# Question related to the Lorentz Invariance

1. Sep 1, 2014

### lhcQFT

I have a question related to the Lorentz invariance.

(on the book of Mark Srednicki Quantum Field Theory, page 35 prob. 2.9 c)

There are representations of $\Lambda$ and S.

In order to show that result of problem, I use number of two ways.

1. I expanded $\Lambda$ to infinitesimal form using Taylor expansion.
2. Just original shape of exponential form.

When I inserted values of S^12, the first method gives me that the value is 0.(by specific element, that is rho = 0, tau =1) But the second method gives me that the value is 1.

The correct answer is 0. because most of components are 0 except for middle row and middle column(2x2). I can't understand this discrepancy. Please explain this error.

(Sorry about that I can't use Latex. because I'm a novice of this web site..)

2. Sep 1, 2014

### ChrisVer

you don't have to use infinitesimal expansions... Just use the series form for the exponential:

$e^{A} = \sum_{n} \frac{A^{n}}{n!}$

I use $\hbar =1$ although it doesn't matter to state since you can also rename $S^{12} \equiv \frac{1}{\hbar} S^{12}$ and using the appropriate form for $S^{12}$ repr, as I do below - no hbar, all to keep exponential argument dimensionless. Also some expressions might look terrible, but if you are accustomed to these calculations, you can jump them - i just did everything almost explicitly because I don't know you.

$e^{-i \theta S^{12}} = \sum_{n} \frac{(-i \theta S^{12})^{n}}{n!}$

$e^{-i \theta S^{12}} = 1_{4 \times 4} + \sum_{p=1} (-i)^{2p} (S^{12})^{2p} (\frac{\theta}{2p!})^{2p}+\sum_{p=0} (-i)^{2p+1} (S^{12})^{2p+1} (\frac{\theta}{2p+1!})^{2p+1}$

Now $[(S^{12})^{2}]^{\rho}_{\omega} = \sum_{\sigma} (S^{12})^{\rho}_{\sigma} (S^{12})^{\sigma}_{\omega} = \sum_{\sigma} (-i)^{2} (g^{1 \rho} \delta^{2}_{\sigma}- g^{2 \rho} \delta^{1}_{\sigma}) (g^{1 \sigma} \delta^{2}_{\omega}- g^{2 \sigma} \delta^{1}_{\omega})= - \sum_{\sigma}[ g^{1 \rho} \delta^{2}_{\sigma} g^{1 \sigma} \delta^{2}_{\omega} - g^{1 \rho} \delta^{2}_{\sigma}g^{2 \sigma} \delta^{1}_{\omega}- g^{2 \rho} \delta^{1}_{\sigma}g^{1 \sigma} \delta^{2}_{\omega}+ g^{2 \rho} \delta^{1}_{\sigma}g^{2 \sigma} \delta^{1}_{\omega}]$

Now from this -due to summing over the delta kroenecker and metric is diagonal- the 1st and last term vanish, while the other two survive... So you have [doing the sum of sigma]:

$[(S^{12})^{2}]^{\rho}_{\omega} = g^{1 \rho} g^{22} \delta^{1}_{\omega}+ g^{2 \rho} g^{11} \delta^{2}_{\omega}$

From here you see that the squared $S^{12}$ is diagonal (try putting $\rho \ne \omega$ ) and has entries 1 at $\rho = \omega = 1,2$. I'll drop 12 of S since we are only caring about them, and with $S$ I'll mean the matrix of $S^{12}$

$S^{2} = \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$

So every power $S^{2k} = S^{2},~ (k>0)$. This can also be seen without doing any calculations if you could have noticed the similarity of $S$ to the Pauli matrice in a block form. What about the $S^{2k+1}$ ? Well you just have to multiply $S^{2}$ with the $S$ matrix...
$S \cdot S^{2} = -i \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & -i & 0\\ 0 & i & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix} = S$

And finally you can find the exponential:

$e^{-i \theta S} = 1_{4 \times 4} + S^{2} \sum_{p=1} (-i)^{2p} (\frac{\theta}{2p!})^{2p}+ S \sum_{p=0} (-i)^{2p+1} (\frac{\theta}{2p+1!})^{2p+1}= 1_{4 \times 4} + S^{2} \sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p}- i S \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1}$

which in matrix form is:

$e^{-i \theta S} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1+\sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p} & - \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 0\\ 0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 1+\sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p} & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$

That's the $\Lambda$ you have.... just see the series expansions of cos and sin in your expression...

$e^{-i \theta S} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p!})^{2p} & - \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 0\\ 0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p!})^{2p} & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & \cos \theta & - \sin \theta & 0\\ 0 & \sin \theta & \cos \theta & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$

Last edited: Sep 1, 2014
3. Sep 1, 2014

### lhcQFT

Ah.. Thank you for your guide. I make sense how to calculate. :-)