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QED - correctly expressing a sum over spin-polarizations

  1. Jul 27, 2010 #1
    1. The problem statement, all variables and given/known data

    You have spinors that you designed in a computer algebra system. You’re trying to verify Casimir’s trace-trick for a single anti-spinor fermion-line...which you don’t understand all that well. You successfully verify the trace to be equal to:

    [tex]{\mathop{\rm Tr}\nolimits} \left( {{\gamma _\mu }(pslas{h_{v'}} - {m_{v'}}){\gamma _0}{{({\gamma ^\mu })}^\dag }{\gamma _0}(pslas{h_\nu } - {m_\nu })} \right) = 4\left( {{{({p_{\nu '}})}^\mu }{{({p_\nu })}^\nu } + {{({p_\nu })}^\mu }{{({p_{\nu '}})}^\nu } + {g^{\mu \nu }}({m^2} - {p_{\nu '}} \bullet {p_\nu })} \right)[/tex]

    …which is presented in one of Griffiths’s examples for, again, a fermion line (I think in Griffiths’s example it’s electron/muon scattering). Due to your success of verifying the trace: you feel pretty OK: “maybe my trace is correct as-coded”.

    But: that sum over spins is hostilely-problematic. Your code simply returns incorrect results. Which brings me to my question: can you check all my code for the problem? Kidding. Actually: my question is: I suspect I may be expanding the sum incorrectly, which means my problem is not “code” but “physics-understanding”. Is the following expansion correct? (see "Attempt...")

    2. Relevant equations

    Casimir’s trick, appearing in Griffiths “Intro to Elementary Particles”, is as follows:
    [tex]{\frac{1}{4}\sum\limits_{ \pm {\textstyle{1 \over 2}}, \pm {\textstyle{1 \over 2}}} {\left( {\bar v{\gamma _\mu }v'} \right)\left( {\bar v'{\gamma _0}{{({\gamma ^\mu })}^\dag }{\gamma _0}v} \right)} = {\mathop{\rm Tr}\nolimits} \left( {{\gamma _\mu }(pslas{h_{v'}} - {m_{v'}}){\gamma _0}{{({\gamma ^\mu })}^\dag }{\gamma _0}(pslas{h_\nu } - {m_\nu })} \right)}[/tex]

    3. The attempt at a solution

    [tex]\begin{array}{l}
    \sum\limits_{ \pm {\textstyle{1 \over 2}}, \pm {\textstyle{1 \over 2}}} {\left( {{{\bar v}_s}({{{\bf{\vec p}}}_0}){\gamma _0}{{v'}_{s'}}({\bf{\vec p}})} \right)\left( {{{\bar v'}_{s'}}({\bf{\vec p}}){\gamma _0}{{({\gamma ^0})}^\dag }{\gamma _0}{v_s}({{{\bf{\vec p}}}_0})} \right)} = \sum\limits_{ \pm {\textstyle{1 \over 2}}, \pm {\textstyle{1 \over 2}}} {\left( {{{\bar v}_s}({{{\bf{\vec p}}}_0}){\gamma _0}{{v'}_{s'}}({\bf{\vec p}})} \right)\left( {{{\bar v'}_{s'}}({\bf{\vec p}}){\gamma _0}{v_s}({{{\bf{\vec p}}}_0})} \right)} \\
    = \left( {{{\bar v}_{ + 1/2}}({{{\bf{\vec p}}}_0}){\gamma _0}{{v'}_{ + 1/2}}({\bf{\vec p}})} \right)\left( {{{\bar v'}_{ + 1/2}}({\bf{\vec p}}){\gamma _0}{v_{ + 1/2}}({{{\bf{\vec p}}}_0})} \right) + \left( {{{\bar v}_{ + 1/2}}({{{\bf{\vec p}}}_0}){\gamma _0}{{v'}_{ - 1/2}}({\bf{\vec p}})} \right)\left( {{{\bar v'}_{ - 1/2}}({\bf{\vec p}}){\gamma _0}{v_{ + 1/2}}({{{\bf{\vec p}}}_0})} \right) \\
    + \left( {{{\bar v}_{ - 1/2}}({{{\bf{\vec p}}}_0}){\gamma _0}{{v'}_{ + 1/2}}({\bf{\vec p}})} \right)\left( {{{\bar v'}_{ + 1/2}}({\bf{\vec p}}){\gamma _0}{v_{ - 1/2}}({{{\bf{\vec p}}}_0})} \right) + \left( {{{\bar v}_{ - 1/2}}({{{\bf{\vec p}}}_0}){\gamma _0}{{v'}_{ - 1/2}}({\bf{\vec p}})} \right)\left( {{{\bar v'}_{ - 1/2}}({\bf{\vec p}}){\gamma _0}{v_{ - 1/2}}({{{\bf{\vec p}}}_0})} \right) \\
    \end{array}[/tex]

    Maybe you need to see the explicit spinors, which are below. If not, please disregard....they may be complicated reading that is for naught.

