# I Spinor representation of Lorentz transformations

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1. Mar 14, 2017

### 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.

2. Mar 14, 2017

### 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 $\Sigma^{ij}$!

3. Mar 15, 2017

### samalkhaiat

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

$$-\frac{i}{2}\omega_{\mu\nu}\Sigma^{\mu\nu} = \begin{pmatrix} \frac{1}{2}\left( - \vec{\beta} + i \vec{\theta} \right) \cdot \vec{\sigma} & 0 \\ 0 & \frac{1}{2}\left( \vec{\beta} + i \vec{\theta} \right) \cdot \vec{\sigma} \end{pmatrix} .$$

4. Mar 15, 2017

### Frank Castle

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)?

5. Mar 15, 2017

### Frank Castle

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?!

6. Mar 15, 2017

### samalkhaiat

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 $S_{R}(\Lambda)$.

7. Mar 15, 2017

### Frank Castle

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