I realize that the imaginary evolution could help us to find the ground state for a system. However, I very puzzled why it works, and what the principle is back up there? I have done some searching on this, but all I could find was people using this but not talking about why. Please explain this part to me in detail.


1 Answer 1


Let's start with the usual time evolution in quantum mechanics to set the scene. It is governed by Schrödinger's equation (limiting the discussion to 1D for simplicity, it is trivial to extend the argument to higher dimensions): $$ i\hbar\frac{\partial\psi(x,t)}{\partial t}=\hat{H}\psi(x,t) $$ When the Hamiltonian is time independent (i.e. when the potential is time independent), then it is straight-forward to solve the above equation to find the time dependence of $\psi(x,t)$. What you need to do is first solve the eigenvalue equation for the Hamiltonian: $$ \hat{H}\psi_n(x)=E_n\psi_n(x), $$ where $\psi_n(x)$ are the eigenstates and $E_n$ the energy eigenvalues. Second, you need to expand the wave function at an initial time (say $t=0$) in terms of the energy eigenstates: $$ \psi(x,0)=\sum_nc_n(0)\psi_n(x), $$ where $c_n(0)$ are the expansion coefficients that you can find by calculating the overlap between $\psi(x,0)$ and the energy basis eigenstates. Third, the wave function at a later time $t$ is given by: $$ \psi(x,t)=\sum_nc_n(0)e^{-iE_nt\hbar}\psi_n(x). $$ The time dependence is such that each energy eigenstate $\psi_n(x)$ "oscillates" at a frequency proportional to the corresponding energy eigenvalue $E_n/\hbar$.

Next, going back to your question, let's consider a change of variables $\tau=it$. You can think of $\tau$ as "imaginary time". Applying this change of variables to the Schrödinger equation we obtain: $$ -\hbar\frac{\partial\psi(x,\tau)}{\partial\tau}=\hat{H}\psi(x,\tau). $$ Again, as $\hat{H}$ is time independent, the dependence on $\tau$ can be solved in the same way in which the dependence on $t$ was solved above, and we obtain: $$ \psi(x,\tau)=\sum_nc_n(0)e^{-E_n\tau/\hbar}\psi_n(x). $$ You can now see that the function $\psi(x,\tau)$ at imaginary time $\tau$ is no longer obtained by an "oscillating" superposition of energy eigenstates, but instead by an "exponentially decaying" superposition of energy eigenstates. Furthermore, the exponential decay rate is proportional to $E_n/\hbar$.

What have we accomplished by making this variable change from $t$ to $\tau$? Consider the limit of large $\tau$: $$ \psi(x,\tau\gg1)\simeq c_0(0)e^{-E_0\tau}\psi_0(x). $$ In this limit, the ground state $n=0$ is "projected out" of the initial state, because the corresponding exponential decay is the slowest one. Therefore, by evolving the system in "imaginary time", we can obtain the ground state of the Hamiltonian $\psi_0(x)$ as the long imaginary time limit.

Will this always work? This will only work if when you expand the initial state in terms of energy eigenstates, there is some contribution by the ground state. If not, the long imaginary time evolution will instead lead to the lowest energy state present in the initial expansion.

An area in which imaginary time evolution is used is in one of the most accurate computational methods to solve the Schrödinger equation for solids, diffusion quantum Monte Carlo. In this method, the imaginary time Schrödinger equation is solved stochastically as a diffusion equation, and the ground state of the system is projected out.

  • 4
    $\begingroup$ This is very clear and extremely helpful! Thank you so much! $\endgroup$
    – Noah Ren
    Commented Jun 8, 2020 at 8:04
  • 1
    $\begingroup$ Very nice, I haven't read this explanation in any of the other dozens answers about imaginary time on this site that just basically answer with another equation. It's an interesting trick with the t=inf*i limit. I wish there was some more insight into the arguments about partition functions that creep up in these discussions as well.. $\endgroup$
    – BjornW
    Commented Oct 8, 2021 at 20:55

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.