Understanding how to derive the Feynman rules out of the path integral

[JD_PM]

TL;DR Summary

I want to understand how to derive the Feynman rules out of the functional integral by means of working out the easiest potential $V(\phi(x))=\lambda \frac{\phi^3}{3!}$

I am studying interacting scalar fields (from Osborn) using the path integral approach.

We define the functional integral \begin{equation*}
Z[J] := \int d[\phi] e^{iS[\phi] + i\int d^d x J(x) \phi(x)} \tag{1}
\end{equation*}

The idea is to differentiate $Z[J]$ with respect to $J$ and end up defining correlation functions. We can define this integral by a perturbation expansion. This can be expressed in terms of Feynman diagrams, and for each diagram there is an amplitude given by the Feynman rules.

I see that, formally, $(1)$ takes the form

\begin{align*} Z[J] &= \exp\left(\frac{i}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \\ &\times \exp\left( i \int d^d x (-V(\phi(x))+J(x)\phi(x))\right) \Big|_{\phi=0} \end{align*}Then Osborn states to "expand this integral to get the perturbation expansion" and then he goes straight to explain the Feynman rules.

My issue is that I do not see how this perturbative expansion leads to write down the Feynman rules.

Could you please explain me the easiest case I could find, $V(\phi(x))=\lambda \frac{\phi^3}{3!}$? In other words, how to expand the following integral perturbatively

\begin{align*} Z[J] &= \exp\left(\frac{i}{2} \int d^d x d^d y \frac{\delta}{\delta \phi(x)} \Delta_F (x-y) \frac{\delta}{\delta \phi(y)}\right) \times \\ &\times \exp\left( i \int d^d x (-\lambda \frac{\phi^3}{3!}+J(x)\phi(x))\right) \Big|_{\phi=0} \tag{2} \end{align*}

So that I see how to establish the Feynman rules (i.e. to find out what the factor of (…) for a $\phi^3$ is etc), what kind of integrals (at least 1-loop) are associated to which diagrams and find all connected one- and two-loop graphs which contribute to $\langle \phi(x_1) \phi(x_2) \rangle$ and $\langle \phi(x_1) \phi(x_2) \phi(x_3)\rangle$

Please note that , to do so, we should only do an inspection of $(2)$ and not the explicit full expansion.

If you find an easier non-trivial potential to work with, please do so instead of using $\phi^3$

Source: Osborn notes, section 2.2. Interacting Scalar Field Theories

Thank you

 

[vanhees71]

That seems to be an interesting alternative to the usual approach to just calculate
$$Z_0[J]=\exp \left (-\frac{\mathrm{i}}{2} \int_{\mathbb{R}^4} \mathrm{d}^4 x \int_{\mathbb{R}^4} \mathrm{d}^4 y \Delta_{\text{F}}(x-y) J(x) J(y) \right).$$
and then use
$$Z[J]=\exp \left [-\mathrm{i} \int_{\mathrm{R}^4} \mathrm{d}^4 x V\left (-\mathrm{i} \frac{\delta}{\delta J(x)} \right ) \right] Z_0[J]$$
to derive the Dyson series for $N$-point functions. This functional approach is pretty straight ford then derive the Feynman rules by expanding the exponential and calculating the corresponding derivatives. At the end of course it boils down that the derivatives obey the same rules as the Wick theorem in the operator formalism. For details, using a diagrammatic approach using this formalism, see
sects. 4.5ff in

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

 

[JD_PM]

That seems to be an interesting alternative to the usual approach

Nice! If you do not mind, let us follow it

My knowledge on this topic is very limited so please let me work out the easiest example I could find in detail. That is: $\phi^3$ theory and the 2-point correlation function

I want to understand how the perturbation expansion really works so let me show all steps up to first order. (Hence we expect to get no loops here).

