# Fermion contraction

1. Nov 23, 2007

### jostpuur

Directly from the Peskin & Shroeder, page 63:

$$S_F(x-y) = \int\frac{d^4p}{(2\pi)^4}\frac{i(\displaystyle{\not}p + m)}{p^2-m^2+i\epsilon} e^{-ip\cdot(x-y)}$$

I'm slightly confused with the notation with the contractions. Things like $\overline{\psi}(x)\psi(y)$ and $\psi(x)\overline{\psi}(y)$ get written carelessly although it doesn't really make sense to put psi bar on the right. On the other hand the propagator itself is a 4x4 matrix, so when I try to make sense out of this, this is the only conclusion I've succeeded to come up with: The contraction should be carried out always with fixed indexes of the fermion operators, and then we choose the corresponding element from the matrix in the propagator. That means, that when $a,b\in\{1,2,3,4\}$, then

$$\textrm{contraction}(\psi_a(x),\overline{\psi}_b(y)) = (S_F(x-y))_{ab} = \int\frac{d^4p}{(2\pi)^4}\frac{i(\displaystyle{\not}p +m)_{ab}}{p^2-m^2+i\epsilon} e^{-ip\cdot(x-y)}$$

Is this correct?

If it was correct, what happens when the psi bar is on the left? What is

$$\textrm{contraction}(\overline{\psi}_a(x),\psi_b(y)) ?$$

Last edited: Nov 23, 2007
2. Nov 23, 2007

### Mr.Brown

ah the point here is that the spinor with the bar represents the creation of a particle at the point coordinate of the spinor, while the spinor without bar stands for annihilation.
So if you interchange the bar you gotta make sure the the zero-component of the coordinate of the spinor is changed as well to not violate causality :)

3. Nov 23, 2007

### blechman

There's nothing wrong with the order of spinors, as long as you are taking an "outer product", that is - the spinor indices are not contracted with each other. Then both products are allowed, and your interpretation of the propogator as a matrix in "spinor" space is correct.

4. Nov 24, 2007

### jostpuur

Peskin & Shroeder write things like this

$$\left\{\begin{array}{l} \langle 0|\psi(x)\overline{\psi}(y)|0\rangle\quad\quad\quad \textrm{for}\;x_0>y_0\\ -\langle 0|\overline{\psi}(y)\psi(x)|0\rangle\quad\quad\quad\textrm{for}\;x_0<y_0\\ \end{array}\right.$$

I don't think that quantity is complex number for $x_0<y_0$ and 4x4 matrix for $x_0>y_0$. That is just confusing notation for something else.

But my problem seems to be solved. When going through the lecture notes, I encountered (probably not for the first time) an equation, that said clearly that

$$\textrm{contraction}(\psi_a(x_1),\overline{\psi}_b(x_2)) = -\textrm{contraction}(\overline{\psi}_b(x_2), \psi_a(x_1)).$$

This removed the need for guessing quite well.

Talking about contractions. How should they be written with latex?

5. Nov 24, 2007

### Mr.Brown

i guess it´s \contraction{A}{B}

but doesn´t work on the forum i guess it needs ams package :(

$$\contraction{B}{C}$$

6. Nov 24, 2007