Finite potential well

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The finite potential well (also known as the finite square well) is a simple problem from quantum mechanics. It is an extension of the infinite potential well, in which a particle is confined to a box, but one which has finite - not infinite - potential walls. This means unlike the infinite potential well, there is a probability associated with the particle being found outside of the box. The quantum mechanical interpretation is unlike the classical interpretation, where if the total energy of the particle is less than potential energy barrier of the walls it cannot be found outside the box. In the quantum interpretation, there is a non-zero probability of the particle being outside the box even when the energy of the particle is less than the potential energy barrier of the walls (because of quantum tunnelling).

1 The particle in a 1-dimensional box
2 See also
3 References

[edit] The particle in a 1-dimensional box

For the 1-dimensional case on the x-axis, the time-independent Schrödinger equation can be written as:

$-\frac{\hbar^2}{2 m} \frac{d^2 \psi}{d x^2} + V(x) \psi = E \psi \quad (1)$

where
$\hbar = \frac{h}{2 \pi}$
$h \,$ is Planck's constant
$m \,$ is the mass of the particle
$\psi\,$ is the (complex valued) wavefunction that we want to find
$V\left(x\right)\,$ is a function describing the potential at each point x and
$E\,$ is the energy, a real number.

For the case of the particle in a 1-dimensional box of length L, the potential is zero inside the box, but rises abruptly to a value Γ at x = -L/2 and x = L/2. The wavefunction is considered to be made up of different wavefuctions at different ranges of x, depending on whether x is inside or outside of the box. Therefore the wavefunction is defined such that:

$\psi = \begin{cases} \psi_1, & \mbox{if }x<-L/2\mbox{ (the region outside the box)} \\ \psi_2, & \mbox{if }-L/2<x<L/2\mbox{ (the region inside the box)} \\ \psi_3 & \mbox{if }x>L/2\mbox{ (the region outside the box)} \end{cases}$

[edit] Inside the box

For the region inside the box V(x) = 0 and Equation 1 reduces to:

$-\frac{\hbar^2}{2 m} \frac{d^2 \psi_2}{d x^2} = E \psi_2$

Letting

$k = \frac{\sqrt{2mE}}{\hbar}$

the equation becomes

$\frac{d^2 \psi_2}{d x^2} = -k^2 \psi_2$

This is a well studied differential equation and eigenvalue problem with a general solution of:

$\psi_2 = A \sin(kx) + B \cos(kx)\quad$

Hence:

$E = \frac{k^2 \hbar^2}{2m}$

Here, A and B can be any complex numbers, and k can be any real number.

[edit] Outside the box

For the region outside of the box, since the potential is constant, V(x) = Γ and Equation 1 becomes:

$-\frac{\hbar^2}{2 m} \frac{d^2 \psi_1}{d x^2} = ( E - \Gamma) \psi_1$

There are two possible families of solutions, depending on whether E is less than Γ (the particle is bound in the potential) or E is greater than Γ (the particle is free).

For a free particle, E > Γ, and letting

$\kappa=\frac{\sqrt{2m(E - \Gamma)}}{\hbar}$

produces

$\frac{d^2 \psi_1}{d x^2} = -\kappa^2 \psi_1$

with the same solution form as the inside-well case:

$\psi_1 = C \sin(\kappa x) + D \cos(\kappa x)\quad$

This analysis will first focus on the bound state, where Γ > E. Letting

$\alpha = \frac{\sqrt{2m(\Gamma - E)}}{\hbar}$

produces

$\frac{d^2 \psi_1}{d x^2} = \alpha^2 \psi_1$

where the general solution is exponential:

$\psi_1 = Fe^{- \alpha x}+ Ge^{ \alpha x} \,\!$

Similarly, for the other region outside the box:

$\psi_3 = He^{- \alpha x}+ Ie^{ \alpha x} \,\!$

Now in order to find the specific solution for the problem at hand, we must specify the appropriate boundary conditions and find the values for A , B , F , G , H and I that satisfy those conditions.

[edit] Finding wavefunctions for the bound state

Solutions to the Schrödinger equation must be continuous, and continuously differentiable. These requirements are boundary conditions on the differential equations previously derived.

In this case, the finite potential well is symmetrical, so symmetry can be exploited to reduce the necessary calculations.

Summarizing the previous section:

where we found $\psi_1, \psi_2 \,\!$ and $\psi_3 \,\!$ to be:

$\psi_1 = Fe^{- \alpha x}+ Ge^{ \alpha x} \,\!$

$\psi_2 = A \sin(kx) + B \cos(kx)\quad$

$\psi_3 = He^{- \alpha x}+ Ie^{ \alpha x} \,\!$

We see that as x goes to $-\infty$ , the F term goes to infinity. Likewise, as x goes to $+\infty$ , the I term goes to infinity. As the wave function must be finite for all x, this means we must set F = I = 0, and we have:

$\psi_1 = Ge^{ \alpha x} \,\!$

and

$\psi_3 = He^{- \alpha x} \,\!$

Next, we know that the overall $\psi \,\!$ function must be continuous and differentiable. In other words the values of the functions and their derivatives must match up at the dividing points:

$\psi_1(-L/2) = \psi_2(-L/2) \,\!$		$\psi_2(L/2) = \psi_3(L/2) \,\!$
$\frac{d\psi_1}{dx}(-L/2) = \frac{d\psi_2}{dx}(-L/2) \,\!$		$\frac{d\psi_2}{dx}(L/2) = \frac{d\psi_3}{dx}(L/2) \,\!$