\begin{align*}
&\left.e^{\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}}e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&=\left[1+\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}+\cdots \right]\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&=\left[1+\frac{i}{2}\int d^d x\int d^d y \frac{\delta}{\delta\phi(x)}\Delta_F(x-y)\left( -\frac{\lambda}{2}i\phi^2(y)+iJ(y)\right) +\cdots\right]\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&=\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}+\frac{i}{2}\int d^d x\int d^d y\,\Delta_F(x-y)\left( -\lambda i\phi(y)\delta(x-y)\right)\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&\hphantom{=}+\frac{i}{2}\int d^d x \int d^d y\,\Delta_F (x-y)\left( -\frac{\lambda}{2}i\phi^2(y)+iJ(y)\right)\left( -\frac{\lambda}{2}i\phi^2(x)+iJ(x)\right)\left.e^{i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&\hphantom{=}+\cdots
\\
&=1+\frac{i}{2}\times 0+\frac{i}{2}\int d^d x \int d^d y\,(iJ(x))\Delta_F(x-y)(iJ(y))+\mathcal{O}(\Delta_F^2)
\end{align*}

Thus, up to first order, we deal with one propagator only and the external legs $iJ(x), iJ(y)$

It makes sense then that the 2-point correlation function up to first order yields

 

[JD_PM]

OK so far.

Let us complicate things a bit :) . Let us study the exact same problem but up to second order. Performing a completely analogous computation I get (I am not sure if this is correct; if not, please let me know and I will show all steps)

\begin{align*}
&\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right)\times \exp\left(i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0} \\
&=1+\frac{i}{2}\times 0+\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y)) \\
&+ \frac{(i)^2}{4}\int d^d x \int d^d y \int d^d t \int d^d \xi\,(iJ(x))\Delta_F(x-y)(iJ(y)) (iJ(t))\Delta_F(t-\xi)(iJ(\xi)) + \mathcal{O}(\Delta_F^3)
\end{align*}

Thus, up to first order, we deal with two propagators and the external legs $iJ(x), iJ(y), iJ(t), iJ(\xi)$

Here I start to get lost, as when I checked Osborn notes, he shows a diagram with a loop

Why do we get a diagram with such a form? How can we deduce it from knowing that we have with two propagators and the external legs $iJ(x), iJ(y), iJ(t), iJ(\xi)$

 

[vanhees71]

Hm, the diagram you look for must of course be a contribution at $\mathcal{O}(\lambda^2)$ with only two $J$'s, i.a., the diagram is
$$\propto \int \mathrm{d}^4 z \int \mathrm{d}^4 w J(x) \Delta_F(x-z) \Delta_F^2(z-w) \Delta_F(w-y) J(y).$$ So you must have forgotten some terms in the derivatives.

 

[JD_PM]

Hm, the diagram you look for must of course be a contribution at $\mathcal{O}(\lambda^2)$ with only two $J$'s, i.a., the diagram is
$$\propto \int \mathrm{d}^4 z \int \mathrm{d}^4 w J(x) \Delta_F(x-z) \Delta_F^2(z-w) \Delta_F(w-y) J(y).$$ So you must have forgotten some terms in the derivatives.

Thank you very much for giving the answer!

As I suspected, this computation was not going to be straightforward . I am struggling a bit on how to deal with squared derivatives.

I will only focus on second order terms. This is what I have done

\begin{align*}
&\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right) \exp\left(i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0}
\\
&=\left[\cdots +\frac{(i)^2}{4}\int d^{2d} z \int d^{2d} w \left(\frac{\delta}{\delta \phi(z)}\right)^2\Delta_F^2(z-w) \left(\frac{\delta}{\delta \phi(w)}\right)^2+\cdots \right]\left.e^{i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&=\left[1+\frac{(i)^2}{4}\int d^{2d} z\int d^{2d} w \left(\frac{\delta}{\delta\phi(z)}\right)^2\Delta_F^2(z-w)\left(\frac{\delta}{\delta\phi(w)}\right)\left( -\frac{\lambda}{2}i\phi^2(w)+iJ(w)\right) +\cdots\right] \\
&\times \left.e^{i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\end{align*}

Am I correct so far?

I know you might not have time to check computations so I will keep trying till I get it

 

[JD_PM]

Am I correct so far?

No I am NOT!

The square of an integral is not the double integral of the square of the integrand i.e.

$$\left[\int dx f(x) \right]^2=\int dx f(x) \int dy f(y) \neq \int d^{2x} f^2(x)$$

Let's start over.

 

[PeterDonis]

Let us study the exact same problem but up to second order.

You're not studying "the exact same problem" if you change the number of external legs. With only two legs you are studying the propagator; with four legs you are studying scattering. These are different physical processes that lead to different sets of Feynman diagrams at each order and you shouldn't mix them up.

