Squeezing in the interaction and Heisenberg picture

Question

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.

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.