\partial_\mu\phi+A_\mu F_{\mu\nu}\phi[/tex]
    [tex]=(D_\mu F_{\mu\nu})\phi+F_{\mu\nu}\partial_\mu\phi[/tex] <-- (this is wrong: see edit below)

    Where did I go wrong in the maths? If you are right, then wouldn't the maths tell me so?


    Never mind! I now realized where I went wrong.
    The last step in this post is wrong. I must add and subtract [itex]F_{\mu\nu}A_\mu\phi[/itex], so that
    [tex]=(\partial_\mu F_{\mu\nu})\phi+F_{\mu\nu}\partial_\mu\phi+A_\mu F_{\mu\nu}\phi-F_{\mu\nu}A_\mu\phi+F_{\mu\nu}A_\mu\phi[/tex]
    [tex]=(\partial_\mu F_{\mu\nu}+[A_\mu,\,F_{\mu\nu}])\phi+F_{\mu\nu}(\partial_\mu+A_\mu)\phi[/tex]
    [tex]=(D_\mu F_{\mu\nu})\phi+F_{\mu\nu}D_\mu\phi[/tex]

    So, samalkhaiat is right: the covariant derivative obeys the product rule, and hence the identity in the first post is shown to be true. Case closed.

    Thanks, all!
    Last edited: Jan 24, 2010
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