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# 物理代考|量子场论代考QUANTUM FIELD THEORY代考|PHYS5125

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## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Feynman rules for scalar QED

Expanding out the scalar QED Lagrangian we find
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^2-\phi^{\star}\left(\square+m^2\right) \phi-i e A_\mu\left[\phi^{\star}\left(\partial_\mu \phi\right)-\left(\partial_\mu \phi^{\star}\right) \phi\right]+e^2 A_\mu^2|\phi|^2 .$$
We can read off the Feynman rules from the Lagrangian. The complex scalar propagator is
$$=\frac{i}{p^2-m^2+i \varepsilon} \text {. }$$
This propagator is the Fourier transform of $\left\langle 0\left|\phi^{\star}(x) \phi(0)\right| 0\right\rangle$ in the free theory. It propagates both $\phi$ and $\phi^{\star}$, that is both particles and antiparticles at the same time – they cannot be disentangled.
The photon propagator was calculated in Section 8.5:
$$\sim m \sim=\frac{-i}{p^2+i \varepsilon}\left[g_{\mu \nu}-(1-\xi) \frac{p_\mu p_\nu}{p^2}\right],$$
where $\xi$ parametrizes a set of covariant gauges.

Some of the interactions that connect $A_\mu$ to $\phi$ and $\phi^{\star}$ have derivatives in them, which will give momentum factors in the Feynman rules. To see which momentum factors we get, look back at the quantized fields:
\begin{aligned} \phi(x) & =\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p e^{-i p x}+b_p^{\dagger} e^{i p x}\right), \ \phi^{\star}(x) & =\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p^{\dagger} e^{i p x}+b_p e^{-i p x}\right) . \end{aligned}

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|External states

Now we know the vertex factors and propagators for the photon and the complex scalar field. The only thing left in the Feynman rules is how to handle external states. For a scalar field, this is easy – we just get a factor 1 . That is because a complex scalar field is just two real scalar fields, so we just take the real scalar field result. The only thing left is external photons.
For external photons, recall that the photon field is
$$A_\mu(x)=\int \frac{d^3 k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_k}} \sum_{i=1}^2\left(\epsilon_\mu^i(k) a_{k, i} e^{-i k x}+\epsilon_\mu^{i \star}(k) a_{k, i}^{\dagger} e^{i k x}\right) .$$
As far as free states are concerned, which is all we need for $S$-matrix elements, the photon is just a bunch of scalar fields integrated against some polarization vectors $\epsilon_\mu^i(k)$. Recall that external states with photons have momenta and polarizations, $|k, \epsilon\rangle$, so that $\left\langle 0\left|A_\mu(x)\right| k, \epsilon_i\right\rangle=\epsilon_\mu^i(k) e^{-i k x}$. This leads to LSZ being modified only by adding a factor of the photon polarization for each external state: $\epsilon_\mu$ if it is incoming and $\epsilon_\mu^{\star}$ if it is outgoing.

For example, consider the following diagram:
where $k^\mu=p_1^\mu+p_2^\mu$. The first polarization $\epsilon_\mu^1$ is the polarization of the photon labeled with $p_\mu^1$. It gets contracted with the momenta $p_2^\mu+k^\mu$ which come from the $-i e A_\mu\left[\phi^{\star}\left(\partial_\mu \phi\right)\right.$ $\left.-\left(\partial_\mu \phi^{\star}\right) \phi\right]$ vertex. The other polarization, $\epsilon_\mu^4$, is the polarization of the photon labeled with $p_\mu^4$ and contracts with the second vertex.

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Feynman rules for scalar QED

$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^2-\phi^{\star}\left(\square+m^2\right) \phi-i e A_\mu\left[\phi^{\star}\left(\partial_\mu \phi\right)-\left(\partial_\mu \phi^{\star}\right) \phi\right]+e^2 A_\mu^2|\phi|^2 .$$

$$=\frac{i}{p^2-m^2+i \varepsilon} \text {. }$$

$$\sim m \sim=\frac{-i}{p^2+i \varepsilon}\left[g_{\mu \nu}-(1-\xi) \frac{p_\mu p_\nu}{p^2}\right],$$

\begin{aligned} \phi(x) & =\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p e^{-i p x}+b_p^{\dagger} e^{i p x}\right), \ \phi^{\star}(x) & =\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p^{\dagger} e^{i p x}+b_p e^{-i p x}\right) . \end{aligned}

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|External states

$$A_\mu(x)=\int \frac{d^3 k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_k}} \sum_{i=1}^2\left(\epsilon_\mu^i(k) a_{k, i} e^{-i k x}+\epsilon_\mu^{i \star}(k) a_{k, i}^{\dagger} e^{i k x}\right) .$$

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