# Are there more than 3 dynamical pictures of quantum mechanics?

There are 3 well known dynamical pictures of quantum mechanics: the Schrödinger picture, the Heisenberg picture and the interaction picture.

In above wikipedia article, their connection is nicely summarized in the following table:

What is done physically in each of these can be summarized as follows:

Each of them is used frequently and has different advantages. However, if you look at the physical summary, there is really (at least) one missing, that is rarely mentioned in undergraduate courses:

• "Interaction picture #2": free time-dependence in the states, interaction time-dependence in the operators

So here are the questions:

1. Is the "interaction picture #2" relevant in quantum theory?
2. If so, where is it used and how is it useful?
3. If so, why on earth does nobody ever talk about it?
4. Are there posssibly even more useful dynamical pictures of quantum theory?

To anticipate a nitpick: you could probably interpret "interaction picture #2" as a normal interaction with a redefinition of the free and interacting Hamiltonian. However, I would argue that this defeats the point in many cases, since the free Hamiltonian often needs to be something simple for many cases. So swapping it for a complicated interaction Hamiltonian is a bit of the cheat and does not comply with the physical notion of each picture summarized above either.

I am adding my own answer on a few of the points, since this was the solution to the discussion in chat. However, it is by no means complete and other answers are (needless to say) more than welcome!

1. Yes!
2. "Interaction picture #2" is frequently used when one considers quantum Langevin dynamics and quantum stochastic proccesses, such as for example the input-output formalism (see e.g. 1). It is useful there, for example, when developing perturbation theory based on operators or operator scattering theory.
3. In this field people usually just call it the interaction picture. However, it is funny to point out that this terminology is in some sense inconsistent with the textbook and wikipedia definition of the interaction picture.
4. There are only two places that you can shift the free and interacting time dependences into: states and operators. That means, the number 4 should be it for the useful dynamical pictures of quantum mechanics.

Despite the length, this is going to be a little bit of an incomplete answer. I think it could be filled out by giving a better explanation of why/when/how we work in interaction frames/pictures. Also an example (detuned Rabi oscillations is a great example) would be extremely helpful.

## Born Rule

The predictive power of quantum mechanics comes entirely from the Born rule. We can measure statistics of observables. If we have two observables $$O_1$$ and $$O_2$$ then we can measure things like

\begin{align} \langle O_1 \rangle, \langle O_2\rangle, \langle O_1^2\rangle, \langle O_1O_2\rangle, \langle O_1O_2^2O_1\rangle \end{align}

Quantum mechanics allows us to predict these expectations through the born rule. Essentially we have, for any operator $$O$$, that

\begin{align} \langle O \rangle = \langle \psi|O|\psi\rangle \end{align}

## Time Dependence

Here no time dependence has been spoken of. Let's now talk about time dependence. I'll just give away the answer here. It can be put forth as a fundamental postulate of quantum mechanics that for every system there is a unitary operator* $$T(t)$$ which has the property that

\begin{align} \langle O(t)\rangle = \langle \psi|T^{\dagger}(t)OT(t)|\psi\rangle \end{align}

That is we can see how the measured statistics of $$O$$ vary as a function of time.

## Frames/Pictures

We are now in a position to talk about pictures. The interpretation of the above equation is something like: "If the system is in state $$|\psi\rangle$$ at $$t=0$$ then the expectation of operator $$O$$ at time $$t$$ is given by the above equation". Note I don't have to talk about the operator or state evolving as a function of time. It is just the expectation evolving. I will call this the "frame agnostic representation". I will subscript these states and operators with a $$0$$ to indicate this.

\begin{align} \langle O(t)\rangle = \langle \psi_0|T_0^{\dagger}O_0T_0|\psi_0\rangle \end{align}

I am going to make a distinction between frames and pictures of quantum mechanics. As far as I can tell I've made up this distinction so you shouldn't put too much stock in it outside the scope of this post. A picture is a special type of frame. What is a frame? Above I have described the frame agnostic frame. You can move from the agnostic frame into frame $$F$$ as follows. $$V$$ and $$W$$ are Unitary operators

\begin{align} |\psi_F\rangle &= V|\psi_0\rangle\\ O_F &= W^{\dagger}O_0W\\ T_F &= W^{\dagger}T_0V^{\dagger} \end{align}

We can translate the above formula for $$\langle O(t) \rangle$$ into the new frame and find

\begin{align} \langle O(t) \rangle &= \langle\psi_F|VV^{\dagger}T_F^{\dagger}W^{\dagger}WO_FW^{\dagger}WT_FVV^{\dagger}|\psi_F\rangle\\ &=\langle \psi_F|T_F^{\dagger}O_F T_F|\psi_F\rangle \end{align}

So the special property of a frame is that it preserves the Born rule for calculating expectation values of operators. The difference between a general frame and the agnostic frame is that in a general frame we talk about time evolution of the kets and operators whereas in the agnostic frame the state is to be thought of as fixed, something like the initial state and a fixed time and the operator is also fixed.

