# I Scattering, 4 point correlator, number of distinct Feynman diagrams

1. Jan 2, 2017

### binbagsss

In order to compute the scattering probability that two particles of type 1 (associated to $\phi_1(x)$) which come from the far past with the momenta $p1$ and$p2$, to scatter and evolve into two particles of type 2 (associated to $\phi_2(x)$) with the momenta $p3$ and $p4$ , I am going to apply the momentum space scattering rules, I'm just first of all trying to establish the number of distinct Feynman diagrams of $<0|T( \phi_1(x_1)\phi_1(x_2)\phi_2(x_3)\phi_2(x_4))|0>$, where the interacting Hamiltonian is: $H_{int}= g/4 \phi_1^{2}(x)\phi_2^{2}(x)$

QUESTION:

The solution lists these two Feynman diagrams only , see attachment 'dia 1'

But I am also getting this diagram, see attachment 'dia 2'

Where solid lines are associated to $\phi_1(x)$, dotted to $\phi_2(x)$, and $z$ and $w$ are the internal variables I am integrating over.

Is there a reason this should be excluded? I know that 'bubble/vacuum' diagrams are excluded - disconnected diagrams with no connection to external points, but I'm unsure here since it's connected...

On another note, my lecture notes describe the contribution from the $g^{0}$ term as 'trivial' since it describes non-interacting particles. so yeh in this case it's $x1, x2$ and $x3, x4$ contracted since you can not contract different fields, however, what is meant by 'trivial' in this sense? Because I know that diagrams that have no connections to external points are not included since they are vacuum contributions and will just cancel out anyway, however here it is a diagram solely between external points, so whilst it doesn't describe any interaction, it still needs to be included right?

There is the 'amputate external legs' rule which in my lecture notes is stated as equivalent to 'moving propagators to external points'...does this cover it? I don't really understand this rule? If so, or anyway, any links or help on this appreciated, ta.

Last edited: Jan 2, 2017
2. Jan 4, 2017

bump

3. Jan 5, 2017

### binbagsss

anyone?

Last edited: Jan 5, 2017
4. Jan 5, 2017

### Staff: Mentor

Sometimes you have to be patient. Nobody gets paid for posting here.

I assume you mean it lists these two diagrams as the only ones having two vertexes. There is, of course, one diagram with one vertex, which is the way the process as a whole would look if there were no quantum effects involved.

If the two particle types are distinguishable, this isn't a valid diagram, because the w vertex doesn't have two legs of type 1 and two legs of type 2; it has four legs of type 1.

If, OTOH, you were talking about the $\phi^4$ theory, where this scattering is from two particles of the same type into two particles of the same type (but possibly with different distribution of momentum between the two), then I think your third diagram would be a valid diagram with two vertices.

5. Jan 6, 2017

### binbagsss

Apologies yes I meant the two vertexes diagrams.

we have just been introduced to these diagrams
and have not been told anything about what you mention here about the w vertex must have two legs of type 1 aand type 2..
we are asked to draw these using the momentum space scatter feynman diagram rules.

The only rule I haven't used because I don't understnad it is the amputate external legs...move propagators to external points...could this possibly cover it? be the same as what you are saying at all....?

many thanks

6. Jan 6, 2017

### Staff: Mentor

It's in the formulas you wrote down, which have two fields, $\phi_1$ and $\phi_2$, which you said describe particles of type 1 and type 2. Each field appears twice (squared in the Hamiltonian, and two factors in the time ordered product), which means each vertex has two legs of each type.

That just means that when you're writing down the amplitude, you don't need to integrate over the variables associated with the external legs. You only need to integrate over internal lines.

7. Jan 6, 2017

### Staff: Mentor

It depends. If you define the initial and final conditions as you did--two particles of type 1 go in, and two particles of type 2 come out--then there is no possible no-vertex diagram (which is what you are describing), because the only way for particles of type 1 to turn into particles of type 2 is through a vertex.

