Why in the interaction picture time evolution of state vector only considers time time dependent part of Hamiltonian and and while calculating expectation value in interaction picture unperturbed part of Hamiltonian is used? And while representing the state vector in interaction picture ,time propagator only considers time independent part of Hamiltonian. Can some one please explain the time evolution of state vectors and operators in the interaction picture with equation.

$$ |\psi_{I}(t)\rangle = U^{\dagger}|\psi_{s}(t)\rangle $$ where time propagator operator is $$ U(t-t_{0}) = e^{-iH_{0}(t-t_{0})/h} $$

In the first equation shouldn't they mention U rather than its complex conjugate? And why they have only used time independent part of Hamiltonian?where $H=H^{0}+H(t)$. Same is the case for the expectation value of interaction picture, it also considers unperturbed part. How to deal with perturbed part?

  • $\begingroup$ What reference are you looking at for this? $\endgroup$ Oct 3, 2018 at 12:37
  • $\begingroup$ I am reading Zetlli book for this $\endgroup$
    – herry
    Oct 3, 2018 at 13:05
  • $\begingroup$ This is an interesting question but it's not clear specifically what you are asking and it seems like you are saying some contradictory things. To answer you question with equations like you ask for us to do it would be helpful if you could clarify your question with equation. See math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Jagerber48
    Oct 3, 2018 at 13:06
  • $\begingroup$ Its hard to visualize interaction picture as compared to Heisenberg and Schrodinger picture where I can imagine rotation of coordinates and vectors. What are the key points regarding basis and hamiltonian in this picture? $\endgroup$
    – herry
    Oct 3, 2018 at 13:08
  • $\begingroup$ If you just want a visualization, the interaction picture can be viewed as a mix between the Schrodinger and Heisenberg picture. In the Schrodinger picture the state vectors have time dependence, in the Heisenberg picture the operators have time dependence. In the interaction picture both are true. Analogous to a rotating reference frame that does not rotate along with the state vector. Of course the analogy is not perfect, but this is what more formal equations are for. $\endgroup$ Oct 3, 2018 at 13:25

2 Answers 2


The state vector in the interaction picture $|\psi_I\rangle$ is defined to be $U_0^\dagger |\psi_S\rangle$, where $|\psi_S\rangle$ is the state vector in the Schrodinger picture, and $U_0=e^{-iH_0(t-t_0)/\hbar}$ with $H=H_0+H_1(t)$ as you say$^*$. If you don't like the complex conjugate in the equation you can instead write it as $|\psi_S\rangle=U_0|\psi_I\rangle$ if you want. But this is where you can start with the interaction picture and derive other useful relationships from there. In other words, this is a relationship between the state vectors within each picture. It is not a full representation of the interaction picture itself.

This also answers "why are we just using $H_0$ and not the full $H$? Once again, this is how we are defining the state vector in this picture in terms of the state vector in the Schrodinger picture. The time dependent part of the Hamiltonia does come into play in the interaction picture though. I will give some of this now.

You have to keep in mind that the Hamiltonian $H$ above is really the Hamiltonian in the Schrodinger picture. If we define $H_{0,I}=U_0^\dagger H_0 U_0=H_0$ (since $U_0$ and $H_0$ commute) and $H_{1,I}=U_0^\dagger H_1 U_0$, then we can get the time evolution of our state vector in the interaction picture.

If we use these definitions and start with the time dependent Schrodinger equation, you can show that $$i\hbar\frac{\partial}{\partial t}|\psi_I\rangle=H_{1,I}|\psi_I\rangle$$ (If you don't like how we just defined these operators in this way, substitute the state vector into the TDSE first and see that defining the operators in this way ends up being very useful).

This is what should help with your confusion. $H_{1,I}=U_0^\dagger H_1 U_0=e^{iH_0(t-t_0)/\hbar}H_1e^{-iH_0(t-t_0)/\hbar}$ depends on both $H_0$ and $H_1(t)$, so we still have both parts of the original Hamiltonian.

This shows how really any operator transforms from the Schrodinger picture to the interaction picture: $$A_I=U_0^\dagger A_S U_0$$ But, once again, just because we are not using $H_1$ here does not mean it is not important. This is just how we relate the two pictures (interaction and Schrodinger). This is then why the expectation values do not depend on $H_1$ as you point out. It is just an artifact of how we define the interaction picture to be "relative to" the Schrodinger picture.

$^*$ The Hamiltonian $H=H_0+H_1(t)$ is viewed in the "Schrodinger picture". Now in your comments and earlier questions we discussed that operators in the Schrodinger picture have no time dependence. This was somewhat loose speaking. What we should have said was that there is no time dependence due to unitary transformations. Operators in the Schrodinger picture can still have a time dependence if something is physically changing. And example of this is if we have a particle in a time dependent electric field. The problem is simpler to solve if we can break the Hamiltonian into a sum time-independent and time-dependent, although this does not have to be the case.

