Why does dimensional regularization respect the Ward identity?

[center o bass]

It is often stated that this is the case, but I have often wondered if it is a general statement or just something that we observe to be the case when calculating the relevant loop corrections. Can it be proven generally? Is it somehow easy to see?

 

[Jolb]

Dimensional regularization is a fairly general mathematical technique. You can always use it for calculations in QFT. As far as I know, in QFT, the technique basically just amounts to introducing extra parameters (regulators) to change an integral originally in some weird coordinate system into a simple euclidean integral [basically the infinitesimal elements and the integration bounds get retrofitted by the introduction of the regulator].

The Ward identity, on the other hand, is a result of QFT. So it only holds in special circumstances.

You might want to look in Peskin and Schroeder, they have a nice little discussion about dimensional regularization which uses the Gamma function very sneakily.

Edit: Maybe you can't always use it: if you're being very creative, you may have to check whether the regulator you introduce is going to cause something fishy to happen. But in Peskin and Schroeder, you'll see they're doing a very innocuous mathematical manipulation.

 

[center o bass]

Dimensional regularization is a general mathematical technique. You can always use it for calculations in QFT. As far as I know, in QFT, the technique basically just amounts to introducing extra parameters (regularizers) to change an integral originally in some weird coordinate system into a simple euclidean integral [basically the infinitesimal elements get fixed by the introduction of the regularizer].

The Ward identity, on the other hand, is a result of QFT. So it only holds in special circumstances.

I agree, but the ward identity holds when one uses dimensional regularization, while it does not when using for example a simple cutoff regulator.

When searching the net I found that this was due to the fact that integrals in a general dimensional respects a translational substitution; i.e that

$$\int d^{d} p f(p+q) = \int d^{d} p f(p).$$

This is not true for a cut-off regulator and since the proof of ward identity involves such a substitution, dimensional regularization respects it, while a cutoff regulator does not.