I am going through Gardiner & Zoller's 'Quantum Noise', and following the derivation of the Langevin equations in terms of input/output operators. Let the bath Hamiltonian be:

$$H_B=\hbar\int_{-\infty}^\infty d\omega\omega b^\dagger(\omega)b(\omega)$$

with interaction term:

$$H_I=\int_{-\infty}^\infty d\omega k(\omega)[b^\dagger(\omega)c-c^\dagger b(\omega)]$$

for some system operator $c$ and function $k(\omega)$ which we later take to be constant (first Markov approximation). Suppose we wish to derive the equations of motion for arbitrary system operator $a$. We then define the input operator to be (for some initial time $t_0$):

$$b_{in}(t)=\frac{1}{\sqrt{2\pi}}\int d\omega e^{-i\omega (t-t_0)}b_{t_0}(\omega)$$

I understand that this gives us a nice final form for the Langevin equation:

$$\dot{a}=-\frac{i}{\hbar}[a,H_S]-[a,c^\dagger]\left[ \frac{\gamma}{2}c+\sqrt{\gamma}b_{in}(t) \right]+\left[ \frac{\gamma}{2}c^\dagger+\sqrt{\gamma}b_{in}^\dagger \right](t)[a,c]$$

However, given a situation we then tend to swap $b_{in}$ for whatever input we have to the system. For example if the system is a two level atom being illuminated by a laser, we let $b_{in}$ represent the input laser field. It isn't clear to me at all from the definition why we can do that.

Similarly we define the output operator: $$b_{out}(t)=\frac{1}{\sqrt{2\pi}}\int d\omega e^{-i\omega(t-t_1)}b_{t_1}(\omega)$$ and claim that in general, we measure the output field $b_{out}$. Why is that, and what in particular do we have to measure to be viewing this field?

  • $\begingroup$ Could you add a page number for where Gardiner and Zoller represent the input laser as $b_{in}$? Usually the interaction with a laser is a coherent process so it should not be considered noise in this way. $\endgroup$ Commented Jun 30, 2015 at 7:45

1 Answer 1


It is much easier to look at this problem in the frequency domain:

$$b_\mathrm{in/out}(\omega) := 1/\sqrt{2\pi} \int dt e^{i\omega t} b_\mathrm{in/out}(t) = e^{i\omega t_{0/1}} b(t_{0/1}, \omega)$$

is the interaction picture operator of the bath evaluated at time $t_{0/1}$.

If one then sends $t_{0/1}\to \mp \infty$, $b_\mathrm{in/out}(\omega)$ is essentially the initial/final frequency content of the bath field.

See also my answer to this question: https://physics.stackexchange.com/a/534741/101770

  • $\begingroup$ i dont think your answer makes it clear whats going on. also why would taking t-> infinity make it the final frequency content of the bath, and how does that matter to the question at hand? $\endgroup$ Commented Mar 14 at 7:06

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