I define a picture to be a special frame which has the property that $$T_P = T_F = 1$$ We see that this means

\begin{align} T_P &= W^{\dagger} T_0 V^{\dagger} = 1\\ &\rightarrow WV = T_0 \end{align}

In a picture we have the property that

\begin{align} \langle O(t) \rangle = \langle \psi_P | O_P |\psi_P\rangle \end{align}

That is, all of the time dependence is included in either the kets or the operators.

## Schrodinger/Heisenberg pictures

Ok with that machinery under our belts we are in a very good place to discuss the usual pictures of quantum mechanics. It is very obvious that the Schrodinger picture arises if we choose $$V_S = T_0$$ and $$W_S = 1$$. In that case we get

\begin{align} |\psi_S\rangle &= T_0|\psi_0\rangle\\ O_S &= O_0\\ T_S &= 1 \end{align}

All time dependence is on the kets.

The Heisenberg picture arises if we choose $$V=1$$ and $$W=T_0$$

\begin{align} |\psi_H\rangle &= |\psi_0\rangle\\ O_H &= T_0^{\dagger} O_0 T_0\\ T_H &= 1 \end{align}

Here all time dependence is on the operators.

## Aside on Hamiltonians

The interaction pictures/frames will be related to splitting the time evolution operator into two parts. Before discussing this an aside on how the Hamiltonian is related to the time-evolution operator. First an aside on Hamiltonians. Consider

$$|\psi_F\rangle = T_F |\psi_0\rangle$$

We can take the time derivative of this to get a differential equation for $$|\psi_F\rangle$$.

$$\frac{d}{dt}|\psi_F\rangle = \frac{d}{dt}T_F |\psi_0\rangle$$

It can be proven (exercise for the reader) that for any unitary operator $$T$$ there is a Hermitian operator which satisfies

\begin{align} \frac{d}{dt} T = -iH_T T \end{align}

We call this $$H_T$$ the Hamiltonian for the system. We get the Schrodinger equation.

$$\frac{d}{dt}|\psi_F\rangle = -iH|\psi_F\rangle$$

There is a one to one relationship between Hamiltonians and time evolution operators.

## To split the time evolution operator

Suppose we have a time evolution operator $$T_0$$ and corresponding Hamiltonian

$$H_T = H_X + H_Y$$

Where $$H_X$$ and $$H_Y$$ are the Hamiltonians corresponding to time evolution operators $$X$$ and $$Y$$ in the way mentioned above. Suppose also that $$H_X$$ is simple in the sense that if we had this Hamiltonian alone we could solve the problem. One might naively think we have $$T_0 = XY$$ but as we will see, if $$[H_X, H_Y]\neq 0$$ then this is not the case. It is a close guess though.

I will leave it as an exercise for the reader or for another question to prove that if $$H_T = H_X + H_Y$$ then we can write

$$T_0 = X\tilde{Y}$$

Where $$\tilde{Y}$$ has the property that its corresponding Hamiltonian is

$$H_{\tilde{Y}} = X^{\dagger}H_Y X$$

This transformation looks obtuse at first however in certain example (such as periodically driven Hamiltonians) sometimes the transformed Hamiltonian $$H_{\tilde{Y}}$$ is easier to solve than $$H_Y$$. For example $$H_{\tilde{Y}}$$ might be time independent even if $$H_Y$$ is time-dependent**. Note that $$H_{\tilde{Y}}$$ is the operator $$H_Y$$ expressed in a frame with $$W=X$$. That is, it is $$H_Y$$ in the $$X$$ frame.

\section{Interaction Pictures/Frames} Now I can finally talk about the interaction pictures and interaction frames. We have seen that if we have a system with Hamiltonian

$$H_T = H_X + H_Y$$

that we can express the time evolution operator as

$$T_0 = X\tilde{Y}$$

For later reference this can also be written

$$T_0 = \tilde{Y}\tilde{Y}^{\dagger}X\tilde{Y} = \tilde{Y}X_{\tilde{Y}}$$

Where this $$X_{\tilde{Y}}$$ is $$X$$ in the frame defined by $$\tilde{Y}$$.

In the traditional ket interaction picture we choose $$V=\tilde{Y}$$ so the the interesting dynamics is put onto the kets and $$W=X$$ so that the boring dynamics is put onto the operators. The time evolution for the kets will be given by (KP stands for ket interaction picture

$$\frac{d}{dt}|\psi_{KP}\rangle = -iH_{\tilde{Y}}|\psi_{KP}\rangle$$

It is good that it evolves under $$H_{\tilde{Y}}$$ because ideally $$H_{\tilde{Y}}$$ is in some way simpler than $$H_Y$$.

## Operator Interaction Picture

This is interaction picture #2 referred to in the original question and answer. In this picture we choose $$W=\tilde{Y}$$ and $$V=X_{\tilde{Y}}$$. In this picture the interesting dynamics is put on the operators while the boring dynamics is somehow put on the kets. Note that (another exercise for the reader, ignore explicit time dependece of $$O$$) the time evolution of the operators will be given by (OP stands for operator interaction picture)

$$\frac{d}{dt}O_{OP} = -i[O_{OP}, H_{\tilde{Y}}]$$

Again it is good that $$O_{OP}$$ evolves under $$H_{\tilde{Y}}$$ because this is supposedly simplified compared to $$H_Y$$.

