I have some confusion regarding squeezing (or in general, quantum optics interactions) and the picture in which this is described. Let's assume we have two free fields described by $\hat{a}$ and $\hat{b}$, where we can assume that the pump field $\hat{b}$ is classical. The Hamiltonian should be something like: \begin{equation} \hat{H}=E_{pump} + \hbar\omega(\hat{a}^\dagger\hat{a}+1/2)+\frac{ir}{2}(\hat{a}^{\dagger 2}-\hat{a}^2) \end{equation} Such Hamiltonian can be divided in two parts, a free Hamiltonian $\hat{H}_0$ and an interacting Hamiltonian $\hat{H}_I$.

For what I understand, evolution due to such Hamiltonians is commonly described using the Dirac or interaction picture, where the states evolve according to the interaction Hamiltonian and the operators evolve with the free Hamiltonian. The problem is, that every time I encounter squeezing, the only Hamiltonian mentioned is and Hamiltonian of the type: \begin{equation} \hat{H}=\frac{ir}{2}(\hat{a}^{\dagger 2}-\hat{a}^2) \end{equation} as for example in A. I. Lvovsky, Squeezed light, arXiv:1401.4118. The author at page 4, then states that in the Heisenberg picture the operators evolve according to said Hamiltonian. But in the Heisenberg picture operators should evolve according to the TOTAL Hamiltonian, which includes the free Hamiltonian.

I calculated the SPDC Hamiltonian and indeed I could check that: \begin{equation} \frac{i}{\hbar}\int_0^t \hat{H}_I dt \propto \frac{r}{2}(\hat{a}^{\dagger 2}-\hat{a}^2) \end{equation} so I can see how this is used in the interaction picture to make the state evolve as: \begin{equation} |\psi(t)\rangle = e^{\frac{i}{\hbar}\int_0^t \hat{H}_I(t)}|\psi(0)\rangle \end{equation} My problem stems from the fact that everybody uses this Hamiltonian to make the OPERATORS evolve, claiming we are dealing with the Heisenberg picture.

I fail to connect the dots. Does someone have some insight?

EDIT: As it has been asked, $\hat{H}_I $ is the interaction Hamiltonian in the interaction picture and it has the form of \begin{equation} \hat{H}_I(t)\propto\frac{r}{2}(\hat{a}^{\dagger 2}e^{(2\omega-\omega_p)}-\hat{a}^2e^{-(2\omega-\omega_p)}) \end{equation} and it basically accounts for the energy conservation. This Hamiltonian and the derivation can be found in Gerry, Knight - Introduction to Quantum Optics.

  • $\begingroup$ can you maybe clarify what is $H_I$. $\endgroup$ Commented Aug 4, 2023 at 3:16
  • $\begingroup$ @ZeroTheHero Hi! I edited my question adding the interaction Hamiltonian :) $\endgroup$
    – Luthien
    Commented Aug 4, 2023 at 13:34

1 Answer 1


The free Hamiltonian is not explicitly accounted for because it doesn't influence the dynamics of the interaction. In the Schrödinger picture the free evolution simply results in a phase shift, and in the phase-space picture this corresponds to a rotation of the quadratures. One usually therefore (sometimes implicitly) considers a rotating reference frame where the free evolutions of the quadrature operators are trivial.

We can motivate this more formally though. The free Hamiltonian results in the following time evolutions of the creation and annihilation operators:

$$ \dot{\hat{a}} = \frac{i}{\hbar}[\hbar\omega(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}),\hat{a}] = i\omega(\hat{a}^{\dagger}\hat{a}\hat{a}-\hat{a}\hat{a}^{\dagger}\hat{a})=-i\omega\hat{a}\\ \dot{\hat{a}}\vphantom{\hat{a}}^{\dagger} = \frac{i}{\hbar}[\hbar\omega(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}),\hat{a}] = i\omega(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}-\hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a})= i\omega\hat{a}^{\dagger} $$

