# Bound state problem

## Homework Statement

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.

## Homework Equations

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

## 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:

mfb
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.

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.
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?

mfb
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.

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.
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?

mfb
Mentor
but as V=infinity at x=0 then $\psi=0$at x=0,so F=0.
Why should the wave function be zero there?

Everything before that line was fine.

Why should the wave function be zero there?

Everything before that line was fine.
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?

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

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

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?

mfb
Mentor
for x=0 $\left.\frac{d\psi}{dx}\right|_{x=0}=0$
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.

• 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$E= - a^2V_{0}^2m/2\hbar^2$

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

Hint:solve Schrodinger's equation outside the potential for E<0
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
Show that it admits a bound state of energy $E=-ma^2V_{0}/2\hbar^{2}$.Are there any other bound states?
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.

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

• E∫ε−εψ(x)dx=0
Hi
Why this integral is equal to zero?(in post #11)

mfb
Mentor
The wave function is bound and the size of the integration range goes to zero in the limit ##\epsilon \to 0##.

but why then the approach of letting the integration of
∫ε−ε(d2ψdx2)dx
equals to 2mU/h^2 led to fine answer?
shouldn't it also be equal to zero?

mfb
Mentor
The second derivative is not bound.

But why the second derivative is not bound?
It goes to zero for both sides.

vela
Staff Emeritus
Homework Helper
When you integrate the second derivative, you get
$$\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx = \psi'(\varepsilon)-\psi'(-\varepsilon).$$ If ##\psi'## is continuous, then
$$\lim_{\varepsilon\to 0^+} \psi'(\varepsilon) = \lim_{\varepsilon\to 0^+} \psi'(-\varepsilon)$$ so that
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx = 0.$$ But that's not what you have in this problem. The delta function potential results in a discontinuity in ##\psi'## so
$$\lim_{\varepsilon\to 0^+} \psi'(\varepsilon) \ne \lim_{\varepsilon\to 0^+} \psi'(-\varepsilon)$$ and
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx \ne 0.$$

When you integrate the second derivative, you get
$$\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx = \psi'(\varepsilon)-\psi'(-\varepsilon).$$ If ##\psi'## is continuous, then
$$\lim_{\varepsilon\to 0^+} \psi'(\varepsilon) = \lim_{\varepsilon\to 0^+} \psi'(-\varepsilon)$$ so that
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx = 0.$$ But that's not what you have in this problem. The delta function potential results in a discontinuity in ##\psi'## so
$$\lim_{\varepsilon\to 0^+} \psi'(\varepsilon) \ne \lim_{\varepsilon\to 0^+} \psi'(-\varepsilon)$$ and
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx \ne 0.$$
Yes, this makes it zero. But it is still bound doesn't it?
Besides the antiderivative of the wave function is not continuous at zero but the integral is equal to zero.

mfb
Mentor
The second derivative diverges as ##\epsilon \to 0##. It is not bound (as function of epsilon).

The second derivative diverges as ##\epsilon \to 0##. It is not bound (as function of epsilon).
but for x<0 ##\psi(x)=\frac {\sqrt {2mU}} {2\hbar}*e^{\frac {\sqrt{2mU}} {\hbar}x}##
for x>0##\psi(x)=\frac {\sqrt {2mU}} {2\hbar}*e^{-\frac {\sqrt{2mU}} {\hbar}x}##

The second derivative of the wave function will converge to##(1/2) \frac {\hbar} {\sqrt {2mU}}## when ##\epsilon \to 0## in both cases x>0 and x<0.
and when ##x \to \infty## or ##-\infty## the second derivative of the wave function will converge to zero.

mfb
Mentor
Sure, but not for x=0.

So I conclude that a wave function to be bound it is a must to be continuous, and once it is bound the definite integral of it from ##-\epsilon \to \epsilon## will approach zero if epsilon approaches zero(even if its antiderivative is not continuous like the case of this problem) .