- #1
Daschm
- 4
- 0
Hi,
After having solved some problems I encountered by using Google and often being linked to threads here, I finally decided to register, especially because I sometimes have problems for which I don't find solutions here and now want to ask them by myself :)
Like the following: I am currently doing my first calculations on loop diagrams working with Peskin/Schroeder and I try to regularize the QED Electron-Vertex-function by dimensional regularisation (I want to solve Problem 7.2: Calculating Z1 and Z2 by dimensional regularisation and manual cutoff and seeing whether Z1=Z2 is fulfilled or not).
I did something, and it seems to be wrong, and I don't understand why.
Looking at the numerator:
[itex]\bar{u}(p^\prime)({\not} k \gamma^\mu {\not} k^\prime + m^2 \gamma^\mu - 2 m (k+k^\prime)^\mu)u(p)[/itex]
I used the usual 4-dimensional contraction and anticommutation rules + Dirac equation to get to the final formula where you have only [itex]\gamma^\mu[/itex] and [itex]\sigma^{\mu \nu}q_\nu[/itex] with certain coefficients. Afterwards I rewrote everything in d dimensions and used the general formulae for dimensional regularisation in order to calculate the integrals. I crosschecked my solution with one I found on the internet (http://www-personal.umich.edu/~jbourj/peskin/homework 6.pdf) and I basically get the same result, except for an additional 2! He gets it because he uses the d-dimensional contraction rules, ending up with an additional factor of [itex](2-\epsilon)^2[/itex], whose mixed term cancels a [itex]\frac{1}{2} \frac{1}{\epsilon}[/itex] to get a finite value of 2.
Now I am a bit confused: Peskin Schroeder also gives the d-dimensional contraction formulae so I understand how I am supposed to get the same result. And it seems to be necessary to contract in d dimensions if I want to regularize it in that way. But I don't understand why it is forbidden to start doing the Dirac algebra in 4 dimensions and then, if I end up with an divergent integral, doing dimensional regularisation, put everything in d dimensions and calculate it and put d=4 at the end. [Doing this, I won't get this peculiar 2 I talked about earlier]
I mean: I see that the two ways give different results, so they can't be equivalent, but it confuses me because the step going into d dimensions is just a temporary aid to see where the divergence occurs and how to renormalize it, but at the end I will set d=4. So intuitively it should not matter at which point I start to generalize and at which point I will specify again, at least for the finite part, shouldn't it?
What I am asking for is maybe an explanation / argument why it matters if I contract in d or in 4 dimensions. And do I really have to calculate/contract+commute everything in d dimensions if I already know that in the end I want to do dimensional regularisation? Even though I just do it to extract the divergent part [itex]1/\epsilon[/itex]?
After having solved some problems I encountered by using Google and often being linked to threads here, I finally decided to register, especially because I sometimes have problems for which I don't find solutions here and now want to ask them by myself :)
Like the following: I am currently doing my first calculations on loop diagrams working with Peskin/Schroeder and I try to regularize the QED Electron-Vertex-function by dimensional regularisation (I want to solve Problem 7.2: Calculating Z1 and Z2 by dimensional regularisation and manual cutoff and seeing whether Z1=Z2 is fulfilled or not).
I did something, and it seems to be wrong, and I don't understand why.
Looking at the numerator:
[itex]\bar{u}(p^\prime)({\not} k \gamma^\mu {\not} k^\prime + m^2 \gamma^\mu - 2 m (k+k^\prime)^\mu)u(p)[/itex]
I used the usual 4-dimensional contraction and anticommutation rules + Dirac equation to get to the final formula where you have only [itex]\gamma^\mu[/itex] and [itex]\sigma^{\mu \nu}q_\nu[/itex] with certain coefficients. Afterwards I rewrote everything in d dimensions and used the general formulae for dimensional regularisation in order to calculate the integrals. I crosschecked my solution with one I found on the internet (http://www-personal.umich.edu/~jbourj/peskin/homework 6.pdf) and I basically get the same result, except for an additional 2! He gets it because he uses the d-dimensional contraction rules, ending up with an additional factor of [itex](2-\epsilon)^2[/itex], whose mixed term cancels a [itex]\frac{1}{2} \frac{1}{\epsilon}[/itex] to get a finite value of 2.
Now I am a bit confused: Peskin Schroeder also gives the d-dimensional contraction formulae so I understand how I am supposed to get the same result. And it seems to be necessary to contract in d dimensions if I want to regularize it in that way. But I don't understand why it is forbidden to start doing the Dirac algebra in 4 dimensions and then, if I end up with an divergent integral, doing dimensional regularisation, put everything in d dimensions and calculate it and put d=4 at the end. [Doing this, I won't get this peculiar 2 I talked about earlier]
I mean: I see that the two ways give different results, so they can't be equivalent, but it confuses me because the step going into d dimensions is just a temporary aid to see where the divergence occurs and how to renormalize it, but at the end I will set d=4. So intuitively it should not matter at which point I start to generalize and at which point I will specify again, at least for the finite part, shouldn't it?
What I am asking for is maybe an explanation / argument why it matters if I contract in d or in 4 dimensions. And do I really have to calculate/contract+commute everything in d dimensions if I already know that in the end I want to do dimensional regularisation? Even though I just do it to extract the divergent part [itex]1/\epsilon[/itex]?