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 Ψ ( x ) \Psi (x) .

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

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

The potential varies as follows:

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

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

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

Let E 1 E_1 , E 2 E_2 , and E 3 E_3 be the lowest, second lowest, and third lowest allowable values of E E , subject to the constraint E > 0 E > 0 . Determine the following ratio:

E 3 E 2 E 2 E 1 \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.

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2 solutions

Karan Chatrath
Apr 7, 2020

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

  • A value of E = 0.1 E=0.1 was initially considered.
  • The ODE was solved numerically assuming an arbitrary value for d Ψ N d x ( π ) \frac{d \Psi_N}{dx}(-\pi)
  • The solution was plotted and the value of Ψ N ( π ) \Psi_N(\pi) was computed.
  • The Energy was incrementally increased and the above steps were performed again till such time Ψ N ( π ) 0 \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 E vs. the value of Ψ \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 E . The allowable values of E E are those for which the graph crosses the horizontal axis. The code for making the graph is attached.

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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

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

Steven Chase - 1 year, 2 months ago

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Interesting. Running the V = 0 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

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