As for the usage of $U_0$ versus $U_0^\dagger$, this is purely a definition. If you want to put something "more physical" with this, you could say that $|\psi_S\rangle=U_0|\psi_I\rangle$ means that $|\psi_S\rangle$ is a time evolved state of $|\psi_I\rangle$ if the Hamiltonian was just $H_0$. But this is not the actual system in question, since $H=H_0+H_1(t)$

  • $\begingroup$ How H can be schrodinger picture hamiltonian? Because it also contain H(t) where as in Schrodinger picture operators are time independent. Secondly I am not so clear about using U dagger in the first equation. ?Because propagator is defined by U not by U dagger $\endgroup$
    – herry
    Oct 3, 2018 at 14:36
  • $\begingroup$ Can you please also explain the physical meaning of these equations of the interaction picture that what the tell us? abut energies about eigen vectors and about time dependent and independent part $\endgroup$
    – herry
    Oct 3, 2018 at 14:44
  • $\begingroup$ @herry Operators can still have a time dependence in the Schrodinger picture when something physical has the time dependence (for example, a time dependent electric field). The time evolution of the Hamiltonian here is not due to the QM formalism (i.e. not a unitary transformation), but rather something actually physically changing. As for your second issue, once again, this is just how things are defined for the interaction picture. If you want to, use $|\psi_S\rangle=U_0|\psi_I\rangle$ $\endgroup$ Oct 3, 2018 at 14:44
  • $\begingroup$ @herry Once again, just like between the Schrodinger and Heisenberg pictures, the interaction picture is does not give anything different physically. The eigenvalues of observables are the same in all pictures. I can add more information about the Hamiltonian though. $\endgroup$ Oct 3, 2018 at 14:45
  • $\begingroup$ What is the physical meaning to first equation in my post? In case of Heisenberg picture we can say that we implement time propagator to see the time evolution of a state. Does the U dagger changes its meaning in interaction picture? $\endgroup$
    – herry
    Oct 3, 2018 at 14:47

You should get comfortable converting between thinking about the unitary time evolution operator $\hat{U}(t)$ and the Hermitian Hamiltonian $\hat{H}(t)$. The two are related by ($\hbar = 1$)

$$ \frac{d}{dt}\hat{U}(t) = -i\hat{H}(t)\hat{U}(t) $$

If $\hat{H}(t)$ is time independent we can integrate this equation without worrying about non-commutativity of operators and find

$$ \hat{U}(t) = \exp(-i\hat{H} t) $$

We can take either $\hat{U}$ or $\hat{H}$ to be fundamental and derive one from the other. I personally think of $\hat{U}$ as being the fundamental thing so I'll talk about things that way.

Suppose we have a system whose Hamiltonian has two parts:

$$ \hat{H}_T = \hat{H}_X + \hat{H}_Y $$

Suppose $\hat{H}_X$ is a well understood, probably time-independent, Hamiltonian while $H_Y$ is more complicated and possibly time-dependent.

There are unitary operators corresponding to $\hat{H}_X$ and $\hat{H}_Y$ which satisfy

\begin{align} \frac{d}{dt}\hat{X} &= -i\hat{H}_X \hat{X}\\ \frac{d}{dt}\hat{Y} &= -i\hat{H}_Y \hat{Y}\\ \end{align}

These are the corresponding time evolution operators for $\hat{H}_X$ and $\hat{H}_Y$. The question is how does $\hat{U}$ (which gives rise to the total Hamiltonian $\hat{H}_T$) relate to $\hat{X}$ and $\hat{Y}$?

To investigate this we see what happens if we write $\hat{U}$ as a product of two unitary operators $\hat{A}$ and $\hat{B}$. We can also define Hamiltonian $\hat{H}_A$ and $\hat{H}_B$ which are related to the time derivatives of $\hat{A}$ and $\hat{B}$.

\begin{align} \hat{U} &= \hat{A}\hat{B}\\ \frac{d}{dt}\hat{U} &= \left(\frac{d}{dt}\hat{A}\right)\hat{B} + \hat{A}\frac{d}{dt}\hat{B}\\ &=-i\left(\hat{H}_A\hat{A}\hat{B} + \hat{A}\hat{H}_B\hat{B} \right)\\ &= -i\left(\hat{H}_A + \hat{A}\hat{H}_B\hat{A}^{\dagger}\right)\hat{U}\\ &= -i\left(\hat{H}_X + \hat{H}_Y \right)\hat{U} \end{align}

