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In studying the the interaction of a single mode electromagnetic field in a coherent state with a two state atomic system (initially in its ground state), the problem is reduced to evaluating the following expectation value:

$$\langle 1,n| \left[\frac{2}{\hat\Omega^2}\left(\hat{H} - \omega\left(\hat{N_e}-\frac{1}{2}\right)\right)\cos(\hat{\Omega} t - 1)\right]|n , 1\rangle$$

Where $|1\rangle$ and $|n\rangle$ are the ground atomic state and a number state, respectively. $\hat{\Omega}$ depends on both atomic and field operators, as does the hamiltonian $\hat{H}$ and $\hat{N_e}$.

For each of the factors in this expectation value, the evaluation of the expectation values have already been calculated, some trivial, others involving a taylor expansion.

My problem is evaluating the above product of operators. I can't simply pass the states through the operators as each operator depends on atomic and field states.

Would anyone be able to point me in the right direction? Ideally there would be a rule that allows me to replace each operator in the product with their expectation values... I feel like the fact that $\hat{\Omega}$ commutes with $\hat{H}$ and $\hat{N_e}$ is important, but still don't know how to proceed.

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If $\hat H$, ${\hat N}_e$, and $\hat \Omega$ all commute, the simplest idea is to take their common eigenstates as a basis for the identity decomposition. If $$ {\hat H} |\lambda\rangle = E_\lambda |\lambda\rangle\;, \;\;\; {\hat N}_e |\lambda\rangle = n_\lambda |\lambda\rangle\;,\;\;\; {\hat \Omega} |\lambda\rangle = \Omega_\lambda |\lambda\rangle $$ then you have $$ \langle 1, n| \frac{2}{{\hat\Omega}^2}\left[\left(\hat{H} - \omega\left(\hat{N_e}-\frac{1}{2}\right)\right)\cos\left(i\hat{\Omega} \tau -1 \right)\right] |n , 1\rangle = \\ =\sum_\lambda{\langle 1, n| \frac{2}{{\hat\Omega}^2}\left[\left(\hat{H} - \omega\left(\hat{N_e}-\frac{1}{2}\right)\right)\cos\left(i\hat{\Omega} \tau -1 \right)\right] |\lambda\rangle \langle \lambda|n , 1\rangle }\\ = \sum_\lambda{ \frac{2}{\Omega_\lambda^2}\left[\left(E_\lambda - \omega\left(n_\lambda-\frac{1}{2}\right)\right)\cos\left(i\Omega_\lambda \tau -1 \right)\right] |\langle \lambda|n , 1\rangle}|^2 $$

If $\hat H$, ${\hat N}_e$, and $\hat \Omega$ do not necessarily commute, or the above doesn't prove tractable:

From the ${\hat \Omega}^{-2}$ and cosine factors I'd guess your expression comes from a double time integral of the form $$ \sim \int_0^t{dt' \int_0^{t'}{d\tau \langle 1,n| \left(\hat{H} - \omega\left(\hat{N_e}-\frac{1}{2}\right)\right)e^{i\hat{\Omega} \tau} |n , 1\rangle}} $$ and its complex conjugated. Or if it doesn't, you can tuck in the ${\hat \Omega}^{-2}$ into something like this.

Then if you have the $\hat H$, $\hat N_e$, and $\hat \Omega$ in terms of field creation and annihilation ops, one idea is to split the operator under the expectation value into a power series of those, and rearrange each term in anti-normal order, so as to get something like $$ \langle 1,n| \left(\hat{H} - \omega\left(\hat{N_e}-\frac{1}{2}\right)\right)e^{i\hat{\Omega} \tau} |n , 1\rangle = \sum_{j, k}{\langle 1,n| {\hat a}^k \;{\hat F}_{jk}(\text{atomic ops})\; ({\hat a^\dagger})^j |1, n\rangle} $$ The point is that you can handle the individual terms using an (overcomplete) decomposition of identity in field coherent states $|\alpha\rangle$ , ${\hat a} |\alpha \rangle = \alpha |\alpha\rangle$, $\langle \alpha | {\hat a}^\dagger = \alpha^* \langle \alpha |$, $$ I = \int{d\alpha |\alpha \rangle \langle \alpha |} $$ which gives $$ \langle 1,n| {\hat a}^k \;{\hat F}_{jk}(\text{atomic ops})\; ({\hat a^\dagger})^j |1, n\rangle = \int{d\alpha \langle 1,n| \;{\hat a}^k\; |\alpha \rangle \;{\hat F}_{jk} (\text{atomic ops})\; \langle \alpha |\;({\hat a^\dagger})^j\; |1, n\rangle } = $$ $$ = \left[ \int{d\alpha \;\alpha^k (\alpha^*)^j \;| \langle \alpha | n\rangle |^2 } \right] \langle 1|{\hat F}_{jk}(\text{atomic ops})|1 \rangle $$ with $$ | \langle \alpha | n\rangle |^2 = e^{-|\alpha|^2} \frac{|\alpha|^{2n}}{n!} $$

Depending on what coefficients you get from $\langle 1|{\hat F}_{jk}(\text{atomic ops})|1 \rangle$, you can experiment with the integrals and the series to simplify or even collapse everything into something tractable.

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You might want to consider inserting the unit operator: $$ \hat 1= \sum_{n_1n_2} \vert n_1 n_2\rangle\,\langle n_1 n_2\vert $$ or a complete set of eigenvectors between your operators. Also, since $\hat \Omega$ commutes with $\hat H$ and $\hat N$, you can act move it to the right.

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  • $\begingroup$ I had tried that for the atomic state complete set, $\hat{1} = |1\rangle\langle 1| + |2\rangle\langle 2| $. It didn't seem to get me anywhere. In your consideration, what are $n_1$ and $n_2$ ? are they number states and atomic states? $\endgroup$ – Adam Prior Mar 2 '17 at 11:19
  • $\begingroup$ @AdamPrior well... I was using your notation so maybe you can clarify what is $\vert 1,n\rangle$? $\endgroup$ – ZeroTheHero Mar 2 '17 at 14:17
  • $\begingroup$ It is the tensor product of $|1\rangle$ and $|n\rangle$ $\endgroup$ – Adam Prior Mar 3 '17 at 16:33
  • $\begingroup$ so then my $\vert n_1n_2\rangle$ would be the tensor $\vert n_1\rangle\vert n_2\rangle$. Presumably the $n_1$ part would be the atomic state and the $n_2$ part would be the number state (if I understand your notation correctly). $\endgroup$ – ZeroTheHero Mar 3 '17 at 19:43

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