Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'd like to know whether we need to take the trace of fermion loops in effective theory in the same way that we need to do so for renormalizable theories. At first thought, it seems obvious that should need to since we are dealing with fermions, but I recently saw a calculation and the professor did not bother to take the trace. Here the particular problem at hand:

The theory at hand is, \begin{equation} {\cal L} = \bar{\psi} ( i \partial _\mu \gamma ^\mu - m ) \psi + \frac{1}{2} ( \partial _\mu \phi ) ^2 - \frac{1}{2} M ^2 \phi ^2 + g \phi \bar{\psi} \psi \end{equation} At low energies we have, \begin{equation} {\cal L} = \bar{\psi} ( i \partial _\mu \gamma ^\mu - m ) \psi + \frac{ a }{ M } ( \bar{\psi} \psi ) ^2 + ... \end{equation} Doing matching between the theories at tree level we find that,$ a = g ^2 $. We then move on to calculate the diagram,

$\hspace{6cm}$enter image description here

which the professor wrote as (omitting the external legs), \begin{equation} \frac{ i g ^2 }{ M ^2 } \int \frac{ \,d^4k }{ (2\pi)^4 } \frac{ k_\mu\gamma^\mu + m }{ k ^2 - m ^2 } \end{equation} Not taking the trace of the diagram changes the final result by a factor of $ 4 - \epsilon $ (using dim-reg). On the one hand, this makes sense since we didn't need to take a trace in the high energy theory where the corresponding diagram is

$\hspace{6cm}$enter image description here

on the other hand these are fermions after all.

Is the professor just being sloppy here or do you not need to take the trace for some reason I am missing?

For more context please see my notes under Effective Field Theory in this link (Eq. 4.6)

share|improve this question
    
Do you mean the trace over the spinor components ? (It is not clear.) If so, I don't think that should be the case. The diagram contributes to the self-energy of the fermions, it thus has spinor components (e.g. coming from the $\gamma^\mu$). You should also see than from the effective four fermions interaction. –  Adam Mar 14 at 1:08
    
Yes, I do mean a trace over spinor components. I find it strange that we don't need to as when you first learn about Feynman rules for fermions we are taught to take the trace over the spinor index when you have a fermion loop. What makes four-fermion interaction different? –  JeffDror Mar 14 at 1:30
    
I hope my answer helps ! –  Adam Mar 14 at 1:54

1 Answer 1

up vote 1 down vote accepted

You don't need to trace over the spinor indices because the self-energy has some: $\Sigma_{ab}$.

But first, let's see why there are traces in the `usual' case of QED. In this case, the fermion-photon index contains a matrix $\gamma^\mu_{ab}$. (I will omit all sign, factor $i$ and $e$, etc.) Therefore, the QED diagram equivalent to the second diagram of the OP's question contains (G_{cd} is the fermionic propagator) $$\gamma^\mu_{ac}G_{cd}\gamma^\nu_{db}, $$ where repeated indices are summed over (this is the usual trace over the spinorial indices) and there should also be a photon propagator.

Now, let's see what happens with the $g\phi\bar\psi\psi$ vertex. There is no matrix $\gamma^\mu_{ab}$, but instead the trivial $\delta_{ab}$. Therefore, the trace over spinor indices of the second diagram is trivial: $$g^2\delta_{ac}G_{cd}\delta_{db}=G_{ab}, $$ and we recover the formula of the OP's professor.

Finally, what about the four-fermions interaction? In this case, the vertex is given by $\frac{a}{M}\delta_{ab}\delta_{cd}$ where the left indices correspond to the $\bar\psi$'s and the right indices to the $\psi$'s. The first diagram gives $$(\delta_{ab}\delta_{cd}+\delta_{ac}\delta_{db})G_{cd} ,$$ and the trace is once again trivial.

share|improve this answer
    
Thanks, that makes sense! I guess that statement that you need to take the trace for fermion loops is assuming 3 point interaction. –  JeffDror Mar 14 at 13:10
    
@JeffDror: I would say that you always have to trace over fermionic indices (since the interaction in the Lagrangian is always Lorentz invariant), but sometimes it is far more trivial than in QED! –  Adam Mar 14 at 13:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.