User talk:Agostino981

Quantum Electrodynamics

To construct the electromagnetic interaction Lagrangian, we start with a free-fermion Lagrangian

$$L_0=\bar{\phi}(i\gamma^\mu \partial_\mu-m)\phi$$

To illustrate the entire Lagrangian, we decompose as follows

$$ L=L_0 + L_I $$

Feynman Diagrams

When we construct a Feynman Diagram, we obtain not just the products, but also the probability amplitude and the operator of the process. To illustrate how Feynman Diagrams facilitate computation, here is a few common examples of quantum electrodynamics.

Consider Møller scattering,

it is pretty straightforward that the operator governing the process needed S-Matrix expansions and it is expected the resultant electrons can have two different configuration of momenta.

To begin, we start with

$$ S^{(2)}(2e^-\rightarrow 2e^-)=S_a+S_b $$

We expect the scattering will have two variables, t-channel and u-channel, so our final operator will be the superposition of two operators.

First, we consider t-channel, where

$$ t=(p_1-p'_1)^2=(p_2-p'_2)^2 $$

Obviously, our operator will be

$$ S_a=-e^2 \int d^4x_1 d^4x_2 N[(\bar{\phi}^-_{p'_1}\gamma^\alpha \phi^+_{p_1})_{x_1}(\bar{\phi}^-_{p'_2}\gamma^\beta \phi^+_{p_2})_{x_2}]iD_{F\alpha\beta}(x_1-x_2) $$

where $$\gamma^\alpha$$ are the gamma matrices, $$N[...]$$ denotes normal ordering and $$D_{F\alpha\beta}(x_1-x_2)$$ is the photon propagator.

Similarly, we would expect the u-channel, where

$$ u=(p_1-p'_2)^2=(p_2-p'_1)^2 $$,

will closely resemble t-channel and thus, the other operator will be

$$ S_b=-e^2 \int d^4x_1 d^4x_2 N[(\bar{\phi}^-_{p'_2}\gamma^\alpha \phi^+_{p_1})_{x_1}(\bar{\phi}^-_{p'_1}\gamma^\beta \phi^+_{p_2})_{x_2}]iD_{F\alpha\beta}(x_1-x_2) $$

So, here it is, the operator governing the Møller scattering has been constructed.

To illustrate how Feynman Diagrams facilitate construction of Feynman amplitude, consider electron scattering by an external field.

To put it simple, we read the diagram from left to the right.

$$ M^{(0)}=ie_0 \bar{u}(p')\gamma^\nu A _{e\nu}(p'-p) u(p) $$

and that the operator expected will be

$$ S_e^{(1)}=ie\int d^4 x \bar{\psi}^-(x) \gamma^\nu A_{e\nu}(x) \psi^+(x) $$

So, to this point, you may wonder, what does the operator and the amplitude do?

Well, we can obtain the transition of particles from state to state. Consider the above example, as we have already constructed our operator and amplitude already. The S-Matrix elements for the transition

$$ |i\rangle=|e^- p\rangle \rightarrow |f\rangle=|e^- p'\rangle $$

It is expected that the uncontracted operators of positive frequency will act on $$|i\rangle$$ to give $$|0\rangle$$ and negative frequency ones will, in turn, act on $$|0\rangle$$ to give $$|f\rangle$$ to complete the process described.

As we will only consider electrons in the example, I will not list the positron ones here, although they are equally important.

$$\psi^+(x)|e^- p\rangle = |0\rangle (\frac{m}{VE_p})^{1/2} u(p) e^{-ipx}$$

$$\bar{\psi}^- |0\rangle = \sum |e^- p\rangle(\frac{m}{VE_p})^{1/2} \bar{u}(p) e^{ipx}$$

where the summation is over spins, states and momenta.

One finds

$$\langle f|S^{(1)}_e|i \rangle = (2\pi) \delta^{(4)}(E'-E) (\frac{m}{VE_p})^{1/2}(\frac{m}{VE_{p'}})^{1/2} M $$

where $$M$$ is given above. Please note the replacement

$$(2\pi)^4\delta^{(4)}(p'+k-p)\rightarrow (2\pi)\delta(E'-E)$$

As the photon is from an external source, the momentum is not conserved, so instead, we consider energy.