Propagator of massless Weyl field

[Siupa]

I have this Lagrangian for a free massless left Weyl spinor, so it’s just the kinetic term, that can be written embedding the field into a larger Dirac spinor and then taking the left projector in this way:
$$i \bar{\psi} \cancel{\partial} P_L \psi$$
Srednicki says that the momentum space propagator can be immediately read off as
$$-P_L \frac{\cancel{p}}{p^2}$$
I don't understand how this can be immediately guessed just by looking at that kinetic term. The propagator should be the inverse of the operator between the fields in the quadratic term, but $P_L$ being a projector can't be inverted

 

[malawi_glenn]

Srednicki says

Which page?
I searched for "immediately" in the pre-draft of his book http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

The propagator is not strictly an "inverse", it is a green's function

 

[Siupa]

Which page?
I searched for "immediately" in the pre-draft of his book http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

The propagator is not strictly an "inverse", it is a green's function

Chapter 75, "Chiral gauge theories and anomalies", page 458. It says "We can easily read the Feynman rules off of..." (the word "immediately" was not present, apologies).

I know I can take the long way and compute the 2-point correlator / Green's function, but the wording "easily read off" suggests there's a shortcut or rule of thumb for writing it down just by looking at the quadratic term.

Now, the quick way I know of doing this just by looking at the quadratic term is to take the inverse of the operator sandwiched between the fields. But this here in my opinion fails, because $P_L$ is not invertible.

How does he do it then if not in this way?

 

[malawi_glenn]

You mean because $P_L P_L \neq I$?

I will get back to you later, need to reread that chapter.

 

[Siupa]

You mean because $P_L P_L \neq I$?

Not only that, $P_L^{-1}$ isn't even defined since projectors aren't invertible (the kernel is not trivial)

 

[malawi_glenn]

Have you done the propgator yourself the "hard way"? I.e performing the fourier transform

 

[vanhees71]

It's a bit misleading. You can formulate the Weyl field as the corresponding 2D spinor (there are two types of Weyl spinors, representing massless spin 1/2 particles, left-handed and right-handed ones). Then no projection operator occurs. What's done here is to use Dirac fields and only consider the left-handed part. The propagator is defined to act on the effectively two-component spinor $\hat{P}_{\text{L}} \psi$, where $\psi$ is a Dirac spinor (or "bispinor").

 

[Siupa]

Have you done the propgator yourself the "hard way"? I.e performing the fourier transform

I didn't, I hoped I could understand how to write it down immediately without doing the formal computation

 

[Siupa]

It's a bit misleading. You can formulate the Weyl field as the corresponding 2D spinor (there are two types of Weyl spinors, representing massless spin 1/2 particles, left-handed and right-handed ones). Then no projection operator occurs. What's done here is to use Dirac fields and only consider the left-handed part. The propagator is defined to act on the effectively two-component spinor $\hat{P}_{\text{L}} \psi$, where $\psi$ is a Dirac spinor (or "bispinor").

I understand all this, but how does one arrive at that final form of the propagator based on these considerations? For example, how does one know if the $P_L$ needs to be to the right or to the left of the gamma matrix in the end? (I guess it should be on the left by inspection if one checks that $O^{-1}O$ acutally equals the identity on the subspace of left-handed Dirac spinors)

 

[vanhees71]

It should be to the left, because you have to project to the left-handed piece of $p_{\mu} \gamma^{\mu} P_L \psi$. The free Weyl-spinor equation for the left-handed field reads
$$\mathrm{i} (\partial_0 - \vec{\nabla} \cdot \vec{\sigma}) \psi_L=0.$$
The corresponding propagator fulfills
$$\mathrm{i} (\partial_0 - \vec{\nabla} \cdot \vec{\sigma}) G(x)=-\delta(x).$$
Translated to Fourier space
$$(p^0-\vec{p} \cdot \vec{\sigma}) \tilde{G}=1$$
and thus
$$\tilde{G}=\frac{p^0+\vec{p} \cdot \vec{\sigma}}{p^2+\mathrm{i} 0^+}.$$
Often one defines
$$(\sigma^{\mu})=(\hat{1}_2,\vec{\sigma}), \quad \bar{\sigma}^{\mu}=(\hat{1}_2,-\vec{\sigma}).$$
Then you have
$$\tilde{G}=\frac{p_{\mu} \sigma^{\mu}}{p^2+\mathrm{i} 0^+}.$$
For the right-handed field you just have to interchange $\sigma^{\mu}$ with $\bar{\sigma}^{\mu}$.

 

[Siupa]

It should be to the left, because you have to project to the left-handed piece of $p_{\mu} \gamma^{\mu} P_L \psi$. The free Weyl-spinor equation for the left-handed field reads
$$\mathrm{i} (\partial_0 - \vec{\nabla} \cdot \vec{\sigma}) \psi_L=0.$$
The corresponding propagator fulfills
$$\mathrm{i} (\partial_0 - \vec{\nabla} \cdot \vec{\sigma}) G(x)=-\delta(x).$$
Translated to Fourier space
$$(p^0-\vec{p} \cdot \vec{\sigma}) \tilde{G}=1$$
and thus
$$\tilde{G}=\frac{p^0+\vec{p} \cdot \vec{\sigma}}{p^2+\mathrm{i} 0^+}.$$
Often one defines
$$(\sigma^{\mu})=(\hat{1}_2,\vec{\sigma}), \quad \bar{\sigma}^{\mu}=(\hat{1}_2,-\vec{\sigma}).$$
Then you have
$$\tilde{G}=\frac{p_{\mu} \sigma^{\mu}}{p^2+\mathrm{i} 0^+}.$$
For the right-handed field you just have to interchange $\sigma^{\mu}$ with $\bar{\sigma}^{\mu}$.

Thank you!