# Bound state problem

1. Mar 2, 2016

1. The problem statement, all variables and given/known data
Consider $V(x)=-aV_{0}δ(x)$. Show that it admits a bound state of energy $E=-ma^2V_{0}/2\hbar^{2}$.Are there any other bound states?Hint:solve Schrodinger's equation outside the potential for E<0, and keep only the solution that has the right behavior at infinity and is continuous at x=0.

Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x=0. Calculate the change in the slope at equate it to $\int_{-ε}^{ε}( \frac{d^2\psi}{dx^2}) dx$
(where ε is infinitesimal) determined from Shrodinger's equation.

2. Relevant equations
Shrodinger equation in X basis : $\frac {d^2\psi}{dx^2}+(2m/\hbar)(E-V)\psi=0$.

3. The attempt at a solution
First how can we prove that $E=-ma^2V_{0}/2\hbar^{2}$ admits a bound state, shall I substitute it in the shrodinger equation then prove that the obtained wave function goes to zero when x goes to infinity?
For are there any other bound states?
I tried to solve the Shrodinger's equation with E<0 and outside the potential, what I got is this :
Let E=-A where A is a positive integer.
$\frac {d^2\psi}{dx^2}+(2m/\hbar)(E)\psi=0$
$\frac {d^2\psi}{dx^2}-(2m/\hbar)(A)\psi=0$
$\psi=Ae^{-\sqrt{2mA/\hbar^2}x}+Be^{\sqrt{2mA/\hbar^2}x}$
$Be^{\sqrt{2mA/\hbar^2}x}$ blows up to zero when x goes to infinity, so B=0.
what we finally get is $\psi=Ae^{-\sqrt{2mE/\hbar^2}x}$. I reached here and I don't know how to complete.
If anybody can help.Thanks.

Last edited: Mar 2, 2016
2. Mar 2, 2016

### Staff: Mentor

x can go both to positive and negative infinity, you have to consider both cases separately.
Then consider x=0 separately again, with the given hint.

Schrödinger (or Schroedinger), by the way.

3. Mar 3, 2016

Let E=-U where U is a positive number.
Take case where x<0
the solution is$\psi=Ae^{-\sqrt{2mU/\hbar^2}x}+Be^{\sqrt{2mU/\hbar^2}x}$
as x goes to -infinity the first part will blow up so A must be zero.
when x>0
the solution is $\psi=Ce^{-\sqrt {2mU/ \hbar^2}x}+De^{\sqrt{2mU/ \hbar^2} x }$
as x goes to +infinity the next part will blow up so D=0.

now for case x=0
the solution is $\psi=E e^{-\sqrt {2m (U+V)\hbar^2}x}+Fe^{\sqrt{2m(U+V)/ \hbar^2} x}$
here when x=0 V=infinity but 0*infinity is not defined. So how we can determine the wave function here?

4. Mar 3, 2016

### Staff: Mentor

The wave function has to be continuous.
You don't need x-dependent functions to consider the case x=0 as the wave function there has a single value only.

5. Mar 4, 2016

Let $\psi$at x=0 be a number called it F.
for x<0 $\psi=Be^{\sqrt{2mU/\hbar^2}x}$
for x>0$\psi=Ce^{-\sqrt{2mU/\hbar^2}x}$
As the wave function must be continuous at x=0,then we have $\lim_{x\rightarrow 0^{-} } {B e^ {\sqrt {2mU/ \hbar ^ 2} x }}=$ $\lim_{x\rightarrow 0^{+} }{Ce^{- \sqrt {2mU/ \hbar^2} x }}=$ $\psi(0)=F$
then B=C=F.
using the normalization condition at $-\infty<x<0 and 0<x<+\infty$
we have as B=C $C\int_{-\infty}^{0^-} e^{\sqrt{2mU/ \hbar^2}x}dx +C\int_{0^+}^{+\infty}e^{-\sqrt{2mU/ \hbar^2}x}dx=1$
this will lead that C=$\sqrt{2mU}/2\hbar$
but as V=infinity at x=0 then $\psi=0$at x=0,so F=0.
and in this case U must be zero!!!as C=B=0.
which means that there is no bound states,so what goes wrong?

