I am currently trying to get my head around the input-output formalism.

In describing the input-output formalism (link) , Gardiner and Collett take ladder operators in the Heisenberg picture and apply the Fourier transform to define the input operator.

Specifically, they start with ladder operator $ b(\omega)$, for a mode of frequency $\omega$ which takes the value $b_0(\omega)$ at time $t_0$ and then define the input field $b_{in}(t)$ as (for $t_0 <t$):

$$b_{in}(t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty d\omega e^{-i\omega(t-t_0)}b_0(\omega) \, .$$

I am confused as to how to interpret this. What frequency does the mode now correspond to if frequency has been 'integrated out'? Also is the operator $b_0(\omega)$ not already in the time domain, since it is specified at $t_0$? Could we not write $b$ as a function of $t$ and $\omega$ eg. as $b(t, \omega) = b_0(\omega)$ for $t=0$?

I would appreciate if anyone could clear this up. I suppose my specific questions would be:

  1. How do we interpret the Fourier transform of an operator?

  2. Why do we do this in the case of the input-output formalism?

  3. What does $b_{in}(t) $ correspond to physically?

  4. Any other hints/tips for thinking about this kind of problem.

  • $\begingroup$ The link is behind a paywall. However I would say that the Fourier transform defines $b(\omega)$ since, as you also state, $b(t)$ is defined via the Heisenberg evolution. $\endgroup$
    – lcv
    Nov 28, 2019 at 12:10

2 Answers 2


I will try to give some intuitive explanations, so long answer incoming...

Physical picture, Hamiltonian and operators

Let us first have a look at the bath model considered by Gardiner and Collett in the reference provided by the OP {1} $$ H_B = \hbar \int_{-\infty}^{\infty} d\omega \omega b^\dagger(\omega)b(\omega).$$ In addition to this bath Hamiltonian Gardiner-Collett also consider an interaction of the bath with a "system", but for the most part of the question this is irrelevant.

This Hamiltonian is a typical quantum optics Hamiltonian, featuring a continuum of modes with bosonic ladder operators $b(\omega)$. What do these modes mean? They are simply degrees of freedom of the field, that is they create/destroy an excitation in a particular spatial field distribution. For example, in one dimensional free space, each of them corresponds to a plane wave at frequency $\omega$. We could for example consider the state $b^\dagger(\omega)|0\rangle$, which is a single excitation (e.g. photon) in the plane wave at frequency $\omega$. In a more general context, they can correspond to different bath degrees of freedoms that are not plane waves.

What is not explicitly written here is that the $b$-operators are time dependent. You can easily see this by computing their equations of motion (for the non-interacting bath Hamiltonian above) $$ \frac{d}{dt} b(\omega) =\frac{i}{\hbar}[H_B, b(\omega)] = -i\omega b(\omega).$$

Without interaction, we should therefore really write $$ b(\omega) = b(\omega, t) = e^{-i\omega (t-t_0)}b(\omega, t_0).$$

Physically, this feature is easy to understand. The state $b^\dagger(\omega, t)|0\rangle$ mentioned above corresponding to a photon excitation in a plane wave mode simply has a harmonically oscillating phase, as we expect also from classical electromagnetism.

So far, this is just a clarification of the notation and what Hamiltonian we are actually looking at.

Frequency and time domain for Heisenberg picture operators

The above clarification illustrates the role of time $t$ and "frequency" $\omega$ (which is really an energy label for the modes) in the Hamiltonian and operators. The input-output formalism works in the Heisenberg picture, so a physical observable could be, for example, $\langle b(\omega, t) \rangle$, which is the photon amplitude at time $t$ in the mode of frequency $\omega$. In analogy, $\langle b^\dagger(\omega, t) b(\omega, t) \rangle$ is the photon number at time $t$ and frequency $\omega$ (counting per unit frequency, see also this question).