 

[JD_PM]

Alright, so based on @PeterDonis comment, I think that the best is to first write down which terms we expect to get within each order of perturbation theory (for $\phi^3$ and the two-point correlation function) before going through the computation

• Zeroth order ($\mathcal{O}(\lambda^0)$)

We just get $1$, no propagator. I guess there is no diagram for $\mathcal{O}(\lambda^0)$

• First order ($\mathcal{O}(\lambda^1)$)

We get the propagator $\Delta_F(x-y)$, with two external legs $iJ(x), iJ(y)$. The diagram is shown at #3

• Second order ($\mathcal{O}(\lambda^2)$)

We get two propagators, $\Delta_F(x-y)$ and $\Delta_F(z-w)$, with four external legs $iJ(x), iJ(y), iJ(z), iJ(w)$. I still do not know how to draw this diagram.So following the pattern one sees that

• nth order ($\mathcal{O}(\lambda^n)$)

We get $n$ propagators with $2n$ external legs.

At this point I have some questions

Hm, the diagram you look for must of course be a contribution at $\mathcal{O}(\lambda^2)$ with only two $J$'s, i.a., the diagram is
$$\propto \int \mathrm{d}^4 z \int \mathrm{d}^4 w J(x) \Delta_F(x-z) \Delta_F^2(z-w) \Delta_F(w-y) J(y).$$ So you must have forgotten some terms in the derivatives.

1) Mmm here we have $4$ propagators so I would say this term corresponds to $\mathcal{O}(\lambda^4)$, not to $\mathcal{O}(\lambda^2)$. Is this a typo? .

2) I see we only have $2$ $J$'s, instead of the expected $8$ (based on the above's pattern). Is there a restriction that modify the number of $J'$s? I've read about a rule called "Euler's formula"

\begin{equation*}
L=I-V + 1
\end{equation*}

Where $L$ is the number of loops, $I$ is the number of internal lines and $V$ is the number of vertices.

 

[PeterDonis]

based on @PeterDonis comment, I think that the best is to first write down which terms we expect to get within each order of perturbation theory

Apparently you didn't read my comment carefully, since you are still changing the number of external legs.

What you need to do first, before anything else, is to decide what physical process you are going to study. That decision determines the number of external legs. For example, if you are going to study the simple process of one particle going from place to place, you have two external legs. If you are going to study the process of two particles scattering off each other, you have four external legs. Or you can even study the "vacuum-vacuum" process, with zero external legs. (This is what you would study, for example, if you were trying to estimate "the energy density of the vacuum", say to try to predict a theoretical value for the cosmological constant.)

These are not different orders of perturbation theory. They are different physical processes. You do perturbation theory separately, order by order, for each process.

For example, if you are studying simple propagation (one particle going from place to place), the single zero order diagram is just the straight line: one propagator, two external legs, no vertices. There is no first order diagram for the $\phi^3$ theory for this case because it is impossible to have a diagram with two external legs and one vertex. (I suggest thinking carefully about this until you realize why; it should help you to understand how the diagrams in general are constructed.) There is only one second order diagram: two external legs, a loop in the middle, and each external leg going into its own vertex on the loop, so two vertices. (Again, I suggest thinking carefully about this until you realize why this is the only possible second order diagram.)

If you are studying scattering, you have four external legs, and there are now two zero order diagrams (i.e., diagrams with no vertices). There are still no first order diagrams. (Can you see why?) There are now three second order diagrams. (Again, can you see why?)

You need to take a step back and rethink your entire approach in the light of the above.

 

[JD_PM]

@PeterDonis extremely helpful comment, I really appreciate it

What you need to do first, before anything else, is to decide what physical process you are going to study. That decision determines the number of external legs.

I am interested in studying $2-$point and $3-$point functions, so two different physical processes (the former with two external legs, the latter with $3$), for $\phi^3$ theory.

I do think that the key of this problem is understanding how to argue the following comments

For example, if you are studying simple propagation (one particle going from place to place), the single zero order diagram is just the straight line: one propagator, two external legs, no vertices. There is no first order diagram for the $\phi^3$ theory for this case because it is impossible to have a diagram with two external legs and one vertex. (I suggest thinking carefully about this until you realize why; it should help you to understand how the diagrams in general are constructed.)

