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Questions from Peskin and Schroeder

  1. Jun 29, 2015 #1

    MathematicalPhysicist

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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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  3. Jun 29, 2015 #2

    vanhees71

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    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
     
  4. Jun 29, 2015 #3

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    On which pages exactly do they discuss this issue?

    Thanks.
     
  5. Jun 29, 2015 #4

    ChrisVer

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    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.
     
  6. Jun 29, 2015 #5

    haushofer

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    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.
     
  7. Jun 29, 2015 #6

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    Thanks guys, your help is appreciated.
     
  8. Jun 29, 2015 #7

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    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.
     
  9. Oct 13, 2015 #8

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    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}##?
     
    Last edited: Oct 13, 2015
  10. Oct 15, 2015 #9

    ChrisVer

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    I guess the boost is taken along the z-axis ?
     
  11. Oct 15, 2015 #10

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    @ChrisVer yes the boost is taken over 3-direction, but how do you arrive at this matrix from ##\omega_{\mu \nu} S^{\mu \nu}##?
     
  12. Oct 15, 2015 #11

    ChrisVer

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    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].
     
  13. Oct 15, 2015 #12

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    @ChrisVer that's what I also thought, but wasn't sure.
     
  14. Oct 17, 2015 #13

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    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.
     
  15. Oct 29, 2015 #14

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  16. Oct 29, 2015 #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.
     
  17. Oct 29, 2015 #16

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  18. Oct 30, 2015 #17

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    @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?
     
  19. Oct 30, 2015 #18

    CAF123

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    @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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