Normalize the even wave functions for the finite square well

[broegger]

I'm trying to normalize the even wave functions for the finite square well. The wave function is:

$$\psi(x)= \begin{cases} Fe^{\kappa x} & \text{for } x< a\\ D\cos(lx) & \text{for } -a\leq x \leq a\\ Fe^{-\kappa x} & \text{for } x> a \end{cases}$$

How can I determine D and F? When I set

$$\int_{-\infty}^{\infty}|\psi(x)|^2dx = 1$$,​

I obtain an equation in the two unknown amplitudes D and F. I could apply some boundary conditions to get more equations, but it gets rather complicated and I know there is an easier way...

 

[James R]

We require that the wave function is continuous, so:

$$Fe^{-\kappa a} = D \cos(la)$$

I don't think you can avoid using this, along with the normalisation condition.

 

[marlon]

Also, include the continuity-demand for the FIRST derivative of the wavefunctions at the "walls" of the well.

marlon

 

[dextercioby]

That $$e^{\kappa x}$$ is honestly nonnormalizable...And one more thing,the interval for this function should be:
$$(-\infty,-a)$$

Daniel.

 

[marlon]

And one more thing,the interval for this function should be:
$$(-\infty,-a)$$

Daniel.

Which is indeed the case

marlon

 

[broegger]

Also, include the continuity-demand for the FIRST derivative of the wavefunctions at the "walls" of the well.

marlon

Do I need this? The continuity of $$\psi$$, as James R says, gives me two equations in two unknowns.

 

[dextercioby]

How did u get those initial functions...?I mean,why aren't there 3 (a priori) different amplitudes,e.g. F,G and H...?

A little remark.I said earlier that $$e^{\kappa x}$$ was not normalizable.It would have been the case for the $$(a,+\infty)$$ interval.Sure,in your case,because of the negative values that "x" takes,it is VERY NORMALIZABLE...

Daniel.

 

[broegger]

How did u get those initial functions...?I mean,why aren't there 3 (a priori) different amplitudes,e.g. F,G and H...?

Because of the symmetry of the finite square well (the well is centered at x = 0), I assume that $$\psi$$ must be symmetric (even or odd).

 

[dextercioby]

Yes,well,then that's it...Though the fact that,by imposing continuity along the real axis,you determine completely the wave function and by checking the normalizability,(i think) it will not hold...

There's something fishy.Check Davydov,Cohen-Tannoudji or Flügge...

Daniel.

 

[polyb]

broegger,

When you are normalizing over the full interval, does it look something like this:

$$\int_{-\infty}^{-a}|\psi(x)|^2dx + \int_{-a}^{a}|\psi(x)|^2dx + \int_{a}^{\infty}|\psi(x)|^2dx = 1$$

Between that and the boundary conditions stated by James R, the unknowns should be pretty easy to solve.