I was going to argue it in terms of conservation of momentum. However, let us not do so and assume we are working in configuration space. By definition, an external leg is represented by a vertex with one outgoing line. Hence, if we were to add another external leg we would necessarily need to add another vertex. Do you agree?

There is only one second order diagram: two external legs, a loop in the middle, and each external leg going into its own vertex on the loop, so two vertices. (Again, I suggest thinking carefully about this until you realize why this is the only possible second order diagram.)

Mmm again really interesting. I would argue this one based on Euler's formula. We have the number of vertices fixed so $L = I -2 +1$. A negative number of loop (I think) does not make sense so at least we need to start with $I=1$. In that case $L =0$ and we recover the zeroth order. If we set $I=2$ we get $L=1$, which is precisely the diagram you pointed out i.e.

I would say this is the unique diagram for $\mathcal{O}(\lambda^2)$ because is the only one satisfying Euler's formula.

Mmm but I think you will not be completely satisfied with my answer, given that Euler's formula only works with connected diagrams. So please feel free to add more insight

Actually, why should not we go a bit further? Let's say we want to find diagrams for $\mathcal{O}(\lambda^4)$. I would again make use of Euler's formula; we know that $L = I -4 +1$. So given that there can only be 2 external legs, the only diagram I see that satisfies $L = I -4 +1$ for $\mathcal{O}(\lambda^4)$ is $I = 5$ i.e.

Do you agree so far?

If you are studying scattering, you have four external legs, and there are now two zero order diagrams (i.e., diagrams with no vertices). There are still no first order diagrams. (Can you see why?) There are now three second order diagrams. (Again, can you see why?)

Please let us first ensure the above is correct. I will of course argue these then.

Again, thank you VERY much. I honestly feel I have learned a lot of conceptual understanding after going through your post.

 

[PeterDonis]

By definition, an external leg is represented by a vertex with one outgoing line.

No. An external leg is an external leg. It does not need to end on a vertex.

Hence, if we were to add another external leg we would necessarily need to add another vertex. Do you agree?

No. You can have a propagator line going from one external leg to the other with no vertex in between. Go back and re-read my description of the zero order diagram for the case of two external legs.

In fact, your reasoning is obviously wrong just from the fact that there can be zero-order diagrams at all. A zero-order diagram, by definition, has no vertex in it. Your reasoning, if correct, would imply that it's impossible to have any diagram with no vertex in it.

I would say this is the unique diagram for $\mathcal{O}(\lambda^2)$ because is the only one satisfying Euler's formula.

I'll have to consider this and the rest of your post. At a glance, your statement of Euler's formula doesn't look correct for the zero order case; the only $V = 0$ diagram has $I = 0$ (no internal lines), but your formula says it should have $L = 1$ (one loop), and it doesn't.

 

[PeterDonis]

Do you agree so far?

As far as whether the diagram you gave is the only possible fourth order diagram, no, I don't agree. There is at least one obvious fourth order diagram (a connected one, so the issue I'll mention in the next paragraph doesn't apply) that you have missed.

In fact, now that you've reminded me that we should also consider disconnected diagrams, I'm not even sure there is only one second order diagram.

 

[JD_PM]

At a glance, your statement of Euler's formula doesn't look correct for the zero order case; the only $V = 0$ diagram has $I = 0$ (no internal lines), but your formula says it should have $L = 1$ (one loop), and it doesn't.

I think this is because the diagram for the zero order case is disconnected, for which Euler's formula does not work. Euler's formula works for connected diagrams only.

For the sake of simplicity, it is OK for me if we only focus on connected diagrams.

I will wait until you post your next reply EDIT: Well, let me add the following

There is no first order diagram for the $\phi^3$ theory for this case because it is impossible to have a diagram with two external legs and one vertex.

In $\phi^3$ we need every vertex to be connected to three lines. Hence a diagram with two external legs with only two vertices is not allowed.

 

[PeterDonis]

the diagram for the zero order case is disconnected

No, it isn't. It just has one line connecting the two external legs, and nothing else. How is that disconnected?

"Disconnected" means there are two or more "pieces" that aren't connected to each other by any lines. For example, if you took the zero order diagram for two external legs and added some vacuum-vacuum diagram as a separate "piece", that would be a disconnected diagram for the case of two external legs, at whatever order the vacuum-vacuum diagram was.

In $\phi^3$ we need every vertex to be connected to three lines. Hence a diagram with two external legs with only two vertices is not allowed.

