Solution for a Single Particle in an Infinite Quantum Well - Darin

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The particle in a box problem is a classic way of understanding yet another difference between the classical and quantum worlds. Imagine a box whose sides represent infinite potential, so no particle has enough energy to climb them and escape. Now say we put in an ordinary ball with some initial velocity and it bounces between the walls. The ball is no more likely to be in one place than another. But by solving the Schrödinger equation we'll see that a quantum particle isn't so egalitarian.
<a title="Benjamin D. Esham, Public domain, via Wikimedia Commons" href=""></a><img width="256" alt="Infinite potential well" src="">

Mathematical Setup

[math]\displaystyle{ \frac{-\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi = E \Psi }[/math]
To describe the state of the quantum particle in the potential well we will find a function [math]\displaystyle{ \Psi }[/math] that satisfies the barrier conditions of the well and the Schrödinger equation (above), a differential equation.

Constructing [math]\displaystyle{ \Psi (x) }[/math]

For the region [math]\displaystyle{ x \lt 0 }[/math] the wave function must be zero because the particle cannot overcome an infinite potential barrier. Similarly for [math]\displaystyle{ x \gt L }[/math] .
Now, we just need to construct the solution/wave-function [math]\displaystyle{ \Psi }[/math] for the region [math]\displaystyle{ 0 \lt x \lt L }[/math] where the potential is zero. Let's rewrite the Schrödinger equation a bit to see what we're dealing with.
[math]\displaystyle{ \frac{\partial^2 \Psi}{\partial x^2} = - \frac{2m}{\hbar^2} E \Psi }[/math]
We can clearly see that a second derivative of [math]\displaystyle{ \Psi }[/math] is just a constant times the same function [math]\displaystyle{ \Psi }[/math]. This should remind you of sins and cosines or exponentials, such as
[math]\displaystyle{ sin(kx) }[/math] or [math]\displaystyle{ cos(kx) }[/math] where [math]\displaystyle{ k = \frac{2m}{\hbar^2} }[/math]. You could use complex exponentials instead of sins and cosines as well, but it turns out in this situation sins and cosines are slightly more convenient. A good exercise would be to use complex exponentials ([math]\displaystyle{ \Psi (x) = A e^{ikx} + B e^{-ikx} }[/math]) and verify that you get the right answer. Remember that differential equations may have several particular solutions and the general solution is a super position of them . So, [math]\displaystyle{ \Psi (x) = A sin(kx) + B cos(kx) }[/math], but this doesn't satisfy the boundary conditions. Plug in 0 and L and see that [math]\displaystyle{ \Psi }[/math] is not zero as it should be. This is because quantum particles have discrete energy levels so there is one more condition we must take into account, which is that for sin, [math]\displaystyle{ k = \frac{n \pi}{L} }[/math] and for cos, [math]\displaystyle{ k = \frac{n \pi}{2L} }[/math]. Plug in the boundary conditions to verify that [math]\displaystyle{ \Psi(0) = \Psi(L) = 0 }[/math]. Therefore, [math]\displaystyle{ \Psi (x) = A sin(\frac{n \pi x}{L}) + B cos(\frac{n \pi x}{2L}) }[/math].

We can find A and B by focusing on the region where [math]\displaystyle{ x = 0 }[/math]. [math]\displaystyle{ \Psi (0) = B = 0 }[/math], so [math]\displaystyle{ \Psi (x) =A sin(\frac{n \pi x}{L}) }[/math]. We also know that the wave function must be normalized.
[math]\displaystyle{ 1 = \int_{0}^L |\Psi|^2 dx = \int_{0}^L |A sin(\frac{n \pi x}{L})|^2 dx = \int_{0}^L A^2 sin^2(\frac{n \pi x}{L}) dx = A^2 [\frac{x}{2} - \frac{cos(\frac{2n \pi x}{L})}{4}]_{0}^L = A^2 (\frac{L}{2} - \frac{cos(2n \pi)}{4}) = A^2 (\frac{L}{2}) = 1 }[/math]. Therefore, [math]\displaystyle{ A = \sqrt{\frac{2}{L}} }[/math] and [math]\displaystyle{ \Psi = \sqrt{\frac{2}{L}} sin(\frac{n \pi x}{L}) }[/math] for n = 1, 2, 3, ... And this is the final normalized wave equation for a quantum particle in a 1D infinite potential well.

Note: We could have plugged in [math]\displaystyle{ x = L }[/math] at the beginning instead of 0, but we would be left with the equation [math]\displaystyle{ 0 = 0 }[/math], which doesn't tell us anything new (I hope).

Another Example

Imagine a semi-infinite well in which one wall is a finite potential barrier and another is infinite. What would be the wave-function of a quantum particle in this potential well?
<img src="">
You have all the tools to do this, but the solution is below for you to check your answer.


Region x < 0:
[math]\displaystyle{ \Psi_1 = 0 }[/math]

Region 0 < x < L:
[math]\displaystyle{ \frac{\partial^2 \Psi}{\partial x^2} = - \frac{2m}{\hbar^2} E \Psi, k = \frac{2m}{\hbar^2} E, \Psi_2 = A sin{kx} + B cos{kx}. }[/math]
Region x > L:
[math]\displaystyle{ \frac{\partial^2 \Psi}{\partial x^2} = \frac{2m}{\hbar^2} (V_0 - E) \Psi, q = \frac{2m}{\hbar^2} (V_0 - E), \Psi_3 = C e^{-iqx} }[/math]
Continuity: [math]\displaystyle{ \Psi_1 (0) = \Psi_2 (0) =\gt 0 = B }[/math]
[math]\displaystyle{ \Psi_2 (L) = \Psi_3 (0L) =\gt A sin{kL} = Ce^{-iqL} }[/math]
Smoothness: [math]\displaystyle{ \frac{d \Psi_2}{dx} (L) = \frac{d \Psi_3}{dx} (L) =\gt Ak cos{kL} = -iqC e^{-iqL} =\gt Ak cos{kL} = -iqAsin{kL} =\gt tan{kL} = \frac{ik}{q} }[/math]


The particle in a box problem is a classic way of understanding yet another difference between the classical and quantum worlds. We saw that while a classical particle would be no more likely to be at one position over another, a quantum particle indeed is more likely to be found at certain places in the potential well.


Devices exploiting quantum wells are especially prominent in optoelectronics. the quantum-well problem interested physicists because it is one of the only problems in quantum mechanics that can be solved analytically without approximations.

See also

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