I'm currently simulating a one-dimensional double Gaussian well potential numerically, and have been asked to find parameters corresponding to the overlap in potential and wave functions from my determined bound energy states.

Using the time-independent Schrödinger equation along with a potential of the form

$$ V(x) = -V_0 \left( e^{-(x-x_0)^2} + e^{-(x+x_0)^2} \right), $$

for some $V_0$ and $x_0$, I'm successfully able to simulate the system and get the bound states, but I've also been asked to use a supposed generalisation of the Hamiltonian in matrix form:

$$\begin{pmatrix} \epsilon_0 - A & W\\ W & \epsilon_0 - A\end{pmatrix}, $$

(where $\epsilon_0$ is the bound state energy at infinite separation, $A$ represents the overlap in potential wells, and $W$ represents the overlap of the wave functions in the two wells), with eigenvalues $\epsilon_{1,2} = \epsilon_0 - A \pm W$.

My task is to determine $W$. I assumed at first it was as easy as taking the difference of the first two bound states (ground and first excited) and dividing that by two, but supposedly $W$ is supposed to vary with energy states (which makes sense, since higher energy wave functions have larger width), but I honestly do not understand how to determine this as a function of energy level - for the system I'm analysis, there are a total of three bound states, and my supposed method would only yield the answer for two of them. Could the first two be the same, and the third different?

  • $\begingroup$ This looks like it falls into the "homework-and-exercises" category, so I'm adding that tag. However, your question seems quite a reasonable one at first glance. I would have expected this kind of analysis to be applied to a system with a higher barrier, and smaller overlap between the wells. Then, there would be several pairs of states lying below the barrier energy, and the question would make more sense. So, is this just a question of choosing more suitable values of $x_0$ and $V_0$? $\endgroup$
    – user197851
    Commented Dec 30, 2018 at 11:07
  • $\begingroup$ @LonelyProf The specific task had pre-determined values $V_0 = 4.0$ and $x_0 = 1.1, 1.3, & 1.5$ (in rewritten, dimensionless units), all of which yield three bound state - the first two of which are incredibly close in energy. Would a deeper well yield more of these pairs, centered around the corresponding bound states of the single well? $\endgroup$
    – shaqman
    Commented Dec 30, 2018 at 11:24

1 Answer 1


In the spirit of not putting answers in the comments section, I will offer my suggestion here.

It seems that the given parameters only produce three bound states, of which the lowest two are a close pair (the energy gap being associated with tunelling).

The first question is "how many bound states are there for the single-well Gaussian potential". It's most likely a finite number (see e.g. S Nandi Amer J Phys, 78, 1341 (2010) also available as a preprint). This will give an upper limit to the number of tunnelling state pairs that one can see. If there are only two or three single-well bound states, it may be necessary to adjust $V_0$ to increase the number, to make it possible to go on and investigate tunneling splitting as a function of energy.

Then I would tend to experiment by increasing $x_0$ beyond the values quoted in the OP comment, so as to increase the barrier height and width without affecting the single-well energies. Then, yes, I would expect to see more of the single-state energies combining to give those tunneling pairs. For large barrier height and width, the interaction between the wells will be just a small perturbation, producing very close pairs of symmetric and antisymmetric combinations of states.

Off the top of my head, I don't know of a reliable way to predict the dependence of the number of tunneling state pairs on these two parameters. To a first approximation, we need the barrier height to be larger than the corresponding single-well energy, of course. (There are plenty of methods for approximately predicting the splitting between pairs).

  • $\begingroup$ This gave me enough background to finish the rest of the exercise, thank you! $\endgroup$
    – shaqman
    Commented Dec 30, 2018 at 12:49

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