# Is There an Identity for Different Vectors in Dimensional Regularization?

• I
• Elmo
Elmo
In dimensional regularization I have seen this relation ##k^{\mu}k^{\nu}=\frac{1}{D}g^{\mu\nu}k^2## but this seems to hold for same types of four vectors k. Is there any similar identity for different vectors like ##k^{\mu}p^{\nu}=\frac{1}{D}g^{\mu\nu}k.p## ?

You will have to put that into context, since it is not true in general. For example, take any ##v \perp k##, then your identity implies
$$0 = k^\mu (k\cdot v) = \frac 1D v^\mu k^2$$

Last edited:
Elmo said:
In dimensional regularization I have seen this relation ##k^{\mu}k^{\nu}=\frac{1}{D}g^{\mu\nu}k^2## but this seems to hold for same types of four vectors k. Is there any similar identity for different vectors like ##k^{\mu}p^{\nu}=\frac{1}{D}g^{\mu\nu}k.p## ?
Why would it? The left hand side is not even symmetric in mu and nu.

haushofer said:
Why would it? The left hand side is not even symmetric in mu and nu.
are you referring to the first one or the second ? Because the first one is very often found in most QFT textbooks. I am asking about the second one.

Elmo said:
Because the first one is very often found in most QFT textbooks.
No it isn’t, because it simply isn’t true and it cannot be true, as already explained.

Hepth said:
so you can write it as a completely symmetric object, using the metric tensor.
No, that is nonsense. The metric has full rank and ##k^\mu k^\nu## doesn’t. Two tensors can both be symmtric without being proportional.

Hepth said:
I think others aren't answering correctly.
No, the others are correct. If it were true in general that ##k^{\mu}k^{\nu}=\frac{1}{D}g^{\mu\nu}k^{2}##, then it should be true in particular for a null vector with length zero: ##k^2=0##. But that results in ##k^{\mu}k^{\nu}=0##; i.e., a null vector must vanish identically. Nonsense!

Hepth said:
I think others aren't answering correctly.

is true because

so you can write it as a completely symmetric object, using the metric tensor.

Then solve for C by multiplying each side by another metric tensor to contract the indices. And two metric tensors multiplied by eachother, then taking the trace is just the dimensionality D. Solve for C.

You can do similar things for two different vectors constructing a symmetric and antisymmetric combination.

so solve for C and plug back in.

and if i remember correctly you can do an antisymmetric one as well using sigma:

contract this with another sigma, and move indices to get the trace (-1) and do the matrix math to get a result for C in terms of D/i/constants.
I have to look at this stuff more clearly, but as others noted this identity has some problems for null vectors. I guess the tricky step is to bluntly say "the tensor product of k with itself is symmetric, hence must be proportional to the metric". That's a natural guess, but not at all guaranteed of course.

Edit I guess TS refers to e.g. eqn. 6.46 of Peskin&Schroeder, where the vector l is defined right after eqn. 6.43 and the identity is used in an integral. How would this identity serve in the photon propagator?

haushofer said:
and the identity is used in an integral
(My emphasis)
This is the crucial thing that I tried to hint the OP towards:
Orodruin said:
You will have to put that into context, since it is not true in general.
The identity as presented by the OP is evidently false. However, there is an identity when you put both expressions inside an appropriate integral.

Yes. I think this is reminiscent about how people get confused about identities concerning Dirac delta distributions without mentioning the integral.

You're all right, I'm wrong completely. I think I'm remembering some identities for building up operator expectation values that restricts what the momenta can be, or what vector or tensor objects can be as a result of like an HQET operator. Obviously the null vector kills these identities.

Like if your operator is <DmuDnu + DnuDmu> then your expectation value can only be proportional to certain available structures.
$$<\phi|\frac{1}{2}\left(D^{\mu}D^{\nu} + D^{\nu}D^{\mu}\right)|\phi> = \alpha g^{\mu \nu} + \beta k^{\mu} k^{\nu}$$
$$<\phi|D^{2}|\phi> = \alpha D_{dim} + \beta k^{2}$$

and things like that. I'll go clear out my post so noone has wrong info.

Greg Bernhardt

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