Quantum Question: Why k^2<0 in QFT?

[latentcorpse]

You are right, this contraction has to be included. But this will gove a so called vacuum, bubble or reducible diagram. iF you consider it in this specific diagram you will have the bosons going straight through plus two separate circles which correponds to fermions created and annihilated again. But first of all, this diagram will give an infinite contribution and second it would also happen in vacuum and would give an infinite contribution to the vacuum. So the practical view is to simply ignore such diagrams. It can also be shown using the partition function that this treatment is correct. The bubble diagram give no physical measurable effect.

Hi, I have one or two other small things in these notes that I was wondering if you could help me with.

In the 1st line of eqn 5.6, this only works if we assume that $u$ and $u^dagger$ commute, but it doesn't say anywhere in the notes that this is true!

For example, if we expand the commutator, we will get something like $b b^\dagger u u^\dagger - b^\dagger b u^\dagger u$ and in order for this to equal $[b, b^\dagger] u u^\dagger$ then the $u$ and $u^\dagger$ must commute. Why is this?

And secondly, in 4.128, why doesn it not make sense to consider something like $\Sigma_{r,s} u^r(\vec{p}) \bar{u}^s(\vec{p})$? What are these indices r and s and what do they mean? Are these the spinor indices?

Thanks again!

 

[betel]

The us are no operators but spinors. nevertheless there is a difference which one comes first, similar to vectors and their transpose. In this case this is taken care of by the indices which pick out exactly one component and sums over them. similar to writing maxtrix vector product using the indices, then you can commute them.

Yes, the r,s are spinor indices. And you only consider the one sum, because this one appears in the diagrams.

 

[latentcorpse]

The us are no operators but spinors. nevertheless there is a difference which one comes first, similar to vectors and their transpose. In this case this is taken care of by the indices which pick out exactly one component and sums over them. similar to writing maxtrix vector product using the indices, then you can commute them.

Yes, the r,s are spinor indices. And you only consider the one sum, because this one appears in the diagrams.

Hmmmm... I don't see how the indices work to tell us which way round the spinors should come?

Could you write it out or elaborate perhaps?

Thanks!

 

[betel]

He was a bit sloppy in the notation. He takes the commutator of the field spinor with its hermitian conjugate. You should notice that this is not well defined, as the first is a 4x4 matrix, whereas the second one is a scalar. What he really wanted to compute was the commutator of one component of each, as in eq. 5.3. The spinors will then commute because you have only simple numbers as explained in more detail below. So to be correct you should put an index alpha or beta on the u or v.
The rest should go through unchanged (I haven't checked) and you would have to take the alpha-beta component of all the $$(\not p-m)\gamm^0$$ combinations at the end. If I should explain in more detail pls let me know.

In standard linear algebra you can write e.g. the scalar product using the vectors.
$$(\vec v,\vec u)=\vec v^\dagger \vec u\neq \vec u\vec v^\dagger$$
Here the order is important.
Using indices the order does not matter
$$(\vec v,\vec u)=\sum_i v_i^*u_i = \sum_i u_i v_i^*$$
In exactly the same way it works here.