Inverse of infinitesimal Lorentz transformation

[Gene Naden]

I am working through Lessons in Particle Physics by Luis Anchordoqui and Francis Halzen. The link is https://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1271v2.pdf. I am on page 21. Between equations (1.5.53) and (1.5.54), the authors make the following statement:
$S^\dagger ( \Lambda ) = \gamma ^0 S^{-1} ( \Lambda ) \gamma ^0$

Here $S$ is the unitary transformation corresponding to a Lorentz transformation and
$\Lambda$ is the Lorentz transformation.

They give the following definition for $S$:
$S = 1 - \frac{i}{2} \omega_{\mu\nu} \Sigma^{\mu\nu}$
where $\Sigma^{\mu\nu}$ is defined as $\frac{i}{4} [ \gamma^\mu, \gamma^\nu]$
My first question has to do with $S^{-1}$: Can I invert $S$ by changing the sign of the $\omega$?
Second, can I assume that $(\gamma^\mu)^\dagger = \gamma^\mu$
I see that this is true in the "standard" representation for $\gamma^\mu$. Is it true in general?

As always, thanks.

 

[Gene Naden]

So I went ahead with the assumption that $S^{-1}=1+\frac{i}{2}\omega_{\mu\nu} \Sigma^{\mu\nu}$. This led me to the conclusion that, for
$S^\dagger ( \Lambda ) = \gamma ^0 S^{-1} ( \Lambda ) \gamma ^0$ to be true, it must be the case that
$[\gamma^\mu,\gamma^\nu] = \gamma^0[\gamma^\mu,\gamma^\nu]\gamma^0$
So I tried to prove this and failed. Anyway, the equality looks false because multiplying before and after by $\gamma^0$ changes the sign for a lot of values of $\mu$ and $\nu$.

Any help would be appreciated.

 

[Sorcerer]

I can’t help you because this is way above my level, but I can say (because I’ve gone through the algebra and seen it despite it being “obvious”) that in the Freshmen version of relativity, if you start with x = γ(x’+vt’) and solve for x’, you will get x’ = γ(x - vt). (as long as you don’t forget to substitute for t’ as you go).

 

[samalkhaiat]

S^\dagger ( \Lambda ) = \gamma ^0 S^{-1} ( \Lambda ) \gamma ^0

Here S is the unitary transformation corresponding to a Lorentz transformation and
\Lambda is the Lorentz transformation.

How can S be unitary when S^{\dagger} = \gamma^{0} S^{-1}\gamma^{0} \neq S^{-1}? The matrix S(\Lambda) cannot be unitary because Lorentz group is non-compact.

They give the following definition for S:
S=1 - \frac{i}{2} \omega_{\mu\nu} \Sigma^{\mu\nu}
where \Sigma^{\mu\nu} is defined as \frac{i}{4} [ \gamma^\mu, \gamma^\nu]

No, this expression for \Sigma is not a definition, because it can be proved.

My first question has to do with S. Can I invert S by changing the sign of the \omega?

Yes, because S(\Lambda)S(\Lambda^{-1}) = S(\Lambda \Lambda^{-1}) = 1 \ \Rightarrow \ S^{-1}(\Lambda ) = S( \Lambda^{-1}). So, if you write \Lambda = 1 + \omega, then to first order you have \Lambda^{-1} = 1 - \omega. Also, if you choose S(\Lambda ) = 1 - (i/2) \omega \cdot \Sigma, then S^{-1}(\Lambda) = 1 + (i/2) \omega \cdot \Sigma.

Second, can I assume that (\gamma^\mu)^\dagger = \gamma^\mu

No, you cannot “assume” that. The relation (\gamma^{\mu})^{\dagger} = \gamma^{0} \gamma^{\mu} \gamma^{0} holds in any representation.

 

[Gene Naden]

Thank for the "tough love." I see that S is not unitary. The relation $(\gamma^{\mu})^{\dagger} = \gamma^{0} \gamma^{\mu} \gamma^{0}$ seems to be the key. The authors use a special symbol to denote that the formula for $\Sigma$ is a definition. I appreciate your quick response.

 

[Gene Naden]

Yes that solves it. Thanks again.