Question related to the Lorentz Invariance

[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..)

 

[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}<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> \end{bmatrix}

So every power S^{2k} = S^{2},~ (k&gt;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...
<br /> S \cdot S^{2} = -i \begin{bmatrix}<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 1 &amp; 0\\<br /> 0 &amp; -1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> \end{bmatrix} <br /> <br /> \begin{bmatrix}<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 1 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 1 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> \end{bmatrix} =<br /> <br /> \begin{bmatrix}<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -i &amp; 0\\<br /> 0 &amp; i &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> \end{bmatrix} = S<br /> <br />

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} =<br /> \begin{bmatrix}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 1+\sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p} &amp; - \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} &amp; 0\\<br /> 0 &amp; \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} &amp; 1+\sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p} &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 1\\<br /> \end{bmatrix} <br />

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

e^{-i \theta S} =<br /> \begin{bmatrix}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; \sum_{p=0} (-1)^{p} (\frac{\theta}{2p!})^{2p} &amp; - \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} &amp; 0\\<br /> 0 &amp; \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} &amp; \sum_{p=0} (-1)^{p} (\frac{\theta}{2p!})^{2p} &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 1\\<br /> \end{bmatrix} <br /> <br /> =<br /> \begin{bmatrix}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; \cos \theta &amp; - \sin \theta &amp; 0\\<br /> 0 &amp; \sin \theta &amp; \cos \theta &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 1\\<br /> \end{bmatrix} <br /> <br />

 

[lhcQFT]

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