# QED via path integrals

1. Jan 31, 2008

### Reggaerules

Does anyone understand how weinberg in his QFT book does QED via path integrals: specifically, how does he integrate out the adjoint dirac spinor?

2. Jan 31, 2008

### Jim Kata

look at pages 411 and 412, if that doesn't answer you maybe, I can try to help you a little more

3. Jan 31, 2008

### Reggaerules

I have gone through the entire chapter, but nothing seems to help me. In the development of the spinor path integral, Weinberg requires d \psi^dagger (d p_m as he calls in the general formulation) in the integral measure, but, when he does QED, this does not appear in the integral measure. If I can understand this, I will be ready for Vol. 2. I just can't wait.

Last edited: Jan 31, 2008
4. Jan 31, 2008

### Jim Kata

I'll give you an example look at equation (9.617) he didn't write them out for notation purposes, but there in there.

Example: Say to the one loop order I wanted to calculate the photon propagator/ ignoring renormalization terms in the Feynman gauge. Using equation (9.617) I'd have an integral like

$$\begin{gathered} - i\Delta _{\mu \tau } \left( {x,y} \right) = \left\langle {T\{ A_\mu (x)A_\tau (y)} \right\rangle = \hfill \\ \frac{{\int {\left[ {\prod\limits_{z,m} {dq_m \left( z \right)} } \right]\left[ {\prod\limits_{z,m} {dp_m \left( z \right)} } \right]\left[ {\prod\limits_{z,m} {da_m (z)} } \right]} a_\mu (x)a_\tau (y)\exp \left( {i[ - \frac{1} {2}\int {d^4 zd^4 wa^\xi (z)a^\zeta (w)D_{z\xi ,w\zeta } - \sum\limits_{r,s} {p_r } } (z)q_s (w)K_{rz,sw} ]} \right)}} {{\int {\left[ {\prod\limits_{z,m} {dq_m \left( z \right)} } \right]\left[ {\prod\limits_{z,m} {dp_m \left( z \right)} } \right]\left[ {\prod\limits_{z,m} {da_m (z)} } \right]} \exp \left( {i[ - \frac{1} {2}\int {d^4 zd^4 wa^\xi (z)a^\zeta (w)D_{z\xi ,w\zeta } - \sum\limits_{r,s} {p_r } } (z)q_s (w)K_{rz,sw} ]} \right)}} \hfill \\ \end{gathered}$$

where

$$q_m (x) = \psi _m (x)$$

$$p_m (x) = - [\bar \psi (x)\gamma ^0 ]_m$$

$$K_{mx,ny} = \left( {\gamma ^0 \left( {\gamma ^\mu \frac{\partial } {{\partial x^\mu }} + m + ie\gamma ^\tau A_\tau (x)\delta ^4 (x - y) - i\varepsilon )} \right)} \right)$$ and

$$D_{x\mu ,y\nu } = \left[ {\eta _{\mu \nu } \frac{{\partial ^2 }} {{\partial x^\rho \partial x_\rho }}\delta ^4 (x - y) + i\varepsilon } \right]$$

Now, try to follow what Weinberg does on page 412. I'm going to integrate over the positions and momentums of the fermion fields at same time, and the field independent determinants will cancel out in the ratio leaving me with

$$- i\Delta _{\mu x,\tau y} = \left\langle {T\{ A_\mu (x)A_\tau (y)} \right\rangle = \frac{{\int {\prod\limits_{z,m} {da_m (z)a_\mu (x)a_\tau (y)\exp \left( {i - \frac{1} {2}\int {d^4 zd^4 wa^\xi (z)a^\zeta (w)D_{z\xi ,w\zeta } } } \right)\exp \left( {\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }} {n}} Tr(F^{ - 1} G)^n } \right)} } }} {{\int {\prod\limits_{z,m} {da_m (z)\exp \left( {i - \frac{1} {2}\int {d^4 zd^4 wa^\xi (z)a^\zeta (w)D_{z\xi ,w\zeta } } } \right)\exp \left( {\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }} {n}} Tr(F^{ - 1} G)^n } \right)} } }}$$

where
$$F^{ - 1} (x,y) = \int {\frac{{d^4 k}} {{(2\pi )^4 }}} \frac{{ - \gamma ^0 }} {{i\gamma ^\mu k_\mu + m - i\varepsilon }}e^{ik \cdot (x - y)}$$

and

$$G(x,y) = ie\gamma ^0 \gamma ^\mu a_\mu (x)\delta ^4 (x - y)$$

Now I said I was only going to do this to one loop order, so I'll neglect all terms n>2 in my
sum in my exponent. I also don't have to worry about the n=1 term since it is tadpole and would break the symmetry of charge conjugation if I included it so I'll let you solve to one loop order, but you should get equation (11.2.1)

5. Jan 31, 2008

### Reggaerules

Wow, you are much more proficient in this than I am. But, I think my questions remains....mind you, I don't have any problems with his conclusions.

As you pointed out, you integrated the positions and momentums of the fermion fields at same time. Weinberg just integrates out the momenta, leaving behind only the position spinor. Thats what I am having a problem with. He does this again in Volume 2. I thought he just left it out for notational compactness, but when I e-mailed him, he replied saying that its doable.

I am thinking that this is some type of Lagrangian version for the spinor path integral. Any help will be much appreciated.

6. Jan 31, 2008

### Jim Kata

Hmm, I've never seen them integrated one at a time, and I've never tried it. What part of volume II are you talking about chapter 17?

7. Feb 1, 2008

### Reggaerules

Last line of page 17, volume 2. Also the measure in 15.4.16 does not have any adjoint spinor terms. I am just afraid to move on (since I am studying QFT by myself). Not sure why Weinberg does this consistently. If I understand it for the simple case of QED (I haven't yet started Vol. 2), I am sure that I could extend the notion.

8. Feb 2, 2008

### Reggaerules

Am I thinking correctly? Does Weinberg explain the disappearence of the adjoint spinors?