Position space Feynman rules

Last updated Dec 29, 2024

• Onepass

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 pushing the $\Box$ to the correlator (Boundary terms vanish because field vanishes at $\infty$) and use the Schwinger-Dyson equations $\Box_x \braket{\phi_x\phi_y} = -i\delta_{xy}$ for vanishing interaction $$\braket{\phi_1\phi_2} = i\int d^4xD_{x1}\Box_{x}\braket{\phi_x\phi_2}=\int d^4xD_{x1}\delta_{x2} = D_{12}$$ This was of course expected since this is the definition of the Feynman propagator

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.

• The result for the 4-point function can already be interpreted in a diagramatic way, drawing lines for each Feynman propagator

For an interacting theory we can solve n-point-functions similarly, here we get additional terms from the Schwinger-Dyson equations, since $\mathcal{L}^\prime_\text{int}$ is not zero.

Example: $\phi^3$ theory

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. this time for the second corrdinate $2$, intoducing a second integral $dy$: $$\braket{\phi_1\phi_2} = D_{12} - \frac{g}{2}\int d^4x d^4y D_{x1}D_{y2} \Box_y \braket{\phi^2_x\phi_y},$$ 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: $$\braket{\phi_1\phi_2}=D_{12} + ig\int d^4x D_{1x}D_{2x}\braket{\phi_x}-\frac{g^2}{4}\int d^4x d^4y D_{x1}D_{2y}\braket{\phi^2_x\phi^2_y}$$ The $\braket{\phi_x}$ and $\braket{\phi_x^2\phi_y^2}$ can both be expanded using Schwinger-Dyson again. At some point one can stop at a desired order in $g$. E.g. terminating at $g^2$ we get $$\braket{\phi_1\phi_2} = D_{12} - g^2 \int d^4xd^4y \left(\frac{1}{2}D_{1x}D^2_{xy}D_{y2} + \frac{1}{4}D_{1x}D_{xx}D_{yy}D_{y2} + \frac{1}{2}D_{1x}D_{2x}D_{xy}D_{yy}\right)$$

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 $g$ (corresponding to the coefficient of $i\mathcal{L}^\prime_\text{int}[\phi]$). 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)

• In 4 we can drop the $1/n!$ because we first get a factor $n$ from the derivative $\mathcal{L}^\prime$. If we then fix the incoming line we have $n-1$ lines. We can permute the fields for these lines leading to the same diagram with different assignments of the flieds to the lines. we have $n-1$ fields giving $(n-1)!$ permutations. Since all these permutations give the same diagram with the same contribution we have a factor $(n-1)!$. Therefore in total the factor $1/n!$ cancels.
• However sometimes permuting the outgoing lines leads to the exact same diagram (not with different field assignments but really the exact same diagram), e.g. when two outgoing lines at a vertex connect back to each other to form a loop. In this case the two permutations are the same diagram so we should not double count this diagram! This leads to the notion of the symmetry factor.

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.

We differ between:

1. Free fields $\phi_0(\vec{x},t)$:

• Evolve with the free Hamiltonian $H_0$: $$i\partial_t\phi_0(\vec{x},t) = [\phi_0(\vec{x}),H_0]$$
• The free Hamiltonian is time independent
• Free fields correspond to the Interaction picture (Time evolution of $H_0$ in the field operators, time evolution of $V$ in the states of the theory)

1. Ineracting fields $\phi(\vec{x},t)$: (Heisenberg picture)

• Evolve with the full Hamiltonian $$i\partial_t\phi = [\phi,H(t)]$$
• The full Hamiltonian can be time dependent.
• Interacting field correspond to the Heisenberg picture (All time evolution of $H=H_0 + V$ in the field operators)

At time $t_0$ the free fields are equal to the interacting fields.

The solution of the time evolution equation for the interacting fields can be written as $$\phi(\vec{x},t) = S^\dagger(t,t_0)\phi(\vec{x},t_0)S(t,t_0),$$ where $S(t,t_0)$ is the time evolution operator. Since $\phi(\vec{x},t)$ satisfies the the Heisenberg equation with the full Hamiltonian we can derive an equation for $S(t,t_0)$: $$i\partial_t S(t,t_0)=S(t,t_0)H(t).$$ Since at $t_0$ the free and interacting fields are equal, we can find a relation between $\phi(\vec{x},t)$ and $\phi_0(\vec{x},t)$ by “evolving $\phi_0$ backwards in time using $H_0$” until $t_0$ and then using the equation for the time evoultion of $\phi(\vec{x},t)$ in terms of $S$: $$\phi(\vec{x},t)=S^\dagger(t,t_0)e^{-H_0(t-t_0)}\phi_0(\vec{x},t)e^{iH_0(t-t_0)}S(t,t_0).$$ With this equation we can define the operator $U(t,t_0)$, which relates the free and interacting fields at the same time $t$ $$U(t,t_0)=e^{iH_0(t-t_0)}S(t,t_0),$$ so that $$\phi(\vec{x},t)=U^\dagger(t,t_0)\phi_0(\vec{x},t)U(t,t_0).$$Using the evolution 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

