Finite Quantum Well (Part 1)

Classical Mechanics Level 5

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:

$V (x) = 1 \,\,\,\,\,\,\,\, x < -1 \\ V (x) = 0 \,\,\,\,\,\,\,\, -1 \leq x \leq 1 \\ V (x) = 1 \,\,\,\,\,\,\,\, x > 1$

Assume that the wave function must satisfy the following constraints:

1) The wave function and its spatial derivative must be continuous for all $x$
2) The integral of the squared modulus of the wave function over all space must be finite
3) The wave function must be even (ex: $\Psi(x) = \Psi(-x))$

Given the constraints, what is the smallest allowable energy $E$ for this system?

Note: It also looks to me like this is the only $E$ value which satisfies the constraints

The answer is 0.546247.

1 solution

Mark Hennings
Sep 21, 2019

Note that $\Psi'' = (1-E)\Psi$ for $|x| > 1$ . Since $\Psi$ is square-integrable, we deduce that $E < 1$ , so that $E = 1-k^2$ for some $k > 0$ . But then $\Psi'' = k^2\Psi$ for $|x| > 1$ , while $\Psi'' + (1-k^2)\Psi = 0$ for $|x| < 1$ . Since $\Psi$ is even, we must have $\Psi(x) \; = \; \left\{ \begin{array}{lll} Ae^{-k|x|} & \hspace{1cm} & |x| > 1 \\ B\cos(\sqrt{1-k^2}x) & & |x| < 1 \end{array}\right.$ Since both $\Psi$ and $\Psi'$ must be continuous, we deduce that $Ae^{-k} \; = \; B\cos(\sqrt{1-k^2}) \hspace{2cm} kAe^{-k} \; = \; B\sqrt{1-k^2} \sin(\sqrt{1-k^2})$ and these conditions allow for nonzero solutions $A,B$ provided that $0 < k < 1$ and $k \; = \; \sqrt{1-k^2}\tan(\sqrt{1-k^2})$ so that $0 < E < 1$ and $\sqrt{1-E} \; = \; \sqrt{E}\tan\sqrt{E}$ Solving this equation numerically, we find that the only solution to this equation is $E = \boxed{0.546247}$ .

I analyzed the equation in a similar way and deduced that the form of the solution outside the interval $-1 \le x\le 1$ must be a decreasing exponential. Matching the boundary conditions at the point of potential discontinuity leads to the answer. This bit is fine. What I did was go the numerical route as well. I tried an arbitrary set of boundary conditions (I would have said initial conditions if the independent variable was time) and solved the equation using the value of energy obtained and I see that the wave function diverges.

I think this is because the general solution to the equation when $E<1$ and when $\mid x \mid >1$ is:

$\psi = c_1 e^{(1-E)x} + c_2 e^{-(1-E)x}$

Assuming that the general solution is only a decreasing exponential just does not seem to be a very rigorous approach to the problem. I would like to know your thoughts on this @Mark Hennings @Steven Chase

Karan Chatrath - 1 year, 8 months ago

We are looking fo what are called bound state solutions , namely ones for which the function $\Psi$ is square-integrable. Square-integrability tells us that the most general solution of the differential equation must be of the form $\Psi = \left\{\begin{array}{lll} A e^{-\sqrt{1-E}x} & \hspace{1cm} & x>1\\ B \cos\sqrt{E}x + C\sin\sqrt{E}x & & -1<x<1\\ D e^{\sqrt{1-E}x} && x<-1 \end{array}\right.$ with $E<1$ . If $E>1$ then $\Psi$ would be oscillatory for large $x$ , and we would not get square-integrability. We could apply continuity of $\Psi$ and its derivative at $1$ and $-1$ to find the relationships between $A,B,C$ and $D$ , and also derive an equation that $E$ must satisfy so that the only solution is not just $A=B=C=D=0$ .

This question makes things a bit easier by looking for even solutions.

However, the important assumption is square-integrability, which is a standard requirement of QM.

A different family of solutions would occur when $E>1$ , and $\Psi$ would be linear combinations of $e^{\pm ix\sqrt{E-1}}$ for $|x|>1$ . These equations would model beams of electrons of fixed momentum either being reflected by or transmitted through the potential well of the problem.

Mark Hennings - 1 year, 8 months ago

@Mark Hennings Sir can you please upload the solution of https://brilliant.org/problems/two-rings-attracting/ with a understandable explanation as always. Please sir. Your solution of every complex question impresses me. Please?

A Former Brilliant Member - 1 year, 7 months ago

Indeed, as Mark said, the growing exponential term outside the well has to be thrown out to ensure that the integral of $|\Psi|^2$ is finite

Steven Chase - 1 year, 8 months ago

Thank you for the comments. My knowledge of QM is low but these exercises are helping me to build on it. So is there a way to numerically ensure that the solution has a finite L2 norm? I ask this as I could not validate the result by simply numerically solving the diff equation.

Karan Chatrath - 1 year, 8 months ago

For the infinite well, we could disregard the requirement for continuity in the derivative, since that would force trivial solutions ( $\Psi = 0$ everywhere). It was also virtually guaranteed that the integral of $|\Psi|^2$ would be finite, due to the finite integration interval. So the brute numerical methods worked very well for that case. Here, I believe it is much more difficult to brute search for a solution. Consequently, I did most of this one analytically.

Steven Chase - 1 year, 8 months ago

@Steven Chase – Solving for bound state solutions and/or electron beam solutions for piecewise constant potentials is a standard part of any course in QM...

Mark Hennings - 1 year, 8 months ago

Finite Quantum Well (Part 1)

The answer is 0.546247.

1 solution

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