[Quantum Optics] Scully and Zubairy Section 6.2

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theuselessone
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Homework Statement



I've been reading through Quantum Optics by Scully and Zubairy and have been stuck in section 6.2 getting from the definition of the atomic inversion given by equation (6.2.20, pg 199)

[itex]W(t)=\sum_{n}\left[|c_{a,n}(t)|^2-|c_{b,n}(t)|^2\right][/itex]

to the atomic inversion for the Jaynes-Cummings model Hamiltonian between a single-mode quantized field and a single two-level atom given by (6.2.21, pg 200)

[itex]W(t)=\sum_{n=0}^{\infty}\rho_{nn}(0)\left[\frac{\Delta^2}{\Omega_n^2}+\frac{4g^2(n+1)}{\Omega_n^2}\cos(\Omega_n t)\right][/itex]

where [itex]|c_n(0)|^2=\rho_{nn}(0)[/itex] is the initial probability of n-photons. The coefficient [itex]|c_{a,n}(t)|^2[/itex] is the probability of being in atomic state [itex]\left|a\right\rangle[/itex] and photonic state [itex]\left|n\right\rangle[/itex] (similarly for [itex]|c_{b,n}(t)|^2[/itex]). In the book, they solve for the case of the atom initially in the excited state [itex]\left|a\right\rangle[/itex] such that [itex]c_{a,n}(0)=c_n(0)[/itex] and [itex]c_{b,n+1}(0)=0[/itex].

Homework Equations



Computing [itex]|c_{a,n}(t)|^2[/itex] and [itex]|c_{b,n}(t)|^2[/itex] using equations (6.2.16) and (6.2.17), respectively (or reading ahead to 6.2.18), I get

[itex]|c_{a,n}(t)|^2=\rho_{nn}(0)\left[\cos^2(\frac{\Omega_n t}{2})+\frac{\Delta^2}{\Omega_n^2}\sin^2(\frac{\Omega_n t}{2})\right][/itex]

and

[itex]|c_{b,n}(t)|^2=\rho_{n-1,n-1}(0)(\frac{4g^2n}{\Omega_{n-1}^2})\sin^2(\frac{\Omega_{n-1}t}{2})[/itex]

The Attempt at a Solution



Plugging these into (6.2.20), I find

[tex] \begin{align}<br /> W(t)&=\sum_{n=0}^\infty\rho_{nn}(0)\left[\cos^2(\frac{\Omega_n t}{2})+\frac{\Delta^2}{\Omega_n^2}\sin^2(\frac{\Omega_n t}{2})\right]-\sum_{n=0}^\infty\rho_{n-1,n-1}(0)(\frac{4g^2n}{\Omega_{n-1}^2})\sin^2(\frac{\Omega_{n-1}t}{2}) \\<br /> &=\sum_{n=0}^\infty\rho_{nn}(0)\left[\cos^2(\frac{\Omega_n t}{2})+\frac{\Delta^2}{\Omega_n^2}\sin^2(\frac{ \Omega_n t}{2})\right]-\sum_{n=-1}^\infty\rho_{nn}(0)(\frac{4g^2(n+1)}{\Omega_{n}^2})\sin^2(\frac{\Omega_{n}t}{2})<br /> \end{align}[/tex]

And that's where I get stuck... So far, the only justification I've come up with is that [itex]\rho_{nn}(0)=0[/itex] for [itex]n<0[/itex] since there shouldn't be less than zero photons. I'm able to get the desired equation (6.2.21) if the second sum runs from [itex]n=0[/itex] to [itex]n=\infty[/itex] instead of starting from [itex]n=-1[/itex].

Thanks in advance for any help.

Note: It's showing \Omeg[itex]a_n t[/itex] instead of [itex]\Omega_n t[/itex] for me in the last equation above.
 
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It's probably because your lines of LaTeX are so long. I tried to fix it, but was unsuccessful. The align bits made it so that I couldn't break up the long TeX expression. Once the expressions get up around 70 characters, they don't render right.