Required math: calculus
Required physics: Schrödinger equation
References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.40.
As a sort of hybrid of the finite and infinite square wells, suppose we have the potential
Here, and . Note that for a bound state, .
Using the continuity of at :
At : continuity of gives
Continuity of gives
Combining the last two gives
We can now define the auxiliary variables
The equation to be solved for is therefore:
This is a transcendental equation which can be solved graphically or numerically. The graph looks like this, with in green and in red:
[The vertical green lines are the asymptotes of ; I couldn’t find a way to get Maple not to draw them.]
As can be seen from the graphical solution, there are 3 bound states. We can solve the equation numerically (using Maple’s ‘fsolve’, for example) to find the three solutions and corresponding energies.
From small to large , we get
For the highest energy, we can find the probability that the particle is outside the box (that is, its position is outside the range ).
From the boundary conditions, we find
so the integral of the square modulus of the wave function over its entire range is
The probability of the particle being outside the box is therefore the second term in the above expression divided by the entire expression:
Using the value of given above, we can get values for and in terms of . Plugging these into the above expression and evaluating (using Maple) gives
Note that this answer is very sensitive to the value of . Even a minor rounding error will cause a significant difference in the value of . For example, rounding to 7.96 gives .
It’s interesting that for this barely bound state, the probability of the particle being outside the box is actually greater than that for being inside.
We can do similar calculations for the other two energies, and find:
Since these energies are lower, the probability of being found outside the well is much smaller, as we would expect.