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 ) .
The quantities V and E are the potential energy and total energy, respectively. E is a positive constant.
− d x 2 d 2 Ψ ( x ) + V ( x ) Ψ ( x ) = E Ψ ( x )
The potential energy varies as follows:
V ( x ) = ⎩ ⎪ ⎨ ⎪ ⎧ ∞ , x 2 , ∞ , x < − π − π ≤ x ≤ π x > π
The boundary conditions on Ψ ( x ) are:
Ψ ( x ) = { 0 , x ≤ − π 0 , x ≥ π
Let E 1 be the smallest allowable non-zero value of E , and let E 2 be the second smallest value. What is E 2 − E 1 ?
Note: You will likely need a computer to solve this. This is essentially a quantum harmonic oscillator with additional boundary constraints.
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@Steven Chase Take a look on last 1 hour notification
I like the way you present your solution in such a beautiful way.
I am reading up how to solve the given differential equation analytically as closed form solutions are available, but to solve this problem, I relied on numerical methods.
d x 2 d 2 Ψ = ( x 2 − E ) Ψ Ψ ( 0 ) = 0 ; d x d Ψ ( 0 ) = 1
The initial derivative of Ψ ( x ) can be taken as anything. The solution is simply scaled in proportion to the initial derivative. E is varied in small increments and the allowable energies are those for which:
∣ Ψ ( π ) ∣ < ϵ Where ϵ is a small numeric tolerance like 1 0 − 3 , for example. Essentially, we search for the solutions which satisfy the non-imposed boundary condition Ψ ( π ) = 0 . The following plot illustrates the first five allowable solutions for the quantum harmonic oscillator.
I also do not know how to interpret when the quantum solution starts to agree with the classical solution. How would I visualize such an agreement? In classical oscillators, there isn't any notion of a 'wave function'.
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@Karan Chatrath has explained how to tackle this problem conceptually. I like to make a plot of Ψ ( π ) vs. E . The energies at which the graph crosses the horizontal axis ( Ψ ( π ) = 0 ) are the allowed energies, given initial conditions (for each E ) of Ψ ( − π ) = 0 and d x d Ψ ( − π ) = 1 .
Since this problem was formulated in terms of the time-independent S E , there is no time-evolution of the wave function. The wave function for the lowest energy state E 1 is symmetric about x = 0 and it has a single peak at x = 0 , meaning that the particle is most likely to be found near the origin, where the potential is minimized. This seems fairly analogous to the low-energy classical picture. A low-energy classical oscillator has to remain near the location of minimum potential. At high energy, a classical oscillator moves back and forth over a wide range, and is not confined to a particular location. With high E , the quantum wave function has many peaks spread throughout the well.
I have plotted Ψ 2 ( x ) for a high value of E ( E = 9 3 . 5 6 ) . You can see that the probability density is more uniform. On a large scale, there is no region of the well that is strongly preferred over any other. However, there is a slight preference for the high-potential regions. I think this is because a classical oscillator has low speed at high potential, and high speed at low potential. Therefore, the classical oscillator spends more of its time in the high potential regions.