## Interaction frames

The above answers part of the question about 4 instead of 3 interaction pictures. However, I want to draw attention to a point which has been a sticking point for me. I don't think anyone actually does calculations in the way I have shown above. That is, by solving the easy dynamics on either the operators or the kets and then solving the hard dynamics on the other one and then combining the two. No, what is rather done is that one simply shunts the easy dyanmics "out of the way" solves the hard dynamics on either the kets (in textbooks for example) or the operators (in Langevin equation based quantum optics, for example) and then just forgets about the easy dynamics!

The question is then how, in principle, would you go about adding the easy dyanmics back into the problem? The answer follows.

## Ket Interaction Frame

In the Ket interaction frame the hard dynamics are put into the kets while the easy dynamcis is left in the time evolution operator. We choose $$V_{KF}=\tilde{Y}$$ and $$W_{KF}=1$$ which leaves $$T_{KF} = X$$. This gives

\begin{align} \langle O(t) \rangle = \langle \psi_{KF}|\tilde{Y}^{\dagger} X^{\dagger} O X \tilde{Y}|\psi_{KF}\rangle \end{align}

How would this expression be solved? Well one would work in some computation basis. One would then diagonalize $$H_{\tilde{Y}}$$ in this basis. One would then know how to express $$\tilde{Y}|\psi_{KF}\rangle$$ in terms of the computation basis. Next, since $$X$$ is solved that means we have already diagonalized $$H_X$$ in terms of the computation basis so then it is easy to calculate $$X\tilde{Y}|\psi_{KF}\rangle$$. The problem is then essentially solved if we know the action of $$O$$ on the computation basis.

## Operator interaction frame

As in the case for the ket interaction frame the problem will be solved in two steps. First a differential equation involving $$H_{\tilde{Y}}$$ will be solved to determine operator evolution under the difficult part of the Hamiltonian. Then, since the action of $$H_X$$ on the original operators is known this can subsequently be applied to solve the problem.

Here we take $$V_{OF} = 1$$, $$W_{OF}=\tilde{Y}$$ leaving $$T_{OF} = X_{\tilde{Y}}$$ We write

\begin{align} \langle O(t) \rangle = \langle \psi_{OF}|X_{\tilde{Y}}^{\dagger}\tilde{Y}^{\dagger}O_{OF}\tilde{Y}X_{\tilde{Y}}|\psi_{OF}\rangle \end{align}

Here we work inside out.

$$\tilde{Y}^{\dagger}O_{OF}\tilde{Y}$$

will result in a differential equation involving the commutator of $$O_{OF}$$ with $$H_{\tilde{Y}}$$ that can be solved to express the operator in terms of the operators at initial times. Next the $$X_{\tilde{Y}}$$ time evolution can be applied.

## Many pictures/frames

So we see there are many pictures/frames in quantum mechanics. I have named at least 7 here. The agnostic frame, the Schrodinger picture, the Heisenberg picture and the operator and ket interaction pictures and frames.

1. Is the "interaction picture #2" relevant in quantum theory?

Yes.

1. If so, where is it used and how is it useful?

it is used in quantum optics applications. Particularly when Heisenberg or Langevin operator formalism is being utilized.

1. If so, why on earth does nobody ever talk about it?

As @Wolpertinger points out in their answer, This interaction picture is often implicitly referred to in quantum literature. I don't know why no one has explicitly called out this distinction. The closest I've seen to a discussion in the literature is in Wiseman and Milburn's Quantum Measurement and Control in Appendix A.1.3. I'm not sure why the existence of this frame is not more publicized. I think part of the answer is that in practice it typically just amounts to multiplying something by $$e^{\pm i \omega t}$$ and everything is taken care by just talking about moving in or out of the rotating frame. Speaking in those terms is actually much more straightforward than keeping track of all of the adjoints and different frames etc. that I have outlined above. I'm not sure if what I've given is the most concise statement of the different options involved and clearly (given by the lack of clarity above) I have some residual confusion about the calculational differences between what I call the interaction picture vs interaction frame. I'd be happy to see some further clarification in the literature.

1. Are there posssibly even more useful dynamical pictures of quantum theory?

I think what I have described above distinguishing the interaction pictures from the interaction frames might qualify as an affirmative answer to this question. If not then I might suggest the possibility that there might be frames in which you can split a Hamiltonian into 3 or more parts to get something useful and there may be some sort of novel picture happening there. The idea of Floquet physics also comes to mind. It may be useful to move into a frame that has some oscillatory behavior that doesn't just split the time evolution operator in two but which somehow simplifies the time evolution operator's action on either the kets and/or the operators.

*Recall that Unitary means $$U^{\dagger}U = UU^{\dagger} = 1$$.

**This is where an example could help

edit: Note that there is really a lot more to be said here that might clarify the situation. For example, one could talk about how this can be interpreted as moving in and out of a "rotating frame". One could also say more about what transformations are done to move in between different frames. All of this might clarify the purpose of the somewhat vexing presence of the weird operators $$\tilde{Y}$$ and $$X_{\tilde{Y}}$$.