- #71
DrDu
Science Advisor
- 6,380
- 983
This weekend I managed to take a quick look on Feynman's calculation.
He calculates the probability amplitude to find circularly polarized plane wave photons, i. e., helicity eigenstates.
For this kind of photons, electric field is [tex] \langle \pm, k_z|\mathbf{E}|0 \rangle=E_0 (1, \pm i, 0)^T \exp(ik_z z) [/tex] (don't ask me, to which helicity the plus and minus refers) for k in z-direction. Other directions can be obtained by rotating first around the z-axis, which introduces only a phase factorn and then around the y-axis ( the rotation matrix has row vectors something like [tex] (\cos(\theta), 0, -\sin(\theta), (0, 0, 0), (\sin(\theta), 0, \cos(\theta)) [/tex]), the relevant vector becomes (up to the phase factor) [tex] (\cos(\theta), \pm i, \sin(\theta))^T [/tex].
The transition is due to the [tex] \mathbf{E}\cdot \mathbf{d}[/tex] coupling in the hamiltonian in the electric dipole approximation. The transition dipole moment has to be calculated with the J=3/2, M_J=3/2 L=1 M_L=1 state and the ground state with J=1/2 M_J=1/2, L=0, M_L=0. As we saw already, spin doesn't change as the operator d only acts on the orbital part. Hence the matrix elements can easily be evaluated from the spherical harmonics. The corresponding matrix element is [tex] \langle 3/2, 3/2| \mathbf{d}|1/2, 1/2\rangle=d_0 (1, i, 0)^T [/tex] (again it could be -i instead of i). The scalar product of the electric field and transition dipole moment is then proportional to [tex]1 \pm \cos(\theta)[/tex] depending on the polarization.
He calculates the probability amplitude to find circularly polarized plane wave photons, i. e., helicity eigenstates.
For this kind of photons, electric field is [tex] \langle \pm, k_z|\mathbf{E}|0 \rangle=E_0 (1, \pm i, 0)^T \exp(ik_z z) [/tex] (don't ask me, to which helicity the plus and minus refers) for k in z-direction. Other directions can be obtained by rotating first around the z-axis, which introduces only a phase factorn and then around the y-axis ( the rotation matrix has row vectors something like [tex] (\cos(\theta), 0, -\sin(\theta), (0, 0, 0), (\sin(\theta), 0, \cos(\theta)) [/tex]), the relevant vector becomes (up to the phase factor) [tex] (\cos(\theta), \pm i, \sin(\theta))^T [/tex].
The transition is due to the [tex] \mathbf{E}\cdot \mathbf{d}[/tex] coupling in the hamiltonian in the electric dipole approximation. The transition dipole moment has to be calculated with the J=3/2, M_J=3/2 L=1 M_L=1 state and the ground state with J=1/2 M_J=1/2, L=0, M_L=0. As we saw already, spin doesn't change as the operator d only acts on the orbital part. Hence the matrix elements can easily be evaluated from the spherical harmonics. The corresponding matrix element is [tex] \langle 3/2, 3/2| \mathbf{d}|1/2, 1/2\rangle=d_0 (1, i, 0)^T [/tex] (again it could be -i instead of i). The scalar product of the electric field and transition dipole moment is then proportional to [tex]1 \pm \cos(\theta)[/tex] depending on the polarization.