However, this theory also supports a "scattering" interaction, where one particle of each type (1 and 2) goes in and one particle of each type comes out. The two-vertex diagrams for this look exactly the same as the two you saw in the solution, all you need to do is switch which axis is time and which axis is space. But for this case there is also a no-vertex diagram with just two straight lines, one for the type 1 particle and one for the type 2 particle. This is possible because in this case neither particle needs to change type, since there is one of each in both the initial and final states.

8. Jan 6, 2017

### Staff: Mentor

Oops--I just realized on looking again that this is not true. You drew the diagram so the loop at the w vertex is a dotted line, which means it's a type 2 loop (if I'm reading the diagrams correctly) on a vertex that's on a type 1 line. So the vertex in fact does have two legs of type 1 and two legs of type 2 (the latter two legs form the loop), and should be allowed.

There should also be another possible "loop" diagram where we have a vertex on a type 2 line with a type 1 loop attached to it (i.e., draw it on one of the dotted legs and have a solid loop). This should also be allowed by the rules as given.

I suspect that the reason these diagrams don't appear in the solution is that they involve loops with only one vertex, and might be viewed as "corrections" to the one-vertex diagram (the one you would get if you just removed the extra loop) rather than "new" two-vertex diagrams. But if that's the reason, it seems like an arbitrary distinction to me.

9. Jan 6, 2017

### binbagsss

So we also have the rules of conservation at a vertex and to integrate over the undetermined momenta only.
So both of these rules are basically just when you integrate over everything, including the external lines you get deltas.
Is integrate over internal lines not the same as the rule integrate over undetermined momenta, since the conservation at a vertex applies to the momenta going in and out of the vertex, i.e. external variables? so the only undetermined momenta remaining are the internal momenta..?

thanks.

10. Jan 6, 2017

### Staff: Mentor

Not just that, but the same deltas for every single diagram with the same external legs. Everybody just agreed that there is no point in writing all that down when what we're really interested in is the differences between the different diagrams for the same external legs.

Yes. Basically you have to integrate over any momenta that are not constrained by conservation laws.

11. Jan 6, 2017

### binbagsss

so the amputation rule, the integrate over all undetermined momenta, and conservation at a vertex are all basically the same rule?

12. Jan 6, 2017

### Staff: Mentor

The first and the last are, yes. Integrating over all undetermined momenta adds something new, though, because that is what leads to the integrals diverging--i.e., giving infinite answers if you really integrate over all undetermined momenta (i.e., allow the undetermined momenta to be arbitrarily large or arbitrarily small). So now you have to figure out how to handle that, since the actual physical probabilities for different interactions obviously aren't infinite. This is where "regularization" and "renormalization" and things like that come in.

13. Jan 6, 2017

### binbagsss

Oh thanks, I have just seen an example of this divergence in Peskin due to external momenta being on shell , a conservation has resulted in $\frac{-i}{p_1^{2}-m^{2} }$ and on shell says $p_{1}^{2}=m^2$.... I have a question however, tong and this text give the propagator in the Feynman rules as $\frac {i}{p^{2}-m^{2}}$ but in my lecture notes it has the signs instead as $\frac{-i}{p^{2}+m^{2}}$ so for this same example I would not get $\frac{1}{0}$... do various texts use different signs like this in the propagator? How is the divergence for such an example apparent in the latter sign choice case? thanks.

14. Jan 6, 2017

### Staff: Mentor

Not that I've seen. The $i / \left( p^2 - m^2 \right)$ version is the one I'm used to seeing for the propagator of a scalar field. I would need to see more context from your lecture notes to understand where $-i / \left( p^2 + m^2 \right)$ is coming from.

15. Apr 15, 2017

### binbagsss

oh the mass-squared on relation is either $m^{2}=\pm p^{2}$ consistent with the sign choice in the propagator, i.e. so this integral will blow-up on shell

16. Apr 15, 2017

### Staff: Mentor

What sign choice in the propagator?

An integral with $p^2 + m^2$ in the denominator won't blow up at all, regardless of what sign choice you make. I'm confused as to what you are referring to.