How to derive Eq. (3.110) in Peskin's book?

  • Thread starter Thread starter physlover
  • Start date Start date
  • Tags Tags
    Book Derive
Click For Summary
SUMMARY

The discussion focuses on deriving Eq. (3.110) from Peskin's Quantum Field Theory (QFT) book, specifically using the transformation relation between spinors under Lorentz transformations. The participant attempts to simplify the derivation by considering a boost along the axis-1 and a rotation around the axis-3, applying the Lorentz transformations as outlined in Eqs. 3.20 and 3.21. The derivation leads to a discrepancy in the left-hand portion of the spinor, indicating a potential error in the application of the transformation laws. The discussion highlights the importance of correctly applying the transformation properties of spinors in QFT.

PREREQUISITES
  • Understanding of Lorentz transformations in Quantum Field Theory
  • Familiarity with spinor notation and properties
  • Knowledge of the Dirac equation and its solutions
  • Proficiency in mathematical techniques for handling complex equations
NEXT STEPS
  • Review the derivation of spinor transformations in Peskin's QFT book, particularly Eqs. 3.20 and 3.21
  • Study the application of the Lorentz transformation to the Dirac spinor
  • Explore the implications of first-order term approximations in spinor calculations
  • Investigate common pitfalls in deriving equations from transformation laws in QFT
USEFUL FOR

Students and researchers in theoretical physics, particularly those studying Quantum Field Theory and the properties of spinors, will benefit from this discussion.

physlover
Messages
1
Reaction score
0
Hello,

I'm having trouble to derive Eq. (3.110) in Peskin's QFT book. In the book, it's said to use
[tex]u(\Lambda^{-1} \tilde{p})=\Lambda^{-1}_{1/2} u(\tilde{p})[/tex]
But I ran into trouble to derive this. Here is my try:

The equation is equivalent to
[tex]u(p)=\Lambda^{-1}_{1/2} u(\Lambda p)[/tex]
To make things simpler, I consider only a boost in axis-1 and a rotation in axis-3. In this way, the Lorentz transform on the momentum is given by Eq. 3.20 and 3.21 in the book, i.e.,
[tex]\tilde{p}^0 = p^0+\beta p^1[/tex]
[tex]\tilde{p}^1 = p^1+\beta p^0 -\theta p^2[/tex]
[tex]\tilde{p}^2 = p^2+\theta p^1[/tex]
[tex]\tilde{p}^3 = p^3[/tex]
Using Eq.3.37 in the book, and working on the left-hand portion of the spinnor
[tex]\left[ \Lambda^{-1}_{1/2} u(\tilde{p}) \right]_L =(1+i\theta\cdot \sigma/2+\beta\cdot \sigma/2) \sqrt{\tilde{p}\cdot \sigma} \xi[/tex]
[tex]=\sqrt{(1+i\theta \sigma^3+\beta \sigma^1)\left[ (p^0+\beta p^1)-(p^1+\beta p^0 -\theta p^2)\sigma^1-(p^2+\theta p^1)\sigma^2 - p^3\sigma^3 \right]} \xi[/tex]
Keeping the 1st order terms, I have
[tex]\left[ \Lambda^{-1}_{1/2} u(\tilde{p}) \right]_L = \sqrt{p\cdot \sigma +i\theta (p^0\sigma^3-p^3) -i\beta(p^2\sigma^3-p^3\sigma^2)} \xi[/tex]
This is a little different from the left-hand portion of spinnor [tex]u(p)[/tex].

Could someone help me find out what's wrong with the above derivation?

Thanks.
 
Physics news on Phys.org
Not sure I can help you, but the same law was used before while building solutions of Dirac equation in order to get u(P) from u(P-rest frame).
 

Similar threads

Replies
5
Views
4K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
4
Views
1K
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 32 ·
2
Replies
32
Views
4K