Contracting operators in Wick diagrams

1. Mar 15, 2013

jackson1

Hi, I'm currently going through Ticciati's book along with the notes from Sidney Coleman's course and I have a question pertaining to Wick diagrams/expansion of S.
In their example (section 4.3 of Ticciati and lecture 9 in Coleman's notes) they never seem to contract the adjoint nucleon field with itself nor the nucleon field with
itself; i.e., you only see contractions of $\psi^\dagger (x_1)$ with $\psi (x_2)$ while you do see contractions of the meson field with itself, $\phi(x_1)$ contracted with $\phi(x_2)$. Why is this so,
why not contract the adjoint nucleon field with itself, why not contract the nucleon field with itself? Also, what happens if you contract two appropriate fields at the
same spacetime point, e.g., $\psi^\dagger (x_1)$ contracted with $\psi(x_1)$? It appears, from the formula for their contraction, that if you contract the two the resulting integral
should be

$$\lim_{\epsilon \rightarrow 0}\int \frac{d^4 k}{(2\pi)^4} e^{-ik\cdot (x-x)}\frac{i}{k^2 - m^2 +i\epsilon} = \lim_{\epsilon \rightarrow 0}\int \frac{d^4 k}{(2\pi)^4} \frac{i}{k^2 - m^2 +i\epsilon}$$
which, I believe, is $-1/2m$. Finally, does anyone know how to include the contraction symbols in latex? Thanks for your time.

2. Mar 15, 2013

Bill_K

Wick's theorem tells us how to convert a time-ordered product into a normal-ordered product. A contraction occurs each time the interchange of two operators results in a nonzero commutator/anticommutator. But for a fermion field, {ψ(x), ψ(x')} = {ψ(x), ψ(x')} = 0.

3. Mar 15, 2013

Chopin

This is perfectly possible--the resulting Feynman graph is called a "tadpole". However, for three-point interactions, momentum conservation dictates that the edge leading into the tadpole will carry 0 momentum, and renormalization can be used to remove their contribution to the scattering amplitude, so one generally doesn't worry about them.

As I recall, there is no easy way to do it. I think Peskin and Schroeder have a section on their website or something where they discuss how they did it in their book, but that's just a dim recollection of mine.

Last edited: Mar 15, 2013
4. Mar 15, 2013

jackson1

@ Bill_K - Duh, thank you for pointing out the P.B. relations for the complex field.
@Chopin - Thanks. Could you do the same for all odd-term interactions?

5. Mar 15, 2013

Chopin

Not all odd-term interactions, but all diagrams or sub-diagrams with a single external line. These diagrams are higher-order contributions to the vacuum expectation value of the field, $\langle 0 | \phi | 0 \rangle$. This term is 0 in a noninteracting theory, so that condition is maintained in an interacting theory by making it a renormalization condition. You said you're looking in Coleman's notes--search for the term "tadpole" and you should see a discussion of it in the renormalization section later on.

If you're just going through all of this for the first time, it's probably not worth worrying about too much right now. Just proceed under the assumption that making those contractions is a perfectly acceptable thing to do, and later on you'll learn why in practice you don't have to worry about it.

Last edited: Mar 15, 2013
6. Mar 15, 2013

jackson1

Ok, I see that later on (lecture 16) he shows that the total contribution of the tadpoles, possibly for a specified theory, is zero - just skimmed ahead and read a few sentences so I'm not sure of the proof, but I'm sure it's exactly as you mentioned. Just to make sure I'm understanding what you're saying, suppose $H_I = g \psi^\dagger \psi \phi$ then at order $g^2$ I will have an operator $\overline{\psi^\dagger(x_1) \psi(x_1)} : \phi(x_1)\psi^\dagger(x_2)\psi(x_2)\phi(x_2) :$, among many others, but I'll see later on that this term can be ignored?

7. Mar 15, 2013

Chopin

Correct. Roughly, you'll end up defining a new object $A$, called a counterterm, that you'll add to your set of diagrams, such that its first-order term $A_1$ will be defined as $A_1 + \overline{\psi^\dagger(x_1)\psi(x_1)} = 0$.

So then, in addition to the term that you mentioned, you'll also have another term $A_1: \phi(x_1)\psi^\dagger(x_2)\psi(x_2)\phi(x_2):$ to work with, meaning that when you add all the diagrams up, you'll have $\overline{\psi^\dagger(x_1)\psi(x_1)} : \phi(x_1)\psi^\dagger(x_2)\psi(x_2)\phi(x_2): + A_1: \phi(x_1)\psi^\dagger(x_2)\psi(x_2)\phi(x_2): = (\overline{\psi^\dagger(x_1)\psi(x_1)} + A_1) : \phi(x_1)\psi^\dagger(x_2)\psi(x_2)\phi(x_2): = 0$

8. Mar 15, 2013

jackson1

Thank you so much.

9. Mar 16, 2013

Chopin

Incidentally, since you're using Coleman's notes...it may interest you to know if you don't already that Harvard recorded his lectures of that course one year, and a while back they made the videos publicly available online. You can find them all at http://www.physics.harvard.edu/about/Phys253.html. I'm an armchair physicist only, so going through those videos has been my primary source of education on the topic. I highly recommend them--in addition to being a fantastic lecturer, the man was downright hilarious.