Consider a particle with energy E moving towards a potential barrier of height U0 and width a:
-----------------------|||||||----------------------- E
_______________________|||||||_______________________ x
                       | -a- |
Using Schrödinger's (time-independent, one-dimensional) Equation, we can solve for the wave function of the particle (using h for h-bar):
   - h22ψ
  --------_ + U(x)ψ = Eψ

The potential U(x) is divided into three parts:

U(x) = { 0 : x < 0,
         U0: 0 < x < a,
         0 : x > a     }

In order to solve for ψ, the wave function of the particle, we also divide it into three parts: ψ0 for x < 0, ψ1 for 0 < x < a, and ψ2 for x > a. Astute readers will notice at this point that the potential is the same for ψ0 and ψ2 -- these two wave functions ought, then, to look at least somewhat similar. As we shall see, they will have the same wavelength but different amplitudes. Since U = 0 for both ψ0 and ψ2, they each take the same form as the wave function for a free particle with energy E, or:

ψ(x) = A*ei*k0x + B*e-ik0x  (where k0 = √(2*m*E/h2) )

The first portion of this equation corresponds to a wave moving rightwards while the second portion corresponds to a wave moving to the left. Or they would, had we folded in time-dependence (see note at the bottom). In order to make our lives easier, it is necessary to think a little bit about what is actually physically happening in this system. Our particle is approaching the potential barrier from the left, moving rightwards. When it hits the potential barrier, common sense says that at least some of the time, the particle will bounce off the barrier and begin moving leftwards. From this, we know that ψ0 contains both the leftward (reflected particle) and rightward (incident particle) portions of the wave function. As the other nodes in this writeup explain, when the particle hits the potential barrier, in addition to bouncing off some of the time, some of the time it will pass through. So we know that ψ2 has at least the rightward-moving component. But there is nothing in the experimental setup that would cause the particle to begin moving towards the left once it has passed through the potential barrier, so we can deduce that the leftward-moving component of ψ2 has an amplitude of zero.

Now, to deal with the particle while it is inside the barrier. Common sense would suggest that the particle can never actually exist within the barrier, (let alone cross over it). Physically, however, we know for sure that a particle can, in certain circumstances, pass through the barrier, so common sense would suggest that if it exists on both sides of the barrier, it must also exist within the barrier. But how on earth are we supposed to observe a particle while it is inside a potential barrier? The answer is that while we can't observe the particle inside the potential barrier, the mathematical properties of the wave function suggests that it does in fact exist while it is inside the barrier.

Since the only thing that matters in physics is relative potential, we can pretend like the particle, while it is inside the potential barrier, isn't in a potential of U0, but rather simply has an energy of E - U0 = - (U0 - E)  (since U0 > E). As before, then, the equation for this situation the wave equation with a wave number (k) of √(2*m*E/h2). In this case however, the particle has negative energy (tis a very good thing we can't physically observe the particle while it is inside the barrier, since negative energies can't exist), so it has an imaginary wave number, k1 = i√(2*m*(U0-E)/h2).

We now know enough to write out all three parts of the wave equation:

ψ(x) = {
      A*ei*k0x + B*e-ik0x : x < 0
      C*e-ik1x + D*eik1x   : 0 < x < a
      E*eik0x           : x > a

The wave function and its first derivative have to continuous over all x ∈ R. We can use these boundary conditions to get four relationships among the constants (ψ0(0) = ψ1(0), ψ0'(0) = ψ1'(0), ψ1(a) = ψ2(a), and ψ1'(a) = ψ2'(a)). Actually solving for the constants is impossible given just these conditions (five unknowns but only four equations), but we can find the probability that the particle reflects off the barrier, and the probability that it tunnels through the barrier. Recall that the probability function of a particle with wave function ψ is

P(x) = |ψ(x)|2

Since we know that the first portion of ψ0 (with amplitude A) represents the incident particle, and the second portion (with amplitude B) represents the reflected particle, the ratio of the two wave functions |B|2/|A|2 is the fraction of the time that the incident particle will reflect off the barrier. Similarly, the ratio |E|2/|A|2 is the fraction of the time that the particle will tunnel through the barrier. After a bit of extraordinarily ugly algebra (don't try this at home), we find that:

            |E|2/|A|2  =    -----------------------
                            1 + 1/4 ---------------
                                      E (U0 - E)

It shouldn't be too hard to convince yourself that since the particle has to do something after hitting the barrier, the probability that it will reflect off is just 1 - |E|2/|A|2. This probability decreases exponentially with a (since sinh(x) = (ex - e-x)/2), so the largest factor in determining tunneling probability is the width of the potential barrier. tdent notes that since the probability also depends exponentially on k1, there's a large dependence on the difference between the barrier height and the energy of the particle, but since the dependence on (U0 - E) is under a square root, this still has less of an effect than a.

Note: For time-independent potentials (∂U/∂t = 0), the time-dependent solution to the Schrödinger equation is just ψ(x)*e-iωt, where ψ is the time-independent wave function and ω = E/h. So, the time-dependent form of the solution ends up looking like:


As t increases then, for the first part of the function to remain constant x must increase and for the second part to remain constant x must decrease. So the first portion of the equation represents a wave travelling towards increasing x (the right), and the second portion represents a wave travelling towards decreasing x (the left).

From personal notes, Modern Physics by Kenneth Krane, and (for the solution to |E|2/|A|2).