Delta function antisymmetric potential problem

[frankcastle]

Homework Statement

particle of mass m is subjected to antisymmetric delta-function potential V(x) =V'Delta(x+a)-V'Delta(x-a) where V'>0
Show that there is only one bound state, and find its energy

Homework Equations

Assuming free particle eqn for x<-a for particle incident from -ve infinity,
psi(x)=Acos(kx)+Bsin(kx) k=(2mE*(4pi)sqrd/(h)sqrd)^1/2

The Attempt at a Solution

I know how to go about the solution for a regular square potential. How does one account for the delta function nature of this potential well?

 

[Avodyne]

Have you seen the problem with a single delta-function, V(x) = -V' delta(x) ? It seems unlikely that you would be assigned this more complicated problem if the simpler one were not already presented. And if not, you should start with it as a warm-up problem. It also has one bound state.

 

[frankcastle]

Oh this isn't an assignment problem, I found it in some text while I was preparing for some QM coursework. I have not come across the single potential question though.

Will it act as an infinite wall? But since the thickness is negligible there must be some flux leakage right?

 

[frankcastle]

What if there is an infinite wall to the left of a single delta function Vdel(x-a)
How will the treatment change as compared to a simple delta function alone?

I obtained one set of boundary conditions by equating wave function on either side of the delta step.
I think the other set can be obtained by integrating the Time independent S.E on either side of the step at a by delx(tends to zero)

am i on the right track?

 

[Avodyne]

The way to treat a delta-function potential, mathematically, is to take the time-independent Schrodinger equation,

$$-{\hbar^2\over 2m}\psi''(x) - V_0 a\,\delta(x)\psi(x) = E\psi(x)$$

where I've called the coefficient of the delta function $$V_0 a$$ because it has dimensions of energy times length. Now, let $$\varepsilon$$ be a small length, and integrate across the delta function location,

$$\int_{-\varepsilon}^{+\varepsilon}dx\left[-{\hbar^2\over 2m}\psi''(x) - V_0 a\,\delta(x)\psi(x)\right] = E\int_{-\varepsilon}^{+\varepsilon}dx\,\psi(x)$$

We get

$$-{\hbar^2\over 2m}\Bigl[\psi'(+\varepsilon) -\psi'(-\varepsilon)\Bigr]- V_0 a\psi(0) \simeq 2\varepsilon E\psi(0)$$

Taking the limit $$\varepsilon\to 0$$, we get

$$-{\hbar^2\over 2m}\Bigl[\psi'(0^+) -\psi'(0^-)\Bigr]- V_0 a\psi(0) = 0$$

This shows that the first derivative of $$\psi$$ is discontinuous across the location of the delta function. And if the derivative has a finite discontinuity, then the function itself is continuous.

So now we know how to match the solutions on either side of the delta function.

 

[frankcastle]

I understand what you are saying,
one question about delta potentials,
can VoDel(x-a) be written as VoaDel(x)?

also in the integrals, wouldn't the limits be a-e and a+e ?

Also I don't understand how to get the second set of boundary conditions,
do I use the last equation you have provided to compute the second set?

How does the infinite well on the left of the delta function of potential affect the wave functions ?

 

[Avodyne]

I understand what you are saying,
one question about delta potentials,
can VoDel(x-a) be written as VoaDel(x)?

No.

also in the integrals, wouldn't the limits be a-e and a+e ?

Yes!

Also I don't understand how to get the second set of boundary conditions,
do I use the last equation you have provided to compute the second set?

Not sure what you mean by "second set". The wave function is continuous at the delta function, and the first derivative is discontinuous, but with a discontinuity given by the equation.

How does the infinite well on the left of the delta function of potential affect the wave functions ?

Again, not sure what you're referring to here. By "infinite well", do you mean the delta function potential itself, or something else?

 

[frankcastle]

no, in the case that there is another square potential barrier of infinite height to the left of the delta function, say at the origin of the X axis, how will that affect the treatment of this problem?