Questions from Peskin and Schroeder

In summary, the conversation discusses the presence of the factor ##\Lambda_{1/2}## in the transformation of the spinor field, which is necessary due to the spinor nature of the field. The transformation of the field is similar to that of the Maxwell equations, where all components must be transformed. The conversation also discusses the exponential matrix used in the boost transformation of the spinor field, and how it can be verified as a solution to the Dirac equation. The identity ##(p\cdot \sigma)(p\cdot \bar{\sigma})=p^2=m^2## is also discussed and proven.
  • #1
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On page 42 of Peskin, at the bottom they say that the next transformation should follow:

##[i\gamma^\mu\partial_\mu - m ]\psi (x) \rightarrow [i\gamma^\mu(\Lambda^{-1})^\nu_\mu \partial_\nu - m ] \Lambda_{1/2} \psi (\Lambda^{-1}x)##

But why does the factor ##\Lambda_{1/2}## appear there? shouldn't it be without this factor?
Thanks in advance.
 
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  • #2
No, there must be this factor, because the spinor field is not a scalar but (as the name suggests) a spinor. More precisely it's a bispinor or Dirac-spinor, and thus ##\Lambda_{1/2}## belongs to the ##(1/2,0) \oplus (0,1/2)## representation of the Lorentz group. Unfortunately Peskin-Schroeder is not very explicit in the group-theoretical aspects of the whole game. To really understand, what's behind the fields, I recommend

Sexl, Roman U., Urbandtke, Helmuth K.: Relativity, Groups, Particles, Springer, 2001
 
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  • #3
On which pages exactly do they discuss this issue?

Thanks.
 
  • #4
Well you can even see that in Srednicki's book...eg eq 34.1
The spinor gets transformed by a Lorentz transformation, as well as its argument.
 
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  • #5
In a coordinatetransformation you should transform everything. The psi transforms nontrivially under a coordinatetransformation. Hence your factor Lambda_{1/2}. Compare to the Maxwell equations; performing a coordinate transformation transforms the field A.
 
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  • #6
Thanks guys, your help is appreciated.
 
  • #7
Wait a minute, let's make it rigorous.
we have ##i\gamma^\mu \partial_\mu -m ] \psi(x) ##, so what is transformed is ##i\gamma^\mu \partial_\mu \rightarrow i\gamma^\mu (\Lambda^{-1})^\nu _\mu \partial_\nu## and ##\psi(x)\rightarrow \Lambda_{1/2} \psi(\Lambda^{-1}x)## is this right?

And we get this annoying factor since it's a transformation of spinors, right? In scala transformation we don't have this factor.
 
  • #8
I have another question from Peskin's, on page 46 they say:"
Now apply the same boost to ##u(p)##. According to equations (3.26) and (3.30) which are respcetively: ##S^{0i}=-\imath /2
\begin{bmatrix}
\sigma^i& 0\\
0& -\sigma^i \end{bmatrix}##

and ##\Lambda_{1/2} = \exp (-\imath /2 \omega_{\mu \nu} S^{\mu \nu}## we get:
[tex]u(p) = \exp(-1/2 \eta
\begin{bmatrix}
\sigma^3& 0\\
0& -\sigma^3 \end{bmatrix}) \sqrt{m}
\begin{bmatrix}
\xi\\
\xi \end{bmatrix}[/tex]

I don't understand how did we get the last boost? How to infer this exponential matrix? Shouldn't we take: ##\Lambda_{1/2}\sqrt{m} (\xi , \xi)^T##, what happened to the other terms in ##\omega_{\mu \nu} S^{\mu \nu}##?
 
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  • #9
I guess the boost is taken along the z-axis ?
 
  • #10
@ChrisVer yes the boost is taken over 3-direction, but how do you arrive at this matrix from ##\omega_{\mu \nu} S^{\mu \nu}##?
 
  • #11
I think that the generators of boosts are the [itex]S^{0i}[/itex]... as a result all the [itex]\omega_{\mu \nu}[/itex] (the transformation parameters) that don't match to this must be zero.
Then taking only the 3-direction of boosts [itex]\omega_{03}\ne0 \equiv \eta[/itex].
 
  • #12
@ChrisVer that's what I also thought, but wasn't sure.
 
