Poincare algebra and its eigenvalues for spinors

[Markus Kahn]

Homework Statement

Show that for
$$W^\mu = -\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}M^{\nu\rho}P^{\sigma},$$
where $M^{\mu\nu}$ satisfies the commutation relations of the Lorentz group and $\Psi$ is a bispinor that transforms according to the $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ representation of the Lorentz group we have
$$W^2\Psi = -\frac{1}{2}\left(\frac{1}{2}+1\right)m^2\Psi.$$

Homework Equations

This is a follow up question to this question : https://www.physicsforums.com/threads/spin-matrices-and-field-transformations.967878/

The Attempt at a Solution

For the purpose of leaving out unnecessary calculations let's assume it is clear that we can write
$$W^\mu = -\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}S^{\nu\rho}P^{\sigma},$$
where $S^{\mu\nu}= \frac{i}{4}[\gamma^\mu,\gamma^\nu]$ and $\gamma$ satisfying the Clifford algebra. My attempt to find $W^2$ goes as follows
$$\begin{align*}W_\mu W^\mu &= -\frac{1}{16}\varepsilon_{\mu\nu\rho\sigma}\varepsilon^{\mu abc}S^{\nu\rho}S_{ab}\partial^{\sigma}\partial_c = -\frac{1}{16}\eta^{ad}\eta^{be}\eta^{cf}\varepsilon_{\mu\nu\rho\sigma}\varepsilon^{\mu}{}_{def}S^{\nu\rho}S^{ab}\partial^{\sigma}\partial^c\\
&= \frac{1}{16}\eta^{ad}\eta^{be}\eta^{cf} \begin{vmatrix}\eta_{\nu d} & \eta_{\nu e} & \eta_{\nu f} \\ \eta_{\rho d} & \eta_{\rho e} & \eta_{\rho f} \\ \eta_{\sigma d} & \eta_{\sigma e} & \eta_{\sigma f} \end{vmatrix}
S^{\nu\rho}S^{ab}\partial^{\sigma}\partial^c\end{align*}$$
At this point things start to get a bit ugly... If one calculates the determinant the first term takes the form $\eta_{\nu d}\eta_{\rho e}\eta_{\sigma f}$ so I thoguht I'd try fist just calculating this one and then worry about the rest. So we get
$$\frac{1}{16}\eta^{ad}\eta^{be}\eta^{cf} \eta_{\nu d}\eta_{\rho e}\eta_{\sigma f}S^{\nu\rho}S^{ab}\partial^{\sigma}\partial^c = \frac{1}{16}S^{ab}S^{ab}\partial^c\partial^c$$
and form here on I'm lost... What exactly am I supposed to do with this expression and how can this result in a scalar value when applied to the bispinor?

 

[jambaugh]

Just a bit of starter advice. Instead of reversing all indices in $W_\mu$ simply lower the index with the metric, or rather raise the index since more properly:
$$W^\mu = \eta^{\mu\lambda} \varepsilon_{\lambda \nu \rho \sigma} S^{\nu\rho}P^{\sigma}$$

There appear to be other places where your raised and lowered indices don't match up. I'm not clear where your factor fo 1/16 came from. Should you be replacing our S's with gammas?

A bit of helpful notation you may find is to write $\gamma^{\mu\nu} = 1/2(\gamma^\mu\gamma^\nu - \gamma^\nu\gamma^\mu)$ to distinguish spin rep from Lie algebra element.

You should work within the clifford algebra which will allow you to algebraically resolve all those index identities and commutation relations.

Let's see...
Assuming a diagonal basis, $\gamma^{\alpha\beta} = \gamma^\alpha\gamma^\beta$, likewise $\gamma^{\alpha\beta\epsilon}=\gamma^\alpha\gamma^\beta\gamma^\epsilon$
and $\gamma^{\alpha\beta\epsilon\phi}=\gamma^\alpha\gamma^\beta\gamma^\epsilon\gamma^\phi$
but in 4 dimensions this is proportional to the top element $\gamma^5=\gamma^1\gamma^2\gamma^3\gamma^4$. In particular:
$$\gamma^{\alpha\beta\epsilon\phi}=\varepsilon^{\alpha\beta\epsilon\phi}\gamma^5$$
You can also use the fact that the expansion: $\varepsilon_{\alpha\beta\mu\nu}\gamma^\alpha\gamma^\beta\gamma^\mu\gamma^\nu$ will be a sum of zero terms plus the 4! copies of $\gamma^5$
This let's you absorb these Levi-Civita symbols into the clifford algebra of the gamma matrices.
Look at the wikipedia page on Gamma Matrices for the list of identities. With these you should be able to reduce your expression down to simplified form.

