# Delta well + infinite barrier -> bound state

maverick280857
Hi,

I'm trying to understand the quantum mechanical solution to this potential:

$$V(x) = \left\{\begin{array}{cc}\infty & \mbox{ for } x < 0,\\-\lambda\delta(x-d) & \mbox { for } x > 0\end{array}\right.$$

A particle of mass m is constrained to move on the half straight line $\{x \in \mathbb{R}: x > 0\}$ under this potential. I want to examine the condition for existence of bound states, and its dependence on $\lambda$ and $d$, where both $\lambda$ and $d$ are positive.

For x < 0, $\psi(x) = 0$

For 0 < x < d, $\psi(x) = A\sin(kx)$ where $k = \sqrt{2mE}/\hbar$.

For x > d, $\psi(x) = Ce^{ikx} + De^{-ikx}$.

Enforcing continuity at x = d, $A sin(kd) = Ce^{ikd} + De^{-ikd}$.

Integrating Schrodinger's equation over a small interval around $x = d$

$$-\frac{\hbar^2}{2m}\int_{d-\epsilon}^{d+\epsilon}\frac{d^2 \psi}{dx^2} dx - \lambda\int_{d-\epsilon}^{d+\epsilon}\delta(x-d)\psi(x) dx = E\int_{d-\epsilon}^{d+\epsilon}\psi(x) dx$$

But I don't get a condition relating $\lambda$ and $d$ for the existence of a bound state. What have I missed?

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## Answers and Replies

Jivesh
There might not be an explicit relation between \\lambda and d, if that what's you are trying to find. But you do have the relation that the difference of the derivatives to the left and right of the peak is equal to the negative of \\lambda times 2m, divided by hbar^2. Comparing the real and imaginary part, you might get a relation between the trigonometric ratios of d and \\lambda. I think the energy levels might have the discrete band structure, as is found for the case of a solid, with the potential represented by the dirac comb.

For a bound state, the wavefunction has to decay exponentially for x>d.

maverick280857
There might not be an explicit relation between \\lambda and d, if that what's you are trying to find. But you do have the relation that the difference of the derivatives to the left and right of the peak is equal to the negative of \\lambda times 2m, divided by hbar^2. Comparing the real and imaginary part, you might get a relation between the trigonometric ratios of d and \\lambda. I think the energy levels might have the discrete band structure, as is found for the case of a solid, with the potential represented by the dirac comb.

Okay, I found the source of this problem and it asks me to find a relation between d and lambda for the existence of at least one bound state.

maverick280857
For a bound state, the wavefunction has to decay exponentially for x>d.

Right so that means k becomes imaginary.

yes, and either C or D becomes 0.

maverick280857
yes, and either C or D becomes 0.

Okay just to clarify matters...

For 0 < x < d, since the wavefunction is constrained to be 0 for x <= 0, it should have the form

$$\psi(x) = A\sin(kx)$$

where $k = \sqrt{\frac{2mE}{\hbar^2}}$ for E > 0. This makes k real.

For x > d, the wavefunction should have the form

$$\psi(x) = B e^{-kx}$$

for the same k. Now, what I want to understand is...what is the definition of a bound state here? Suppose the delta well didn't exist. Then, the "bound"
state would just be $\psi(x) = A\sin(kx)$ (or in other words, no bound state, as its just an infinite step at x = 0). With the delta well in place at x = d, what really is a bound state for the composite potential? In other words, how is an E > 0 solution a bound state?

All solutions to the Schroedinger equation which are normalizable are bound states.
The other solutions, like sin(kx) in the free case are not normalizable and belong to the continuum. Whether one considers non-normalizable solutions as admissible solutions at all depends on the formalism. In the orthodox view a la von Neumann, they don't belong to the Hilbert space at all and are non-admissible. In the rigged Hilbert space approach of Bogoliubov which accomodates naturally the Dirac bra-ket formalism, they are ok.

maverick280857
All solutions to the Schroedinger equation which are normalizable are bound states.

Got it. So, are you saying that the solutions I listed -- the sin(kx) in (0, d) and the exponentially decaying wave for (d, \infty) are bound states for E > 0? The reason for my confusion in defining a "bound state" and distinguishing between E > 0 and E < 0 stems from the fact that we have a barrier which goes to +\infty and a well which goes to -\infty.

maverick280857
Here's my working.

$$\psi(x) = \left\{\begin{array}{cc}A\sin(kx) & 0 < x < d\\Ce^{-kx} & x > d\end{array}\right.$$

1. Continuity at x = d:

$$A\sin(kd) = Ce^{-kd}$$

2. Integration of Schrodinger's equation around x = d

$$\sin(kd) + \cos(kd) = \frac{2m\lambda}{\hbar^2 k}\sin(kd)$$

where $k = \sqrt{\frac{2mE}{\hbar^2}}$ and E > 0.

This gives (if my algebra is correct) the "condition relating $\lambda$ and $d$ for the existence of a bound state" as

$$\tan(kd) = \frac{1}{\frac{2m\lambda}{\hbar^2 k}-1}$$

Is this the desired condition, even though it involves another parameter (E)?