The time evolutions of the quadrature operators are:

$$ \begin{align} \dot{\hspace{-0.6pt}\hat{X}} &= \frac{\dot{\hat{a}} + \dot{\hat{a}}\vphantom{\hat{a}}^{\dagger}}{\sqrt{2}} = \frac{-i\omega\hat{a}+i\omega\hat{a}^{\dagger}}{\sqrt{2}} = \omega\frac{\hat{a}-\hat{a}^{\dagger}}{i\sqrt{2}} = \omega\hat{P}\\ \dot{\hspace{-2pt}\hat{P}} &= \frac{\dot{\hat{a}} - \dot{\hat{a}} \vphantom{\hat{a}}^{\dagger}}{i\sqrt{2}} = \frac{-i\omega\hat{a}-i\omega\hat{a}^{\dagger}}{i\sqrt{2}} = -\omega\frac{\hat{a}+\hat{a}^{\dagger}}{\sqrt{2}} = -\omega\hat{X} \end{align} $$

This system of differential equations has a solution (for suitable initial conditions):

$$ \begin{align} \hat{X}(t) &= \sin(\omega t)\\ \hat{P}(t) &= \cos(\omega t) \end{align} $$

You can now see how some initial state evolving freely simply rotates in phase space, and since we don't care about this rotation we choose to view the states in this rotating frame.

Update: you asked for a more formal argument. We don't want to solve the dynamics in the stationary frame, because in this frame the squeezing angle as well as the states are continuously rotating and the description is much more complicated. However, we can do the transformation to the rotating frame formally and see how the free Hamiltonian vanishes. See also this excellent answer.

Let the Hamiltonian be

$$ \hat{H}(t) = \hat{H}_0 + \hat{H}_I(t), $$

with the free, trivially time-dependent part being $H_0$. The time-evolution operator for the free time evolution is

$$ \hat{U} = \mathrm{exp}\left[-\frac{i}{\hbar} \int_{t_0}^t \hat{H}_0 d\tau\right]. $$

Let $\hat{T} = \hat{U}^{\dagger}$ be the transformation to the rotating frame. Note that

$$ \frac{\partial \hat{T}}{\partial t} = \frac{i}{\hbar}\hat{H}_0\hat{T} = \frac{i}{\hbar}\hat{T}\hat{H}_0. $$

In the Heisenberg picture the transformation $\hat{T}$ acts on the annihilation operator as

$$ \hat{T}\vphantom{T}^{\dagger} \hat{a} \hat{T} = \hat{b}, $$

and we want to find the time evolution of the transformed operator $\hat{b}$:

$$ \tag{1} \frac{\partial \hat{b}}{\partial t} = \frac{\partial}{\partial t} \hat{T}\vphantom{T}^{\dagger} \hat{a} \hat{T} = \dot{\hat{T}}\vphantom{T}^{\dagger} \hat{a} \hat{T} + \hat{T}\vphantom{T}^{\dagger} \dot{\hat{a}} \hat{T} + \hat{T}\vphantom{T}^{\dagger} \hat{a} \dot{\hat{T}}. $$

Using the Heisenberg evolution $\dot{\hat{a}} = \frac{i}{\hbar}[\hat{H},\hat{a}]$ the middle term can be re-written as

$$ \hat{T}\vphantom{T}^{\dagger} \dot{\hat{a}} \hat{T} = \frac{i}{\hbar}\hat{T}\vphantom{T}^{\dagger}(\hat{H}\hat{a}-\hat{a}\hat{H})\hat{T} = \frac{i}{\hbar} (\hat{T}\vphantom{T}^{\dagger}\hat{H}\hat{T}\hat{T}\vphantom{T}^{\dagger}\hat{a}\hat{T}- \hat{T}\vphantom{T}^{\dagger}\hat{a}\hat{T}\hat{T}\vphantom{T}^{\dagger}\hat{H}\hat{T}) = \frac{i}{\hbar}[\hat{H}\vphantom{H}',\hat{b}], $$ where $\hat{H}\vphantom{H}'=\hat{T}\vphantom{T}^{\dagger}\hat{H}\hat{T}$.

