---
abstract: 'The question raised by \[Bastin and Martin 2003 J. Phys. B: At. Mol. Opt. Phys. [**36**]{}, 4201\] is examined and used to explain in more detail a key point of our calculations. They have sought to rebut criticisms raised by us of certain techniques used in the calculation of the off-resonance case. It is also explained why this result is not a problem for the off-resonance case, but, on the contrary, opens the door to a general situation. Their comment is based on a blatant misunderstanding of our proposal an d as such is simply wrong.'
address: |
    $^1$Mathematics Department, Faculty of Science, South Valley University, 82524 Sohag, Egypt.\
    $^2$Mathematics Department, Faculty of Science, Al-Azhar University, Nasr City, Cairo, Egypt. 
author:
- 'Mahmoud Abdel-Aty$^{1,}$, A.-S. F. Obada$^2$'
title: '**Reply to Comment on “Quantum inversion of cold atoms in a microcavity: spatial dependence”** '
---

[**Submitted to:**]{}

In our paper \[2\] we have used the Hamiltonian (in the mesa mode case)

$$\hat{H}=\frac{p_{z}^{2}}{2M}+\frac{\Delta }{2}\sigma_{z}+\omega (a^{\dagger
}a+\frac{1}{2}\sigma_{z})+\lambda f(z)\{\sigma ^{-}a^{\dagger }+a\sigma
^{+}\}.  \label{1}$$

Let us write equation (1) in the following form (in the mesa mode case $%
f(z)=1)$ $$\begin{aligned}
\hat{H} &=&\frac{p_{z}^{2}}{2M}+\hat{V}  
\nonumber 
\\
\hat{V} &=&\frac{\Delta }{2}\sigma_{z}+\omega (a^{\dagger }a+\frac{1}{2}%
\sigma_{z})+\lambda \{\sigma ^{-}a^{\dagger }+a\sigma ^{+}\}.\end{aligned}$$ It is easy to show that in the $2\times 2$ atomic-photon space the eigenvalues and eigenfunction of the interaction Hamiltonian $\hat{V}$ ($%
\hat{V}|\Phi_{n}^{\pm }\rangle =E_{n}^{\pm }|\Phi_{n}^{\pm }\rangle )$ are given by \[3\], $$E_{n}^{\pm }=\omega (n+\frac{1}{2})\pm \sqrt{\frac{\Delta ^{2}}{4}+\lambda
^{2}(n+1)},$$ $$\begin{aligned}
|\Phi_{n}^{+}\rangle &=&\cos \theta_{n}|n+1,g\rangle +\sin \theta
_{n}|n,e\rangle,  
\nonumber 
\\
|\Phi_{n}^{-}\rangle &=&-\sin \theta_{n}|n+1,g\rangle +\cos \theta
_{n}|n,e\rangle,\end{aligned}$$ where $$\theta_{n}=\tan ^{-1}\left( \frac{\lambda \sqrt{n+1}}{\sqrt{\frac{\Delta
^{2}}{4}+\lambda ^{2}(n+1)}-\frac{\Delta }{2}}\right),\qquad$$ We write the wave function $|\Psi (z,t)\rangle =\sum\limits_{n}C_{n}^{\pm
}(z,t)|\Phi_{n}^{\pm }\rangle .$ Then using the total Hamiltonian (1) we have $$\begin{aligned}
\hat{H}|\Psi (z,t)\rangle  &=&\sum\limits_{n}\left( \frac{p^{2}}{2M}%
+V\right) C_{n}^{\pm }(z,t)|\Phi_{n}^{\pm }\rangle   
\nonumber 
\\
&=&\sum\limits_{n}\frac{p^{2}}{2M}C_{n}^{\pm }(z,t)|\Phi_{n}^{\pm }\rangle
+\sum\limits_{n}VC_{n}^{\pm }(z,t)|\Phi_{n}^{\pm }\rangle   
\nonumber 
\\
&=&\sum\limits_{n}\frac{p^{2}}{2M}C_{n}^{\pm }(z,t)|\Phi_{n}^{\pm }\rangle
+\sum\limits_{n}C_{n}^{\pm }(z,t)E_{n}^{\pm }|\Phi_{n}^{\pm }\rangle  
\nonumber 
\\
&=&\sum\limits_{n}\left( \frac{p^{2}}{2M}C_{n}^{\pm }(z,t)+E_{n}^{\pm
}C_{n}^{\pm }(z,t)\right) |\Phi_{n}^{\pm }\rangle ,\end{aligned}$$ because of the orthonormality of the wavefunctions $|\Phi_{n}^{\pm }\rangle $ then $$i\frac{\partial }{\partial t}C_{n}^{\pm }=\left( -\frac{1}{2M}\frac{\partial
^{2}}{\partial z^{2}}+E_{n}^{\pm }\right) C_{n}^{\pm },$$ with no coupling even in the presence of the detuning (equation (13) in BM comment \[1\]).

It may be worthwhile for the authors to consult some papers that have been published previously (see for example Refs. \[4-6\]) where the detuning has been considered and similar results have been obtained. To be more precisely:

-   We have used the interaction picture, so that the term $(n+\frac{1%
    }{2})\omega $ does not appear, it can be used as a phase only. Bearing in mind the case of mesa mode is being treated in our paper i.e. $f(z)=1$.

