# Delta well + infinite barrier -> bound state

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?

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

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.

reilly
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.