Physical interpretation of V and A couplings

[bomanfishwow]

Hi,

I'm trying to remind myself of exactly what, physically, is the difference between V and A couplings. Now, a vector coupling is of the form $$\bar{\psi}\gamma^\mu\psi$$, and axial coupling of the form $$\bar{\psi}\gamma^\mu\gamma^5\psi$$. Thinking in terms of a chiral fermion expanded as:

$$f = \left[\left(\frac{1-\gamma^5}{2}\right) + \left(\frac{1+\gamma^5}{2}\right)\right]\psi$$

and where $$\bar{f} = \gamma^\dagger\gamma^0$$, I assume the difference between the V and A couplings has to do with how the L and R projection operators commute through either $$\gamma^\mu$$ or $$\gamma^\mu\gamma^5$$ from the 'coupling' term, and the $$\gamma^0$$ from the conjugate field term in a given Lagrangian.

However, as $$[\gamma^5,\gamma^5] = 0$$, I don't see how a difference in the chiral treatment between V and A couplings can arise. Am I barking up completely the wrong tree? Any insight welcomed!

 

[Vanadium 50]

You can always redefine what you mean by "V" and "A", as purely vector or purely axial currents conserve parity. It's the mixed terms that generate parity violation.

 

[bomanfishwow]

You can always redefine what you mean by "V" and "A", as purely vector or purely axial currents conserve parity. It's the mixed terms that generate parity violation.

So, a mixed term is something like $$\bar{\psi}\gamma^\mu\left(1-\gamma^5\right)\psi$$? How can it be shown that this term violates parity?

Cheers.

 

[Meir Achuz]

Gamma 5 is a pseudoscalar. Adding a pseudoscalar to a scalar violates parity.
When used in beta decay, the V-A interaction leads to negative helicity for outgoing leptons. This violates parity.

 

[bomanfishwow]

Gamma 5 is a pseudoscalar. Adding a pseudoscalar to a scalar violates parity.

Sure, but I'm interested in showing why this is the case mathematically...

 

[Vanadium 50]

Given $P|\psi> = (-1)^n|\psi>$

Then $P(a + b\gamma^5)|\psi> = aP|\psi> + bP \gamma^5|\psi>$

Which is $\left( (-1)^na + (-1)^{n+1}b \right) |\psi>$, so is only an eigenvalue of parity if $a=0$ or $b=0$.

 

[bomanfishwow]

Ah yes, very elegant. Thanks!