    …in which: (1) primes denote post-collision momentum (in spherical coordinates) (2) no-prime means “before-collision momentum” (my spinor has only z-momentum, which is OK to choose) (3) the “s” subscript means “spin”. That means the following anti-spinors (which represent a neutron, if you want to know):

    [tex]\begin{array}{l}
    {v_{ + 1/2}}({{{\bf{\vec p}}}_0}) = \left[ {\begin{array}{*{20}{c}}
    0 \\
    { - {\textstyle{{{Q_z}} \over {2\sqrt {{\textstyle{1 \over 2}}{E_D} + m} }}}} \\
    0 \\
    { - \sqrt {{\textstyle{1 \over 2}}{E_D} + m} } \\
    \end{array}} \right]{v_{ - 1/2}}({{{\bf{\vec p}}}_0}) = \left[ {\begin{array}{*{20}{c}}
    { - {\textstyle{{{Q_z}} \over {2\sqrt {{\textstyle{1 \over 2}}{E_D} + m} }}}} \\
    0 \\
    {\sqrt {{\textstyle{1 \over 2}}{E_D} + m} } \\
    0 \\
    \end{array}} \right] \\
    {{v'}_{ + 1/2}}({\bf{\vec p}}) = \left[ {\begin{array}{*{20}{c}}
    { - {\textstyle{P \over {\sqrt {E + m} }}}{e^{ - {\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\
    {{\textstyle{P \over {\sqrt {E + m} }}}\cos ({\textstyle{1 \over 2}}\theta )} \\
    {\sqrt {E + m} \cdot {e^{ - {\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\
    { - \sqrt {E + m} \cdot \cos ({\textstyle{1 \over 2}}\theta )} \\
    \end{array}} \right]{{v'}_{ - 1/2}}({\bf{\vec p}}) = \left[ {\begin{array}{*{20}{c}}
    {{\textstyle{P \over {\sqrt {E + m} }}}\cos ({\textstyle{1 \over 2}}\theta )} \\
    {{\textstyle{P \over {\sqrt {E + m} }}}{e^{{\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\
    {\sqrt {E + m} \cdot \cos ({\textstyle{1 \over 2}}\theta )} \\
    {\sqrt {E + m} \cdot {e^{{\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\
    \end{array}} \right] \\
    {{\bar v}_{ + 1/2}}({{{\bf{\vec p}}}_0}) = \left[ {\begin{array}{*{20}{c}}
    0 & { - {\textstyle{{{Q_z}} \over {2\sqrt {{\textstyle{1 \over 2}}{E_D} + m} }}}} & 0 & {\sqrt {{\textstyle{1 \over 2}}{E_D} + m} } \\
    \end{array}} \right] \\
    {{\bar v}_{ - 1/2}}({{{\bf{\vec p}}}_0}) = \left[ {\begin{array}{*{20}{c}}
    { - {\textstyle{{{Q_z}} \over {2\sqrt {{\textstyle{1 \over 2}}{E_D} + m} }}}} & 0 & { - \sqrt {{\textstyle{1 \over 2}}{E_D} + m} } & 0 \\
    \end{array}} \right] \\
    {{\bar v'}_{ + 1/2}}({\bf{\vec p}}) = \left[ {\begin{array}{*{20}{c}}
    { - {\textstyle{P \over {\sqrt {E + m} }}}{e^{{\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} & {{\textstyle{P \over {\sqrt {E + m} }}}\cos ({\textstyle{1 \over 2}}\theta )} & { - \sqrt {E + m} \cdot {e^{{\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} & {\sqrt {E + m} \cdot \cos ({\textstyle{1 \over 2}}\theta )} \\
    \end{array}} \right] \\
    {{\bar v'}_{ - 1/2}}({\bf{\vec p}}) = \left[ {\begin{array}{*{20}{c}}
    {{\textstyle{P \over {\sqrt {E + m} }}}\cos ({\textstyle{1 \over 2}}\theta )} & {{\textstyle{P \over {\sqrt {E + m} }}}{e^{ - {\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} & { - \sqrt {E + m} \cdot \cos ({\textstyle{1 \over 2}}\theta )} & { - \sqrt {E + m} \cdot {e^{ - {\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\
    \end{array}} \right] \\
    \end{array}[/tex]
     
  2. jcsd
  3. Jul 28, 2010 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    It's not clear what you are trying to do. Are you just trying to prove that the sum over spins on the LHS is equal to the trace on the RHS?

    That should be easy enough. You don't need the explicit spinors; you just need the orthogonality and completeness relations (and the anticommutation relations for the Dirac matrices, of course). Then you can compute the sum. If you just make all the spinor indices explicit, then you should see (after applying the completeness relation) that all the indices are contracted in such a way to give the trace on the LHS.

    As for actually evaluating the trace itself, you just need the Dirac algebra. I don't have my Griffiths or Peskin & Schroeder here with me, but somewhere in one (or both) of those books is a section that gives you a bunch of traces of products of Dirac matrices. Just break it down and do it step-by-step.

    I've certainly done it wrong myself before, so don't sweat it...it just takes some careful algebra.
     
  4. Jul 28, 2010 #3
    Oh, my question was: "Is the following expansion correct?". I just wanted to know if I did the expansion correctly.

    Also: another question came up: can you put opposite momenta in the arguments of the spinors? E.g., initial vs. final momenta VS. final vs. initial momenta?
     
  5. Jul 28, 2010 #4
    ...oops. more clarification: did i carry out the sum right? if i didn't, that would explain why my computer code is behaving badly.
     
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