I think you mean only one vertex.

 

[PeterDonis]

I've read about a rule called "Euler's formula"

\begin{equation*}
L=I-V + 1
\end{equation*}

Where $L$ is the number of loops, $I$ is the number of internal lines and $V$ is the number of vertices.

Where have you read this? The usual presentations of Euler's formula talk about vertices, edges, and faces, and the formula is $V - E + F = 2$.

 

[JD_PM]

Where have you read this? The usual presentations of Euler's formula talk about vertices, edges, and faces, and the formula is $V - E + F = 2$.

Osborn notes, page 24

PS: I have been studying the $2-$point function for $\phi^3$. I will share it to check my understanding

 

[PeterDonis]

Osborn notes, page 24

Hm. His definition of "connected" vs. "disconnected" graphs seems to match mine (see the discussion and examples on p. 26--all of the examples there have four external legs, not two, but the pattern seems clear). But it is still the case that the zero order diagram we have been using for the case of two external legs does not appear to satisfy his version of Euler's formula, even though it is connected by his definition. So I am still not sure about that formula.

 

[PeterDonis]

I am still not sure about that formula.

To be clear, what I'm not sure about is not Euler's formula for planar graphs itself; that's a well known mathematical theorem. What I'm not sure about is how to count loops, vertices, and internal lines in Feynman diagrams in order to correctly apply the formula to them. It's possible that the formula simply does not apply to the zero order 2-external leg diagram, with $I = V = 0$ (since by the formula that diagram should have one loop), or in general to diagrams with no internal lines and/or no vertices.

 

[JD_PM]

It's possible that the formula simply does not apply to the zero order 2-external leg diagram, with $I = V = 0$ (since by the formula that diagram should have one loop)

Yes, I was going to reply a similar idea: I suspect Euler's formula does not apply to non-interacting diagrams (i.e. diagrams coming from the free Lagrangian density).

 

[JD_PM]

Thank you very much @PeterDonis & @vanhees71, I think I got it!

Let us assume that Euler's formula applies only to interacting terms in the Lagrangian density.

Let's rewrite #9 and write down all possible one-loop, two-loop connected diagrams (and the zeroth order i.e. the propagator) for the $2-$point function in $\phi^3$

• Zeroth order ($\mathcal{O}(\lambda^0)$)

We just get one diagram: the propagator.

• First order ($\mathcal{O}(\lambda^1)$)

There is no diagram. In $\phi^3$, we need every vertex to be connected to three lines. Hence a diagram with two external legs and only one vertex is not allowed.

• Second order ($\mathcal{O}(\lambda^2)$)

A connected one-loop diagram

This is the unique diagram because there is no other way of connecting two external legs with two vertices, which have (each) three lines associated to it.

• Third order ($\mathcal{O}(\lambda^2)$)

There is no contribution

• Fourth order ($\mathcal{O}(\lambda^2)$)

There are two connected two-loop diagrams:

If I am not mistaken, those are all one-loop, two-loop connected diagrams for the $2-$point function in $\phi^3$.

 

[JD_PM]

A bit of an extra question: I have been trying to obtain all one-loop, two-loop connected diagrams for the $3-$point function in $\phi^4$. However, I did not come up with any.

Let us focus on the lowest order: the idea is take three external legs and connect them to one vertex, which has 4 lines connected to it. However, I do not see how to do so.

 

[nrqed]

Thank you very much for giving the answer!

As I suspected, this computation was not going to be straightforward . I am struggling a bit on how to deal with squared derivatives.

I will only focus on second order terms. This is what I have done

\begin{align*}
&\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right) \exp\left(i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0}
\\
&=\left[\cdots +\frac{(i)^2}{4}\int d^{2d} z \int d^{2d} w \left(\frac{\delta}{\delta \phi(z)}\right)^2\Delta_F^2(z-w) \left(\frac{\delta}{\delta \phi(w)}\right)^2+\cdots \right]\left.e^{i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\\
&=\left[1+\frac{(i)^2}{4}\int d^{2d} z\int d^{2d} w \left(\frac{\delta}{\delta\phi(z)}\right)^2\Delta_F^2(z-w)\left(\frac{\delta}{\delta\phi(w)}\right)\left( -\frac{\lambda}{2}i\phi^2(w)+iJ(w)\right) +\cdots\right] \\
&\times \left.e^{i \int d^d t \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)}\right|_{\phi=0}
\end{align*}

Am I correct so far?

