Infinite square well with attractive potential

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Homework Statement


We have an infinite square well potential of width 2L centered at the origin, with an attractive delta function potential V0δ(x) at the origin, with the properties
[tex]V_0<0, -V_0>\frac{\hbar^2}{mL^2}[/tex]

Determine the conditions for a negative energy bound state.
There are a few other parts to the question, but I do not have the sheet at the moment.

Homework Equations


Schrodinger Equation

The Attempt at a Solution


In the absence of the delta function, we get
[tex]-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}=E\psi[/tex]

This differential equation has character polynomial given by
[tex]x^2+\frac{2mE}{\hbar^2}=0, x=\pm\frac{i\sqrt{2mE}}{\hbar}[/tex]

The solution is then
[tex]\psi=Asin\left(\frac{\sqrt{2mE}}{\hbar}x\right)+Bcos\left(\frac{\sqrt{2mE}}{\hbar}x\right)[/tex]

Using the boundary condition that the wave function is zero at ±L and normalizing the wave function, I get
[tex]\psi=L^{-\frac{1}{2}}cos\left(\frac{n\pi x}{2L}\right), L^{-\frac{1}{2}}sin\left(\frac{n\pi x}{2L}\right)[/tex]

Where the cosine solution is for odd n and the sine solution is for even n. Also, the energy spectrum is given by
[tex]E_n=\frac{n^2\pi^2\hbar^2}{8mL^2}[/tex]

For any positive integer n. This is the solution for the infinite well without the delta function, the full Schrodinger equation should be
[tex]-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V_0\delta(x)\psi=E\psi[/tex]

I thought about integrating both sides over the width of the well to eliminate the delta function and get
[tex]-\frac{\hbar^2}{2m}\int_{-L}^{L}\frac{\partial^2\psi}{\partial x^2}dx+V_0\psi(0)=E\int_{-L}^{L}\psi dx[/tex]

For even n this is trivial, but for odd n it seems to give me that V0=0, which is not true, nor very helpful. Obviously the bound state will most likely be the ground state n=1, so I thought it could be
[tex]E_1<-V_0, \frac{\pi^2\hbar^2}{8mL^2}<-V_0[/tex]

However, I have a feeling that I am supposed to work in the conditions imposed on V0 somehow. Is the work so far on the right track, or have I missed something important? There are a few more parts to the question, but this is all I could remember without the sheet near me.
 

Answers and Replies

  • #2
gabbagabbahey
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I thought about integrating both sides over the width of the well to eliminate the delta function and get
[tex]-\frac{\hbar^2}{2m}\int_{-L}^{L}\frac{\partial^2\psi}{\partial x^2}dx+V_0\psi(0)=E\int_{-L}^{L}\psi dx[/tex]

For even n this is trivial, but for odd n it seems to give me that V0=0, which is not true, nor very helpful.
Actually, it gives you [itex]V_0\psi(0)=0[/itex], and since [itex]V_0\neq 0[/itex], [itex]\psi(0)=[/itex]___?

You should not be surprised by this result; the wavefunction is always zero in regions where the potential is infinite. The effect of this is simply to divide the well in two.
 
  • #3
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Given the solutions I got for the wavefunction, [itex]\psi(0)=0[/itex] for even n, and [itex]\psi(0)=L^{-\frac{1}{2}}[/itex], for odd n. But since V0 is non zero, in order for [itex]V_0\psi(0)=0[/itex], it would imply [itex]\psi(0)=0[/itex]. However, wouldn't this make the wavefunction discontinuous for odd n at the origin?

Would this still have the effect of splitting the potential into two regions even if it is attractive? It makes sense if the delta function has a positive coefficient. Visually, this problem should look like a square well with a large dip toward [itex]-\infty[/itex] at the origin. This is what would produce the negative energy bound state I'm looking for.
 
  • #4
gabbagabbahey
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No, your solutions are invalid. The fact that [itex]\psi(0)=0[/itex] provides an extra boundary condition which you must apply to the general solution to Schroedinger's equation in inside [itex]|x|\leqL[/itex]...make sense?
 
  • #5
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With the additional condition that [itex]\psi(0)=0[/itex], If I apply that to my general solution from above, since cosine zero is never zero, then B=0 and the solution is just
[tex]\psi=Asin\left(\frac{\sqrt{2mE}}{\hbar}x\right)[/tex]

Applying the boundary conditions at the walls of the well and normalizing tells me
[tex]\psi(x)=L^{-\frac{1}{2}}sin\left(\frac{n\pi}{L}x\right), E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}[/tex]

Is the incorrect solution simply because I did not apply the condition at x=0 to my first solution, or do I have to solve the differential equation leaving the delta potential term intact? If it's the latter, I don't think I can still solve the differential equation by finding the roots of the characteristic equation.

We derived the solution for an attractive delta potential in class, but that was without the infinite potential walls at ±L. In that case solutions involved the exponential function, but those can't satisfy the boundary conditions at the walls of the well, so I don't think that example will be of much help.
 
  • #6
gabbagabbahey
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As I said earlier, the delta function effectively divides your well into two halves, giving you an new boundary at [itex]x=0[/itex]. Your original solution was incorrect because it failed to take this boundary and its corresponding boundary condition into account.
 