PS -- The question also says that an observer can detect the existence of the potential by bouncing back waves from the wall at x = 0, of the form sin(kx) which will acquire a phase shift when the delta well is present, and look like sin(kx + phi). If this is to be believed, the exponentially decaying solution should be replaced with the complex exponential wave, don't you think?

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I didn't check your calculations but I'd rather expect that the bound state has E<0 in that problem as the continuum of free states still starts at E=0.

Btw, instead of sin and cos you should also use sinh and cosh.

maverick280857
Btw, instead of sin and cos you should also use sinh and cosh.

Why? Is that because E < 0? And why is that so?

PS -- The question also says that an observer can detect the existence of the potential by bouncing back waves from the wall at x = 0, of the form sin(kx) which will acquire a phase shift when the delta well is present, and look like sin(kx + phi). If this is to be believed, the exponentially decaying solution should be replaced with the complex exponential wave, don't you think?

The ansatz in post #1 will yield the correct answer for scattering states.
For the bound state, as E<0, k will be imaginary and if you write exp(-kx) on the right, you should also write sinh (kx).

maverick280857
Ok, here is my revised working

$$k = \sqrt{-2mE/\hbar^2}$$

where E < 0

For 0 < x < d

$$\psi(x) = A\sinh(kx) + B\cosh(kx)$$

as $\psi(0) = 0$, B = 0.

for x > d,

$$\psi(x) = C\sinh(kx) + D\cosh(kx)$$

Integrating Schrodinger's equation,

$$k(C-A)\cosh(kd) + \left(kD + \frac{2m\lambda}{\hbar^2}A\right)\sinh(kd) = 0$$

which gives (please correct me if I'm wrong)

$$C = A$$
$$kD + \frac{2m\lambda}{\hbar^2}A = 0$$

So,

$$\psi(x) = \left\{\begin{array}{cc}A\sinh(kx) & \mbox{ for } 0 < x < d\\A\sinh(kx) -\frac{2m\lambda}{\hbar^2 k}A\cosh(kx) &\mbox{ for } x > d\end{array}\right.$$

But this doesn't give me any condition connecting $\lambda$ and $d$. Also, how can I write the wavefuntion as sin(kx + phi) when the delta barrier is present? The question suggests I should do so, to get to a transcendental equation for phi.

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sigh, for x>d an independent combination of sinh and cosh is not allowed, as I thought I had explained before as these functions diverge when x goes to infinity. Only for D=-C, that is, for exp (-kx), a normalizable solution does result.

maverick280857
sigh, for x>d an independent combination of sinh and cosh is not allowed, as I thought I had explained before as these functions diverge when x goes to infinity. Only for D=-C, that is, for exp (-kx), a normalizable solution does result.

Oops, I'm sorry...the last step I wrote here is wrong...I figured out the problem btw, and was just going to post the correction. As usual, I had messed up because of 'k'. You're right of course, I wrote sinh() and cosh() but on my notebook I still had the old sin() and cos(). Everything works out now...including the condition on lambda and d, which is

$$\frac{2m\lambda d}{\hbar^2} > 1$$

(If this condition is satisfied, there is at least one bound state.)

As for the second part of the question, what they're actually proposing is that we bounce waves back from the wall at x = 0 (when lambda = 0). That is a scattering problem, and so has nothing to do with the bound states that we just computed. I was confused by the language because the question refers to the solution of the first part.

PS -- A bad day :-/

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maverick280857
DrDu, can you help me understand this second part properly. I am stuck somewhere.

An observer at $x = \infty$ can infer the existence of $V(x)$ by bouncing back waves from the wall at $x = 0$. For $\lambda = 0$, the wavefunction with energy $E = \hbar^2 k^2/2m$ would be $\psi(x) = \sin(kx)$. But with $\lambda \neq 0$, the wavefunction acquires a phase $\phi$ so that $\psi(x) = \sin(kx + \phi)$ at large x.

I have to find an equation determining $\phi$. In this case, the wavefunction is $\psi(x) = B\sin(kx + \phi)$ (for E > 0 and real k) for large x. Can this wavefunction be used as that near the delta well? And what is the wavefunction for $0 < x < d$ when the delta well is present? Can that just be $A\sin(kx)$?

You can use that ansatz for x>d and for 0<x<d, A sin kx is ok.

maverick280857
You can use that ansatz for x>d and for 0<x<d, A sin kx is ok.

Ok, so if I do that,

continuity at x = d gives

$$A\sin(kd + \phi) = B\sin(kd)$$

and integrating Schrodinger's equation around x = d gives

$$\frac{\hbar^2 k}{2m}\cos(kd + \phi) + \left(\lambda-\frac{\hbar^2 k}{2m}\cot(kd)\right)\sin(kd + \phi) = 0$$

This equation can be solved 'exactly' for $\tan(kd + \phi)$ and hence for $\phi$. But the questions asks for a 'transcendental equation' for $\phi$.