I know you might not have time to check computations so I will keep trying till I get it

I am not sure if you are still interested in getting the rules from the path integral, but the key point is that when you square you will have to apply
\begin{align*}
\frac{(i)^2}{4}\int d z \, dw \, dx \, dy \,\left(\frac{\delta}{\delta\phi(z)}\right)\Delta_F(z-w)\left(\frac{\delta}{\delta\phi(w)}\right)
\left(\frac{\delta}{\delta\phi(x)}\right)\Delta_F(y-x)\left(\frac{\delta}{\delta\phi(y)}\right)
\end{align*} on the exponential.

A bit of an extra question: I have been trying to obtain all one-loop, two-loop connected diagrams for the $3-$point function in $\phi^4$. However, I did not come up with any.

Let us focus on the lowest order: the idea is take three external legs and connect them to one vertex, which has 4 lines connected to it. However, I do not see how to do so.

Right, there is no contribution to the three point function in $\phi^4$ theory in the first order in the coupling constant (so with a single vertex).

 

[PeterDonis]

If I am not mistaken, those are all one-loop, two-loop connected diagrams for the $2-$point function in $\phi^3$.

I can think of one other two vertex, one loop diagram for second order:

Have one vertex on the line between the two external legs. The third line coming out of this vertex (the first two are going to the legs) goes to a second vertex which has a loop hanging off of it. There are no other vertexes on the loop.

This diagram appears to obey the Feynman rules for $\phi^3$ theory.

 

[JD_PM]

I am not sure if you are still interested in getting the rules from the path integral, but the key point is that when you square you will have to apply
\begin{align*}
\frac{(i)^2}{4}\int d z \, dw \, dx \, dy \,\left(\frac{\delta}{\delta\phi(z)}\right)\Delta_F(z-w)\left(\frac{\delta}{\delta\phi(w)}\right)
\left(\frac{\delta}{\delta\phi(x)}\right)\Delta_F(y-x)\left(\frac{\delta}{\delta\phi(y)}\right)
\end{align*} on the exponential.

I am, but decided to first gain conceptual understanding. The result of the second-order computation can be found here.

Right, there is no contribution to the three point function in $\phi^4$ theory in the first order in the coupling constant (so with a single vertex).

Thank you for point this out! Then let's focus on the three point function in $\phi^4$ at $\mathcal{O}(\lambda^2)$.

For some reason, I do not see how to pictorially combine three external legs and connect them to two vertices (which each has 4 lines connected to it). I have been looking for books with an explanation on that kind of diagrams but did not find any.

Peskin & Schroeder work with $\phi^4$, but only with the 2-point function (unless I am missing it). Might you please give me a hint on how to draw such contribution or point me at bibliography?

 

[JD_PM]

• Third order ($\mathcal{O}(\lambda^2)$)

There is no contribution.

• Fourth order ($\mathcal{O}(\lambda^2)$)

Typo, I of course meant

• Third order ($\mathcal{O}(\lambda^3)$)

There is no contribution.

• Fourth order ($\mathcal{O}(\lambda^4)$)

I can think of one other two vertex, one loop diagram for second order:

Have one vertex on the line between the two external legs. The third line coming out of this vertex (the first two are going to the legs) goes to a second vertex which has a loop hanging off of it. There are no other vertexes on the loop.

This diagram appears to obey the Feynman rules for $\phi^3$ theory.

Thank you @PeterDonis, I missed that one!

Oh, so

There is only one second order diagram: two external legs, a loop in the middle, and each external leg going into its own vertex on the loop, so two vertices. (Again, I suggest thinking carefully about this until you realize why this is the only possible second order diagram.)

It turns out there is not only one second order diagram but two (?). What was your reasoning #9 that made you conclude that there was only one possible second order diagram?

 

[nrqed]

I am, but decided to first gain conceptual understanding. The result of the second-order computation can be found here.
Thank you for point this out! Then let's focus on the three point function in $\phi^4$ at $\mathcal{O}(\lambda^2)$.

For some reason, I do not see how to pictorially combine three external legs and connect them to two vertices (which each has 4 lines connected to it). I have been looking for books with an explanation on that kind of diagrams but did not find any.

