# QED Path Integrals: Weinberg QFT Book Explained

• Reggaerules
In summary, Weinberg does QED via path integrals, but requires d \psi^dagger (d p_m as he calls in the general formulation) in the integral measure. However, when he does QED, this does not appear in the integral measure.
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?

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

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.

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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)

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.

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?

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.

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

## What is a QED path integral?

A QED path integral is a mathematical tool used in quantum field theory that allows for the calculation of amplitudes and probabilities of particle interactions. It involves summing over all possible paths that a particle can take between two points in spacetime.

## How does the Weinberg QFT book explain QED path integrals?

The Weinberg QFT book provides a thorough and rigorous explanation of QED path integrals, starting with the basics of quantum mechanics and building up to the more complex concepts of quantum field theory. It also includes examples and exercises to help readers understand and apply the theory.

## Why are QED path integrals important in quantum physics?

QED path integrals are important because they provide a way to calculate the probabilities of particle interactions in quantum field theory. This allows for the prediction and understanding of various phenomena, such as the behavior of subatomic particles and the properties of quantum fields.

## What are some challenges in understanding QED path integrals?

One of the main challenges in understanding QED path integrals is the use of complex mathematical concepts, such as functional integrals and Feynman diagrams. It also requires a solid understanding of quantum mechanics and quantum field theory, which can be difficult for those without a strong background in physics.

## How can QED path integrals be applied in research and experimentation?

QED path integrals have numerous applications in research and experimentation, particularly in high-energy physics and particle accelerators. They are used to calculate the probabilities of particle interactions and to make predictions about the behavior of particles in various physical systems.

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