Quantum Well - Squared Cosine Potential

Classical Mechanics Level pending

A particle in a one-dimensional quantum well is governed by a variant of the time-independent Schrodinger equation, expressed in terms of wave function $\Psi (x)$ .

The quantities $V$ and $E$ are the potential energy and total energy, respectively.

$-\frac{d^2}{d x^2} \, \Psi (x) + V(x) \, \Psi (x) = E \, \Psi (x)$

The potential varies as follows:

$V (x) = \infty \,\,\,\,\,\,\,\, x < -\pi \\ V (x) = \cos^2 x \,\,\,\,\,\,\,\, -\pi \leq x \leq \pi \\ V (x) = \infty \,\,\,\,\,\,\,\, x > \pi$

The boundary conditions on $\Psi (x)$ are:

$\Psi (x) = 0 \,\,\,\,\,\,\,\, x \leq -\pi \\ \Psi (x) = 0 \,\,\,\,\,\,\,\, x \geq \pi$

Let $E_1$ , $E_2$ , and $E_3$ be the lowest, second lowest, and third lowest allowable values of $E$ , subject to the constraint $E > 0$ . Determine the following ratio:

$\frac{E_3 - E_2}{E_2 - E_1}$

Note: Only apply the Schrodinger equation within the finite potential region

The answer is 2.87.

2 solutions

Karan Chatrath
Apr 7, 2020

This problem was treated purely numerically. Steps performed are as follows:

A value of $E=0.1$ was initially considered.
The ODE was solved numerically assuming an arbitrary value for $\frac{d \Psi_N}{dx}(-\pi)$
The solution was plotted and the value of $\Psi_N(\pi)$ was computed.
The Energy was incrementally increased and the above steps were performed again till such time $\Psi_N(\pi) \approx 0$ was satisfied. This is the other boundary condition. The values of energy for what this happened are the allowable values of energy for this system.
It was also observed that the energy values are independent of initial derivative.

A plot showing the wave functions at different allowable energy levels is as follows:

Steven Chase
Apr 7, 2020

I used the same procedure outlined by @Karan Chatrath . What we want to do is make a graph of $E$ vs. the value of $\Psi$ at the right end of the well, assuming that we initialize properly at the left end of the well for each value of $E$ . The allowable values of $E$ are those for which the graph crosses the horizontal axis. The code for making the graph is attached.

import math

x1 = -math.pi
x2 = math.pi

Emax = 10.0

dE = Emax/(10.0**4.0)
dx = (x2-x1)/(10.0**4.0)

E = dE

print "E Psi(x2)"

while E <= Emax:  # sweep E over a range

    Psi = 0.0
    Psid = 1.0   # initial derivative chosen arbitrarily

    x = x1       # for each E, initialize at left end of well

    V = (math.cos(x))**2.0

    Psidd = (V-E)*Psi

    while x <= x2:  # for each E, determine Psi(x2) at right well end
                    # by integrating from left well end

        Psi = Psi + Psid * dx
        Psid = Psid + Psidd * dx

        V = (math.cos(x))**2.0

        Psidd = (V-E)*Psi

        x = x + dx

    print E,Psi # plot E vs. Psi(x2)

    E = E + dE

Very nice! From the looks of the plot, the gap between successive allowable energies increases as energy increases.

Karan Chatrath - 1 year, 2 months ago

Indeed. I was thinking back to the simple potential-free textbook quantum well. And the energy there is proportional to $n^2$ , where $n = 1,2,3, etc.$ . The gaps between energy levels increase there too, as a result of the $n^2$ dependence. Just for fun, I ran this same code with $V = 0$ and reproduced the results for the classic quantum well, showing the $n^2$ dependence. The ratio in that case is $\frac{9 - 4}{4 - 1} = \frac{5}{3}$ .

Steven Chase - 1 year, 2 months ago

Interesting. Running the $V=0$ case is a good way of cross-checking one's code cause the results are well known. I say this especially since in the last problem, my code had some errors.

Karan Chatrath - 1 year, 2 months ago

Quantum Well - Squared Cosine Potential

The answer is 2.87.

2 solutions

0 pending reports