From this illustration we can understand that to count all photons, one should integrate over frequency, as is also done in the Hamiltonian. In this context, the operator $$b(t) = \int_{-\infty}^{\infty} d\omega b(\omega, t) = \int_{-\infty}^{\infty} d\omega e^{-i\omega(t-t_0)} b(\omega, t_0)$$

is relevant, where I have neglected factors of $2\pi$. Its expectation value $\langle b(t) \rangle$ would then for example correspond to the time dependent photon amplitude. For an illustration see also my answer to this question, where it is also shown how this relates to forming "wave packet operators" out of the different frequency contributions, in close analogy to Fourier transforms in classical optics.

We are now in a position to answer some of the OP's questions directly

  1. How do we interpret the Fourier transform of an operator?

Think about wave packets. It is just like in classical electromagnetism, only with operators.

What frequency does the mode now correspond to if frequency has been 'integrated out'?

The Fourier transformed operator has no notion of modes any more. I like to think of it as a "wave packet operator", although that term is vague. Your new operator is a superposition of contributions at different frequencies $\omega$. The time dependence of the operators at frequency $\omega$ appears because the excitation of that degree of freedom can vary in time. This is like in classical optics, where you can write your field as a superposition of e.g. plane waves, whose amplitude can vary in time.

Also is the operator $b_0(\omega)$ not already in the time domain, since it is specified at $t_0$? Could we not write b as a function of t and ω eg. as $b(t,ω)=b_0(ω)$ for $t=0$?

We can and have to! However, $b(t,ω)$ only knows about the time dependence of one mode at $\omega$.

Inputs and outputs

Well, so what is the fuss about input and output operators? In this context it is important to note that the time dependence that we derived above only includes the effect of the bath Hamiltonian, that is the bath operator's free evolution in the absence of interactions.

When interactions are present, the definition of input-output operators becomes crucial. Let us consider a frequency space version of the input operator $$b_\textrm{in}(\omega) = e^{i \omega t_0} b(\omega, t_0).$$ Note that $b_\textrm{in}(\omega)$ is not time dependent at all. So what does it mean? It is simply the interaction picture operator given by a boundary condition at $t=t_0$. In particular, with $b_\textrm{in}(\omega)$ given, we can infer the free time evolution of the operator. In practice, one will usually take $t_0 \rightarrow -\infty$, such that one specifies the boundary condition in the infinite past.

The input-operator given by the OP is then simply the time domain version of this operator $$b_\textrm{in}(t) = \int_{-\infty}^{\infty} d\omega e^{-i\omega t} b_\textrm{in}(\omega).$$

This is the same formula as in the question, just rewritten.

  1. Why do we do this in the case of the input-output formalism?

The reason is that you have interactions. For the free theory, defining input-output operators does not do much, it is simply swapping from frequency to time domain and building "wave packet operators". However, for the interacting theory, there are additional time dependencies of the operators given by additional terms in the Hamiltonian (see Gardiner-Collet paper {1}). The interpretation of the input-operator is then that you specify an initial condition of your quantum field in the infinite past. This can for example be interpreted in the context of scattering theory {2-6}. Importantly, there is an input-output relation relating the output operator (in the infinite future) to the input operator (infinite past) and the dynamics of the interacting system operators. In the case considered by Gardiner-Collett {1}, this is

$$ b_\textrm{out}(t) - b_\textrm{in}(t) = \sqrt{\gamma} c(t),$$

where $c(t)$ is a system operator interacting with our bath operators. What this relation tells you is that if you know $c(t)$ (for example from solving a Master equation), you can calculate what output you get in the bath dynamics.

From this relation one can also understand why the input-output operators are defined as they are. If we switch off the interaction by setting $\gamma=0$, we get

$$ b_\textrm{out}(t) = b_\textrm{in}(t),$$

as one would expect for a free field that has no interactions. In words: if you put an excitation in a free theory, it just oscillates without scattering.

