Proving Bound State Exists for V(x)=-aV_0δ(x)

[amjad-sh]

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 Schrödinger'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.

 

[mfb]

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.

Sc[/color]hrö[/color]dinger (or Schroedinger), by the way.

 

[amjad-sh]

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]

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.

 

[amjad-sh]

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 \psiat 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&lt;x&lt;0 and 0&lt;x&lt;+\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=0at 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]

but as V=infinity at x=0 then \psi=0at x=0,so F=0.

Why should the wave function be zero there?

Everything before that line was fine.

 

[amjad-sh]

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]

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

 

[amjad-sh]

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]

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.

 

[amjad-sh]

mmmm ok.
I will use now the Schroedinger equation -\hbar^2 /2m\int_{-ε}^{ε}\psi&#039; &#039;(x)dx+\int_{-ε}^{ε}V(x)\psi(x)dx=E\int_{-ε}^{ε}\psi(x)dx

Then -\hbar^2 /2m[\psi&#039;_{x&gt;0}(0)-\psi&#039;_{x&lt;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 thatU=0 or U=a^2V_{0}^2m/2\hbar^2
and so the two allowed bound states are E=0 andE= - a^2V_{0}^2m/2\hbar^2

 

[mfb]

U=0 leads to a contradiction elsewhere.
Do the units match for the final result? The approach looks fine.

 

[amjad-sh]

Hint:solve Schrödinger'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]

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

 

[amjad-sh]

E∫ε−εψ(x)dx=0

Hi
Why this integral is equal to zero?(in post #11)

 

[mfb]

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

 

[amjad-sh]

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]

The second derivative is not bound.

 

[amjad-sh]

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

 

[vela]

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
$$\lim_{\varepsilon\to 0^+}\int_{-\varepsilon}^\varepsilon \psi''(x)\,dx \ne 0.$$

 

[amjad-sh]

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]

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

 

[amjad-sh]

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]

Sure, but not for x=0.

 

[amjad-sh]

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) .

 

[mfb]

There are bound, not continuous functions.
A continuous function over a compact support has to be bound, however.

and once it is bound the definite integral of it from $-\epsilon \to \epsilon$ will approach zero if epsilon approaches zero

Sure.

 

[amjad-sh]

There are bound, not continuous functions.
A continuous function over a compact support has to be bound, however.Sure.

I in fact posted a thread couple of days concerning a problem somehow similar to this problem but much more complicated.
The difference in that problem is that the wavefunctions are scattering states and not bounded states.
The problem asked to derive the scattering coefficients by explicitly solving the Hamiltonian.
I used the same approach I used for this problem but didn't get an answer.
I would be glad if you can take a look at it: www.physicsforums.com/threads/solving-scattering-problem-including-spin-flip.952575/#post-6035381
some of the detailed steps may be wrong but I want know if my general approach is right.

There are bound, not continuous functions.
A continuous function over a compact support has to be bound, however.

I will appreciate it if you can also send for me a pdf that explains these concepts in details.

 

[George Jones]

But why the second derivative is not bound?

What is the derivative of a step function?

 

[amjad-sh]

What is the derivative of a step function?

the dirac delta function.

 

[George Jones]

What is the derivative of a step function?

the dirac delta function.

Since

Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x=0.

the first derivative has a jump discontinuity at $x=0$. This means that the second derivative (the derivative of the first derivative) involves a Dirac delta function, and, consequently, the second derivative is unbounded.