Feynman Rules on ##\phi^3, \phi^4## theory

[JD_PM]

You have 3 external lines (representing the 3 particles at infinity) and 3 legs of the vertex to connect. Therefore 6 possibilities (3 for the first pair, 2 for the second, and the last one is immediately determined).

OK I get we have 3 external lines but what do you mean by 'legs of the vertex' ?

 

[Gaussian97]

Let's see if I can light your way, in the picture, I draw a first diagram, where you can see the three external lines (the outer ones) and the three legs of the vertex (the three lines touching the point). What you need is to pair these lines (one external with one leg of the vertex), there are six ways to do it, which I have drawn after the $=$ sign.

Note I am using the orientation of the external and internal lines as a way to distinguish them, but in reality, we can adjust these directions to make the six diagrams look exactly the same as the first one.

 

[JD_PM]

Sorry but I am afraid I got lost.

what kind of rules are you following to draw the internal lines? I do not understand why you are allowed to curve them (I was only thinking in terms of straight lines)

 

[Gaussian97]

I'm just connecting the external lines with the legs of the vertex with lines, as always in Feynman diagrams, curved lines or straight lines make no difference, is just a way to make them look nicer. If I write only straight lines then all the pictures look the same and there's no way to see why there are 6 and only 6.
There's no secret, just draw on a paper the external lines and the legs of the vertex (i.e. draw on a paper the picture on the left of the $=$) and then connect them by lines...
In any case, if you are not used to this kind of pictures, is better to forget about the 3!, and take
$$\lambda = -i\lambda_3$$as your Feynman rule (which is what everyone does), and then you will need to draw only one diagram.
Although I must say that this kind of pictures are really useful to determine symmetry factors.

 

[JD_PM]

OK thank you.

Before showing that the term proportional to $\lambda_4$ vanishes, please let me do an overall checking to see if I understand the whole process.

In first order, we have checked in detail the $\phi \to \phi \phi$ process and shown that its Feynman amplitude is $\mathcal{M}= -i\lambda_3$. Let's look at the other first order terms

For $|0\rangle \to \phi \phi \phi$, $\phi \phi \phi \to |0\rangle$ and $\phi \phi \to \phi$ I get the exact same Feynman amplitude i.e. $\mathcal{M}= -i\lambda_3$; What is this telling us?

 

[Gaussian97]

Well, this tells us that the Feynman rule associated with the vertex is independent of the details on how this vertex is connected to the rest of the diagram. For each vertex, you multiply by the same factor, i.e. $-i\lambda_3$.
If this were not true then you couldn't have such a general rule, you should have different vertex factor for different cases, fortunately, it's not the case.

 

[JD_PM]

Just to check my basic understanding of Feynman diagrams; these are the basic (i.e. only first order) processes of our theory (in a very schematic way)

 

[Gaussian97]

Correct

 

[JD_PM]

$$S = \langle \phi\phi | \phi \rangle + i \frac{\lambda_3}{3!} \int d^4 x \langle \phi\phi| :\phi(x) \phi(x) \phi (x): |\phi\rangle + i\frac{\lambda_4}{4!} \int d^4 x \langle \phi\phi | :\phi(x) \phi(x) \phi(x) \phi (x) : | \phi \rangle $$
Now show that the first and last terms vanish and then, from all the terms arising from splitting $\phi = \phi^+ + \phi^-$ look which ones survive.

Let's show that the last term vanishes!

Let's focus on the matrix element $\langle f| :\phi(x) \phi(x) \phi(x) \phi (x) : |i \rangle$; we have

$$\langle 0|b(\vec p_2) b(\vec p_3) b(\vec k_2) b(\vec k_3) \phi (x) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle$$

So we see there's a 'free variable'; that's because we are only dealing with 3 fields (i.e. we are studying the process $\phi \to \phi\phi$) while there are 4 nestled. Then we have

\begin{align*}
&\langle 0|b(\vec p_2) b(\vec p_3) b^{\dagger}(\vec k_2) b^{\dagger}(\vec k_3) (b(\vec k)+b^{\dagger}(\vec k)) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle \\
&= \langle 0|b(\vec p_2) b(\vec p_3) b^{\dagger}(\vec k_2) b^{\dagger}(\vec k_3) b(\vec k) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle + \langle 0|b(\vec p_2) b(\vec p_3) b^{\dagger}(\vec k_2) b^{\dagger}(\vec k_3) b^{\dagger}(\vec k) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle
\end{align*}

In the $\langle 0|b(\vec p_2) b(\vec p_3) b^{\dagger}(\vec k_2) b^{\dagger}(\vec k_3) b(\vec k) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle$ term we observe that $b(\vec k)$ will hit $|0\rangle$ so this term will vanish. Besides, in the $\langle 0|b(\vec p_2) b(\vec p_3) b^{\dagger}(\vec k_2) b^{\dagger}(\vec k_3) b^{\dagger}(\vec k) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle$ term we observe that $b^{\dagger}(\vec k)$ will hit $\langle 0 |$ so this term will vanish.

Thus we indeed get

$$i\frac{\lambda_4}{4!} \int d^4 x \langle \phi\phi | :\phi(x) \phi(x) \phi(x) \phi (x) : | \phi \rangle =0$$

PS: @Gaussian97 again, another really fun and useful discussion! Again, I hope to come across you very soon!

 

[Gaussian97]

$$\langle 0|b(\vec p_2) b(\vec p_3) b(\vec k_2) b(\vec k_3) \phi (x) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle$$

Well, in fact you should study
$$\langle 0|b(\vec p_2) b(\vec p_3) \phi(x)\phi(x) \phi (x) \phi(x) b^{\dagger}(\vec p_1) |0 \rangle$$
but yes, the reason is very obvious you can always move an anihilation operator to the right or a creation to the left, so all the possible terms will vanish.
Indeed, it's a good exercise to prove that any expectation value of the form
$$\left< 0\right|\prod_{i=0}^n\prod_{j=1}^{m_i} b^{(i)}(k_{ij})\left|0\right>,\quad
\text{where}\quad b^{(i)}=\begin{cases}b& \text{ for } i = \dot{2}\\ b^{\dagger}& \text{ for } i \neq \dot{2}\end{cases}$$
and where $n\neq \dot{2}$ (if $n$ is even then the expresion is trivial since there are anihilation operators acting on the vacuum state). This is always zero unless
$$\sum_{i=0}^n(-1)^{i}m_i = 0$$
Then the case
$$\langle 0|b(\vec p_2) b(\vec p_3) b^{\dagger}(\vec k_2) b^{\dagger}(\vec k_3) b(\vec k) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle$$
is just given by $n=3, m_0=2, m_1=2, m_2=2, m_3=1$ and therefore $m_0-m_1+m_2-m_3=1\neq 0$ so that expression must be zero, and similarly
$$\langle 0|b(\vec p_2) b(\vec p_3) b^{\dagger}(\vec k_2) b^{\dagger}(\vec k_3) b^{\dagger}(\vec k) b(\vec k_1) b^{\dagger}(\vec p_1) |0 \rangle$$
is the case for $n=3, m_0=2, m_1=3, m_2=1, m_3=1$ so $m_0-m_1+m_2-m_3=-1$ so obviously is also zero.
You can use this formula to prove that all the other terms in the expansion of $\phi^4$ vanish.