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I Spinor representation of Lorentz transformations

  1. Mar 14, 2017 #1
    I've been working my way through Peskin and Schroeder and am currently on the sub-section about how spinors transform under Lorentz transformation. As I understand it, under a Lorentz transformation, a spinor ##\psi## transforms as $$\psi\rightarrow S(\Lambda)\psi$$ where $$S(\Lambda)=\exp\left(-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu}\right)$$ with $$\Sigma^{\mu\nu}=\frac{i}{4}\left[\gamma^{\mu},\,\gamma^{\nu}\right]=-\Sigma^{\nu\mu}$$ Then, in the Weyl representation we have that $$\Sigma^{0i}=-\frac{i}{2}\left(\begin{matrix}\sigma^{i}&&0\\ 0&&-\sigma^{i}\end{matrix}\right)$$ and $$\Sigma^{ij}=\frac{i}{2}\varepsilon^{ijk}\left(\begin{matrix}\sigma^{k}&&0\\ 0&&\sigma^{k}\end{matrix}\right)$$ Given this, what confuses me is how one ends up with the following left-handed and right-handed transformations: $$S(\Lambda)_{L}=\exp\left(-\frac{\mathbf{\beta}\cdot\mathbf{\sigma}}{2}+i\frac{\mathbf{\theta}\cdot\mathbf{\sigma}}{2}\right) \\ \\ S(\Lambda)_{R}=\exp\left(\frac{\mathbf{\beta}\cdot\mathbf{\sigma}}{2}+i\frac{\mathbf{\theta}\cdot\mathbf{\sigma}}{2}\right)$$ Where does the additional ##i## come from in the spatial rotations term?

    I have read from other sources, that the parameters ##\omega_{\mu\nu}## are defined such that ##\omega_{0i}=\beta_{i}## and ##\omega_{ij}=\varepsilon_{ijk}\theta^{k}##, which are the boost and rotation parameters respectively. Given these, however, I can't arrive at the above expressions. For example, for ##S(\Lambda)_{L}## I obtain
    $$S(\Lambda)_{L}=\exp\left(-\frac{\mathbf{\beta}\cdot\mathbf{\sigma}}{4}+\varepsilon_{ijk}\varepsilon^{ijl}\frac{\theta^{k}\sigma^{l}}{4}\right)=\exp\left(-\frac{\mathbf{\beta}\cdot\mathbf{\sigma}}{4}+\frac{\mathbf{\theta}\cdot\mathbf{\sigma}}{2}\right)$$ where I have used that ##\varepsilon_{ijk}\varepsilon^{ijl}=2\delta_{kl}##.

    Would someone be able to explain this to me as I'm really stuck on this point at the moment.
     
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  3. Mar 14, 2017 #2

    vanhees71

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    The rotations are represented by the usual SU(2) operations on the left and right-handed parts of the Dirac operator. Thus, there's a factor ##\mathrm{i}## too much in your definition of ##\Sigma^{ij}##!
     
  4. Mar 15, 2017 #3

    samalkhaiat

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    Using the antisymmetry of the Lorentz parameters, you can write [tex]-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu} = \frac{1}{4} \omega_{\mu\nu} \gamma^{\mu}\gamma^{\nu} .[/tex] Expanding the summation leads to [tex]-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu} = \frac{1}{4}\omega_{0k}\gamma^{0}\gamma^{k} + \frac{1}{4}\omega_{k0}\gamma^{k}\gamma^{0} + \frac{1}{4}\omega_{ij}\gamma^{i}\gamma^{j} .[/tex] The first two terms are equal, because [itex]\omega_{k0}=-\omega_{0k}[/itex] and [itex]\gamma^{k}\gamma^{0} = - \gamma^{0}\gamma^{k}[/itex]. Thus

    [tex]-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu} = \frac{1}{2} \omega_{0k}\gamma^{0}\gamma^{k} + \frac{1}{4} \omega_{ij}\gamma^{i}\gamma^{j} . \ \ \ (1)[/tex]

    Now, in the chiral representation we have

    [tex]\gamma^{0}\gamma^{k} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^{k} \\ - \sigma^{k} & 0 \end{pmatrix} = \begin{pmatrix} - \sigma^{k} & 0 \\ 0 & \sigma^{k} \end{pmatrix} ,[/tex]

    [tex]\gamma^{i}\gamma^{j} = \begin{pmatrix} 0 & \sigma^{i} \\ - \sigma^{i} & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^{j} \\ - \sigma^{j} & 0 \end{pmatrix} = - i \epsilon^{ijk} \begin{pmatrix} \sigma^{k} & 0 \\ 0 & \sigma^{k} \end{pmatrix} .[/tex]