Hm, isn't tan(kd+phi)=y a transcendental equation for phi?

Btw, I have an extra problem for you.
You found already the k for which for x>d phi is proportional to exp(-kx).
Can you find other solutions of that equation with complex k ?

maverick280857
Hm, isn't tan(kd+phi)=y a transcendental equation for phi?

According to my understanding, it would be if y is a function of phi (http://en.wikipedia.org/wiki/Transcendental_equation). I am confused here because the question says 'the transcendental equation for phi cannot be solved exactly. For small energies, assume $\phi \approx a k$ and find $a$'. We are assuming that $\psi(x) = B\sin(kx + \phi)$ all the way up to $x = d$ (whereas the question gives it as an asymptotic form of $\psi(x)$), and that $\psi(x) = A\sin(kx)$ for $0 < x < d$. What is wrong with these assumptions?

maverick280857
You found already the k for which for x>d phi is proportional to exp(-kx).

I'm not sure I follow you. Can you please restate this statement? The condition for the existence of a bound state is obtained from

$$k = \frac{m\lambda}{\hbar^2}(1-e^{-2kd})$$

when $k = \sqrt{-2mE/\hbar^2}$ for E < 0 (bound states). I argued that this equation has a solution only if the RHS as a function of k has a slope greater than that of the LHS. That's how I got the condition for the existence of a bound state. Is this what you were referring to?

How shall I say? For a potential of infinite range, the statement only holds asymptotically. But for a potential of finite range d, if A exp (ikx)+B exp(ikx) =C sin (kx+phi) holds asymptotically (i.e. abs(A)=abs(B)) then it holds down to x=d.

maverick280857
How shall I say? For a potential of infinite range, the statement only holds asymptotically. But for a potential of finite range d, if A exp (ikx)+B exp(ikx) =C sin (kx+phi) holds asymptotically (i.e. abs(A)=abs(B)) then it holds down to x=d.

Thanks for your reply DrDu. That does clear things up a bit. But how does one get a transcendental equation for $\phi$ which 'cannot be solved exactly'? I seem to have found one which can be solved exactly for $\phi$. They seem to want something like $\phi = f(\phi)$. Do you see something wrong in my reasoning?

As you keep complaining I just tried to solve the problem for the bound state.
so for x<d psi is A sinh kx and for x>d psi is B exp(-kx).
The condition from the delta function is (with the limit epsilon to 0 understood)
$$\frac{\psi'(d+\epsilon)}{\psi(d)}-\frac{\psi'(d-\epsilon)}{\psi(d)}=\frac{2m\lambda}{\hbar^2}$$ or
$$-k - k \coth(kd)=\frac{2m\lambda}{\hbar^2}$$.
That looks quite transcendental and different from what you ever proposed.

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Your post #20 seems to be correct. So the only transcendental content is the arctan.

maverick280857
As you keep complaining I just tried to solve the problem for the bound state.
so for x<d psi is A sinh kx and for x>d psi is B exp(-kx).
The condition from the delta function is (with the limit epsilon to 0 understood)
$$\frac{\psi'(d+\epsilon)}{\psi(d)}-\frac{\psi'(d-\epsilon)}{\psi(d)}=\frac{2m\lambda}{\hbar^2}$$ or
$$-k - k \coth(kd)=\frac{2m\lambda}{\hbar^2}$$.
That looks quite transcendental and different from what you ever proposed.

I wasn't complaining DrDu, appreciate your help :-). But this is what I did to get that condition for the delta function:

$$-\frac{\hbar^2}{2m}\int_{d-\epsilon}^{d+\epsilon}\frac{d^2\psi}{dx^2}dx - \lambda\int_{d-\epsilon}^{d+\epsilon}\psi(x)\delta(x-d) dx = E\int_{d-\epsilon}^{d+\epsilon}\psi(x) dx$$

the rhs = 0 in the limit of epsilon -> 0. This gives the equation I got.

By the way, the equation you wrote in your post does not contain any phase $\phi$. I assume you have solved only the bound state problem, which I have solved as well. I am now referring to the problem I wrote in post #18 below.

Think of the delta function, with positive lambda, as a very narrow, deep potential well. Use your common sense and experience with the potential well to get the appropriate general form of the solution.
Regards,
Reilly Atkinson

maverick280857
Think of the delta function, with positive lambda, as a very narrow, deep potential well. Use your common sense and experience with the potential well to get the appropriate general form of the solution.
Regards,
Reilly Atkinson

Reilly, thanks for your reply, but which part are you referring to?

As I stated in an earlier post, I was able to get the bound state solution. But in post #18, I posed the second part of the problem, which deals with the asymptotic nature of the wavefunction in case of a scattering problem for the same V(x). I solved that and got an equation for the phase shift.

But as far as I can tell, the phase shift can be determined exactly from that equation and unless I have made a mistake in the computation, there is no need to assume that $\phi \approx ak$ and determine $a$ as the question demands.