We drop the subscript $I$ of $V_I$. 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 $U(t_0,t_0)=1$ 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 in $V$ by inserting the zeroth order solution $U(t^\prime, t_0)=1$ to get the first order solution $U(t,t_0) = 1 - i \int^t_{t_0}dt^\prime V(t^\prime),$ then inserting this again, etc… $$U(t,t_0)=1 - i\int^t_{t_0}dt^\prime V(t^\prime) - \int^t_{t_0}dt^\prime \int^{t^\prime}_{t_0} dt^{\prime\prime}V(t^\prime)V(t^{\prime\prime}) + …$$ This can be expressed in terms of time-ordered products by using (since $t^{\prime\prime}<t^\prime$) $$\int^t_{t_0}dt^\prime \int^{t^\prime}_{t_0} dt^{\prime\prime}, V(t^\prime)V(t^{\prime\prime})=\int^t_{t^\prime}dt^{\prime\prime} \int^{t}_{t_0} dt^{\prime} V(t^{\prime\prime})V(t^{\prime})$$ 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

The solution for $U$ is given by the dyson series as time ordering of the exponential $$U(t,t_0) = T\bigg{\exp\left[-i\int^t_{t_0}dt^\prime V_I(t^\prime)\right]\bigg}.$$ Using $V_I = - \int d^3x \mathcal{L}_\text{int}[\phi_0]$ we can rewrite this, evolving from $t_i$ to $t_j$ $$U_{ij} = T\bigg{\exp\bigg[i\int^{t_j}_{t_i}d^4x \mathcal{L}_\text{int}[\phi_0]\bigg]\bigg}.$$ This is true for all times since at $t=t_0$ the free fields equal the interacting ones $\phi_0(\vec{x},t_0)=\phi(\vec{x},t_0)$ and plugging in the definition of $V_I$ we find that we can write it in terms of the time dependent free fields.

We note:

• $U^{-1}{ij} = U^\dagger{ij} = U_{ji}$

• 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: $$\ket{\Omega} \propto \lim_{t\to -\infty}S^\dagger(t,t_0)e^{-H_0(t-t_0)}\ket{0}=U(t_0,-\infty)\ket{0}$$
• However exept at $t=-\infty$ the two vacua $\ket{0}$ and $\ket{\Omega}$ are not the same.

• From this one can derive an expression for correlation functions in the interacting theory

Correlation functions

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

Derivation interacting correlation functions

Using the relation of the vacuum states $$\ket{\Omega}=\mathcal{N}U_{0 -\infty}\ket{0}$$ we can write a correlation function as $$\braket{\Omega|\phi(x_1)…\phi(x_n)|\Omega}\propto\braket{0|U_{\infty 0}U_{01}\phi_0(x_1)U_{10}U_{02}\phi_0(x_2)U_{20}…U_{0n}\phi_0(x_n)U_{n0}U_{0-\infty}|0},$$ where we used the relation between interacting and free fields $\phi(y)=U_{t_y 0}\phi_0(y)U_{0t_y}.$ Next we can rewrite this as $$\braket{0|U_{\infty 1}\phi_0(x_1)U_{12}\phi_0(x_2)U_{23}…U_{(n-1)n}\phi_0(x_n)U_{n-\infty}|0}.$$ Here we assumed that $t_1 > … > t_n$, therefore this is equally valid for a timeordered correlation function. $$\braket{\Omega|T{\phi(x_1)…\phi(x_n)}|\Omega}\propto\braket{0|T{U_{\infty 1}\phi_0(x_1)U_{12}…\phi_0(x_n)U_{n-\infty}|0}.$$ Now we can use that every $U_{t_it_j}$ are themselfs time ordered products only containing times between $t_i$ and $t_j$, all operators are already in time order! If they already are in time order although they are stading insinde a $T{…}$, we can rearrange them however we like. E.g. We can push all $U$s to the right side giving: $$\braket{0|T{\phi_0(x_1)…\phi_0(x_n)U_{\infty,-\infty}}|0}.$$

Since $\braket{\Omega|\Omega}=1$ we know that the normalization constant $\mathcal{N}=\braket{0|U_{\infty,-\infty}|0}^{-1}$. Therefore $$\braket{\Omega|T{\phi(x_1)…\phi(x_n)|0}=\frac{\braket{0|T{\phi_0(x_1)…\phi_0(x_n)U_{\infty,-\infty}|0}}{\braket{0|U_{\infty,-\infty}|0}},$$ where we can insert the definition of $U$ in terms of $V_I$. Then we can relate $V_I$ to the interaction by noting:

• 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 $$V_I(t) = \int d^3x \frac{g}{3!}\phi_0(\vec{x},t)^3$$

Note:

• This is manifestly lorentz invariant, since $\mathcal{L}_\text{int}$ is a Lorentz scalar.

