Spinor product in Peskin-Schroeder problem 5.3

  • Context: Graduate 
  • Thread starter Thread starter Manu_
  • Start date Start date
  • Tags Tags
    Product Spinor
Click For Summary
SUMMARY

The discussion centers on problem 5.3 from Peskin and Schroeder, specifically the proof of the Fierz identity involving left-handed and right-handed spinors. The identity is expressed as $$ \bar{u}_{L}(p_{1}) \gamma^{\mu} {u}_{L}(p_{2}) [\gamma_{\mu}]_{ab} = 2 [u_{L}(p_{2})\bar{u}_{L}(p_1) +u_{R}(p_{1})\bar{u}_{R}(p_2) ]_{ab} $$, with the Dirac matrix M defined by $$ [M]= \left( \frac{1-\gamma^{5}}{2} \right) \gamma_{\mu} V^{\mu} + \left( \frac{1+\gamma^{5}}{2} \right) \gamma_{\mu} W^{\mu} $$. The user seeks clarification on the identification of $V^{\mu}$ and $W^{\mu}$ as specific spinor products. Additionally, the discussion touches on defining a spinor product for right-handed spinors, $$s(p_1,p_2)=\bar{v}_{R}(p_1) u_{L}(p_2)$$.

PREREQUISITES
  • Understanding of Fierz identities in quantum field theory
  • Familiarity with Dirac matrices and spinor notation
  • Knowledge of Peskin and Schroeder's "An Introduction to Quantum Field Theory"
  • Basic concepts of left-handed and right-handed spinors
NEXT STEPS
  • Study the derivation of Fierz identities in quantum field theory
  • Learn about the properties of Dirac matrices and their applications
  • Review the spinor product formalism in quantum electrodynamics
  • Explore advanced quantum field theory textbooks for deeper insights, such as Atkinson's and Radovanovic's works
USEFUL FOR

Graduate students in theoretical physics, quantum field theorists, and anyone studying spinor algebra and Fierz transformations in particle physics.

Manu_
Messages
12
Reaction score
1
Hello,

I am currently stuck on problem 5.3 (c) about spinor products in PS, where one needs to prove the Fierz identity:
$$ \bar{u}_{L}(p_{1}) \gamma^{\mu} {u}_{L}(p_{2}) [\gamma_{\mu}]_{ab} = 2 [u_{L}(p_{2})\bar{u}_{L}(p_1) +u_{R}(p_{1})\bar{u}_{R}(p_2) ]_{ab} $$
They say that a Dirac matric M satisfies:
$$ \gamma^{5} [M]=-[M]\gamma^{5}$$
hence should be of the form:
$$ [M]= \left( \frac{1-\gamma^{5}}{2} \right) \gamma_{\mu} V^{\mu} + \left( \frac{1+\gamma^{5}}{2} \right) \gamma_{\mu} W^{\mu} $$

But then, to get the answer, I suppose that:
$$ V^{\mu} = u_{L}(p_{2})\bar{u}_{L}(p_1) $$
$$ W^{\mu} = u_{R}(p_{1})\bar{u}_{R}(p_2) $$
Honestly, I don't see exactly why. Can someone point me out the way to make this identification?

Next, in part (d), we should get an amplitude of the form:
$$ i\mathcal{M} = (-ie)^{2} \bar{v}_{R} (k_{2}) \gamma^{\mu} u_{R} (k_{1}) \frac{-i}{s} \bar{u}_{R}(p_{1})\gamma_{\nu} v_{R}(p_2)$$
Thus, we have terms in u and v. However, all the spinor product formalism has been developed in terms of u. My question is: can one define a spinor product $$s(p_1,p_2)=\bar{v}_{R}(p_1) u_{L}(p_2) $$?

Thanks,
Emmanuel
 
Physics news on Phys.org
All these messy calculations involving Fierz transformations identities can be found in the books by Atkinson's and Radovanovic's.
 
Thanks for the references, MathematicalPhysicist.
But I took a look on these books, and all I saw was the general Fierz identities, and these are already treated in PS.
However, I still don't see where does this result come from. I'm sure it's something basic, but I can't see it...
 
Does anyone else have a suggestion?
Thanks!
 
Hi!
Yes, I have seen this manual, but unfortunately, he states that this is all obvious...
 
  • Like
Likes   Reactions: MathematicalPhysicist
Hi @Manu_ , have you tried the exercises textbook of Atkinson's?
 
Hi!
I took a look in this book and Radovanovic's, but they cover more or less what has been treated in P&S before problem 5.3. They do not mention Fierz identities used with spinor products.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 49 ·
2
Replies
49
Views
7K
  • · Replies 2 ·
Replies
2
Views
7K