I have a puzzle about Schroedinger equation with time-dependent hamiltonian, which is usually used in time-dependent quantum systems.

However, one of the axioms in quantum mechanics postulates that the hamiltonian as the generator of the one-parameter unitary group $U(t)$ is not time-dependent, and the evolution of a quantum state should obey the Schroedinger equation with time-independent hamiltonian.

So why can Schroedinger equation with time-dependent hamiltonian be used without hesitation?

Where does the time-dependent hamiltonian come from, except for the case of interaction picture? And should there be an additional axiom for it?

The key is to recognize which are from axiom and which are from model. I foolishly thought that every Shroedinger equation with a time-dependent Hamiltonian can be derived by starting from that strong-continuity axiom in quantum mechanics. Also, I do not think the above comments answered my question.

  • 4
    $\begingroup$ An emergency? As Heisenberg said, if you don't have time, I don't have the energy ... Ok fine. Are you sure your axiom is not only for the time-independent Schroedinger equation, which describes stationnary states? Do you have more info on that axiom? $\endgroup$
    – fffred
    Aug 23, 2013 at 6:56
  • 4
    $\begingroup$ I would like to see the source of this axiom. I've never heard an axiom say that the Hamiltonian must be time independent. There is no constraint from unitary evolution other than that the Hamiltonian has to be hermitian. $\endgroup$
    – Michael
    Aug 23, 2013 at 6:58
  • $\begingroup$ There exists a one-parameter group $U_t$ of unitary operators that map an intial state $\psi_0$ at the time $t=0$ to the state $\psi_t=U_t\psi_0$ at time t. The operator $U_t$ is of the form $U_t=e^{-itH}$. If $\psi_0\in D(H)$, the $\psi_t$ is a solution of the Schroedinger equation $d_t\psi_t=-iH\psi_t$. That's the axiom. $\endgroup$
    – user28662
    Aug 23, 2013 at 7:12
  • $\begingroup$ I'm guessing here from the standpoint of knowing finite dimensional Lie theory - for an operator belonging to the identity-connected component of the group $U(N)$ you can always find an $H$ s.t. $U = \exp(i\,H\,t_{end})$ ($\exp$ maps the Lie algebra surjectively in this special case) but the path through the Lie group doesn't have do follow the flow $U = \exp(i\,H\,t)$ to get there. Is the author talking about an equivalent path with the same endpoint in the Lie group as a general path maybe? You can certainly construct a flow path to do this ("one-parameter group") ... $\endgroup$ Aug 27, 2013 at 14:39
  • 1
    $\begingroup$ ...where did you get this axiom from - can you cite its source? I think I must be missing something. $\endgroup$ Aug 27, 2013 at 14:40

3 Answers 3


From the given information it's a little hard to deduce what you want. But say you have a time dependent magnetic field interacting with a system of particles. Then the spin state of the system of the particles will be time dependent. Therefore, the Hamiltonian of the system of particles will be time dependent. Usually a time dependent Hamiltonian comes from a non-conserved system (here we observe only the system of particles interacting with the magnetic field, rather than the whole picture, which would also include what is generating the magnetic field). Since we ignore what is generating the magnetic field, then the system is not conserved therefore the Hamiltonian is often time dependent.

That's why it is used without hesistation. It's all about the context of the problem.