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. $$\braket{0|T\big{\phi_0(x_1)\phi_0(x_2)e^{i\int d^4x \mathcal{L}_\text{int}[\phi_0]}|0}=T^{(0)}+T^{(1)}+T^{(2)}+…,$$ where the $T^{(i)}$ are the terms to order $i$ in $g$: $$T^{(0)}=\braket{0|T{\phi_0(x_1)\phi_0(x_2)|0}$$ $$T^{(1)}=\frac{ig}{3!}\int d^4x \braket{0|T{\phi_0(x_1)\phi_0(x_2)\phi_0(x)^3|0}$$ $$T^{(2)}=\left(\frac{ig}{3!}\right)^2\frac{1}{2}\int d^4x \int d^4y\braket{0|T{\phi_0(x_1)\phi_0(x_2)\phi_0(x)^3\phi_0(y)^3|0}$$

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, e.g. $$\braket{0|T{\phi_0(x_1)\phi_0(x_2)\phi_0(x)^3}|0}=\braket{0|T{\phi_+(x_1)\phi_+(x_2)\phi_+(x)^3|0} + \braket{0|T{\phi_-(x_1)\phi_+(x_2)\phi_+(x)^3}|0}+…$$In order to give a finite result every created particle needs to be annihilated again, so the number of $\phi_+$ needs to equal the $\phi_-$. Since the above correlator has an odd number of fields, it has to vanish since no pairing is possible. Moving on to the next term we have $$\braket{0|T{\phi_0(x_1)\phi_0(x_2)\phi_0(x)^3\phi_0(y)^3}|0}=\braket{0|T{\phi_+(x_1)\phi_+(x_2)\phi_+(x)^3\phi_+(y)^3|0}+…$$ The result is calculated by building all possible $\phi_+$ and $\phi_-$ pairs called contractions according to Wicks theorem, which give a Feynman propagator. If we insert a Feynman propagort $D_{xy}$ for every possible contraction we get for the above term: $$\braket{0|T{\phi_0(x_1)\phi_0(x_2)\phi_0(x)\phi_0(x)\phi_0(x)\phi_0(y)\phi_0(y)\phi_0(y)}|0}=$$ $$9D_{12}D_{xx}D_{xy}D_{yy}+6D_{12}D_{xy}$$ $$+18D_{1x}D_{2x}D_{xy}D_{yy}+9D_{1x}D_{2y}D_{xx}D_{yy}+18D_{1x}D_{2y}D_{xy}^2$$ $$+18D_{1y}D_{2y}D_{xy}D_{yy}+9D_{1y}D_{2x}D_{xx}D_{yy}+18D_{1y}D_{2x}D_{xy}^2$$ Here the prefactors arise by counting the number of possible contractions leading to the respective term. Putting this into the formular for the interacting correlation function for $T^{(2)}$ we get $$\braket{\Omega|T{\phi(x_1)\phi(x_2)}|\Omega}=\frac{1}{\braket{0|T{e^{i\int\mathcal{L}_\text{int}}}|0}}\left[D_{12}- g^2\int d^4x\int d^4y \left[\frac{1}{8}D_{12}D_{xx}D_{yy} + \frac{1}{12}D_{12}D_{xy}^3 + \frac{1}{2}D_{1x}D_{2x}D_{xy}D_{yy} + \frac{1}{4}D_{1x}D_{xx}D_{yy}D_{y2} + \frac{1}{2}D_{1x}D_{xy}^2D_{y2}\right]+…\right]$$ We can now connect this expansion and calculation to Feynman diagram, where each term represents a particluar diagram. The feynman rules for the diagrams are the same as before. However we have two additional terms and the normalization factor $\braket{0|T{e^{i\int\mathcal{L}_\text{int}}}|0}$.

The additional terms are referred to as bubble diagrams.

Definition: bubble diagrams

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

These bubble diagram contributions cancel with the normalization $\braket{0|T{e^{i\int\mathcal{L}_\text{int}}}|0}$, 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.

Cancellation of bubble diagrams

We can calculate $\braket{0|T{e^{i\int\mathcal{L}_\text{int}}}|0}$ by expanding the exponential and using wicks theorem again. this gives similar correlation functions as before but without external points: $$\braket{0|T{e^{i\int\mathcal{L}_\text{int}}}|0}=1+\left(\frac{ig}{3!}\right)^2\frac{1}{2}\int d^4x\int d^4y \braket{0|T{\phi_0(x)^3\phi_0(y)^3}|0} + …$$ Calculating the correlation function leads to terms corresponding exactly to the bubbles! (only the disconnected part) One can show that in the nominator, the bubbles always factorize: $$A=A|_\text{no bubbles}\times A|_\text{bubbles}.$$ Now since $$\braket{0|T{e^{i\int\mathcal{L}_\text{int}}}|0}=A|_\text{bubbles}$$ In the total interacting correlation function things cancel: $$\frac{A}{\braket{0|T{e^{i\int\mathcal{L}_\text{int}}}|0}}=A|_\text{no bubbles}$$

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

Position space Feynman rules

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