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 $\gamma$.

Ans:

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

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

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

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

with

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

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 intial state is $|\chi _{0}\rangle =|e,0,0;\phi \rangle$ and final state is $|g,0,0;1_{\lambda _{k},k},1_{\lambda '_{k'},k'}\rangle$. Therefore

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

where to change from state $|e\rangle$ to $|i\rangle$ to $|g\rangle$, we need two momentum operator, so the first term must be zero. And so

${\begin{aligned}&\langle g,0,0;1_{\lambda _{k},k},1_{\lambda '_{k'}}|\chi (t)\rangle \\&\approx {\frac {1}{(i\hbar )^{2}}}\int _{-\infty }^{t}dt'\int _{-\infty }^{t'}dt''\langle g,0,0;1_{\lambda _{k},k},1_{\lambda '_{k'},k'}|{\mathcal {H}}'_{I}(t'){\mathcal {H}}'_{I}(t'')|e,0,0;\phi \rangle \\&={\frac {1}{(i\hbar )^{2}}}\int _{-\infty }^{t}dt'\int _{-\infty }^{t'}dt''\sum _{m=-1,0,1}\langle g,0,0;1_{\lambda _{k},k},1_{\lambda '_{k'},k'}|{\mathcal {H}}'_{I}(t')|i,1,m;1_{\lambda ,k}\rangle \langle i,1,m;1_{\lambda ,k}|{\mathcal {H}}'_{I}(t'')|e,0,0;\phi \rangle \\&=\sum _{m=-1,0,1}f(t,w_{k},w_{k}')\left(\langle g,0,0|\mathbf {p} |i,1,m\rangle \cdot {\hat {\lambda _{k'}^{'*}}}\right)\left(\langle i,1,m|\mathbf {p} |e,0,0\rangle \cdot {\hat {\lambda _{k}^{*}}}\right)\end{aligned}}$

where $f(t,w_{k},w_{k}')$ is independent of the direction of the emitted photons.

Here

${\begin{aligned}{\frac {\mathbf {p} }{m}}&={\frac {i}{\hbar }}[{\mathcal {H}}_{0}^{(at)},\mathbf {r} ]\\\langle i,1,m|{\frac {\mathbf {p} }{m}}|e,0,0\rangle &={\frac {i}{\hbar }}\langle i,1,m|[{\mathcal {H}}_{0}^{(at)},\mathbf {r} ]|e,0,0\rangle \\&={\frac {i}{\hbar }}(E_{i}-E_{e})\langle i,1,m|\mathbf {r} |e,0,0\rangle \\\langle g,0,0|{\frac {\mathbf {p} }{m}}|i,1,m\rangle &={\frac {i}{\hbar }}(E_{g}-E_{i})\langle g,0,0|\mathbf {r} |i,1,m\rangle \end{aligned}}$

Taking away the radial dependent part, ie, take away $|e\rangle ,|i\rangle ,|g\rangle$ parts, and define

${\begin{aligned}r_{+}&=x+iy=rsin\theta e^{i\phi }\\r_{-}&=x-iy=rsin\theta e^{-i\phi }\\x&=(r_{+}+r_{-})/2\\y&=(r_{+}-r_{-})/2i\end{aligned}}$

we know

$\langle l',m'|z|l,m\rangle \neq 0$ only if $m'=m$,

$\langle l',m'|r_{+}|l,m\rangle \neq 0$ only if $m'=m+1$,

$\langle l',m'|r_{-}|l,m\rangle \neq 0$ only if $m'=m-1$

${\begin{aligned}\langle 1,m|\mathbf {r} |0,0\rangle &=\langle 1,m|{\hat {x}}x+{\hat {y}}y+{\hat {z}}z|0,0\rangle \\&={\frac {\hat {x}}{2}}\langle 1,m|r_{+}+r_{-}|0,0\rangle +{\frac {\hat {y}}{2i}}\langle 1,m|r_{+}-r_{-}|0,0\rangle +{\hat {z}}\langle 1,m|z|0,0\rangle \end{aligned}}$

So

${\begin{aligned}\langle 1,-1|\mathbf {r} |0,0\rangle &={\frac {1}{2}}({\hat {x}}+i{\hat {y}})\langle 1,-1|r_{-}|0,0\rangle \\&={\frac {1}{2}}({\hat {x}}+i{\hat {y}})r\int d\Omega sin\theta e^{-i\phi }{Y_{1}^{-1}}^{*}Y_{0}^{0}\\&={\frac {1}{2}}({\hat {x}}+i{\hat {y}})r{\sqrt {\frac {2}{3}}}\end{aligned}}$

${\begin{aligned}\langle 1,1|\mathbf {r} |0,0\rangle &={\frac {1}{2}}({\hat {x}}-i{\hat {y}})\langle 1,1|r_{+}|0,0\rangle \\&={\frac {1}{2}}({\hat {x}}-i{\hat {y}})r\int d\Omega sin\theta e^{i\phi }{Y_{1}^{1}}^{*}Y_{0}^{0}\\&=-{\frac {1}{2}}({\hat {x}}-i{\hat {y}})r{\sqrt {\frac {2}{3}}}\end{aligned}}$

${\begin{aligned}\langle 1,0|\mathbf {r} |0,0\rangle &={\hat {z}}\langle 1,0|z|0,0\rangle \\&={\hat {z}}r\int d\Omega cos\theta {Y_{1}^{0}}^{*}Y_{0}^{0}\\&={\hat {z}}r{\sqrt {\frac {1}{3}}}\end{aligned}}$