  • $\begingroup$ Yes, but why can schroedinger equation be used for a non-conserved system? Merely starting from physical intuition? $\endgroup$
    – user28662
    Aug 23, 2013 at 7:20
  • $\begingroup$ You do know that the time dependent Schrodinger's equation is the generalization of the time independent Schrodinger's equation right? The Hamiltonian is basically characterized as the energy of the wave function. If this energy changes with time, well then our system changes with time. Therefore, we would expect that Schrodinger's equation account for a time dependent system, which is does! $\endgroup$ Aug 23, 2013 at 7:32
  • $\begingroup$ Just like you can use Newton's equations to describe a non-conserved system, so can you use Schrodinger's equation to describe a non-conserved system. There's no special differentiation between the two when you apply the equations. The only difference is in a non-conserved system, you're not describing 'everything' that is happening, whereas in a conserved system you are. This doesn't limit the use of Schrodinger's equation though, just like I outlined in my answer with the magnetic field $\endgroup$ Aug 23, 2013 at 7:33
  • $\begingroup$ Newton's mechanics is based on experience rather than axiom on which quantum mechanics lives. Now that the existing axioms of quantum mechaincs are so sufficient that any additional one cannot be added, then the case for any non-conserved system must have been considered. My puzzle is whether Schroedinger equation with time-dependent hamiltonian can be deduced from the axiom of quantum mechanics, except in interaction picture. $\endgroup$
    – user28662
    Aug 23, 2013 at 8:17
  • 1
    $\begingroup$ Nowhere does this axiom exclude evaluating a system in which it is externally influenced. I think you're really missing the point between differentiating these two systems. In reality if we took the Hamiltonian of the whole system, it probably would be time independent! (although this is getting into a larger debate...is this universe's Hamiltonian time independent?). It should be quite clear that the Hamiltonian changes with time if you are pumping energy into a system and ONLY applying Schrodinger's equation to the observed system having energy pumped into $\endgroup$ Aug 23, 2013 at 8:36

$$ \left\vert\Psi\left(t\right)\right\rangle = {\rm T}\exp\left(-\,{{\rm i} \over \hbar} \int_{t'}^{t}H\left(t''\right)\,{\rm d}t''\right)\left\vert\Psi\left(t'\right)\right\rangle\,, \qquad t > t' $$

  • $\begingroup$ @WetSavannaAnimalakaRodVance Note the time ordering symbol. This is just the short hand notation for the iterative solution of the usual Dyson series, which is fully general. No need for $[H(t_1),H(t_2)]=0$. $\endgroup$
    – Michael
    Aug 30, 2013 at 6:35
  • $\begingroup$ @MichaelBrown Ah, sorry, I missed this: thanks, that makes things much clearer. $\endgroup$ Aug 30, 2013 at 8:05
  • $\begingroup$ Dear Felix @MichaelBrown has put me straight - I was forgetting about the Dyson series (I come from a quantum optics background where I haven't see such things used) so I agree, this is a valid alternative (and equivalent) to the Peano-Baker series. You might like to add some words to it as it looks a bit fearsome, particularly to someone who is seeking to understand axioms. I must say the time ordering operator is a bit fearsome to me anyway, I think I'd like to be in a well lit street when I met him at night. $\endgroup$ Aug 30, 2013 at 8:13

It would be good for you to cite where you have seen the time independent Hamiltonian as an axiom because I, along with several other readers in comments, have not seen this axiom.

However, I can imagine that something like it might be a reasonable axiom in nonrelativistic quantum mechanics, something like:

There exists a basis for the Hilbert space of quantum states of an isolated (i.e. not interacting at all with the outside world) wherein the Hamiltonian is not dependent on time**.

There is actually something like this (although not explicitly stated as an axiom) in a discussion of the Hamiltonian in the "Feynman Lectures on Physics" in chapter 8 "The Hamiltonian Matrix" of the third volume, especially in the sections 8.3 "What are the basis states of the world?" and 8.4 "How states change with time".

Before I go on: hereafter I'm going to treat my state transition operators as though they are finite ($N$) dimensional square matrices so I can speak in the language of finite dimensional Lie theory, where I feel most at home. I'm sure one can generalize to the countably infinite dimensional Hilbert space by putting suitable boundedness and other conditions on the operators concerned into place to broaden my arguments, but I don't have a good enough grip on spectral theory to be sure of myself.

The gist of Feynman's discussion and also my preferred way of thinking about the Schrödinger equation is that the Schrödinger equation asserts the time shift invariance of the description of a quantum system's state when that system is sundered from the rest of the World as well as the linearity and probability-conservation of that description. Nature doesn't care whether I - a monkey in a suit whose forebears evolved only a few hundred thousand years ago - begin my analysis of a system at four o'clock or whether I wait until after my coffee break at half past four - the description I get cannot be essentially different - it's an essentially "Copernican" idea. Our system's state is represented by a state vector $\psi$ in the Hilbert state space and if we further assume linearity (see this question) of our state's evolution then our state vector (now written as a column vector) is going to evolve following some matrix equation: $\psi(t) = \mathbf{U}(t) \psi(0)$, where the state transition matrix $\mathbf{U}(t) \in \mathfrak{U}(N)$ must:

  1. Fulfil $\mathbf{U}(t+s) = \mathbf{U}(t) \mathbf{U}(s) = \mathbf{U}(s) \mathbf{U}(t)$ for any time intervals $t$ and $s$. This is simply our discussion about time shift invariance above. Straight away we know $ \mathbf{U}(t) = \exp(A t)$, for some constant matrix $A \in \mathfrak{u}(N)$ (the Lie algebra of skew-Hermitian matrices) as the exponential is the only continuous function with this time shift invariance property; (see footnote)
  2. Be unitary to conserve norms and thus probabilities.

