User:Crazyjimbo/Draft of Finite potential well

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-Dimensional Finite Potential Well


For the 1-dimensional case on the x-axis, the potential of the finite square well is


 * $$V(x) = \begin{cases}

-V_0, & \textrm{for } -a \le x \le a, \\ 0, & \textrm{for } |x| > a, \end{cases}$$

where a and V0 are positive constants. This potential admits both bound states and scattering states depending on whether E > 0 or E < 0.

Bound States
Bound states occur when E < 0. To solve the Schrödinger equation for this potential, the areas to the left of the well, within the well and to right of the well must be considered separately.

Left of the Well
To the left of the well, where x < -a, the potential is zero and the time independent Schrödinger equation reduces to
 * $$-\frac{\hbar}{2m} \frac{d^2 \psi}{dx^2} = E \psi. \qquad \text{(1)}$$

Setting
 * $$k = \frac{\sqrt{-2mE}}{\hbar},$$

where k is positive since E < 0, the time independent Schrödinger equation can be written as
 * $$\frac{d^2 \psi}{dx^2} = k^2 \psi.$$

This is a well studied differential equation and eigenvalue problem with a general solution of
 * $$\psi(x) = Ae^{-kx} + Be^{kx}$$

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

If this solution is to represent a real world particle it must be normalisable and since e-kx goes to infinity as x goes to infinity in the negative direction, B must be zero. The physically admissible solution to equation (1) is then
 * $$\psi(x) = Ae^{kx}$$

Inside the Well
When -a < x < a, the potential is given by V(x) = V0 and time independent Schrödinger equation is
 * $$-\frac{\hbar}{2m} \frac{d^2 \psi}{dx^2} -V_0\psi = E \psi. \qquad \text{(2)}$$

Setting
 * $$k = \frac{\sqrt{2m(E + V_0)}}{\hbar},$$

the time independent Schrödinger equation can be written as
 * $$\frac{d^2 \psi}{dx^2} = -l^2 \psi.$$

Note that l is real since E > Vmin = -V0 and thus E + V0 > 0.

This equation has a general solution of
 * $$\psi(x) = C \sin(lx) + D \cos(lx),$$

where C and D can be any complex numbers.

Right of the Well
By similar treatment to left of the well, when x > a, the physically admissible solution to equation (1) is
 * $$\psi(x) = Fe^{-kx}$$

Solution
The potential is an even function so the full solutions are either even or odd. For the even solutions, the solution inside the well will be $$\psi(x) = D \cos(lx)$$ and the full solution given by:


 * $$\psi(x) = \begin{cases}

F e^{-kx}, & \textrm{for } x > a, \\ D \cos(lx), & \textrm{for } -a \le x \le a, \\ F e^{kx}, & \textrm{for } x < -a, \end{cases}$$

$$\psi$$ and $$\frac{d \psi}{dx}$$ are required to be continuous at x = a and x = -a and so
 * $$F e^{-ka} = D cos(la),$$

and
 * $$-k F e^{-ka} = -l D sin(la).$$

Dividing, gives
 * $$k = l \tan(la).$$

This equation gives a condition on E but cannot be solved analytically for exact solutions.

Similar analysis gives the odd solution as


 * $$\psi(x) = \begin{cases}

F e^{-kx}, & \textrm{for } x > a, \\ D \sin(lx), & \textrm{for } -a \le x \le a, \\ F e^{kx}, & \textrm{for } x < -a, \end{cases}$$ with
 * $$l = -k \tan(la).$$