Position space Feynman rules

Last updated Aug 5, 2023

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The Schwinger-Dyson equations can be used to calculate correlation functions

If $\mathcal{L}\text{int}=0$ we can directly calculate correlation functions e.g. with the Feynman propagator $D_F(x_i,x_j)=:D{ij}$

2-point function

$$\braket{\phi_1\phi_2}=D_{12}$$

Derivation of 2-point function

Since the Feynman propagator is the greens function it is given by $$\Box_x D_{x1}=-i\delta_{x1},$$ where $\delta_{x1}=\delta^4(x-x_1)$

We can artificially introcude an integral $$\braket{\phi_1\phi_2} = \int d^4x \delta_{x1}\braket{\phi_x\phi_2} = i\int d^4x (\Box_xD_{x1})\braket{\phi_x\phi_2}$$ we partially integrate at this point why do the boundary terms of time integration vanish? and use the Schwinger-Dyson equations $\Box_x \braket{\phi_x\phi_y} = -i\delta_{xy}$ $$\braket{\phi_1\phi_2} = i\int d^4xD_{x1}\Box_{x}\braket{\phi_x\phi_2}=\int d^4xD_{x1}\delta_{x2} = D_{12}$$

4-point function

$$\braket{\phi_1\phi_2\phi_3\phi_4} = D_{12}D_{34} + D_{13}D_{24} + D_{14}D_{23}$$

Derivation of 4-point function

Same as for the 2-point functions, by applying the Schwinger-Dyson equations to $\Box_x \braket{\phi_x\phi_2\phi_3\phi_4}$ after partial integration and resolving the resulting 2-point functions with the result above.

For an interacting theory we solve perturbatively. E.g. take $\mathcal{L}= -\frac{1}{2}\phi\Box\phi + \frac{g}{3!}\phi^3$

For the 2-point function we follow the derivation of the free case by

1. Adding an integral $\int d^4x \delta_{x1}$
2. Replace $\Box_x D_{x1} = -i\delta_{x1}$
3. Integrate partially to get $$\braket{\phi_1\phi_2} = i\int d^4x D_{1x}\Box_x\braket{\phi_x\phi_2}$$ Now when applying the Schwinger-Dyson equations we need to take in the interaction term, which leads to two terms:

• The term involving $\delta_{x2}$ which we get in the free theory
• The interaction term from $\mathcal{L}\text{int}^\prime$ $$\braket{\phi_1\phi_2} = i\int d^4x D{1x}(\frac{g}{2}\braket{\phi^2_x\phi_2}-i\delta_{x2})$$ To treat the interaction term we again go through the steps 1. 2. and 3. $$\braket{\phi_1\phi_2} = D_{12} - \frac{g}{2}\int d^4x d^4y D_{x1}D_{y2} \Box_y \braket{\phi^2_x\phi_y},$$ where we can again use Schwinger-Dyson equations to get a $\delta$-term $\propto g\braket{\phi_x}$ and an interaction term $\propto g^2 \braket{\phi^2_x\phi^2_y}$, which can be treated with Schwinger-Dyson equations again. To get a perturbative result one can terminate at some order of $g$, e.g. $$\braket{\phi_1\phi_2} = D_{12} - g^2 \int d^4xd^4y \left(\frac{1}{2}D_{1x}D^2_{xy}D_{y2} + \frac{1}{4}D_{1x}D_{xx}D_{yy}D_{y2} + \frac{1}{2}D_{1x}D_{2x}D_{xy}D_{yy}\right)$$

Note:

• The new terms correspond to new “diagrams”
• The last three terms are integrated over since we inserted the integrals
• The last three terms correspond to loops

These calculations can be generalized and represented as pictoral diagram, with the Feynman rules:

Position space Feynman rules

To calculate a correlation function $\braket{\phi_1\dotsc\phi_n}$:

Draw diagrams corresponding to the different terms of the perturbative expansion

1. Start with $n$ external points $x_i$ representing the positions of the field in the correlator
2. Draw a line from each point. A line can

• Contract with another line giving a factor of the Feynman propagator between the two endpoints $D_{xy}$
• Split at a new internal vertex giving a factor $ig$ with the coupling $g$. At the new vertex new lines emerge as many as there are fields in the interaction lagrangian

3. The Result is the sum of all diagrams with all lines connected and the position of internal vertices integrated over
4. Drop the $n!$ factor from the interaction term and divide by the symmetry factor (permutations in the interaction give the same diagram)

We can use time dependent perturbation theory: We have $$H = H_0 + V,$$ where the solution of $H_0$ is known, e.g. $H_0$ could be the free hamiltonian and $V$ an interaction term.

