Undergrad Spinor Representation of Lorentz Transformations: Solving the Puzzle

[Frank Castle]

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.

 

[vanhees71]

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 $\Sigma^{ij}$!

 

[samalkhaiat]

Where does the additional $i$ come from in the spatial rotations term?
Would someone be able to explain this to me as I'm really stuck on this point at the moment.

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

-\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)

Now, in the chiral representation we have

\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} ,

\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} .

Substituting these expressions in (1), you get

-\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} .

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

-\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} .

 

[Frank Castle]

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

-\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)

Now, in the chiral representation we have

\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} ,

\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} .

Substituting these expressions in (1), you get

-\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} .

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

-\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} .

Great answer, thanks.
There's just a couple of things that I'm not quite sure about, how did you 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} = - i \epsilon^{ijk} \begin{pmatrix} \sigma^{k} & 0 \\ 0 & \sigma^{k} \end{pmatrix} . 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 \theta^{k} = - \frac{1}{2}\epsilon^{ijk}\omega_{ij} for convenience (absorbing the minus sign)?

 

[Frank Castle]

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 $\Sigma^{ij}$!

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?!

 

[samalkhaiat]

Naively, I get $$\gamma^{i}\gamma^{j} = \begin{pmatrix} 0 & \sigma^{i} \\ - \sigma^{i} & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^{j} \\ - \sigma^{j} & 0 \end{pmatrix} = \begin{pmatrix} -\sigma^{i}\sigma^{j} & 0\\ 0 & -\sigma^{i}\sigma^{j} \end{pmatrix}$$

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

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

Also, do we simple choose that \theta^{k} = - \frac{1}{2}\epsilon^{ijk}\omega_{ij} for convenience?

Yes. You can choose any sign for \theta^{k} (and for \beta^{k}). I picked the minus sign so that my results agree with the expressions you wrote for S_{L}(\Lambda) and S_{R}(\Lambda).

 

[Frank Castle]

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

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

Yes. You can choose any sign for \theta^{k} (and for \beta^{k}). I picked the minus sign so that my results agree with the expressions you wrote for S_{L}(\Lambda) and S_{R}(\Lambda).

Ok great, thanks for your help. I assume what I wrote at the end of post #4 is correct?!