  • #7
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Then is this solution from a previous post correct?
[tex]\psi(x)=L^{-\frac{1}{2}}sin\left(\frac{n\pi}{L}x\right), E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}[/tex]

It satisfies that the wave function is zero at x=0,L,-L, is normalized and satisfies the Schrodinger equation.

Also, the last few parts ask about limits on the binding energy when [itex]-V_0[/itex] is large, and when [itex]-\frac{mLV_0}{\hbar^2}=1+\delta, \delta<<1[/itex] and if the energy is continuous at delta=0. That's nothing difficult once I find the relation between E and V.

Once I have the correct expression for the energy states, I also have the facts that [itex]V_0<0, -V_0>\frac{\hbar^2}{mL}[/itex]. But in what way do I combine these to find the condition of a bound energy state? Presumably it would occur when the energy of the particle is insufficient to escape the attractive well, so E+V<0
 
  • #8
gabbagabbahey
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Then is this solution from a previous post correct?
[tex]\psi(x)=L^{-\frac{1}{2}}sin\left(\frac{n\pi}{L}x\right), E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}[/tex]

It satisfies that the wave function is zero at x=0,L,-L, is normalized and satisfies the Schrodinger equation.
Well, these are eigenstates for this potential, but they aren't really the eigenstates you were asked for now, are they?

Determine the conditions for a negative energy bound state.
:wink:

What is the general solution for [itex]E<0[/itex]? What do you get when you apply your boundary conditions to it? (Remember to consider the regions [itex]-L<x<0[/itex] and [itex]0<x<L[/itex] separately)
 
  • #9
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I solved the equations again, separately for each region and took a few ideas from my notes when thinking about the boundary conditions at the walls of the well. Without going through that whole process, I got:
[tex]\psi_1(x)=A_1sin(k(L-x)), 0<x\leq L[/tex]
[tex]\psi_2(x)=A_2sin(k(L+x)), -L\leq x<0[/tex]

With the usual [itex]k^2=\frac{\sqrt{2mE}}{\hbar}[/itex]. Since the wave function must be continuous at x=0, then we get that [itex]A_2=\pm A_1[/itex]. More specifically, we know [itex]\psi(0)=0[/itex]. (The normalization constant is still [itex]A_1=L^{-1/2}[/itex], but it hasn't been needed for anything yet.)

So if we consider the first case, [itex]A_2=-A_1[/itex], then at x=0
[tex]A_1sin(kL)=-A_1sin(kL)[/tex]

This implies that [itex]kL=n\pi[/itex], and gives the usual result of
[tex]E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}[/tex]

Also, in this case, the derivative of the wave function is continuous at x=0. For the case of [itex]A_2=A_1[/itex], the wave function is still continuous at x=0 for our expression of k, but there is now a discontinuity in the derivative. Integrating over a small region near x=0, this can be described by the relation
[tex]2A_1kcos(kL)=\frac{2mV_0}{\hbar^2}A_1sin(kL)[/tex]

If we let z=kL, then we find a transcendental equation
[tex]tan(z)=\frac{z\hbar^2}{mLV_0}[/tex]

If you look at this graphically, there are an infinite number of solutions that occur where the two functions intersect. Looking at the limit [itex]-V_0\rightarrow\infty[/itex], then [itex]tan(z)=0[/itex], and [itex]kL=n\pi[/itex]. This leads to the usual relation
[tex]E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}[/tex]

In the other limit [itex]-\frac{mLV_0}{\hbar^2}=1+\delta, \delta<<1[/itex], I was thinking that
[tex]tan(z)=\frac{-z}{1+\delta}[/tex]

Then I'm kind of stuck from here. I was thinking that the intersection in this case would occur at a small enough value of z so I could use the approximation
[tex]tan(z)\approx\frac{z}{1-\frac{1}{2}z^2}[/tex]

Then I thought, even if that were the case, it would only apply to the first intersection point.
 
  • #10
gabbagabbahey
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I must apologize, but I think I may have steered you in the wrong direction when I said [itex]\psi(0)=0[/itex]. Looking again at the equation,

[tex]-\frac{\hbar^2}{2m}\int_{-L}^{L}\frac{d^2\psi}{d x^2}dx+V_0\psi(0)=E\int_{-L}^{L}\psi dx[/tex]

It's true that the RHS will be zero since the wavefunction is continuous and [itex]\psi(L)=\psi(-L)=0[/itex], but [itex]\frac{d\psi}{dx}[/itex] need not be continuous, because of the delta function at the center

[tex]\implies \int_{-L}^{L}\frac{d^2\psi}{d x^2}dx\neq \left.\frac{d\psi}{d x}\right|_{-L}^{L}[/itex]

Which was my basis for claiming that [itex]V_0\psi(0)=0[/itex]

Instead, for negative energy states, I think you'll want to write the general solution in the form:

[tex]]\psi_1(x)=A_1\sinh(\kappa x)+B_1\cosh(\kappa x), -L\leq x<0[/tex]

[tex]\psi_2(x)=A_2\sinh(\kappa x)+B_2\cosh(\kappa x), 0< x\leq L[/tex]

where [itex]\kappa\equiv\frac{-2mE}{\hbar^2}[/itex] is real and positive.

Then apply your boundary conditions at [itex]x=\pm L[/itex] and the fact that the wavefunction is continuous at [itex]x=0[/itex], but its derivative has a finite discontinuity there.
 
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