# Clarification of statement on how to evaluate S-matrix elements.

1. Oct 13, 2012

### center o bass

Hi. I'm trying to understand a statement from the book Field Quantization on the evaluation of the S-matrix elements for a certain initial and final state in QED.

The author states, if one is evalutating a matrix element on the form

$$\langle k_1' \lambda_1', \ldots, \bar p_1' \bar s_1', \ldots, p_1' s_1', \ldots |\hat S|p_1s_1, \ldots, \bar p_1 \bar s_1, \ldots, k_1 \lambda_1, \ldots \rangle$$

where (p,s) correspond to momenta and spins of electrons while (k,λ) correspond to momenta and spins of photons, that

"In the evaluation matrix elements of the type above, obviously the only terms of the S-matrix expansion that will contribute are those that contain a 'matching' configuration of field operators in the normal product. To be more specific , for each electron/positron/photon in the initial state a matching annihilation operator is needed,i.e., a factor $\psi^{(+)}/\bar \psi^{(+)}/A_\mu^{(+)}$ and similartly for the final state a creation operator of the type $\psi^{(-)}/\bar \psi^{(-)}/A_\mu^{(-)}$ must appear. If these conditions are not fulfilled the matrix element involving hte states will vanish since unpaired operators are present, which annihilate the vacuum state."

The way I understand this statement is that one has to have exactly as many annhilation operators/creation operators (of the relevant type) in the contributing term as the number of particles (corresponding to relevant operators). But I do not see that it's obvious. How is it obvious? For example; for an electron scattering let

$$|i\rangle = b_{p_2 s_2}^\dagger b_{p_1 s_1}^\dagger |0\rangle , \ \ |f\rangle = b_{p_2' s_2'}^\dagger b_{p_1' s_1'}^\dagger |0\rangle$$

be the initial and final states. According to the statement above one should think that only terms including the operators

$$:\bar \psi^{(-)} \bar \psi^{(-)} \psi^{(+)}\psi^{(+)}:$$
should contribute. But what about terms of the type

$$\bar \psi^{(-)} \psi^{(+)}?$$

Since $\bar \psi^{(-)} \sim b_{p s}^\dagger, \ \ \ \psi^{(+)} \sim b_{p s}$ this would mean that the matrix element would be come something like

$$\langle 0 | b_{p_2' s_2'} b_{p_1' s_1'} (b_{p s}^\dagger b_{q \bar s}) b_{p_2 s_2}^\dagger b_{p_1 s_1}^\dagger |0\rangle$$

with exponential factors and integrations over momenta p and q. Why is it obvious that this vanishes? The creation operator within the parentheses is inhibited to annihilate the vacuum to the left and the annihilation operator is inhibited to annhilate the vacuum to the right. It seems to me that there does not need to exact pairing of operators, but rather at least one creation (annihilation) operator of the relevant type corresponding the creation of a particle in the final (initial) state.

I would be very greatfull if someone could tell me where I mess this up :)

2. Oct 16, 2012

### geoduck

If p1, p2, and p'1 and p'2 are not equal, then the amplitude should be zero.

Your operator conserves momentum, so when p1 becomes p1' it must be true that p1'=p1 or else your matrix elements ends up being zero (you need to annihilate anything you created with the exact same momentum)

3. Oct 17, 2012

### center o bass

How is it that my operator conserves momentum? How do you see that easily? Do you just use the feynman rules in momentum space?

I'm more conserned about the picking out of contributions to different physical processes and how you see, without appealing to the feynman rules which terms contribute from the Wick expansion of the S-matrix. From the argument the author put forward

$$:\bar \psi^{(-)} \bar \psi^{(-)} \psi^{(+)}\psi^{(+)}:$$

shuld be the only set of operators that contribute, because 'obviously' - "the only terms of the S-matrix expansion that will contribute are those that contain a 'matching' configuration of field operators in the normal product."

I do not see why it is obvious, by just looking at the operators, why terms like

$$\bar \psi^{(-)} \psi^{(+)}$$

do not contribute.

Last edited: Oct 17, 2012
4. Oct 17, 2012

### Dickfore

Ignore this. If it had been true, there would be no scattering. As far as momentum being conserved, that's true for the total momentum, not the momentum of each particle separately.

5. Oct 17, 2012

### center o bass

Alright. Thanks, I will. Is my question clear btw? Do you understand what I'm trying to figure out?

6. Oct 17, 2012

### geoduck

You wrote that the operators were built out of fields: this should be enough to conserve momentum (no explicit coordinate dependence).

Take a concrete example: incoming particles with momentum 6 and 4, and outgoing particles with momentum 7 and 3. The amplitude <7,3|S|4,6>=0 for your interactions because you only have two operators. Your annihilation operator can delete the 4 to get

<7,3|,6>

but your creation operator can only create 4 (to conserve momentum), so you get back:

<7,3|4,6>=0

You need at least 4 operators, so that the two annnihilation operators can get rid of 4 and 6:

<7,3|>

And the two creation operators can create 7 and 3 from vacuum:

<7,3|7,3> ≠ 0

This is okay because this conserves momentum, as 4+6=7+3