So the most general state evolution in keeping with the above is $\psi(t) = \exp(-\hbar^{-1} i\, \hat{H}\, t)\,\psi(0)$, where $\hat{H}$ is a constant, Hermitian matrix (this is equivalent to the unitaryhood statement - $i\,\hat{H} \in \mathfrak{u}(N)$). This evolution in turn is equivalent to:

$$i\,\hbar\,\mathrm{d}_t \psi = \hat{H}\,\psi$$

which is the Schrödinger equation: for our purposes here $\hbar$ is simply a constant one pulls out and $i$ is another constant pulled out so that our Hamiltonians are Hermitian (i.e. we have real-valued energies) and not skew Hermitian as one wontedly thinks of in Lie theory (the Lie algebra $\mathfrak{u}(N)$ is the algebra of skew-Hermitian matrices).

OK. So there is a case for an axiom something like you say: but there is the proviso that it is constant only modulo a unitary co-ordinate transformation. Things would be too restrictive and impracticable otherwise. We might just want to look at the problem from another co-ordinate frame: we're always doing such things in Physics! The interaction picture is a good example as you rightly point out if we want to split our isolated system into a "main" subsystem with known behavior and contribution to the Hamiltonian and a "perturbation" system with a perturbing contribution to the Hamiltonian. So, for this or some other reason, we now "roll our co-ordinate reference axes around in the Lie group $\mathfrak{U}(N)$" by some unitary co-ordinate-mapping matrix $\mathbf{V}(t)$ where:

$$\mathrm{d}_t \mathbf{V}(t) = -\frac{i}{\hbar}\,\tilde{K}(t)\,\mathbf{V}(t) = -\frac{i}{\hbar}\,\mathbf{V}(t)\,K(t)$$

where $i\, K(t),\,i\, \tilde{K}(t)\in \mathfrak{u}(N)$ and there are two alternatives because left translation or right translation of neighbourhoods of the identity work equally well to define the Lie group topology). Therefore, we now have our new state $\psi^\prime(t) = \mathbf{V}(t) \psi(t)$ so that:

$$\begin{array}{lcl}\mathrm{d}_t \psi^\prime(t) &=& \mathrm{d}_t(\mathbf{V}(t))\, \psi^\prime(t) + \mathbf{V}(t) \,\mathrm{d}_t \psi^\prime(t) \\ &=& -\frac{i}{\hbar}\left(\mathbf{V}(t)\,K(t)\,\mathbf{V}^{-1}(t) + \mathbf{V}(t)\,\hat{H}(t)\,\mathbf{V}^{-1}(t)\right)\,\psi^\prime(t) \\ &=& -\frac{i}{\hbar} \hat{H}^\prime(t)\,\psi^\prime(t)\end{array}$$

where we can always find an $i\,\hat{H}^\prime(t) \in \mathfrak{u}(N)$ such that $\hat{H}^\prime(t) = \mathbf{V}(t)\,K(t)\,\mathbf{V}^{-1}(t) + \mathbf{V}(t)\,\hat{H}(t)\,\mathbf{V}^{-1}(t)$, as we can understand either because:

  1. We can left translate a $C^1$ path through the identity with tangent $\frac{i}{\hbar}(K + \hat{H})$ there to a $C^1$ path through $\mathbf{V}$ with tangent $\frac{i}{\hbar}\,\mathbf{V}\,(K + \hat{H})$ there and then right translate this path back to a (in general different) $C^1$ path through the identity with the in general different tangent $\frac{i}{\hbar} \hat{H}^\prime = \frac{i}{\hbar}\,\mathbf{V}\,(K + \hat{H}) \mathbf{V}^{-1}$ there; or
  2. More jargonistically, a Lie group acts on its own Lie algebra through conjugation in the group's adjoint representation. So now we have another perfectly valid Schrödinger equation, only this time it is time-varying:

$$i\,\hbar\,\mathrm{d}_t \psi^\prime = \hat{H}^\prime(t)\,\psi^\prime$$

and, as in Michael Brown's comment, the only requirement of $\hat{H}$ is that it must be Hermitian. Of course when we roll around in the Lie group like this, physical observables become themselves time varying, as they do in both the interaction and Heisenberg picture.

