Is proper time renormalization gauge invariant?

Proper Time Renormalization is achieved by putting:

$$\int_0^\infty e^{iat} dt = {1\over ia}$$

Is it true that this is the only kind of normalization that is gauge invariant? If so, why do famous books ignore it even though it has such an important property (gauge invariance)?

Does the reason behind the name have anything to do with this?

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Can I know why down voting? –  TMS Dec 2 '12 at 10:44
Perhaps because it's not clear what you're asking –  David Z Dec 2 '12 at 20:38
I see, will make sure that this not happens again. –  TMS Dec 2 '12 at 21:15
The first step to doing that: remove the phrase "Any suggestions?" from the end of your question, and ask something more specific. "Any suggestions?" is never a valid question here. –  David Z Dec 2 '12 at 21:43
@David: Understood, hope now it's ok. –  TMS Dec 2 '12 at 21:47
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The thing is called the Schwinger proper time parametrization, and it isn't completely ignored--- it's the basis for combining denominators and it's the way to interpret field correlation functions as particle path sums. It isn't a regularization by itself, it is just an alternate representation of the propagator. You write

$${1\over p^2 + m^2 } = \int e^{-(p^2 + m^2)\tau} d\tau$$

or the same thing with an i in the proper place on both sides, and you get a proper time representation. Each separate $\tau$ contribution is the contribution of a path which takes proper time $\tau$, as you can see by Fourier transforming both sides to x (the left hand side is the spatial propagator, the right hand side is an integral over spreading Gaussians).

To make it a regularization, you want to give a cutoff on small proper times, so that the lower integration limit is not zero, or the integral is weighted with a factor that depends on $\tau$. This is a proper perturbative regulator, and it can be gauge invariant, but it isn't the way people do it anymore.

The way people do it today is to regulate the integrals using a factor of ${1\over p^\epsilon}$ in front of every $d^dp$ that appears. This is dimensional regularization, and you can understand it in several equivalent ways:

1. putting a factor of p to a tiny fractional power as I said above
2. continuing the dimension of spacetime away from integer dimensions

But 2 sounds more mysterious than it is, and leads to confusion about what to do with gamma matrices and index sums. You should treat the gamma matrices and index sums as you always do, keeping them 4 dimensional, and then you can do dim-reg consistently.

Anyway, putting this factor in the integrals is tantamount to adding a factor of $\tau^\epsilon$ inside the proper time integral, which effectively cuts off the proper times at short $\tau$. So what you are describing, when it is turned into a regularization, is identical to what everyone does, except phrased less mysteriously than continuing the dimension.

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Thx ron, stupid me with the integral, it was deep night when I posted it. Can you please advice why people don't use it this days? and dose dimensional regularization preferred due to the dimension compactification approach for unification theories? what about gauge invariance? –  TMS Dec 2 '12 at 10:48
@TMS: The Schwinger representation is what people use these days--- I explained how to turn it into dim-reg. All perturbative regulators have a Schwinger interpretation which is easier. Continuing the dimension is a misleading way to say it, because the regulated integrals don't have to reproduce consistent other-dimension theories when you make $\epsilon$ an integer. Dimensional regularization, as any other Schwinger thing, is "gauge invariant" (meaning regulating charged particles doesn't introduce new gauge breaking terms), but remember that you need to fix the gauge on the vector. –  Ron Maimon Dec 2 '12 at 20:30
Already done, thx for answer. –  TMS Dec 2 '12 at 20:34