Chirality, helicity, and electron-photon vertices, and polarised electron beams

[kurros]

I am having some conceptual difficulties here. Let's start with the electron-photon vertex piece of the QED Lagrangian:

$-e\overline{\psi}\gamma^\mu\psi A_\mu$

Now we can write this in the chiral representation as

$-e\overline{\psi}_L\sigma^\mu\psi_R A_\mu - e\overline{\psi}_R\overline{\sigma}^\mu\psi_L A_\mu$

So this implies that every time an electron (in a chirality eigenstate) interacts with a photon then its chirality gets flipped. Ok I'll roll with that for now...

Next, there is helicity. Imagine our electron propagating along, then it softly emits a photon and continues with almost unchanged momentum. It's helicity must be flipped because a unit of angular momentum gets carried off by the photon. If it was a really hard event then ok helicity might not flip, but at least spin projected along some constant direction must flip.

Alright.

Now what if I am at the SLC and I have made myself a polarised electron beam? This means we have a beam of electrons in a known helicity eigenstate right? But how can this beam stay polarised given my above comments? Shouldn't the spins be flipping all over the place each time an interaction with a photon occurs? Which should be all the time?

Since this does not happen I guess I am wrong that every interaction with a photon flips the spin of an electron. So what is wrong with my reasoning?

Also I am not very clear on the benefits of colliding a polarised electron beam with a positron beam, as opposed to an unpolarised beam. I initially thought it must be because different beam polarisations would behave differently under weak interactions, but that is related to chirality, not helicity.

I'm so confused.

 

[kurros]

But ok chirality and helicity become basically the same thing in the high energy limit, so does this mean that if I create two left-handed helicity electron beams and try to collide them then no interactions will take place? This would be a little bit mind blowing for me if it is true, after all they both have electric charge so naively I would expect them to scatter. Does it mean that repulsion within a bunch is similarly suppressed? Similarly is it bad for the SLC polarized e- beam, i.e. do they only scatter off half of the positrons in the unpolarised e+ beam?

 

[blechman]

I am having some conceptual difficulties here. Let's start with the electron-photon vertex piece of the QED Lagrangian:

$-e\overline{\psi}\gamma^\mu\psi A_\mu$

Now we can write this in the chiral representation as

$-e\overline{\psi}_L\sigma^\mu\psi_R A_\mu - e\overline{\psi}_R\overline{\sigma}^\mu\psi_L A_\mu$

So this implies that every time an electron (in a chirality eigenstate) interacts with a photon then its chirality gets flipped. Ok I'll roll with that for now...

Next, there is helicity. Imagine our electron propagating along, then it softly emits a photon and continues with almost unchanged momentum. It's helicity must be flipped because a unit of angular momentum gets carried off by the photon. If it was a really hard event then ok helicity might not flip, but at least spin projected along some constant direction must flip.

Alright.

Now what if I am at the SLC and I have made myself a polarised electron beam? This means we have a beam of electrons in a known helicity eigenstate right? But how can this beam stay polarised given my above comments? Shouldn't the spins be flipping all over the place each time an interaction with a photon occurs? Which should be all the time?

Since this does not happen I guess I am wrong that every interaction with a photon flips the spin of an electron. So what is wrong with my reasoning?

Also I am not very clear on the benefits of colliding a polarised electron beam with a positron beam, as opposed to an unpolarised beam. I initially thought it must be because different beam polarisations would behave differently under weak interactions, but that is related to chirality, not helicity.

I'm so confused.

NO! You made a mistake here. Photons do *NOT* flip chirality! The interaction should have been

$$-e\overline{\psi}_L\sigma^\mu\psi_L A_\mu - e\overline{\psi}_R\overline{\sigma}^\mu\psi_R A_\mu$$

Just like the Kinetic term. It is the MASS term that flips chirality.

 

[kurros]

Oh, well that certainly would make more sense. I must have gotten my gamma matrix in the wrong representation or something. Must go check.

 

[kurros]

Ahh, bugger I see what I did. The gamma matrix in the Weyl basis indeed flips the left and right chirality pieces of the spinor fields, but of course there is a $\gamma^0$ hidden in the $\overline{\psi}$ which flips it back again. Duh.

 

[blechman]

quite so! you do these calculations a million times you don't even think about them anymore...