We again calculate the $\phi^3$ example: $$V(t) = \int d^3x \frac{g}{3!}\phi(\vec{x},t)^3.$$ We integrate over space since we are looking for the hamiltonian and not the density. The time dependence is in the field operators, as we are in the Heisenberg picture, where all time dependence is in operators.

For perturbation theory we change to the Interaction picture, where the time evolution of the free hamiltonian is factored out. At time $t_0$ the free fields are equal to the Heisenberg picture fields. Then:

• The free Hamiltonian $H_0$ is time-independent
• When free fields evolve with $H_0$ as $i\partial_t\phi_0(x,t) = [\phi_0(x),H_0]$ we get $$\phi_0(\vec{x},t) = e^{iH_0(t-t_0)}\phi(\vec{x},t_0)e^{-iH_0(t-t_0)}$$
• So at $t=t_0$ the free fields equal the interacting field $\phi_0(\vec{x},t_0) = \phi(\vec{x},t_0)$
• The full Hamiltonian may be time-dependent
• The Heisenberg picture fields $\phi(x,t)$ evolve with $H(t)$ as $i\partial_t\phi = [\phi(x),H(t)]$ $$\phi(\vec{x},t) = S^\dagger(t,t_0)\phi(\vec{x},t_0)S(t,t_0)$$ with the evolution operator $S(t,t_0)$ as $$i\partial_t S(t,t_0) = S(t,t_0) H(t)$$
• Since at $t_0$ the Heisenberg fields equal the free ones we have $$\phi(\vec{x},t) = U^\dagger(t,t_0)\phi_0(\vec{x},t)U(t,t_0),$$ with $U(t,t_0)=e^{iH_0(t-t_0)}S(t,t_0)$.
• Using the defining equation for $S$ one can derive a differential equation for $U$ $$i\partial_t U(t,t_0) = V_I(t)U(t,t_0),$$ with $V_I(t)=e^{iH_0(t-t_0)}V(t_0)e^{-iH_0(t-t_0)}$ the potential in interaction picture
• Vacuum state in Heisenberg-Picture $\ket{\Omega}$ and in interaction picture $\ket{0}$ are different (vacuum states annihilated by heisenberg / interaction picture operators…)
• At $t=-\infty$ the states are the same, but evolve differently with time. Up to a normalization: $$\ket{\Omega} \propto \lim_{t\to -\infty}S^\dagger(t,t_0)e^{-H_0(t-t_0)}\ket{0}=U(0,-\infty)\ket{0}$$
• From this one can derive an expression for correlation functions

Correlation functions

$$\braket{\Omega|T{\phi(x_1)\dotsc\phi(x_n)}|\Omega}=\frac{\braket{0|T{\phi_0(x_1)\dotsc\phi_0(x_n)\exp[-i\int^\infty_{-\infty}dt V_I(t)]}|0}}{T{\exp[-i\int^\infty_{-\infty} V_I(t)]}}$$

• At $t_0$ the interaction potential can be expressed in terms of free fields $$V(t_0) = \int d^3x \frac{g}{3!} \phi(\vec{x},t_0)^3 = \int d^3x \frac{g}{3!} \phi_0(\vec{x},t_0)^3$$
• Since the free states evolve with $H_0$ we then have $$V_I(t) = \int d^3x \frac{g}{3!}\phi_0(\vec{x},t)^3$$