Peskin & Schroeder work with $\phi^4$, but only with the 2-point function (unless I am missing it). Might you please give me a hint on how to draw such contribution or point me at bibliography?

Indeed, one cannot have a three point function in pure $\phi^4$ theory. This can be seen from the fact that if we use V vertices and we connect three of them to external states, this necessarily leaves an odd number of legs to pair up, which is impossible. So the three point function vanishes, to all order in perturbation theory.

 

[JD_PM]

Right, there is no contribution to the three point function in $\phi^4$ theory in the first order in the coupling constant (so with a single vertex).

Alright, I find no diagram with three external legs and having 4 connected lines for each vertex (i.e. no order in perturbation theory seems to satisfy such rules) .

Does the 3-point correlation function have contributions for $\phi^4$? I start to think that the answer is no.

EDIT: @nrqed has just answer the question!

 

[JD_PM]

• Fourth order ($\mathcal{O}(\lambda^4)$)

There are two connected two-loop diagrams:

View attachment 279728 View attachment 279729

If I am not mistaken, those are all one-loop, two-loop connected diagrams for the $2-$point function in $\phi^3$.

Well, once we include tadpoles, $7$ more contributions to $\mathcal{O}(\lambda^4)$ arise (just as one more arose for $\mathcal{O}(\lambda^2)$). Source: Srednicki QFT, chapter 9

 

[PeterDonis]

What was your reasoning #9 that made you conclude that there was only one possible second order diagram?

I had forgotten about tadpoles.

 

[nrqed]

Well, once we include tadpoles, $7$ more contributions to $\mathcal{O}(\lambda^4)$ arise (just as one more arose for $\mathcal{O}(\lambda^2)$). Source: Srednicki QFT, chapter 9

View attachment 279760

Good! One comment: in particle physics, people often omit the tadpoles. The reason for this is that they always disappear once one renormalizes. The reason is that the integrals over the loop momenta (that flow in the tadpoles) do not contain the external momenta. In the examples here, we then get that the tadpoles are just (divergent ) factors multiplying the two point functions, factors which are independent of the external momenta. After renormalization, these terms get canceled completely. However, when working in condensed matter systems (and at T not equal to zero), these tadpoles do not cancel and are therefore important.

Anyway, I know this is a technical point but I thought I would point it out.

[Moderator's note: An additional off topic comment has been deleted.]

 

[JD_PM]

I am not sure if you are still interested in getting the rules from the path integral, but the key point is that when you square you will have to apply
\begin{align*}
\frac{(i)^2}{4}\int d z \, dw \, dx \, dy \,\left(\frac{\delta}{\delta\phi(z)}\right)\Delta_F(z-w)\left(\frac{\delta}{\delta\phi(w)}\right)
\left(\frac{\delta}{\delta\phi(x)}\right)\Delta_F(y-x)\left(\frac{\delta}{\delta\phi(y)}\right)
\end{align*} on the exponential.

Please let us discuss the up-to-second-order result

\begin{align*}
&\exp\left(\frac{i}{2}\int d^d x \int d^d y \frac{\delta}{\delta \phi(x)}\Delta_F(x-y) \frac{\delta}{\delta \phi(y)}\right)\times \exp\left(i \int d^d z \left( -\frac{\lambda}{3!}\phi^3+J\phi \right)\right)\Big|_{\phi=0} \\
&=1+\frac{i}{2}\int d^d x \int d^d y\,(i(J(x))\Delta_F(x-y)(iJ(y)) + \\
&-\lambda \int d^d z \int d^d w \Delta_F (0) \Delta_F (z-w) \\
&-\frac{1}{4}\int d^d x \int d^d y \int d^d z \int d^d w \Delta_F(x-y) \Delta_F(z-w) J_x J_y J_z J_w + \mathcal{O}(\Delta_F^4)
\end{align*}

I conceptually understood that there are three diagrams for the $2-$function.

My question now is: how could we establish the Feynman rules out such result so that we can obtain such three diagrams algorithmically/systematically?

 

[JD_PM]

Hi @vanhees71 , I just wanted to ping you because I thought you might be interested in #33

 

[vanhees71]

Hi @vanhees71 , I just wanted to ping you because I thought you might be interested in #33

I think I've to do the calculation myself first. That needs a bit of time. I'm not sure whether this approach is in any way simpler than the standard approach though.