Now if we relax the time-shift-invariance constraint, as in Spaderdabomb's answer, we still must have unitary evolution of the state transition matrices to make sure that norms and thus probabilities are conserved. So now our state transition operator still stays in the Lie group $\mathfrak{U}(N)$ but it no longer traces a one-parameter group (i.e. left / right translationally-invariant flow line) but rather traces some general $C^1$ path through the Lie group. Its time derivative must be in the left or right translated Lie algebra so now we have:

$$\mathrm{d}_t \mathbf{U}(t) = -\frac{i}{\hbar}\,\hat{H}(t)\,\mathbf{U}(t) = -\frac{i}{\hbar}\,\mathbf{U}(t)\,\tilde{\hat{H}}(t)$$

for some $\frac{i}{\hbar}\,\hat{H}(t), \frac{i}{\hbar}\,\tilde{\hat{H}}(t)\in \mathfrak{u}(N)$ and we still have a Schrödinger equation:

$$i\,\hbar\,\mathrm{d}_t \psi = \hat{H}(t)\,\psi$$

Note that, as in Spaderdabomb's answer this situation arises when an outside influence, e.g. a magnetic field, can be treated classically so that it begets time varying $c$-valued parameters of the Hamiltonian, in contrast with the complicated case when we represent the magnetic field quantum mechanically. As Spaderdabomb alluded to in the comments, if we extend the system description to include a quantum mechanical model of the magnetic field and its source we in principle have a time invariant Hamiltonian and your axiom holds sway, although this is very in principle and never done owing to the forbidding complexity of such a description.


  1. As you are likely aware: mathematicians define a powerful and beautifully simple abstraction of $\mathbf{U}(t+s) = \mathbf{U}(t) \mathbf{U}(s) = \mathbf{U}(s) \mathbf{U}(t)$ in the idea of a flow.
  2. There are some weird and quite wonderful everywhere discontinuous solutions to the functional equation $\mathbf{U}(t+s) = \mathbf{U}(t) \mathbf{U}(s) = \mathbf{U}(s) \mathbf{U}(t)$ if you're interested in Hewitt and Stromberg, "Real and Abstract Analysis", Springer-Verlag, Berlin, 1965 Chapter 1, §5 but we throw these out as "unphysical".
  3. However: In the case of a semisimple, compact Lie group the weird and wonderful Hewitt and Stromberg behaviour is eliminated because there is only one possible topology for such a Lie group that makes the group a Lie group. Or, otherwise put, there are NO automorphisms of the group (thought of only as an abstract, non-Lie group) that are not also Lie group automorphisms, so ALL automorphisms are also group topology homeomorphisms. This amazing fact was proven by van der Waerden in 1932 (van der Waerden, B. L., Mathematische Zeitschrift 36 pp780 - 786). So we don't even have to worry about assuming continuity: semisimplicity and compactness do the job for us. We nearly have this in the above discussion: $\mathfrak{SU}(N)$ is compact and simple, but $\mathfrak{U}(N)$ has a continuous, Lie group centre $\mathfrak{U}(1)$ and $\mathfrak{U}(N) = \mathfrak{SU}(N) \otimes \mathfrak{U}(1) = \mathfrak{U}(1) \otimes \mathfrak{SU}(N)$). I think this is quite profound from a physics standpoint, since most often we have to make assumptions about the mathematical objects that are very strong (i.e. highly specializing) mathematical assumptions: smoothness, for example. Here is an example where the coarsest of mathematical assumptions (one parameter groups that are not even assumed continuous) yields smoothness, indeed $C^\omega$ (analytic) behavior. Sadly, my gut feeling is that this point is likely the only thing in this post which will not survive the passage to countably infinite dimensions: van der Waerden's stunning little trick relies wholly on finiteness of dimension.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.