- #1
bjnartowt
- 265
- 3
Homework Statement
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:
{\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)
…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...")
Homework Equations
Casimir’s trick, appearing in Griffiths “Intro to Elementary Particles”, is as follows:
{\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)}
The Attempt at a Solution
\begin{array}{l}<br /> \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)} \\ <br /> = \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) \\ <br /> + \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) \\ <br /> \end{array}
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):
\begin{array}{l}<br /> {v_{ + 1/2}}({{{\bf{\vec p}}}_0}) = \left[ {\begin{array}{*{20}{c}}<br /> 0 \\<br /> { - {\textstyle{{{Q_z}} \over {2\sqrt {{\textstyle{1 \over 2}}{E_D} + m} }}}} \\<br /> 0 \\<br /> { - \sqrt {{\textstyle{1 \over 2}}{E_D} + m} } \\<br /> \end{array}} \right]{v_{ - 1/2}}({{{\bf{\vec p}}}_0}) = \left[ {\begin{array}{*{20}{c}}<br /> { - {\textstyle{{{Q_z}} \over {2\sqrt {{\textstyle{1 \over 2}}{E_D} + m} }}}} \\<br /> 0 \\<br /> {\sqrt {{\textstyle{1 \over 2}}{E_D} + m} } \\<br /> 0 \\<br /> \end{array}} \right] \\ <br /> {{v'}_{ + 1/2}}({\bf{\vec p}}) = \left[ {\begin{array}{*{20}{c}}<br /> { - {\textstyle{P \over {\sqrt {E + m} }}}{e^{ - {\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\<br /> {{\textstyle{P \over {\sqrt {E + m} }}}\cos ({\textstyle{1 \over 2}}\theta )} \\<br /> {\sqrt {E + m} \cdot {e^{ - {\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\<br /> { - \sqrt {E + m} \cdot \cos ({\textstyle{1 \over 2}}\theta )} \\<br /> \end{array}} \right]{{v'}_{ - 1/2}}({\bf{\vec p}}) = \left[ {\begin{array}{*{20}{c}}<br /> {{\textstyle{P \over {\sqrt {E + m} }}}\cos ({\textstyle{1 \over 2}}\theta )} \\<br /> {{\textstyle{P \over {\sqrt {E + m} }}}{e^{{\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\<br /> {\sqrt {E + m} \cdot \cos ({\textstyle{1 \over 2}}\theta )} \\<br /> {\sqrt {E + m} \cdot {e^{{\bf{i}}\phi }}\sin ({\textstyle{1 \over 2}}\theta )} \\<br /> \end{array}} \right] \\ <br /> {{\bar v}_{ + 1/2}}({{{\bf{\vec p}}}_0}) = \left[ {\begin{array}{*{20}{c}}<br /> 0 & { - {\textstyle{{{Q_z}} \over {2\sqrt {{\textstyle{1 \over 2}}{E_D} + m} }}}} & 0 & {\sqrt {{\textstyle{1 \over 2}}{E_D} + m} } \\<br /> \end{array}} \right] \\ <br /> {{\bar v}_{ - 1/2}}({{{\bf{\vec p}}}_0}) = \left[ {\begin{array}{*{20}{c}}<br /> { - {\textstyle{{{Q_z}} \over {2\sqrt {{\textstyle{1 \over 2}}{E_D} + m} }}}} & 0 & { - \sqrt {{\textstyle{1 \over 2}}{E_D} + m} } & 0 \\<br /> \end{array}} \right] \\ <br /> {{\bar v'}_{ + 1/2}}({\bf{\vec p}}) = \left[ {\begin{array}{*{20}{c}}<br /> { - {\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 )} \\<br /> \end{array}} \right] \\ <br /> {{\bar v'}_{ - 1/2}}({\bf{\vec p}}) = \left[ {\begin{array}{*{20}{c}}<br /> {{\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 )} \\<br /> \end{array}} \right] \\ <br /> \end{array}