# Spin-1 operator identity

1. Jan 22, 2010

### 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:

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

This is almost equal to RHS, except for that stupid $A_\mu$ term. How the hell do I get rid of it?!?

Last edited: Jan 23, 2010
2. Jan 23, 2010

### ansgar

$$\text{LHS}=(-D_\nu D^2+D^2 D_\nu-2D_\mu F_{\mu\nu})\phi$$

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.

3. Jan 23, 2010

### 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.

Last edited: Jan 23, 2010
4. Jan 23, 2010

### samalkhaiat

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

Last edited: Jan 23, 2010
5. Jan 23, 2010

### 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.

6. Jan 23, 2010

### 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;

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

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

regards

sam

7. Jan 24, 2010

### 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!

Last edited: Jan 24, 2010