Note that above $\hat{A}$ and $\hat{B}$ were arbitrary unitary operators so we can choose them however we like. We see that if we choose $\hat{A} = \hat{X}$ then we have

\begin{align} \hat{H}_A &= \hat{H}_X\\ \hat{H}_B &= \hat{X}^{\dagger}\hat{H}_Y\hat{X} = \hat{H}_{Y,I} \end{align}

We need a new unitary operator $\tilde{Y}$ which gives rise to Hamiltonian $\hat{H}_{Y,I}$

$$ \frac{d}{dt}\tilde{Y} = -i\hat{H}_{Y,I}\tilde{Y} $$

Of course $\tilde{Y}=\hat{B}$

This means that if $\hat{H}_T = \hat{H}_X+\hat{H}_Y$ then we know that $\hat{U}$ can be written

$$ \hat{U} = \hat{X}\tilde{Y} $$

Where $\hat{X}$ and $\tilde{Y}$ give rise to the Hamiltonians indicated above.

Now we require a few conditions to make the working in the interaction picture useful.

  1. $\hat{X}$ represents well understood "simple" time evolution which is previously solved.
  2. $\hat{H}_Y$ is in some sense complicated and not yet solved but the combination $\hat{H}_{Y,I} = \hat{X}^{\dagger}\hat{H}_Y\hat{X}$ is somehow more simple and solvable. For example, $\hat{H}_Y$ may be time dependent while, by luck and cleverness, $\hat{H}_{Y,I}$ is time-independent.

Suppose now that we are working in some complete orthogonal "calculation basis" given by the kets $\{|q_i\rangle\}$. To say that $\hat{X}$ is "solved" means that we know how to act it on any of these kets and get a time-dependent resultant ket. This could be found by solving the Schrodinger equation with the Hamiltonian $\hat{H}_X$.

What the interaction picture says is that, since $\hat{U} = \hat{X}\tilde{Y}$, if we can figure out the action of $\tilde{Y}$ on a ket in the calculation basis, then we can figure out the action of $\hat{U}$ on a ket in the calculation basis.


$$ \hat{X} |q_i\rangle = \sum_{j} c^X_{ij}(t)|q_j\rangle $$

with $c^X_{ij}(t)$ known time-dependent coefficients capturing the time-evolution under $\hat{X}$ or equivalently under $\hat{H}_X$. Suppse we also solve the Schrodinger equation with $\hat{H}_{Y,I}$. then we also know

$$ \tilde{Y} |q_i\rangle = \sum_j c^Y_{ij}(t)|q_j\rangle $$

We "know" tihs formula in the sense that the time-dependent coefficients $c^Y_{ij}(t)$ are known.

Then we can put these two solution together to find

$$ \hat{U}|q_i\rangle = \hat{X}\tilde{Y}|q_i\rangle = \sum_j \hat{X} c^Y_{ij}(t)|q_j\rangle = \sum_{jk} C^X_{kj}(t)C^Y_{ij}(t)|q_k\rangle $$

A few translations to make the abstract stuff I'm talking about here commensurate with what you'll usually see. In comparison with Aaron Stevens' answer

\begin{align} \hat{H}_X &= H_0\\ \hat{H}_Y &= H_1\\ \hat{H}_{Y,I}&=H_{1,I}\\ |\psi_I\rangle &= \hat{X}^{\dagger}|\psi_S\rangle\\ \hat{X} &= U_0\\ \end{align}

One final note: Often when the interaction picture is used authors simply ignore $\hat{X}$ and act as if it doesn't exist and continue on with the problem. This is done for two reasons. 1) the dynamics under $\hat{X}$ are boring and well understood so authors assume that if the reader is worried about those dynamics they can carry out the last step of calculation themselves to return into the non-interaction picture. 2) One gets a lot of intuition about the solution to problems by looking at their solutions in the interaction picture alone. All of this ends up being a bit confusing because it seems like authors are ignoring something at least somewhat important.

Edit: Another point I'd like to add. Suppose the initial state is $|\psi_0\rangle$. Note that this is a ket at a fixed moment in time. It will never change. We can define the Schrodinger picture ket as

$$ |\psi_S\rangle = \hat{U}|\psi_0\rangle = \hat{X}\tilde{Y}|\psi_0\rangle $$

We can see by the above manipulations that

\begin{align} |\psi_I\rangle &= \hat{X}^{\dagger}|\psi_S\rangle = \hat{X}^{\dagger}\hat{U}|\psi_0\rangle = \hat{X}^{\dagger}\hat{X}\tilde{Y}|\psi_0\rangle\\ &= \tilde{Y}|\psi_0\rangle \end{align}

This explains the sense in which "moving into the interaction picture" does the job of "removing or rotating out the time evolution under $\hat{X}$" from the problem.


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