# Questions from Peskin and Schroeder

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1. Jun 29, 2015

### MathematicalPhysicist

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?

2. Jun 29, 2015

### vanhees71

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

3. Jun 29, 2015

### MathematicalPhysicist

On which pages exactly do they discuss this issue?

Thanks.

4. Jun 29, 2015

### ChrisVer

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.

5. Jun 29, 2015

### haushofer

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.

6. Jun 29, 2015

### MathematicalPhysicist

Thanks guys, your help is appreciated.

7. Jun 29, 2015

### MathematicalPhysicist

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. Oct 13, 2015

### MathematicalPhysicist

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:
$$u(p) = \exp(-1/2 \eta \begin{bmatrix} \sigma^3& 0\\ 0& -\sigma^3 \end{bmatrix}) \sqrt{m} \begin{bmatrix} \xi\\ \xi \end{bmatrix}$$

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 $\omega_{\mu \nu} S^{\mu \nu}$?

Last edited: Oct 13, 2015
9. Oct 15, 2015

### ChrisVer

I guess the boost is taken along the z-axis ?

10. Oct 15, 2015

### MathematicalPhysicist

@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. Oct 15, 2015

### ChrisVer

I think that the generators of boosts are the $S^{0i}$... as a result all the $\omega_{\mu \nu}$ (the transformation parameters) that don't match to this must be zero.
Then taking only the 3-direction of boosts $\omega_{03}\ne0 \equiv \eta$.

12. Oct 15, 2015

### MathematicalPhysicist

@ChrisVer that's what I also thought, but wasn't sure.

13. Oct 17, 2015

### MathematicalPhysicist

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?).

14. Oct 29, 2015

### MathematicalPhysicist

Anyone?

15. Oct 29, 2015

### Dr.AbeNikIanEdL

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 $-ip$.

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

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.

16. Oct 29, 2015

### MathematicalPhysicist

17. Oct 30, 2015

### MathematicalPhysicist

@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. Oct 30, 2015

### CAF123

@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 $(\vec\sigma \cdot \vec p)^2 = \vec p^2$