Until $t=0$ a system is in an eigenstate $\psi_0(x)$ of the Hamiltonian $\hat{H}_0$. The time-evolution is the trivial phase factor. Now at $t=0$ the system changes to $\hat{H}$ (you can assume it is a small perturbation away). The old states will start to develop according to the TDSE to the new state $$\psi(t,x)=e^{-\frac{i}{\hbar} \hat{H} t} \psi_0(x) \, .$$ The old state $\psi_0(x)$ can be expanded in terms of the new set of stationary solutions/eigenfunctions $\phi_n(x)$ of $\hat{H}$ $$\psi(t,x)=e^{-\frac{i}{\hbar} \hat{H} t} \sum_{n=0}^\infty c_n \, \phi_n(x)=\sum_{n=0}^\infty c_n \, \phi_n(x) \, e^{-\frac{i}{\hbar} E_n t} \, . \tag{1}$$

Under which conditions does the state $\psi(t,x)$ converge to a stationary state/eigenstate or ground state of the new system as $t\rightarrow \infty$?

From the above expression, one often argues: the larger the energy $E_n$, the stronger this contribution will oscillate or similarly by Wick-Rotation $t\rightarrow -i\tau$ and then let $\tau \rightarrow \infty$ to see that the dominant contribution is actually the ground state $\phi_0(x)$ (or any higher state e.g. $\phi_1(x)$ if $\langle \phi_0 | \psi_0 \rangle=0$) of the new system with energy $E_0$, but this Wick-Rotation has always been somewhat wishy-washy to me without mathematical grounds.

Is there a more correct way to see how the dominant contribution actually arises from the slowest oscillatory term?

What I'm also somewhat confused about is that this can not always be right. For example if $E_n=2\pi \hbar n/T$ (harmonic oscillator), then the new state is actually periodic in $t$ with period $T$, so why can one even argue this way? Or put differently, when is one allowed to argue this way?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Commented Apr 29, 2020 at 0:51

1 Answer 1


Assuming that $\psi_0(x)$ is not also an eigenstate of the full $H$, the short answer to the question "Under which conditions does the state 𝜓(𝑡,𝑥) converge to a stationary state/eigenstate or ground state of the new system as $t\to\infty$?" is roughly "never". As time evolution is unitary under the expansion you presented, then the coefficients $c_n$ are conserved during the time evolution, each just acquiring a phase, and the overlap $|\langle \phi_n | \psi(t) \rangle |^2$ is fixed in time.

The longer is more complicated answer is "it depends". One way is to slightly change the question and ask "how will most of the obersvables look when taken in this state, at long times?" which is to say we are not interested in the state itself, but rather in a set of physical observable quantities that we average over time $O \equiv \lim_{T\to\infty}\frac{1}{T}\int_0^{T}\langle \psi(t)| \hat{O}| \psi(t) \rangle $. Now one can argue that the largest contributions will come from the states with the slowest time oscillations - that is the ground state.

Another way is to allow some sort of dissipation. This, in a way, breaks the unitarity of the time evolution. If the Hamiltonian represent some infinitely large system, and we are interested in local observables, and we take the so-called thermodynamic limit of $L\to\infty$ before we take $t\to\infty$ then we can create an effective source of dissipation in the system, that reflect the fact that in order for the superposition to oscillate we will have to wait times longer than $L/v$ (for some $v$ that is related to the density of states), and therefore we avoid it. There are other ways to create dissipation: for example coupling to another (infinite set of) degrees of freedom, and then tracing them out.

In most physical contexts, this is the case, as most systems are not completely isolated, and we are interested in observables and not the state itself.

  • $\begingroup$ Hey, thanks for your answer. One example of why I ask this question is what happens to the new state $\psi(t,x)$ of the nuclei wave function in the BO approximation when an electronic transition occurs. The electronic transition is fast compared to anything else, so the potential energy surface in which the nuclei move will suddenly change. After some time I expect the new system to be somewhere in its new ground state. What does the new system interact with? Or is it indeed just the observables that make it look like as the oscillations describing it are just a mathematical tool!? $\endgroup$
    – Diger
    Commented Apr 28, 2020 at 10:40
  • $\begingroup$ The interaction between the electrons and the nucleus is the physical mechanismby which this relaxation occur. The nucleus wave function will evolve (slowly) and the electrons will respond to that, this is a mechanism of energy transfer between the two subsystems. Now, thinking of the entire system together it will potentially oscillate among the different states of the new Hamiltonian, but eventually it will leak energy outside (through EM radiation or interaction with neighboring nuclei and electrons, that are outside the scope of the BO approx) until it will stabilize near its ground state $\endgroup$
    – user245141
    Commented Apr 28, 2020 at 10:48
  • $\begingroup$ Does emitting EM radiation mean that the system is somehow interacting with some inherent vacuum background of the quantized EM field? Or is it just some random process that can happen any time (more sooner than later)? I mean as long as EM radiation is not emitted, how can it interact with the EM fields? There are no fields yet... $\endgroup$
    – Diger
    Commented Apr 28, 2020 at 13:20
  • $\begingroup$ are you asking how the system behave in a true experimental setup or how do we treat it theoretically within the BO approximation? $\endgroup$
    – user245141
    Commented Apr 28, 2020 at 13:51
  • $\begingroup$ Feel free to answer both interpretations ;) But in particular I was thinking about its theoretical approximative approach... $\endgroup$
    – Diger
    Commented Apr 28, 2020 at 13:52

Your Answer

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

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