The differential equation for the time evolution operator $U$ is $$i\partial_t U(t, t_0) = V(t)U(t,t_0),$$ which we can integrate with the condition $U(t_0,t_0)$ to $$U(t,t_0) = 1 - i \int^t_{t_0}dt^\prime V(t^\prime)U(t^\prime,t_0).$$ We can solve this perturbatively by inserting the first zeroth order solution $U(t^\prime, t_0)=1$, then inserting the first order solution $U(t,t_0) = 1 - i \int^t_{t_0}dt^\prime V(t^\prime)U(t^\prime,t_0),$ etc… $$U(t,t_0)=1 - i\int^t_{t_0}dt^\prime V(t^\prime) - \int^t_{t_0}dt^\prime \int^t_{t_0} dt^{\prime\prime}V(t^\prime)V(t^{\prime\prime}) + …$$ This can be expressed in terms of time-ordered products by using $$\int^t_{t_0}dt^\prime \int^{t^\prime}_{t_0} dt^{\prime\prime} V(t^\prime)V(t^{\prime\prime})=\int^t_{t^\prime}dt^{\prime\prime} \int^{t}_{t_0} dt^{\prime} V(t^{\prime\prime})V(t^{\prime})$$ and then writing $$ \begin{align} \int^t_{t_0} dt^\prime \int^{t^\prime}_{t_0} dt^{\prime\prime} V(t^\prime)V(t^{\prime\prime} &= \frac{1}{2}\int^t_{t_0}dt^\prime \left[\int^{t^\prime}_{t_0}dt^{\prime\prime} V(t^\prime)V(t^{\prime\prime}) + \int^{t}_{t^\prime}dt^{\prime\prime}V(t^{\prime\prime})V(t^\prime)\right]\\ &= \frac{1}{2}\int^t_{t_0}dt^\prime \int^t_{t_0}dt^{\prime\prime}T{V(t^\prime)V(t^{\prime\prime})} \end{align} $$ So the integrals from the perturbative solution can be expressed as time ordered product with the Dyson series

Dyson series

$$U(t,t_0) = T{\exp\left[-i\int^t_{t_0}dt^\prime V_I(t^\prime)\right]}$$

and with $V_I = - \int d^3x \mathcal{L}_\text{int}[\phi_0]$, finally

$$U_{ij} = T{\exp[i\int^{t_j}_{t_i}d^4x \mathcal{L}_\text{int}[\phi_0]]}$$

$$\braket{\Omega|T{\phi(x_1)\dotsc\phi(x_n)}|\Omega}=\frac{\braket{0|T{\phi_0(x_1)\dotsc\phi_0(x_n)e^{i\int d^4 \mathcal{L}_\text{int}[\phi_0]}}|0}}{\braket{0|T{e^{i\int d^4x \mathcal{L}_\text{int}[\phi_0]}}|0}}$$

We use the derived formulas to again derive $\braket{\Omega|T{\phi(x_1)\phi(x_2)}|\Omega}$ with $\mathcal{L}_\text{int}[\phi] = \frac{g}{3!}\phi^3$.

To do so we expand $\braket{0|T{\phi_0(x_1)\dotsc\phi_0(x_n)e^{i\int d^4 \mathcal{L}_\text{int}[\phi_0]}}|0}$ perturbatively in $g$ by using the expansion of the $\exp$ function.

We now only have to calculate correlation functions of products of free fields!

For that we split the fields $\phi_0(x)=\phi_+(x)+\phi_-(x)$ where $+$ corresponds to the creation operator part and $-$ to the annihilation operator part of the field. $$\phi_+(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2w_p}} a^\dagger_p e^{ipx}$$ $$\phi_-(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2w_p}} a_p e^{-ipx}$$

After this decomposition we have products of creation and annihilation fields. In order to give a finite result every created particle needs to be annihilated again, so the number of $\phi_+$ needs to equal the $\phi_-$. The result is calculated by building all possible $\phi_+$ and $\phi_-$ pairs called contractions according to Wicks theorem, which give a Feynman propagator.

In such calculations bubble diagrams appear.

Definition: bubble diagrams

Bubble diagrams are subgraphs in diagrams without connection to external points.

These bubble diagram contributions cancel with the normalization $\braket{0|Te^{i\int\mathcal{L}_\text{int}}|0}$, so that

$$\braket{\Omega|T{\phi(x_1)\phi(x_2)}|\Omega} = \braket{0|T{\phi_0(x_1)\phi_0(x_2)e^{i\int\mathcal{L}\text{int}}}|0}\text{no bubbles}$$

which means the subgraphs not connected to external points are excluded.

This leads to the same terms and thus Feynman rules as the lagrangian derivation.

The normalization factors for diagrams come from different sources:

• Factor $\frac{1}{m!}$ from the expansion of $\exp(i\mathcal{L}_\text{int})$
• At order $m$ there are $m$ interaction vertices. The positions of these vertices can be distributed in $m!$ ways leading to a Factor $m!$
• Factor $\frac{1}{j!}$ for the conventional normalization of the interaction
• Factor $j!$ for the permutation of identical lines at an interaction vertex