- #1
elmerx25
- 14
- 0
Hello:
A spinless electron can interact with [itex]A^\mu[/itex] only via its charge; the coupling is proportional to [itex](p_{f} + p_{i})^{\mu}[/itex]. An electron with spin, on the other hand, can also interact with the magnetic field via its magnetic moment. This coupling involves the factor [itex]i\sigma^{\mu\nu}(p_{f} - p_{i})[/itex]The relation between the Dirac current and the Klein-Gordon current can be studied as
follows: Define the antisymmetric sigma tensor as:
[tex]i\sigma^{\mu\nu} = \frac{i}{2} (\gamma^{\mu}\gamma^{\nu} - \gamma^{\nu}\gamma^{\mu})[/tex]
And the Gordon decomposition of the Dirac current can be made:
[tex]\bar u_{f}\gamma^{\mu}u_{i} = \frac{1}{2m} \bar u_{f} [(p_{f} + p_{i})^{\mu} + i\sigma^{\mu\nu} (p_{f} - p_{i})_{\nu} ] u_{i}[/tex]
To identify the magnetic moment interaction [itex](-{\mu} . B)[/itex] it suffices to show that:
[tex]\int[-\frac{e}{2m} \bar \psi_{f} i\sigma_{\mu\nu} (p_{f} - p_{i})^{\nu}\psi_{i}] A^{\mu} d^{3}x = \int\psi^{f*}_{A}(\frac{e}{2m}{\sigma} . B) \psi^{i}_{A} d^{3}x[/tex]
Can someone please tell me how I can desmostrate this ecuation?
Thanks.
P.S.: Exercise 6.2 of "Quarks and Leptons. An Introductory Course in Modern Particle Physics - F.Halzem,A.Martin"
A spinless electron can interact with [itex]A^\mu[/itex] only via its charge; the coupling is proportional to [itex](p_{f} + p_{i})^{\mu}[/itex]. An electron with spin, on the other hand, can also interact with the magnetic field via its magnetic moment. This coupling involves the factor [itex]i\sigma^{\mu\nu}(p_{f} - p_{i})[/itex]The relation between the Dirac current and the Klein-Gordon current can be studied as
follows: Define the antisymmetric sigma tensor as:
[tex]i\sigma^{\mu\nu} = \frac{i}{2} (\gamma^{\mu}\gamma^{\nu} - \gamma^{\nu}\gamma^{\mu})[/tex]
And the Gordon decomposition of the Dirac current can be made:
[tex]\bar u_{f}\gamma^{\mu}u_{i} = \frac{1}{2m} \bar u_{f} [(p_{f} + p_{i})^{\mu} + i\sigma^{\mu\nu} (p_{f} - p_{i})_{\nu} ] u_{i}[/tex]
To identify the magnetic moment interaction [itex](-{\mu} . B)[/itex] it suffices to show that:
[tex]\int[-\frac{e}{2m} \bar \psi_{f} i\sigma_{\mu\nu} (p_{f} - p_{i})^{\nu}\psi_{i}] A^{\mu} d^{3}x = \int\psi^{f*}_{A}(\frac{e}{2m}{\sigma} . B) \psi^{i}_{A} d^{3}x[/tex]
Can someone please tell me how I can desmostrate this ecuation?
Thanks.
P.S.: Exercise 6.2 of "Quarks and Leptons. An Introductory Course in Modern Particle Physics - F.Halzem,A.Martin"