Talk:Phy5646

Question: A monoatomic atom undergo spontaneous emission. It changes from an excited state $|e\rangle$ with $l=0$ to an intermediate state $|i\rangle$ with $l=1$ emitting a photon with wave vector $\mathbf {k}$, and then to the ground state $|g\rangle$ with $l=0$ emitting a photon with wave vector $\mathbf {k'}$. Find the probability that the angle between the two wavevectors to be $\theta$.

Ans:

${\mathcal {H}}={\frac {1}{2m}}\left(p-{\frac {e}{c}}A(r)\right)^{2}+V(r)+\sum _{k,{\hat {\lambda }}}\hbar \omega _{k}\left({\hat {a}}_{k{\hat {\lambda }}}^{\dagger }{\hat {a}}_{k{\hat {\lambda }}}+{\frac {1}{2}}\right)$

where $\mathbf {{\hat {A}}(r)} ={\frac {1}{\sqrt {V}}}\sum _{k,\lambda }\left[{\sqrt {\frac {2\pi \hbar }{\omega _{k}}}}c\;\left({\hat {a}}_{k,{\hat {\lambda }}}{\hat {\lambda }}e^{ik\cdot r}+{\hat {a}}_{k,{\hat {\lambda }}}^{\dagger }{\hat {\lambda ^{*}}}e^{-ik\cdot r}\right)\right]$

Using second order time dependent perturbation theory , we can write the wavefunction in Dirac picture as,

$|\chi (t)\rangle \approx |\chi _{0}\rangle +{\frac {1}{i\hbar }}\int _{-\infty }^{t}dt'{\mathcal {H}}'_{I}(t')|\chi _{0}\rangle +{\frac {1}{(i\hbar )^{2}}}\int _{-\infty }^{t}dt'\int _{-\infty }^{t'}dt''{\mathcal {H}}'_{I}(t'){\mathcal {H}}'_{I}(t'')|\chi _{0}\rangle$

where

${\mathcal {H}}'_{I}(t)=e^{{\frac {i}{\hbar }}{\mathcal {H}}_{0}^{(at)}t}\left(-{\frac {e}{mc}}A(r,t)\cdot p+{\frac {e^{2}}{2mc^{2}}}A(r,t)\cdot A(r,t)\right)e^{-{\frac {i}{\hbar }}{\mathcal {H}}_{0}^{(at)}t}$

with

$\mathbf {A(r,t)} ={\frac {1}{\sqrt {V}}}\sum _{k,\lambda }\left[{\sqrt {\frac {2\pi \hbar }{\omega _{k}}}}c\;\left({\hat {a}}_{k,{\hat {\lambda }}}{\hat {\lambda }}e^{ik\cdot r-i\omega _{k}t}+{\hat {a}}_{k,{\hat {\lambda }}}^{\dagger }{\hat {\lambda ^{*}}}e^{-ik\cdot r+i\omega _{k}t}\right)\right]$

Since the system has rotational symmetry, so the internal eigenstate is $|n,l,m\rangle$, where $|n\rangle =|e\rangle ,|i\rangle ,|g\rangle$. The initial photon field is null $|\phi \rangle$. Initially $l=0$, so $m=0$, so

$|\chi _{0}\rangle =|e,0,0;\psi \rangle$.

The final state is $|g,0,0;1_{\lambda ,k},1_{\lambda ',k'}\rangle$.

$\langle e,0,0;\psi |\chi (t)\rangle \approx |\chi _{0}\rangle +{\frac {1}{i\hbar }}\int _{-\infty }^{t}dt'{\mathcal {H}}'_{I}(t')|\chi _{0}\rangle +{\frac {1}{(i\hbar )^{2}}}\int _{-\infty }^{t}dt'\int _{-\infty }^{t'}dt''{\mathcal {H}}'_{I}(t'){\mathcal {H}}'_{I}(t'')|\chi _{0}\rangle$

where