**Required math: calculus **

**Required physics: **Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.30.

The finite square well has the potential

where is a positive constant energy, and is a constant location on the axis.

We’ve seen that since the potential is an even function, we can look for solutions of the Schrödinger equation that are either even or odd. In the even function case, we’ve seen that the solution has the form

Imposing boundary conditions gives us the relations

from which we get the relation between and :

To complete the analysis, we need to normalize the wave function. Since it’s an even function, the integral from 0 to of the square modulus is half the integral from to , so we get:

Doing the integrals gives us

From the first boundary condition above, we have

Solving these last two equations, we get

Using the relation between and along with the trig identity we get

While we’re at it, we can also normalize the odd solution. Here, the solution has the form

The boundary conditions give

with the resulting relation between and :

The normalization integral is

which gives

The first boundary condition gives

Solving these two equations gives

Rewriting the relation between and we get

and inserting this into the solutions gives

The final result is then

where we’ve taken the negative root of in order to satisfy the original boundary conditions above (we could also have taken with a negative sign and with a positive sign).

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JonathanThere is a mistake in the expression for the B the coefficient of the even solutions, the exponential should read exp(kappa*a) and not exp(mu*a).

growescienceFixed now. Thanks.

Guilherme Siqueira GomideI’m pretty sure your constant B has a minus (for the odd solution). If you apply the original boundary condition (not the squared version of it), you get that:

B = -C*sin(mu*a)*exp(kappa*a). That minus is lost when you square both sides.

growescienceQuite right; I’ve fixed it now.

Anonymouswhen you normalize the wave equation, how come you set it equal to 1/2 instead of 1?

growescienceSince the integrand is an even function, the integral from 0 to of the square modulus is half the integral from to .

Bill BenishEq. 14 (the expression for B) is missing.

gwrowePost authorFixed now. Thanks.