**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

We can look for bound states () for this potential. From our analysis of the infinite and finite square wells, we can write down the wave function’s general form:

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

Also,

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.

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alexwhen you found the probability p3, did you have to do that fraction? shouldnt the numerator just be equal to 1 because of normalization?

alexcrap i meant denominator

growescienceSince we didn’t normalize the wave function (that is, we never determined the constant in equation 2) then, yes, we do have to work out both numerator and denominator in equation 28. Actually, if you wanted to normalize the wave function then the denominator determines since

apashanka daswill there be any transmission probability when the particle is having bound state energy level? Please explain.

gwrowePost authorThere is no transmission for a bound state. The wave function does extend into the region where , but it decays exponentially (see equation 2) so its amplitude tends to zero as .

apashanka dassir here the transmission probability is there, although the particle is in the highest energy bound state which is greater than 50 percent chance ,is it consistent with the above?please explain

gwrowePost authorSorry, I don’t understand what you mean.

Bill BenishI believe that constant a in the exponent of equation 2 should be replaced by variable x. Thanks.

gwrowePost authorFixed now. Thanks.