For the first term in the R.H.S. of (1) we have

$$ \dot{\hat{T}}\vphantom{T}^{\dagger} \hat{a} \hat{T} = -\frac{i}{\hbar}\hat{H}_0\hat{T}\vphantom{T}^{\dagger}\hat{a}\hat{T} = -\frac{i}{\hbar}\hat{H}_0 \hat{b} $$

and for the last term:

$$ \hat{T}\vphantom{T}^{\dagger} a \dot{\hat{T}} = \frac{i}{\hbar} \hat{T}\vphantom{T}^{\dagger}a\hat{T}\hat{H}_0 = \frac{i}{\hbar} \hat{b}\hat{H}_0 $$

Putting it together we get

$$ \frac{\partial \hat{b}}{\partial t} = -\frac{i}{\hbar}\hat{H}_0 \hat{b}+ \frac{i}{\hbar}[\hat{H}\vphantom{H}',\hat{b}]+ \frac{i}{\hbar} \hat{b}\hat{H}_0 = \frac{i}{\hbar}[\hat{H}\vphantom{H}',\hat{b}] - \frac{i}{\hbar} [\hat{H}_0,\hat{b}] $$

Since $\hat{T}\hat{H}_0 = \hat{H}_0\hat{T}$ we have

$$ \frac{i}{\hbar}[\hat{H}\vphantom{H}',\hat{b}] = \frac{i}{\hbar}[\hat{T}\vphantom{T}^{\dagger}H_0\hat{T}+\hat{T}\vphantom{T}^{\dagger}\hat{H}_I\hat{T},\hat{b}] = \frac{i}{\hbar} [H_0,\hat{b}] + \frac{i}{\hbar}[\hat{H}\vphantom{H}'_I,\hat{b}] $$

with $\hat{H}\vphantom{H}'_I = \hat{T}\vphantom{T}^{\dagger}\hat{H}_I\hat{T}$. Finally we get that

$$ \frac{\partial \hat{b}}{\partial t} = [\hat{H}\vphantom{H}'_I,\hat{b}], $$

or in other words, in the rotating frame the annihilation operator evolves only under the interaction Hamiltonian.

  • $\begingroup$ Hi! Thank you for your input :) I have some questions though. I'm aware that formally the free Hamiltonian just corresponds to a trivial rotation in phase space. But when we decide to do the evolution in the Heisenberg picture this trivial rotation should be present, am I right? So, are we just implicitly saying that we neglect this trivial rotation and just evolve with the interaction Hamiltonian? Are we doing it in a formal way or are we just "skipping" the uninteresting part? $\endgroup$
    – Luthien
    Commented Aug 4, 2023 at 13:24
  • $\begingroup$ See my updated answer. Solving the dynamics in the stationary frame would be a headache. You can go to the rotating frame formally, but personally I find the hand-wavy argument more insightful than the calculation. I hadn't done it before though so it was interesting. $\endgroup$
    – fulis
    Commented Aug 4, 2023 at 17:43
  • $\begingroup$ Wow, thanks a lot for all this work! So if I understand correctly we just decide to work in the rotating reference frame and "live there", and in this frame I will apply the Heisenberg or Schroedinger picture, just forgetting about the "old frame", am I right? And this would be different from working in the interaction picture? $\endgroup$
    – Luthien
    Commented Aug 4, 2023 at 17:57
  • $\begingroup$ Yes, you just "live" there in most cases. One reason for this is that the rotating frame is actually the one you measure in when doing homodyne detection of light. In this case the rotating frame is defined by the local oscillator you use to measure the light. The interaction picture is what you get when you use this transformation to put the free evolution into the operators, and use the interaction Hamiltonian to evolve the state vector, so I think you could say it's the rotating frame in the Schrödinger picture. $\endgroup$
    – fulis
    Commented Aug 4, 2023 at 18:34
  • $\begingroup$ @Luthien a textbook where I have seen this point clearly is Wiseman & Milburn's "Quantum Measurement and Control". In Appendix A.1.3 at p. 404 they precisely explain how "working in the interaction frame" means "ignoring the boring part" and this is different from working in the interaction picture, as explained in textbook quantum mechanics. $\endgroup$ Commented Jun 20 at 9:03

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