Why Can't I Show the Simple Identity for the Spin-1 Operator in This Paper?

[TriTertButoxy]

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):

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

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

\text{LHS}=(-D_\nu D^2+D^2 D_\nu-2D_\mu F_{\mu\nu})\phi[/itex]<br /> \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<br /> combine underlined terms using identity stated above<br /> =(-[D_\nu,\,D_\mu]D_\mu+D_\mu F_{\mu\nu}-2D_\mu F_{\mu\nu})\phi<br /> then first term is -[D_\nu,\,D_\mu]D_\mu=+F_{\mu\nu}D_\mu, and 2nd and 3rd terms add<br /> =(F_{\mu\nu}D_\mu-D_\mu F_{\mu\nu})\phi<br /> Finally, use product rule in 2nd term: D_\mu(fg)=(D_\mu f)g+f\partial_\mu g.<br /> =F_{\mu\nu}(\partial+A)_\mu\phi-(D_\mu F_{\mu\nu})\phi-F_{\mu\nu}\,\partial_\mu\phi<br /> to get<br /> =F_{\mu\nu} A_\mu \phi-(D_\mu F_{\mu\nu})\phi.<br /> <br /> This is <i>almost</i> equal to RHS, except for that stupid A_\mu term. How the hell do I get rid of it?!?

 

[ansgar]

<br /> \text{LHS}=(-D_\nu D^2+D^2 D_\nu-2D_\mu F_{\mu\nu})\phi<br />

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

??

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.

 

[TriTertButoxy]

There's a typo in my post. The identity should read:

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

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.

 

[samalkhaiat]

[D_\mu,\,D_\nu]=D^2D_\nu-D_\mu D_\nu D_\mu

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

regards

sam

 

[TriTertButoxy]

Whoops! Another typo. The identity should have read
D_\mu F_{\mu\nu}=D^2D_\nu-D_\mu D_\nu D_\mu. I'm fixing this in my original post.
I'm still stuck.

 

[samalkhaiat]

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

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

If it is a c-number function, then

D_{\mu}\Phi = \partial_{\mu}\Phi

In both cases, the covariant derivative is distributive;

<br /> D_{\mu}(F_{\mu\nu}\Phi) = (D_{\mu}F_{\mu\nu})\Phi + F_{\mu\nu}D_{\mu}\Phi<br />

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

regards

sam

 

[TriTertButoxy]

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, A_\mu is not an object that behaves like the derivative.

If my test function were a column vector, shouldn't I have
D_\mu(F_{\mu\nu}\phi)
=(\partial+A)_\mu(F_{\mu\nu}\phi)
=(\partial_\mu F_{\mu\nu})\phi+F_{\mu\nu}\partial_\mu\phi+A_\mu F_{\mu\nu}\phi
=(D_\mu F_{\mu\nu})\phi+F_{\mu\nu}\partial_\mu\phi <-- (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?

--EDIT--

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

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!