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enter image description here I have been trying to understand the input output formalism for a cavity by reading chapter 7 of Walls and Milburn (D. F. Walls and G. J. Milburn. Quantum optics. Springer, 2006.). There is something that I am struggling to understand. This input output formalism describes quantum mechanically the interaction between the internal field of a single-mode cavity, and the extra-cavity fields, namely the cavity inputs and outputs. However, when describing the model for a single-sided cavity, he said explicitly that all the fields considered are propagating in the same direction, the positive x direction. This is the quote right above equation 7.1 of the book: "We will assume that the only modes that are excited have the same plane polarisation and are all propagating in the same direction, which we take to be the positive x-direction." This puzzles me greatly. He surely wants to describe both input and output to the single-sided cavity. If all the fields considered are propagating in the same direction, how can there be both input and output? Shouldn't the input and output fields for a single-sided cavity necessarily be propagating in opposite directions?

In his model, the field operator describing the extra-cavity field is as follows: $$b(x,t)=e^{-i \Omega (t-x/c)} \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} d \omega b(\omega) e^{-i \omega (t-x/c)}.$$ where each $b(\omega)$ is an annihilation operator for an extra-cavity mode with frequency $\omega$ (actually $\omega+\Omega$, where $\Omega$ is the carrier frequency), $x$ is the position, $t$ means time and $c$ is the propagation velocity. His way of describing input and output is as follows, which, because of the above doubt that I have, I find quite hard to understand. He defined a specific combination of the extra-cavity field at an early time $t_0$ as the input and another specific combination of the extra-cavity fields at a late time $t_1$ as the output: $$ a_{IN}(t)=- \frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega e^{-i \omega (t-t_0) b_0(\omega)}, $$ $$a_{OUT}(t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} d \omega e^{-i \omega (t-t_1) b_1(\omega)}.$$ where $b_0(\omega)$ and $b_1(\omega)$ are the time-dependent heisenberg picture operator $b(\omega, t_0)$ and $b(\omega, t_1)$, respectively. This sounds quite natural to me, since I think he probably has the whole thing in mind as a scattering process. The field at a very early time is the input and the field at a very late time is the output. He shows that under some very reasonable approximations one can get two very simple equations $$\dot{a}(t)=-\frac{i}{\hbar} [a(t),H_{SYS}] + \frac{\gamma}{2}a(t) -\sqrt{\gamma}a_{IN}(t),$$ and $$a_{IN}(t)+a_{OUT}(t)= \sqrt{\gamma}(t),$$ which then lead to quite reasonable physical predictions. However, both these input and output field operators are constructed using the mode annihilation operators $b(\omega)$ that he introduced at the beginning, which - he stated explicitly as mentioned above - are all modes propagating in the positive $x$-direction! So whatever interaction these modes may have with the cavity, any kind of propagation have to be in the positive $x$-direction, right? How then can one possibly describe reflection?

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