6. Mar 4, 2016

### Staff: Mentor

Why should the wave function be zero there?

Everything before that line was fine.

7. Mar 4, 2016

OK.
then B=C=F=$\sqrt{2mU}/2\hbar$
and so for x<0 $\psi=(\sqrt{2mU}/2\hbar )* e^{\sqrt{2mU/ \hbar^2}x}$
and so for x>0$\psi=(\sqrt{2mU}/2\hbar)* e^{-\sqrt{2mU/ \hbar^2}x}$
and for x=0$\psi=\sqrt{2mU}/2\hbar$
But this means that all negative energies are allowed, since their is no restriction on negative energy.Is this right?

8. Mar 4, 2016

### Staff: Mentor

The restriction comes from the change of the first derivative that happens at x=0.

9. Mar 5, 2016

for x<0 $\left.\frac{d\psi}{dx}\right|_{x=0}=mU/ \hbar^2$
for x=0 $\left.\frac{d\psi}{dx}\right|_{x=0}=0$
for x>0 $\left.\frac{d\psi}{dx}\right|_{x=0}=-mU/ \hbar^2$
The change of the first derivative at x=0 is $-mU/ \hbar^2$
Now as I calculated the change of slope, I will equate to $\int_{-ε}^{ε} (\frac {d^2\psi}{dx^2})dx$
$\int_{-ε}^{ε} (\frac {d^2\psi}{dx^2})dx=-2mU/ \hbar^2 e^{-\sqrt{2mU/ \hbar^2}ε} +2mU/ \hbar^2$
but I didn't see that equating the change of slope to this integral leads to any restriction in energy?
what is the trick here?

10. Mar 5, 2016

### Staff: Mentor

There is no derivative if your function has a single point only.

The delta potential at x=0 allows to calculate the change of the first derivative. Set it equal to $\frac{2mU}{\hbar^2}$ and you get the constraints on the energy.

11. Mar 6, 2016

mmmm ok.
I will use now the Schroedinger equation $-\hbar^2 /2m\int_{-ε}^{ε}\psi' '(x)dx+\int_{-ε}^{ε}V(x)\psi(x)dx=E\int_{-ε}^{ε}\psi(x)dx$

Then $-\hbar^2 /2m[\psi'_{x>0}(0)-\psi'_{x<0}(0)]-aV_{0}\psi(0)=0 (asE\int_{-ε}^{ε}\psi(x)dx=0)$
so $-\hbar^2/2m[(-2mU/2\hbar^2)-2mU/2\hbar^2]=aV_{0}\psi(0)$
then $U=aV_{0}\sqrt{2mU}/2\hbar$
$U^2=a^2V_{0}^2mU/4\hbar^2$
then $U(U-a^2V_{0}^2m/2\hbar^2)=0$
which means that$U=0 or U=a^2V_{0}^2m/2\hbar^2$
and so the two allowed bound states are $E=0$ and$E= - a^2V_{0}^2m/2\hbar^2$

12. Mar 6, 2016

### Staff: Mentor

Do the units match for the final result? The approach looks fine.

13. Mar 6, 2016

As the hint says, I must suppose E<0 out side the potential, so there will be one bound state which is $E=-a^2V_{0}^2m/2\hbar^2$.I think this is right because it matches what the question wants,as
By the way, it is $E=-a^2V_{0}^2m/2\hbar^2$ and not $E=-a^2V_{0}m/2\hbar^2$ I copied it wrong in the first post.
shall I suppose E<0 for simplicity or things would go wrong if I have dealt with E>0?as the hint restricted me to deal with E<0 out side the potential.

14. Mar 6, 2016

### Staff: Mentor

E>0 cannot be a bound state. There are states with positive energy, but they are not bound.