[Addendum] I forgot to mention you can reposition terms in a product of gamma matrices using the anti-commutator. In general
$$X\alpha\beta Y = X\{\alpha,\beta\}Y-X\beta\alpha Y$$
and of course: $\{\gamma^\alpha,\gamma^\beta\} = -2\eta^{\alpha\beta}$ and $\{\gamma^\alpha,\gamma_\beta\} = -2\delta^\alpha_\beta$

 

[Markus Kahn]

First of all I'd like to thank you for your suggestions! I think I was able to simplify the problem quite a bit thanks to them. So let me just quickly explain what I did.. As you already said, I messed up some of the indices, which lead to wrong results. I now get

$$W_\mu W^\mu = \frac{1}{4}S_{ab}P_cS^{de}P^f \eta^{a\alpha}\eta^{b\beta}\eta^{c\gamma}\varepsilon_{\mu\alpha\beta\gamma}\varepsilon^{\mu}{}_{def}$$

By making use of the relations you mentioned and the one which were provided on Wikipedia I was able to simplify this to
$$\begin{align*}W_\mu W^\mu &= \frac{3}{2}P^\mu P_\mu +\frac{1}{2}S^{\alpha\beta}P^\gamma S_{\beta\gamma}P_\alpha+ \frac{1}{2}S^{\alpha\beta}P^{\gamma}S_{\gamma\alpha}P_\beta\\
&= \frac{3}{2}P^\mu P_\mu +\frac{1}{2}S^{\alpha\beta} S_{\beta\gamma}P^\gamma P_\alpha+ \frac{1}{2}S^{\beta\alpha}S_{\alpha\gamma}P^{\gamma}P_\beta\\
&= \frac{3}{2}P^\mu P_\mu +S^{\alpha\beta} S_{\beta\gamma}P^\gamma P_\alpha\end{align*},$$
where I just literally calculated every term that was generated by the Levi-Civita contraction on its own and then summed all of them up. And this is where I'm currently stuck.. What I have tried so far
$$\begin{align*}S^{\alpha\beta} S_{\beta\gamma} &= \left(\frac{i}{4}\right)^2 [\gamma^\alpha,\gamma^\beta][\gamma_\beta,\gamma_\gamma]\\
&= -\frac{1}{16}\left(\gamma^\alpha\gamma^\beta\gamma_\beta\gamma_\gamma -\gamma^\alpha\gamma^\beta\gamma_\gamma\gamma_\beta -\gamma^\beta\gamma^\alpha\gamma_\beta\gamma_\gamma +\gamma^\beta\gamma^\alpha\gamma_\gamma\gamma_\beta\right)\\
&=-\frac{1}{16}\left(4\gamma^\alpha\gamma_\gamma +2\gamma^\alpha\gamma_\gamma + 2\gamma^\alpha\gamma_\gamma +4\eta^\alpha{}_\gamma\right)\\
&= -\frac{1}{4}(2\gamma^\alpha\gamma_\gamma +\eta^\alpha{}_\gamma)
\end{align*}$$
Inserting this into the above expression gives
$$W_\mu W^\mu=\frac{3}{2}P^\mu P_\mu -\frac{1}{4}(2\gamma^\alpha\gamma_\gamma +\eta^\alpha{}_\gamma)P^\gamma P_\alpha$$
I'm not really sure how to proceed from here.. To get to the solution I need to show that
$$W_\mu W^\mu = \frac{3}{4}P^\mu P_\mu$$
but for this to be true I need
$$\gamma^\alpha\gamma_\gamma P^\gamma P_\alpha=0 \quad\text{and}\quad \eta^\alpha{}_\gamma P^\gamma P_\alpha = P^\gamma P_\gamma,$$
where the second equation is certainly not true.. Can you maybe help me here?

EDIT: Not sure if this is useful in this situation, but we have
$$2\gamma^\alpha\gamma_\gamma = \gamma^\alpha\gamma_\gamma+\gamma^\alpha\gamma_\gamma = [\gamma^\alpha,\gamma_\gamma]+ 2\eta^\alpha{}_\gamma$$

 

[samalkhaiat]

$$W_\mu W^\mu=\frac{3}{2}P^\mu P_\mu -\frac{1}{4}(2\gamma^\alpha\gamma_\gamma +\eta^\alpha{}_\gamma)P^\gamma P_\alpha$$

This is correct and simplifies to [do you know that $\eta^{\mu}_{\nu} = \delta^{\mu}_{\nu}$?]
$$W^{2} = \frac{3}{2} P^{2} - \frac{1}{2} \gamma^{\mu} P_{\mu} \gamma^{\nu} P_{\nu} - \frac{1}{4} P^{2} ,$$ or $$W^{2} = \frac{5}{4} P^{2} - \frac{1}{2} \gamma^{\mu}P_{\mu}\gamma^{\nu}P_{\nu} .$$ Now, if you don't know the relation $\gamma^{\mu} P_{\mu} \gamma^{\nu}P_{\nu} = ( \gamma^{\mu}P_{\mu})^{2} = P^{2}$, substitute $P_{\mu} = i \partial_{\mu}$ in the second term and apply $W^{2}$ that to the Dirac spinor $\Psi$: $$W^{2}\Psi (x) = \frac{5}{4} P^{2}\Psi (x) - \frac{1}{2}i \gamma^{\mu}\partial_{\mu} \left(i \gamma^{\nu}\partial_{\nu}\Psi (x) \right) .$$ Okay, now use $P^{2}\Psi = m^{2}\Psi$ in the first term, and in the second term use the Dirac equation $i\gamma^{\sigma}\partial_{\sigma}\Psi (x) = m \Psi (x)$ twice.