-   The most serious point is that Bastin and Martin have overlooked the formulae for $$\cos 2\theta_{n}\qquad and\qquad \sin 2\theta_{n}.$$ From equation (10) raised in Bastin and Martin comment, it is easy to write $$\tan \theta_{n}=\frac{\lambda \sqrt{n+1}}{\sqrt{\frac{\Delta ^{2}}{4}%
    +\lambda ^{2}(n+1)}-\frac{\Delta }{2}}=\frac{\sqrt{\frac{\Delta ^{2}}{4}%
    +\lambda ^{2}(n+1)}+\frac{\Delta }{2}}{\lambda \sqrt{n+1}}$$ then $$\tan 2\theta_{n}=\frac{\lambda \sqrt{n+1}}{-\frac{\Delta }{2}}$$ Also, it is easy to prove that $$\cos 2\theta_{n}=\frac{-\Delta /2}{\sqrt{\frac{\Delta ^{2}}{4}+\lambda
    ^{2}(n+1)}}\qquad and\qquad \sin 2\theta_{n}=\frac{\lambda \sqrt{n+1}}{%
    \sqrt{\frac{\Delta ^{2}}{4}+\lambda ^{2}(n+1)}}.$$ Once these formulae inserted in equations (18) and (19) of Bastin and Martin comment, we find that, the second terms vanish identically.

Now let us look more carefully at the general case when we take f(z) no longer a constant, i.e. we go beyond the mesa mode case. In this case the orthonormal functions $|\Phi_{n}^{\pm }\rangle $ in the $2\times 2$ system diagonalize the Hamiltonian $V$ and its elements are diagonal in this set of functions with $$\begin{aligned}
V_{n}^{\pm } &=&(n+\frac{1}{2})\omega \pm \sqrt{\frac{\Delta ^{2}}{4}%
+\lambda ^{2}f^{2}(z)(n+1)}  \nonumber \\
\tan 2\theta_{n} &=&\frac{\lambda f(z)\sqrt{n+1}}{-\frac{\Delta }{2}}.\end{aligned}$$ The states $|\Phi_{n}^{\pm }\rangle $ are z-dependent through the trigonometric functions, they satisfy $$\begin{aligned}
\frac{\partial }{\partial z}|\Phi_{n}^{\pm }\rangle &=&\pm |\Phi_{n}^{\pm
}\rangle \frac{d\theta }{dz},  \nonumber \\
\frac{\partial ^{2}}{\partial z^{2}}|\Phi_{n}^{\pm }\rangle &=&\pm |\Phi
_{n}^{\pm }\rangle \frac{d^{2}\theta }{dz^{2}}-|\Phi_{n}^{\pm }\rangle
\left( \frac{d\theta }{dz}\right) ^{2}.\end{aligned}$$ Then $|\Psi (z,t)\rangle $ can be expanded in the form $|\Psi (z,t)\rangle
=\sum\limits_{n}C_{n}^{\pm }(z,t)|\Phi_{n}^{\pm }\rangle $ and it satisfies the Schrodinger equation $$i\frac{\partial }{\partial z}|\Psi (z,t)\rangle =H|\Psi (z,t)\rangle .$$

Hence the coefficients $C_{n}^{\pm }(z,t)$ satisfy the coupled equations $$\begin{aligned}
i\frac{\partial C_{n}^{+}}{\partial z} &=&\left( -\frac{1}{2M}\frac{\partial
^{2}}{\partial z^{2}}+V_{n}^{+}-\left( \frac{d\theta }{dz}\right)
^{2}\right) C_{n}^{+}-\left( 2\frac{\partial C_{n}^{-}}{\partial z}\left( 
\frac{d\theta }{dz}\right) +C_{n}^{-}\left( \frac{d\theta }{dz}\right)
^{2}\right) ,  
\nonumber 
\\
i\frac{\partial C_{n}^{-}}{\partial z} &=&-\left( -\frac{1}{2M}\frac{%
\partial ^{2}}{\partial z^{2}}+V_{n}^{-}-\left( \frac{d\theta }{dz}\right)
^{2}\right) C_{n}^{-}+\left( 2\frac{\partial C_{n}^{+}}{\partial z}\left( 
\frac{d\theta }{dz}\right) +C_{n}^{+}\left( \frac{d\theta }{dz}\right)
^{2}\right) ,\end{aligned}$$ These equations should replace equations (18) and (19) of the comment of \[1\]. But once $f(z)$ is taken to be constant, then $\frac{d\theta }{dz}$ will vanish and we get equations (7) and the results of \[2,3\].

[9]{} Bastin T and Martin J 2003 J. Phys. B: At. Mol. Opt. Phys. [**36**]{}, 4201.

Abdel-Aty M and Obada A-S F 2002 J. Phys. B: At. Mol. Opt. Phys. **35** 807.

Abdel-Aty M and Obada A-S F 2002 Modern Physics Letters B **16** 117.

Battocletti M and Englert B-G, 1994 J. Phys. II France **4** 1939.

Zhang, Z-M, et al 2000 J.Phys.B: At. Mol. Opt. Phys. **33** 2125.

Zhang, Z-M, He L-S 1998 Opt. Commun. **157** 77.
