Infinite Square well with a Finite square well inside

[user3]

Ok here's a potential I invented and am trying to solve:

V =
-Vo in -b<x<b
and 0 in -a<x<-b , b<x<a where b<a
and ∞ everywhere elseI solved it twice and I got the same nonsensical transcendental equation for the allowed energies: \frac{-k}{\sqrt{z_0 - k^2}} \frac{e^{2kb} + e^{2ka}}{e^{2kb} - e^{2ka}} = tan(b \sqrt{z_0-k^2}).

, where k=\frac{\sqrt{-2mE}}{\hbar} and z_0 = \frac{2mV_0}{\hbar^2}

The problem is that when I take the limit as b→a (the ordinary infinite square well) I get only one solution, where I should get infinity many.

So is there something fundamentally wrong with trying to solve this potential? Is it wrong to have an Infinite potential and bury some of it under the 0 (negative potential) ? Note: I am solving it for negative energies(Bound bound states?).

 

[mfb]

How did you solve it? Can you show what you did?

Your choice of k could indicate that you looked at solutions for E<0 only. Is that true? You would expect solutions for positive energies as well.

 

[user3]

Yes that's true. I was looking for the E<0 solutions. Only the even solutions also.

Here's how I solved it:

\psi(x) = C sin(lx) + Dcos(lx)\:\:\:\:\: 0&lt;x&lt;b \:\:\:\:\:[1]

Note: l = \frac{\sqrt{2m(E+V_0)}}{\hbar}

\psi(x) = Ae^{-kx} + Be^{kx}\:\:\:\:\:b&lt;x&lt;a \:\:\:\:\:[2]

since I am looking only for the even solutions, I can use only the cos part in equation [1]
---> \psi(x) = Dcos(lx)\:\:\:\:\:0&lt;x&lt;b \:\:\:\:\:[3]

Also \psi(x)=0 at x = a: Ae^{-ka} + Be^{ka} = 0 ----> A = -Be^{2ka}

that turns equation [2] into \psi(x) = Be^{-kb} (e^{2kb} - e^{2ka}) \:\:\:\:\:[4]

\psi(x) is continuous at x = b: Be^{-kb} (e^{2kb} - e^{2ka}) = Dcos(lx)

that gives us D = \frac{Be^{-kb} (e^{2kb} - e^{2ka})}{cos(lb)}\:\:\:\:\:[5]

Using the fact that \frac{d\psi}{dx} is continuous ,the fact that l = \sqrt{z_0 - k^2}, and equation [5], we get

\frac{-k}{\sqrt{z_0 - k^2}} \frac{e^{2kb} + e^{2ka}}{e^{2kb} - e^{2ka}} = tan(b \sqrt{z_0-k^2})

 

[vela]

Yes that's true. I was looking for the E<0 solutions. Only the even solutions also.

Here's how I solved it:

\psi(x) = C sin(lx) + Dcos(lx)\:\:\:\:\: 0&lt;x&lt;b \:\:\:\:\:[1]

Note: l = \frac{\sqrt{2m(E+V_0)}}{\hbar}

\psi(x) = Ae^{-kx} + Be^{kx}\:\:\:\:\:b&lt;x&lt;a \:\:\:\:\:[2]

since I am looking only for the even solutions, I can use only the cos part in equation [1]
---> \psi(x) = Dcos(lx)\:\:\:\:\:0&lt;x&lt;b \:\:\:\:\:[3]

Also \psi(x)=0 at x = a: Ae^{-ka} + Be^{ka} = 0 ----> A = -Be^{2ka}

that turns equation [2] into \psi(x) = Be^{-kb} (e^{2kb} - e^{2ka}) \:\:\:\:\:[4]

I think you mean $\psi(x) = Be^{-kx}(e^{2kx}-e^{2ka})$. I'd write it differently, however:
$$\psi(x) = -B e^{2ka} e^{-kx} + B e^{kx} = Be^{ka} (-e^{ka}e^{-kx} + e^{-ka}e^{kx}) = 2Be^{ka} \sinh k(x-a) = B' \sinh k(x-a).$$

\psi(x) is continuous at x = b: Be^{-kb} (e^{2kb} - e^{2ka}) = Dcos(lx)

that gives us D = \frac{Be^{-kb} (e^{2kb} - e^{2ka})}{cos(lb)}\:\:\:\:\:[5]

Using the fact that \frac{d\psi}{dx} is continuous ,the fact that l = \sqrt{z_0 - k^2}, and equation [5], we get

\frac{-k}{\sqrt{z_0 - k^2}} \frac{e^{2kb} + e^{2ka}}{e^{2kb} - e^{2ka}} = tan(b \sqrt{z_0-k^2})

Why do you think you're getting only one solution when $b \to a$? It seems to me that if $V_0$ (and hence $z_0$) is big enough, you can get multiple solutions with E<0. The rest of the solutions will be for E>0.

 

[user3]

Yes That's what I expected. I even solved the problem of the Well where the limit has already been taken and It gave me multiple solutions, like expected.

I am sorry, I just realized that I have been taking the limit incorrectly. I now realize the I have a division by 0 when I take the limit. What can I do about that?

 

[mfb]

The 0 should cancel with another expression going to zero. Consider both at the same time in the limit.

 

[user3]

As b→a, the transcendental equation becomes \frac{-k}{\sqrt{z_0 - k^2}} \frac{2e^{2ka}}{0} = tan(a \sqrt{z_0-k^2})


I can't see any other 0's.

 

[vela]

You could simply flip the equation over to avoid the division by 0 so that you have $\cot (a\sqrt{z_0-k^2}) = 0$.