  1. Any other hints/tips for thinking about this kind of problem.

Note that that input-output theory is usually done in the Heisenberg picture. Its useful to get your head around that. For more advanced topics in relation to input-output theory see {6-8} and for applications see in particular {5,9}. The standard textbook references on the topic are {10,11}.

{1} C. W. Gardiner and M. J. Collett Phys. Rev. A 31, 3761 (1985) (journal)
{2} S. Xu and S. Fan Phys. Rev. A 91, 043845 (2015) (journal)
{3} T. Caneva et al New J. Phys. 17 113001 (2015) (journal) (arxiv)
{4} T. Shi, D. E. Chang, and J. I. Cirac Phys. Rev. A 92, 053834 (2015) (journal)
{5} R. Trivedi, K. Fischer, S. Xu, S. Fan, and J. Vuckovic Phys. Rev. B 98, 144112 (2018) (journal) (arxiv)
{6} D. Lentrodt and J. Evers Phys. Rev. X 10, 011008 (2020) (journal)
{7} J. Zhang et al Phys. Rev. A 87, 032117 (2013) (journal)
{8} L. Diósi Phys. Rev. A 85, 034101 (2012) (journal)
{9} R. Trivedi et al Phys. Rev. Lett. 122, 243602 (2019) (journal)
{10} C. Gardiner and P. Zoller Quantum noise (2004) (publisher)
{11} D. F. Walls and G. J. Milburn Quantum optics (2008) (publisher)

  • $\begingroup$ Thanks, this is a fantastic answer! I forgot to check as it is now quite an old question, so apologies for the delay in accepting the answer! $\endgroup$
    – asph
    Apr 20, 2020 at 10:14
  • $\begingroup$ @asph Thanks for the kind words and the accept tick :) Let me know if you have any further questions or suggestions for improving the answer. I also thought that your question was very well phrased btw and it covers some other more vague ones on the network. $\endgroup$ Apr 20, 2020 at 10:29

The notion of ladder operators (creation and annihilation operators) that are functions of frequency and then take on a certain value at a certain time is a bit contradictory. One runs into the uncertainty principle. For each value of $\omega$ the creation operator $b^{\dagger}(\omega)$ would create a state of that frequency that in effect exists forever. The moment on starts to restrict its existence to a finite duration, one ends up needing more frequencies.

So when one defines a time-dependent operator in terms of the inverse Fourier transform of the frequency dependent operators, one effectively specifies the temporal behavior of a state produced by such an operator. One can compare the definition of such a time-dependent operator with the definitions of field operators in quantum field theory. So, one way to look at these operators given in your expression is to interpret them as field operators that are shifted in time.

  • $\begingroup$ Apologies for only commenting now, but I down-voted this answer a while ago, because I manifestly disagree with its main point. The first and second sentence are wrong or at least highly misleading. There is nothing contradictory about frequency and time dependent operators, in fact the underlying field theory used by Gardiner and Collett is defined in terms of such operators. The answer thus seems to be based on a misconception about the structure of the Hamiltonian and what "frequency" is used for here. $\endgroup$ Mar 5, 2020 at 10:40
  • $\begingroup$ @Wolpertinger: sorry but I don't understand your issue here. Are you saying one can have an operator $b^{\dagger}(\omega,t)$ where $\omega$ and $t$ are independent variables? Such a situation would only occur in cases where some nonlinear evolution is incorporated into the ladder operators, which is usually a bad idea. In the context of perturbation theory, it is usually better to do the quantization for a free-field theory and then add the interaction as a perturbation. In that case one would either have a time dependence or a frequency dependence and not both at the same time. $\endgroup$ Mar 5, 2020 at 12:11
  • $\begingroup$ Are you saying one can have an operator b†(ω,t) where ω and t are independent variables? -> Yes. If you look at the Gardiner-Collett paper the free field theory is simply defined in terms of these operators. For details see my answer. $\endgroup$ Mar 5, 2020 at 15:02

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