# Ground state calculation with a time averaged wave function

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• hilbert2
In summary, the conversation discusses a method for calculating the ground state of a quantum system using real time propagation and time averaging. This approach can potentially provide estimates for higher excited states as well. However, it may not be as efficient as the traditional imaginary time method.
hilbert2
Gold Member
TL;DR Summary
Here I describe an unusual way to calculate the ground state of a quantum system, using real time propagation instead of the usual imaginary time calculation.
An idea I was thinking about for the last few days:

You want to calculate the ground state of some system from the Schrödinger eqn ##\hat{H}\left|\psi\right.\rangle = E\left|\psi\right.\rangle##. One way is to choose a trial state ##\left|\psi (t_0 )\right.\rangle## and use the TDSE to propagate it further in imaginary time to obtain

##\displaystyle\left|\psi (t)\right.\rangle = C_0 e^{-iE_0 t/\hbar}\left|0\right.\rangle + C_1 e^{-iE_1 t/\hbar}\left|1\right.\rangle + C_2 e^{-iE_2 t/\hbar}\left|2\right.\rangle + \dots##

where the ##E_j## are the eigenvalues for the eigenstates of ##\hat{H}##, denoted ##\left|j\right.\rangle## and the initial state at time ##t_0## is

##\displaystyle\left|\psi (t_0 )\right.\rangle = C_0 \left|0\right.\rangle + C_1 \left|1\right.\rangle + C_2 \left|2\right.\rangle + \dots##.

If the time coordinate advances along the imaginary time axis instead of real time, the excited state components will eventually become negligible compared to the ground state component.

Now suppose the initial state is propagated along the real time axis instead, and you also calculate the time average of ##\left|\psi\right.\rangle##

##\overline{\left|\psi (t_0 +\Delta t)\right.\rangle} = \frac{1}{\Delta t}\int\limits_{t_0}^{t_0 + \Delta t} \left|\psi (t')\right.\rangle dt'= \frac{1}{\Delta t}\left[C_0 \frac{i\hbar}{E_0}\left(e^{-iE_0 (t_0+\Delta t)/\hbar}-e^{-iE_0 t_0/\hbar}\right)\left|0\right.\rangle\right.##
##\left.+ C_1 \frac{i\hbar}{E_1}\left(e^{-iE_1 (t_0 +\Delta t)/\hbar}-e^{-iE_1 t_0/\hbar}\right)\left|1\right.\rangle + C_2 \frac{i\hbar}{E_2}\left(e^{-iE_2 (t_0 +\Delta t)/\hbar}-e^{-iE_2 t_0/\hbar}\right)\left|2\right.\rangle + \dots\right]##.

then also this will contain less excited state components than the initial state does. This is because the new multipliers ##\displaystyle\frac{i\hbar}{E_j}## have a smaller value for higher integer value ##j##. An exception is when the new time value ##t_0+\Delta t## is one where the exponential multiplier ##\displaystyle e^{-iE_0 t/\hbar}## has circled an integer number of periods in the complex plane and ##\displaystyle e^{-iE_0 (t_0+\Delta t)/\hbar}-e^{-iE_0 t_0/\hbar} = 0##.

The time evolution simulation can also be stopped after a while, when the expectation value ##\left<\hat{H}\right>## calculated for the average state ##\displaystyle\overline{\left|\psi (t)\right.\rangle}## is at a local minimum value and therefore the time average state is likely to be close to the ground state. Then you set that ##\displaystyle\overline{\left|\psi (t)\right.\rangle}## to be the initial state ##\displaystyle\left|\psi (t_0 )\right.\rangle## in a new simulation, to get even more of the excited components removed.

Now, this kind of calculation doesn't immediately seem to have advantages over the imaginary time method, because it can't be converted to a diffusion monte carlo simulation that could be parallelized. However, if you record the time evolution of ##\left|\psi\right.\rangle## and ##\overline{\left|\psi\right.\rangle}## throughout the calculation, it's possible that you can (from the same information) extract an estimate for the first and second excited state as well. Clearly, in a 1d system with symmetric potential energy ##V(x)##, the wavefuntion corresponding to ##\overline{\left|\psi\right.\rangle}## would be closest to an odd function with only one node approximately when

##\displaystyle e^{-iE_0 (t_0+\Delta t)/\hbar}-e^{-iE_0 t_0/\hbar} = 0##

and the ground state component in ##\overline{\left|\psi\right.\rangle}## vanishes.

Last edited:
hilbert2 said:
TL;DR Summary: Here I describe an unusual way to calculate the ground state of a quantum system, using real time propagation instead of the usual imaginary time calculation.

If the time coordinate advances along the imaginary time axis instead of real time, the excited state components will eventually become negligible compared to the ground state component.
I am not certain that your observation for imaginary time would be also applied to ordinary real time evolution.

It doesn't, but if you calculate the average state(wavefunction) on the time interval ##[t_0 , t_0 + \Delta t]## and have a good enough guess for ##E_0##, you can set the value of ##\Delta t## to about half of the value needed for ##e^{-iE_0 t/\hbar}## to go one circle around the origin of ##\mathbb{C}##. Then that time average is a better approximation for the ground state ##\left|0\right.\rangle## and you can use it as the initial state ##\left|\psi (t_0 )\right.\rangle## in a new round of simulation where you also calculate the time averaged state in addition to the temporary state ##\left|\psi (t)\right.\rangle##. Doing this ##n## times will accumulate multipliers of ##\displaystyle\frac{1}{(E_j )^n}## to each component and those become rapidly smaller for excited state components.

Last edited:

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