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# 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Position-space Feynman rules

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## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Position-space Feynman rules

The Schwinger-Dyson equations specify a completely non-perturbative relationship among correlation functions in the fully interacting theory. Some non-perturbative implications will be discussed in later chapters (in particular Sections 14.8 and 19.5). In this section, we will solve the Schwinger-Dyson equations in perturbation theory.

For efficiency, we write $\delta_{x i}=\delta^4\left(x-x_i\right)$ and $D_{i j}=D_{j i}=D_F\left(x_i, x_j\right)$. We will also set $m=0$ for simplicity (the $m \neq 0$ case is a trivial generalization), and $\hbar=1$. With this notation, the Green’s function equation for the Feynman propagator can be written concisely as
$$\square_x D_{x 1}=-i \delta_{x 1}$$
This relation can be used to rewrite correlation functions in a suggestive form. For example, the 2-point function can be written as
$$\left\langle\phi_1 \phi_2\right\rangle=\int d^4 x \delta_{x 1}\left\langle\phi_x \phi_2\right\rangle=i \int d^4 x\left(\square_x D_{x 1}\right)\left\langle\phi_x \phi_2\right\rangle=i \int d^4 x D_{x 1} \square_x\left\langle\phi_x \phi_2\right\rangle,$$
where we have integrated by parts in the last step. This is suggestive because $\square_x$ acting on a correlator can be simplified with the Schwinger-Dyson equations.

Now first suppose we are in the free theory where $\mathcal{L}{\text {int }}=0$. Then the 2-point function can be evaluated using the Schwinger-Dyson equation, $\square_x\left\langle\phi_x \phi_y\right\rangle=-i \delta{x y}$, to give
$$\left\langle\phi_1 \phi_2\right\rangle=\int d^4 x D_{x 1} \delta_{x 2}=D_{12}$$
as expected. For a 4-point function, the expansion is similar:
\begin{aligned} \left\langle\phi_1 \phi_2 \phi_3 \phi_4\right\rangle & =i \int d^4 x D_{x 1} \square_x\left\langle\phi_x \phi_2 \phi_3 \phi_4\right\rangle \ & =\int d^4 x D_{x 1}\left{\delta_{x 2}\left\langle\phi_3 \phi_4\right\rangle+\delta_{x 3}\left\langle\phi_2 \phi_4\right\rangle+\delta_{x 4}\left\langle\phi_2 \phi_3\right\rangle\right} \end{aligned}
Collapsing the $\delta$-functions and using Eq. (7.15), this becomes
\begin{aligned} & \left\langle\phi_1 \phi_2 \phi_3 \phi_4\right\rangle=D_{12} D_{34}+D_{13} D_{24}+D_{14} D_{23} \ & =\overbrace{x_2}^{x_1}+\bullet_{x_2}^{\bullet_3} \bullet_{x_4}^{x_1}+\bullet_{x_2}^{x_3} \bullet_{x_4}^{x_1} \ & \end{aligned}

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

In this section, we reproduce the position-space Feynman rules using time-dependent perturbation theory. Instead of assuming that the quantum field satisfies the EulerLagrange equations, we instead assume its dynamics is determined by a Hamiltonian $H$ by the Heisenberg equations of motion $i \partial_t \phi(x)=[\phi, H]$. The formal solution of this equation is
$$\phi(\vec{x}, t)=S\left(t, t_0\right)^{\dagger} \phi(\vec{x}) S\left(t, t_0\right)$$
where $S\left(t, t_0\right)$ is the time-evolution operator (the $S$-matrix) that satisfies
$$i \partial_t S\left(t, t_0\right)=H(t) S\left(t, t_0\right)$$
These are the dynamical equations in the Heisenberg picture where all the time dependence is in operators. States including the vacuum state $|\Omega\rangle$ in the Heisenberg picture are, by definition, time independent. As mentioned in Chapter 2, the Hamiltonian can either be defined at any given time as a functional of the fields $\phi(\vec{x})$ and $\pi(\vec{x})$ or equivalently as a functional of the creation and annihilation operators $a_p^{\dagger}$ and $a_p$. We will not need an explicit form of the Hamiltonian for this derivation so we just assume it is some time-dependent operator $H(t)$.
The first step in time-dependent perturbation theory is to write the Hamiltonian as
$$H(t)=H_0+V(t)$$
where the time evolution induced by $H_0$ can be solved exactly and $V$ is small in some sense. For example, $H_0$ could be the free Hamiltonian, which is time independent, and $V$ might be a $\phi^3$ interaction:
$$V(t)=\int d^3 x \frac{g}{3 !} \phi(\vec{x}, t)^3$$
The operators $\phi(\vec{x}, t), H, H_0$ and $V$ are all in the Heisenberg picture.
Next, we need to change to the interaction picture. In the interaction picture the fields evolve only with $H_0$. The interaction picture fields are just what we had been calling (and will continue to call) the free fields:
$$\phi_0(\vec{x}, t)=e^{i H_0\left(t-t_0\right)} \phi(\vec{x}) e^{-i H_0\left(t-t_0\right)}=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p e^{-i p x}+a_p^{\dagger} e^{i p x}\right) .$$
To be precise, $\phi(\vec{x})$ is the Schrödinger picture field, which does not change with time. The free fields are equal to the Schrödinger picture fields and also to the Heisenberg picture fields, by definition, at a single reference time, which we call $t_0$.

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Position-space Feynman rules

$$\square_x D_{x 1}=-i \delta_{x 1}$$

$$\left\langle\phi_1 \phi_2\right\rangle=\int d^4 x \delta_{x 1}\left\langle\phi_x \phi_2\right\rangle=i \int d^4 x\left(\square_x D_{x 1}\right)\left\langle\phi_x \phi_2\right\rangle=i \int d^4 x D_{x 1} \square_x\left\langle\phi_x \phi_2\right\rangle,$$

$$\left\langle\phi_1 \phi_2\right\rangle=\int d^4 x D_{x 1} \delta_{x 2}=D_{12}$$

\begin{aligned} \left\langle\phi_1 \phi_2 \phi_3 \phi_4\right\rangle & =i \int d^4 x D_{x 1} \square_x\left\langle\phi_x \phi_2 \phi_3 \phi_4\right\rangle \ & =\int d^4 x D_{x 1}\left{\delta_{x 2}\left\langle\phi_3 \phi_4\right\rangle+\delta_{x 3}\left\langle\phi_2 \phi_4\right\rangle+\delta_{x 4}\left\langle\phi_2 \phi_3\right\rangle\right} \end{aligned}

\begin{aligned} & \left\langle\phi_1 \phi_2 \phi_3 \phi_4\right\rangle=D_{12} D_{34}+D_{13} D_{24}+D_{14} D_{23} \ & =\overbrace{x_2}^{x_1}+\bullet_{x_2}^{\bullet_3} \bullet_{x_4}^{x_1}+\bullet_{x_2}^{x_3} \bullet_{x_4}^{x_1} \ & \end{aligned}

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

$$\phi(\vec{x}, t)=S\left(t, t_0\right)^{\dagger} \phi(\vec{x}) S\left(t, t_0\right)$$

$$i \partial_t S\left(t, t_0\right)=H(t) S\left(t, t_0\right)$$

$$H(t)=H_0+V(t)$$

$$V(t)=\int d^3 x \frac{g}{3 !} \phi(\vec{x}, t)^3$$

$$\phi_0(\vec{x}, t)=e^{i H_0\left(t-t_0\right)} \phi(\vec{x}) e^{-i H_0\left(t-t_0\right)}=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p e^{-i p x}+a_p^{\dagger} e^{i p x}\right) .$$

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