Sine Potential - Energy Quantization

Classical Mechanics Level 4

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

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

− d 2 d x 2 Ψ N ( x ) + V ( x ) Ψ N ( x ) = E Ψ N ( x ) \large{-\frac{d^2}{d x^2} \, \Psi_N (x) + V(x) \, \Psi_N (x) = E \, \Psi_N (x)}

The potential varies as follows:

V ( x ) = ∞ x < 0 V ( x ) = sin ⁡ ( x ) 0 ≤ x ≤ π V ( x ) = ∞ x > π V (x) = \infty \,\,\,\,\,\,\,\, x < 0 \\ V (x) = \sin(x) \,\,\,\,\,\,\,\, 0 \leq x \leq \pi \\ V (x) = \infty \,\,\,\,\,\,\,\, x > \pi

What is the smallest allowable value of energy E E for this system?

Hint: The solution to the previous problem shows how to solve this one. My approach is:

1) Evaluate candidate values of E E over a continuum
2) For each value of E E , start with Ψ N ( 0 ) = 0 \Psi_N(0) = 0 and some arbitrary value for ( d d x Ψ N ) ( 0 ) (\frac{d}{dx} \Psi_N)(0)
3) For each E E , numerically integrate to get the value of the wave function at the other end of the well ( Ψ N ( π ) ) (\Psi_N (\pi))
4) The allowable values of E E are those for which Ψ N ( π ) = 0 \Psi_N (\pi) = 0 . These form a discrete set.


The answer is 1.845.

Steven Chase by Steven Chase , 37, USA

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1 solution

Steven Chase
Aug 23, 2019

The attached code implements the procedure outlined in the problem hint. E ≈ 1.845 E \approx 1.845 is the lowest energy level for which Ψ N ( x f ) = 0 \Psi_N (x_f) = 0 . Other allowed energy levels are shown in the diagram (the values of E E at which the graph crosses the horizontal axis).

Also note that it is very simple to evaluate the wave function numerically, even though it would be difficult (or perhaps impossible) to do by hand. 

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import math

xf = math.pi  # End of well

dE = 10.0/(10.0**4.0)  # Energy increment
dx = xf/(10.0**4.0)    # x increment

E = dE

while E <= 10.0:   # sweep E over a range

    Psi = 0.0      # for every E, initialize Psi and Psi derivative
    Psid = 0.5

    x = 0.0

    V = math.sin(x)
    Psidd = (V-E)*Psi

    ###################################

    while x <= xf:    # for every E, numerically integrate to determine Psi at x = xf

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

        V = math.sin(x)
        Psidd = (V-E)*Psi

        x = x + dx

    ###################################

    print E,Psi   # Plot E against the value of Psi at x = xf
                  # Allowed energy levels correspond to Psi = 0 at x = xf

    E = E + dE

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