Infinite potential well with delta well

[bznm]

I Have tried to solve a problem about infinite potential well with a delta well in the middle, but I haven't the results and so I can't check if the proceeding is wrong...

We have a particle in 1D that can moves only on $[-a.a]$ because of the potential $V(x)=\begin{cases}-\lambda \delta (x), x\in(-a,a)\\ \infty, otherwise\end{cases}$
($\lambda>0)$

I have to find the autofunctions, the expression of energy levels and say if the sprectrum is limited

My attempt solution:

The Schroedinger's Equation is:

$\psi''(x)=\frac{2m}{\hbar^2} (V(x)-E) \psi (x)$
so we have: $\psi''(x)=-\frac{2m}{\hbar^2} E \psi (x)$
I have assumed $E>0$

Then I have translated the axis origin, and the segment [-a, a] now is [0,2a]

We have two wavefunctions:

$\psi_-=A\sin kx +B\cos kx$ for $0<x<a$
$\psi_+=C\sin kx +D\cos kx$ for $a<x<2a$

Conditions:

- from $\psi_- (0)=0 \rightarrow B=0$
- from$\psi_+ (2a) = 0 \rightarrow \tan(2ka)=-D/C$
- from $\psi_- (a) = \psi_+ (a) \rightarrow \tan(ka)=\frac{D}{A-C}$
- from normalization $A^2=C^2+D^2=1$
- from $\psi_+' (a)=\psi'_- (a)-\frac{2mg}{\hbar^2}\psi(a) \rightarrow kC \cos(ka) - Dk \sin(ka)= Ak \cos(ka)-\frac{2mg}{\hbar^2}A\sin(ka)$

I've tried substituting in the last relation the previous relations but I couldn't get to the result...

 

[Orodruin]

I suggest that you do not make the translation to the interval [0,2a]. The original problem is symmetric around x=0 and therefore has eigenfunctions which are either symmetric or anti-symmetric.

 

[bznm]

I have done the translation because my texbook, dealing with the infinite well, explicitly says "In order to avoid different steps for odd and even eigenfunctions, we choose the coordinates origin in one of the extreme points of the segment" and so does in the next steps.. So I have thought to do the same thing... But maybe I can't to do it because of the delta function? Do I have to consider even and odd solution because of the delta?

 

[Avodyne]

You can do it with the shifted interval.

Your normalization condition is wrong. And it won't be needed in any case, it doesn't help you find the spectrum.

Your assumption E>0 is not necessarily correct. You should consider E>0 and E<0 separately.

You should consider the cases $\psi(a)=0$ and $\psi(a)\ne 0$ separately.

Then, your goal is to eliminate $A, B, C, D$ and get an equation involving $ka$ only. The equation you get does not necessarily have an analytic solution, and then needs to be studied graphically. (If you've seen the finite square-well problem, you have probably seen this technique.)

 

[vela]

I have done the translation because my texbook, dealing with the infinite well, explicitly says "In order to avoid different steps for odd and even eigenfunctions, we choose the coordinates origin in one of the extreme points of the segment" and so does in the next steps.. So I have thought to do the same thing... But maybe I can't to do it because of the delta function? Do I have to consider even and odd solution because of the delta?

You can do it either way, but given the symmetry of the problem, I'd stick with the original interval.

One trick to make the math a bit simpler is to use wave functions of the form $\psi_+ = A \sin k(x+a)$ and $\psi_+ = B \sin k(x-a)$. These automatically satisfy the boundary conditions at $x = \pm a$.

 

[bznm]

You should consider the cases ψ(a)=0 and ψ(a)≠0 separately.

if $\psi(a)=0$, $k=\frac{n \pi}{a}$ (and so I obtain the spectrum.. right?)

if $\psi(a) \neq 0$, $tg ka=\frac{D}{A-C}$...

but I can't find the equation that involves only $ka$.. can you help me?

 

[bznm]

One trick to make the math a bit simpler is to use wave functions of the form ψ+=Asink(x+a) and ψ+=Bsink(x−a). These automatically satisfy the boundary conditions at x=±a.

And what about the condition on energy levels?

 

[vela]

What equations do the other two conditions at x=0 give you (for the unshifted well)?

 

[Avodyne]

if $\psi(a)=0$, $k=\frac{n \pi}{a}$ (and so I obtain the spectrum.. right?)

Yes. In the original unshifted interval, these are the eigenvalues corresponding to the odd-parity eigenfunctions. These are the same as in the box without the delta function. This is because these eigenfunctions vanish at the delta function, and so are unaffected by the delta function.

if $\psi(a) \neq 0$, $tg ka=\frac{D}{A-C}$...

but I can't find the equation that involves only $ka$.. can you help me?

You already have

- from $\psi_+ (2a) = 0 \rightarrow \tan(2ka)=-D/C$
- from $\psi_- (a) = \psi_+ (a) \rightarrow \tan(ka)=\frac{D}{A-C}$
...
- from $\psi_+' (a)=\psi'_- (a)-\frac{2mg}{\hbar^2}\psi(a) \rightarrow kC \cos(ka) - Dk \sin(ka)= Ak \cos(ka)-\frac{2mg}{\hbar^2}A\sin(ka)$

Use the first equation to express $D$ in terms of $C$. Plug this into the second equation. Use it to solve for $A$ in terms of $C$. Plug your expressions for $D$ and $A$ into the third equation. Cancel out common factors of $C$. Simplify as much as possible.

 

[bznm]

A lot of thanks!