Anti-commutation of Dirac Spinor and Gamma-5

Click For Summary
SUMMARY

The forum discussion centers on the anti-commutation of the Dirac spinor and the gamma matrix gamma^5 in the context of the interaction Lagrangian $$ \mathcal{L}_{int} = \lambda \phi \bar{\psi} \gamma^5 \psi$$. The user is attempting to derive the second-order scattering amplitude for the process $$\psi\psi \to \psi\psi$$ and is uncertain about the validity of the relation $$\{\gamma^5, \bar{\psi}\} = 0$$. The discussion highlights the implications of this relation on the statistics of the scattering process, particularly in distinguishing between t-channel and u-channel contributions.

PREREQUISITES
  • Understanding of Dirac spinors and their properties
  • Familiarity with quantum field theory concepts, particularly scattering amplitudes
  • Knowledge of Wick's theorem and its application in quantum field theory
  • Proficiency in manipulating gamma matrices and their algebra
NEXT STEPS
  • Research the properties of gamma matrices, specifically the implications of anti-commutation relations
  • Study the derivation of scattering amplitudes in quantum field theory using Peskin and Schroeder's methods
  • Explore Wick contractions in detail and their role in simplifying quantum field calculations
  • Investigate the physical significance of pseudoscalar fields in particle interactions
USEFUL FOR

Particle physicists, quantum field theorists, and graduate students studying advanced topics in quantum mechanics and scattering theory will benefit from this discussion.

Dewgale
Messages
98
Reaction score
9

Homework Statement


Given an interaction Lagrangian $$ \mathcal{L}_{int} = \lambda \phi \bar{\psi} \gamma^5 \psi,$$ where ##\psi## are Dirac spinors, and ##\phi## is a bosonic pseudoscalar, I've been asked to find the second order scattering amplitude for ##\psi\psi \to \psi\psi## scattering. I've been able to get extremely close, but my main sticking point is whether I can anti-commute ##\gamma^5## and ##\bar{\psi}##. Details are below.

Homework Equations


$$S = T( e^{-i \int_{-\infty}^\infty dt H})$$

The Attempt at a Solution


The Hamiltonian density for this Lagrangian density is $$\mathcal{H} = - \mathcal{L} = - \lambda \phi \bar{\psi} \gamma^5 \psi.$$

We therefore have $$ \left< f \right| S \left| i \right> = \left< f \right| T( e^{i \lambda \int d^4x \phi \bar{\psi} \gamma^5 \psi}) \left| i \right>.$$

At second order, this gives
$$\left< f \right| S \left| i \right> \sim \left< f \right|T(\int d^4x \int d^4y \phi(x) \bar{\psi}(x) \gamma^5 \psi(x) \phi(y) \bar{\psi}(y) \gamma^5 \psi(y)) \left| i \right>$$

Let's identify the momenta of the inbound states as ##p,q##, and the outbound states as ##p',q'##. Following Peskin and Schroeder's example for the Yukawa interaction, we can contract the two ##\phi##s to get their Feynman propagator. This leaves us with two cases: where ##\bar{\psi}(x)## and ##\psi(x)## are contracted to the "same" momentum (i.e. q' and q, respectively), and when they are contracted to "opposite" momenta (i.e. p' and q, respectively). These correspond the to t-channel and the u-channel, respectively.

I'm not sure how to write Wick contractions on Physics Forums version of latex, but if we don't write the contractions we have

$$ \left< 0 \right| \hat{a}_{q'} \hat{a}_{p'}\, \bar{\psi}(x) \gamma^5 \psi(x)\, \bar{\psi}(y) \gamma^5 \psi(y) \,\hat{a}_p^\dagger \hat{a}_q^\dagger \left| 0 \right>$$

Recall that the two cases are
(1) ##\bar{\psi}(x)## contracted with ##\hat{a}_{q'}## and ##\psi(x)## contracted with ##\hat{a}_q^\dagger## (the other two are appropriately contracted)
(2) ##\bar{\psi}(x)## contracted with ##\hat{a}_{p'}## and ##\psi(x)## contracted with ##\hat{a}_p^\dagger## (the other two are again appropriately contracted)

We can see that, if we assume ##[\gamma^5, \bar{\psi}]=0## (which isn't true), then the first situation requires two interchanges to "untangle" the contractions, while the second case only requires one. This is the case of the Yukawa interaction, and as each interchange provides a factor of ##(-1)##, it gives the appropriate statistics in that case.

In the second case, no spinors need to cross a ##\gamma^5##, so it's not an issue. However, in the first case, we do need a spinor (##\bar{\psi}(y)##) to cross a ##\gamma^5##.

My question, then, is if ##\{\gamma^5, \bar{\psi}\} = 0## is true? My rational is that
$$
\gamma^5 \bar{\psi} =\gamma^5 \psi^\dagger \gamma^0 \stackrel{?}{=} \psi^\dagger \gamma^5 \gamma^0 = - \psi^\dagger \gamma^0 \gamma^5 = - \bar{\psi} \gamma^5
$$
If it's not true, then I'm not sure how to figure out the statistics for this. It's pretty obvious the net effect will be to put a ##\gamma^5## between each spinor in the numerator of each term, but the relative minus sign between the t-channel and u-channel is important.

Thanks in advance for the help!
 
Last edited:
Physics news on Phys.org
In these situations where you may get confused by strings of products of spinors, it is usually useful to write the indices out. First:
<br /> \bar{\psi}(x) \gamma^5 \psi(x)\, \bar{\psi}(y) \gamma^5 \psi(y) =<br /> \bar{\psi}_a(x) \gamma^5_{ab} \psi_b(x)\, \bar{\psi}_c(y) \gamma^5_{cd} \psi_d(y)<br />
From here, you may temporarily move the \gamma^5 factors out of the expectation value (but keep track of their indices!). Then, order the fermionic degrees of freedom using the canonical anticommutator:
<br /> \{ \psi_a(x) , \bar{\psi}_b(y) \} = \delta^{d}(x - y) \gamma^0_{ab}<br />
(you can move any factors of \gamma^0 out of the expectation value as well). Once you've moved things to their desired place for the given Wick ordering, then you can replace the expectation value with a pair of Dirac propagators, and then realign your propagators/gamma matrices by their indices to get the spinor multiplication which you've generated.
 

Similar threads

Replies
95
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K