    Substituting these expressions in (1), you get

    [tex]-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu} = \omega_{0k} \begin{pmatrix} - \frac{\sigma^{k}}{2} & 0 \\ 0 & \frac{\sigma^{k}}{2} \end{pmatrix} + i \left( - \frac{1}{2}\epsilon^{ijk}\omega_{ij} \right) \begin{pmatrix} \frac{\sigma^{k}}{2} & 0 \\ 0 & \frac{\sigma^{k}}{2} \end{pmatrix} .[/tex]

    Now, if you define the parameters [itex]\beta^{k} = \omega_{0k}[/itex] and [itex]\theta^{k} = - \frac{1}{2}\epsilon^{ijk}\omega_{ij}[/itex], you get

    [tex]-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu} = \begin{pmatrix} \frac{1}{2}\left( - \vec{\beta} + i \vec{\theta} \right) \cdot \vec{\sigma} & 0 \\ 0 & \frac{1}{2}\left( \vec{\beta} + i \vec{\theta} \right) \cdot \vec{\sigma} \end{pmatrix} .[/tex]
     
  5. Mar 15, 2017 #4
    Great answer, thanks.
    There's just a couple of things that I'm not quite sure about, how did you get [tex]\gamma^{i}\gamma^{j} = \begin{pmatrix} 0 & \sigma^{i} \\ - \sigma^{i} & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^{j} \\ - \sigma^{j} & 0 \end{pmatrix} = - i \epsilon^{ijk} \begin{pmatrix} \sigma^{k} & 0 \\ 0 & \sigma^{k} \end{pmatrix} .[/tex] Naively, I get $$\gamma^{i}\gamma^{j} = \begin{pmatrix} 0 & \sigma^{i} \\ - \sigma^{i} & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^{j} \\ - \sigma^{j} & 0 \end{pmatrix} = \begin{pmatrix} -\sigma^{i}\sigma^{j} & 0\\ 0 & -\sigma^{i}\sigma^{j} \end{pmatrix}$$ If I then use ##\left[\sigma^{i},\,\sigma^{j}\right]=2i\varepsilon^{ijk}\sigma^{k}## then I'll just end up with ##\sigma^{i}\sigma^{j}=\sigma^{j}\sigma^{i}+2i\varepsilon^{ijk}\sigma^{k}##?! Are you simply using that $$\omega_{ij}\gamma^{i}\gamma^{j}=\frac{1}{2}\omega_{ij}\left[\gamma^{i},\,\gamma^{j}\right]$$ such that in the Chiral representation $$\omega_{ij}\gamma^{i}\gamma^{j}=\omega_{ij}\begin{pmatrix} -\left[\sigma^{i},\,\sigma^{j}\right] & 0\\ 0 & -\left[\sigma^{i},\,\sigma^{j}\right] \end{pmatrix}=-2i\omega_{ij}\varepsilon^{ijk}\begin{pmatrix} \sigma^{k} & 0\\ 0 & \sigma^{k} \end{pmatrix}$$

    Also, do we simple choose that [itex]\theta^{k} = - \frac{1}{2}\epsilon^{ijk}\omega_{ij}[/itex] for convenience (absorbing the minus sign)?
     
  6. Mar 15, 2017 #5
    Good point, I'd forgotten about the ##i## already in the definition of ##\Sigma^{\mu\nu}##!
    Also, are the choices of for the forms of the parameters somewhat arbitrary? For example, does one choose ##\theta^{k}=-\frac{1}{2}\varepsilon^{ijk}\omega_{ij}## purely for convenience to absorb the extra factor of ##1/2## floating around?!
     
  7. Mar 15, 2017 #6

    samalkhaiat

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    So, [tex]\omega_{ij}\gamma^{i}\gamma^{j} = - \begin{pmatrix} \omega_{ij}\sigma^{i}\sigma^{j} & 0 \\ 0 & \omega_{ij}\sigma^{i}\sigma^{j} \end{pmatrix} .[/tex]

    Now, use the identity [itex]\sigma^{i}\sigma^{j} = \delta^{ij} I_{2} + i \epsilon^{ijk}\sigma^{k}[/itex] together with [itex]\omega_{ij}\delta^{ij} = 0[/itex].
    Remember that this is a rotation in the [itex](ij)[/itex]-plane, i.e., [itex]i \neq j[/itex]. And, in this case, [itex]\sigma^{i}\sigma^{j} = i \epsilon^{ijk}\sigma^{k}[/itex].

    Yes. You can choose any sign for [itex]\theta^{k}[/itex] (and for [itex]\beta^{k}[/itex]). I picked the minus sign so that my results agree with the expressions you wrote for [itex]S_{L}(\Lambda)[/itex] and [itex]S_{R}(\Lambda)[/itex].
     
  8. Mar 15, 2017 #7
    Ok great, thanks for your help. I assume what I wrote at the end of post #4 is correct?!
     
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