# Matrix valued propagators

1. Apr 9, 2009

### Bobhawke

For scalar fields the propagator is just a number that represents the amplitude for a particle to go from one space time point to another.

For fermions, the propagator is matrix valued. What then is the amplitude for a fermion to go from one point to another? How are the elements of the matrix to be interpreted in terms of probability ampltudes?

2. Apr 9, 2009

### malawi_glenn

row-vector times matrix times column vector = number.

3. Apr 9, 2009

### Bobhawke

The fermion propagator involves the dot product of a 4 vector (momentum) and the gamma matrices. Im pretty sure the result is a matrix.

4. Apr 9, 2009

### malawi_glenn

No, since the states are vectors... (4-spinors)

Have you never worked with relativistic quantum identities? What part of my first post did you not understand?

5. Apr 9, 2009

### humanino

The signature :uhh:

6. Apr 9, 2009

### malawi_glenn

shall I change it? :uhh:

But seriously, what is wrong with my explanation?

7. Apr 9, 2009

### blechman

The fermion propogator IS a matrix, Bobhawke is correct. It is an outer product of spinors, not an inner product.

I think the answer he is looking for is $S_{\alpha\beta}\equiv\langle T\psi_\alpha\bar{\psi}_\beta\rangle$ is the amplitude for a fermion of polarization $\alpha$ to propogate to a fermion of polarization $\beta$.

Up to signs and whatnot!

Malawi_glenn: don't change your signature for anyone!

8. Apr 9, 2009

### humanino

I think we agree, there is not much more to this equation than what malawi_glenn was writing in "row-vector times matrix times column vector". It is also written in "the states are vectors... (4-spinors)".

9. Apr 9, 2009

### malawi_glenn

yes, the propagator is matrix, but the amplitude is a number. And that was the question, if the propagator is a matrix, what will happen to the amplitude.

10. Apr 9, 2009

### blechman

I think that is the question I answered. The matrix elements of the propagator are the amplitudes for the polarization states to propagate.

11. Apr 9, 2009

### Bobhawke

Thanks for the replies everyone!

12. Apr 11, 2009

### Hans de Vries

Actually, the propagator is generally the propagation from a source.

So in QED the photon is propagated from the transition current (the interference
pattern caused by an electron in two momentum states) and the electron is
propagated from the interference term $e\gamma^\mu\,A_\mu\psi$

It is indeed the amplitude per polarization state but there are some interesting
details about the interplay between SU(2) and U(1). For instance in a magnetic
field B the energy will be different per polarization state:

$$\binom{~~\exp(-i[E+\Delta E]t)~~}{~~\exp(-i[E-\Delta E]t)~~} ~~=~~ \binom{~~\exp(-i\Delta Et)~~}{~~\exp(+i\Delta Et)~~}~\exp(-iEt})$$

At the RHS the energy is the same for both states but the spinor represents
a precessing spinor around the direction of the magnetic field. Note that:

$$x\uparrow=\binom{1}{1}, ~~y\uparrow=\binom{1}{i},~~x\downarrow=\binom{1}{-1}, ~~y\downarrow=\binom{1}{-i}$$

(up to an overal factor of $1/\sqrt{2}$)

Regards, Hans.