Do definite energy states exist for a single particle when its potential itself changes with time?

I tried solving it and the equations seem to show that they do not exist. But then i am confused as to what energies will be observed when it is measured. What does the expectation value of energy mean in this case?

As a specific example, consider a 1D infinite potential well with $V(x,t)=t$. What energies are observed and with what probability when the system's energy is measured?


2 Answers 2


In this case, the eigenstates of the Hamiltonian are not useful to solve the problem, and one has to work with the Schrödinger equation directly:

$$ i\hbar \, \partial_t \psi(x,t)=\frac{-\hbar^2}{2m}\partial_x^2\psi(x,t)+P\,t\,\psi(x,t) $$ Using a Fourier transform in the variable $x$ you can show that the general solution is $$ \psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \hat{\psi}(k,0)\,\exp\left(\frac{-i\hbar\,k^2\,t}{2m} - \frac{i\,P\,t^2}{2\hbar}+ikx \right)\,dk $$ where $\hat{\psi}(k,0)\in L^2(\mathbb{R})$ is the Fourier transform of the initial condition $\psi(x,0)$. Note that the kernel of the integral operator $$ \exp\left(\frac{-i\hbar\,k^2\,t}{2m} - \frac{i\,P\,t^2}{2\hbar}+ikx \right) $$ is a eigenfuction of the Hamiltonian $\hat{H}(t)$ for all $t$ with eigenvalue $$ E(t) = \frac{\hbar^2k^2}{2m} + Pt $$ This is because $[\hat{H}(t),\hat{H}(t')]=0$ for different times $t$ and $t'$. When you make a measurement at time $t$ your eigenfunction will collapse to the instantaneous eigenvector of $\hat{H}(t)$. Suppose now you have an eigenvector $\psi(x,t_0)$ of $\hat{H}(t_0)$ with eigenvalue $E(t_0)$, how will it evolve? Using the formula from before I get $$ \psi(x,t) = \psi(x,t_0) \, \exp\left(\frac{-i\,E(t_0)\,(t-t_0)}{\hbar} -\frac{i\,P\,(t-t_0)^2}{2\hbar} \right) $$ So $\psi$ will be an instantaneous eigenvector of $\hat{H}(t)$ for all times. In this sense the state is "stationary", since you will always find the state of the system in the corresponding eigenvector, but the value of energy I measure will depend on time since the eigenvalues of the Hamiltonian depend on time too. I'll like to add that for time dependent Hamiltonians with the property $[\hat{H}(t),\hat{H}(t')]=0$ the evolution operator can be written $$ \hat{U}(t_1,t_0) = \exp\left(-\frac{i}{\hbar} \int_{t_0}^{t_1} \hat{H}(t) \, dt \right) $$

  • $\begingroup$ Thanks for the answer, it has made this a little less confusing.Also, can eigenvalues vary with time?I thought they had to be constant?Does it suffice if it is just a number, even if it is not a constant? $\endgroup$
    – SpikeCB
    Sep 7, 2014 at 17:29
  • $\begingroup$ yes, the eigenvalues of the hamiltonian can vary with time, since the operator itself vary with time. It's not a problem at all, the thing is that the value you measure of energy will be different at different times, but will always be eigenvalues of the hamiltonian you have at that time. $\endgroup$
    – Héctor
    Sep 7, 2014 at 17:32
  • $\begingroup$ In classical mechanics when you have a time depend hamiltonian, the energy vary with time and you have no regrets. This is the quantum analogue of that. $\endgroup$
    – Héctor
    Sep 7, 2014 at 17:37

For a particle in a time dependant external potential any Hermitian the Hamiltonian will still have a complete set of eigenstates, and the corresponding eigenvalues will still be the allowed energies of the system. These energies and eigenstates will however generally be time dependant and this means that they are of far less use for solving the TDSE.

For example in the infinite-square-well-with-rising-bottom that you proposed it is pretty simple to show that, for $0< x < L$ \begin{align} \hat{H}\alpha \sin(kx) & = \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + Pt \right]\alpha\sin(kx)\\ & = \left[\frac{\hbar^2}{2m}k^2 + Pt\right]\alpha\sin(kx) \end{align} So $\alpha\sin(kx)$ is an eigenstate of $\hat{H}$ with energy $E(t) = \frac{\hbar^2}{2m}k^2 +Pt$ and $k$ chosen to satisfy the conditions at the edge of the well. However if we try to plug this into the TDSE, writing $\psi(x) = \alpha \sin(kx)$, we find that \begin{align} \imath\hbar \frac{\partial}{\partial t} e^{-\imath \frac{E(t)}{\hbar}t}\psi(x) & = \left(E+\dot{E}t\right)e^{-\imath \frac{E}{\hbar}t}\psi(x) \\ & = \left[\hat{H}+Pt\right]e^{-\imath \frac{E}{\hbar}t}\psi(x) \end{align} So the states of definite energy are not stationary states of the system, as they must have a more complex time dependence than $e^{-\imath \frac{E}{\hbar}t}$.

In general this type of problem cannot be solved analytically. There are results for the limiting behavior for very rapid or very slowly changing potentials but normally these problems are solved using numerical techniques or time dependant perturbation theory.

Also it should be noted that a problem with a time varying potential should not be confused with the Heisenberg picture of QM, which is equivalent to the Schrodinger picture, but with the time dependence moved from the state vectors to the operators and is essentially a change of basis.

  • $\begingroup$ So what happens when energy is measured?Since energy depends in some sense on time(i am still wrapping my head around your answer), does'nt this mean there are no 'definite' energy states, as i recall that stationary and definite energy states mean the same thing?Thanks for the answer, btw.One of the students for whom i am a TA asked me this after the tutorial and i was not sure what happens. $\endgroup$
    – SpikeCB
    Sep 7, 2014 at 16:19
  • $\begingroup$ A stationary state is a state in which no observables change with time, only the phase of the state. A definite energy state is an eigenstate of the Hamiltonian, in exactly the same way that a definite state of any other observable is a eigenstate of its respective operator. If the Hamiltonian is time independent then it follows from the TDSE that a state is stationary if and only if it has a definite energy. For time varying potentials this may not be the case. Definite energy states always exist because the Hamiltonian is Hermitian but I see no reason to assume that stationary states exist. $\endgroup$ Sep 7, 2014 at 16:56
  • $\begingroup$ I have trouble accepting the statement that when E itself is a function of 't', the corresponding state can be called an eigenstate of the hamiltonian, as the eigenvalue itself is not a constant but varies with time.Could you explain this or recommend me a source?Also, i believe the time dependent part of the wavefunction can be calculated by taking the "integral of energy with time" instead of 'Et' in the exponential part, correct me if i am wrong. $\endgroup$
    – SpikeCB
    Sep 7, 2014 at 17:16
  • $\begingroup$ Time is not a quantum mechanical quantity; there is no time operator. Even with a time varying potential, the Hamiltonian is still a purely spatial operator. It is simply a different spatial operator at different times with correspondingly different eigenvalues. Its also worth noting that a system with a time varying potential is necessarily not isolated. Something external is doing work on the system, and this would be expected to change the eigenvalues. And yes you could find the time dependence of definite energy states by integrating the energy. That may not always be practical, however. $\endgroup$ Sep 7, 2014 at 17:50
  • $\begingroup$ That was exactly what i was thinking might be possible, that a time varying potential must mean that the system is not isolated and hence the energy might(must?) change with time.Thanks for validating this. $\endgroup$
    – SpikeCB
    Sep 7, 2014 at 17:54

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