  • #13
On page 46 it's argued that you can verify directly that ##u(p) =
\begin{bmatrix}
\sqrt{p\cdot \sigma} \xi\\
\sqrt{p\cdot \bar{\sigma}} \xi
\end{bmatrix}##
solves the Dirac equation: ##
\begin{bmatrix}
-m& i\sigma \cdot \partial\\
i\bar{\sigma}\cdot \partial & -m \end{bmatrix}
\begin{bmatrix}
\psi_L\\
\psi_R\end{bmatrix} = 0
##
I don't see how exactly, if I plug ##u(p)## back to Dirac equation I get:
##
\begin{bmatrix}
-m\sqrt{p\cdot \sigma} \xi + i\sigma \partial (\sqrt{p\cdot \bar{\sigma}} \xi)\\
i\bar{\sigma}\partial(\sqrt{p\cdot \sigma} \xi) - m \sqrt{p\cdot \bar{\sigma}}\xi\end{bmatrix}## what identity do I miss here and why (I mean why does the identity follow or where does it appear in Peskin's?).

Thanks in advance.
 
  • #14
Anyone?
 
  • #15
I think actually not ##u(p)## is the solution but ##u(p)e^{-ipx}## (compare to eq. 3.45). The ##\partial##'s in your equation will then be replaced by ##-ip##.
The first line then reads

## -m \sqrt{p\sigma}\xi + \sigma p \sqrt{p\bar\sigma}\xi = -m \sqrt{p\sigma}\xi + \sqrt{(p \sigma) (p\sigma) (p\bar\sigma)}\xi ##

With relation 3.51 ## (p \sigma)(p\bar\sigma) = p^2 = m^2 ## it follows that this is zero, and similar for the second line.
 
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  • #16
  • #17
@Dr.AbeNikIanEdL how do I prove this identity (3.51)?

I mean we have: ##p^\mu \cdot \sigma^\mu p^\mu \cdot \bar{\sigma}^\mu = (E-p\cdot \sigma)(E+p\cdot \sigma) = E^2 - (p\cdot \sigma)^2=E^2 - \vec{p}^2 ##

Why is the last equality correct?
 
  • #18
@MathematicalPhysicist: $$(\vec \sigma \cdot \vec p)^2 = \sigma_i p_i \sigma_j p_j = \sigma_i \sigma_j p_i p_j = (2 \delta_{ij} - \sigma_j \sigma_i)p_i p_j = 2 \vec p^2 - (\vec \sigma \cdot \vec p)^2$$ Therefore ## (\vec\sigma \cdot \vec p)^2 = \vec p^2##
 
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1. What is the main purpose of "Questions from Peskin and Schroeder"?

The main purpose of "Questions from Peskin and Schroeder" is to serve as a supplementary resource for students studying quantum field theory. It contains a collection of challenging and thought-provoking questions that aim to deepen the understanding of the concepts and principles discussed in the textbook "An Introduction to Quantum Field Theory" by Michael E. Peskin and Daniel V. Schroeder.

2. Who is the target audience for "Questions from Peskin and Schroeder"?

The target audience for "Questions from Peskin and Schroeder" is primarily graduate and advanced undergraduate students in physics or related fields who are studying quantum field theory. However, it can also be a useful resource for researchers and professionals in the field who want to test their knowledge and understanding of the subject.

3. Are the questions in "Questions from Peskin and Schroeder" based on real-life applications?

Yes, many of the questions in "Questions from Peskin and Schroeder" are based on real-life applications and experiments in quantum field theory. They are designed to provide a deeper understanding of the theoretical concepts and their practical applications in various fields of physics, such as particle physics, condensed matter physics, and cosmology.

4. How can "Questions from Peskin and Schroeder" be used effectively in studying quantum field theory?

"Questions from Peskin and Schroeder" can be used effectively as a supplement to the textbook "An Introduction to Quantum Field Theory" by Peskin and Schroeder. Students can use it to test their understanding of the material, identify areas where they need to improve, and deepen their knowledge through self-study and problem-solving. It can also be used by instructors as a source of additional practice problems for their students.

5. Is "Questions from Peskin and Schroeder" suitable for self-study or is it best used in a classroom setting?

"Questions from Peskin and Schroeder" can be used effectively for both self-study and in a classroom setting. However, it is recommended for students to have a solid understanding of the material covered in the textbook before attempting the questions. In a classroom setting, instructors can use the questions as homework assignments or in-class exercises. For self-study, students can work through the questions at their own pace and use them as a tool for self-assessment and review.

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