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Question related to the Lorentz Invariance

  1. Sep 1, 2014 #1
    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 [itex]\Lambda[/itex] and S.

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

    1. I expanded [itex]\Lambda[/itex] 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. jcsd
  3. Sep 1, 2014 #2

    ChrisVer

    User Avatar
    Gold Member

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

    [itex] e^{A} = \sum_{n} \frac{A^{n}}{n!}[/itex]

    I use [itex] \hbar =1 [/itex] although it doesn't matter to state since you can also rename [itex]S^{12} \equiv \frac{1}{\hbar} S^{12}[/itex] and using the appropriate form for [itex]S^{12}[/itex] 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.

    [itex] e^{-i \theta S^{12}} = \sum_{n} \frac{(-i \theta S^{12})^{n}}{n!}[/itex]

    [itex]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} [/itex]

    Now [itex][(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}][/itex]

    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]:

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

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

    [itex] S^{2} = \begin{bmatrix}
    0 & 0 & 0 & 0\\
    0 & 1 & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 0\\
    \end{bmatrix} [/itex]

    So every power [itex]S^{2k} = S^{2},~ (k>0) [/itex]. This can also be seen without doing any calculations if you could have noticed the similarity of [itex]S[/itex] to the Pauli matrice in a block form. What about the [itex]S^{2k+1}[/itex] ? Well you just have to multiply [itex]S^{2}[/itex] with the [itex]S[/itex] matrix...
    [itex]
    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

    [/itex]

    And finally you can find the exponential:

    [itex]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} [/itex]

    which in matrix form is:

    [itex]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}
    [/itex]

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

    [itex]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}

    [/itex]
     
    Last edited: Sep 1, 2014
  4. Sep 1, 2014 #3
    Ah.. Thank you for your guide. I make sense how to calculate. :-)
     
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