Particle in a one-sided infinite step function plus a delta-function attractive well Given the following potential:
$$
\begin{cases} 
      \infty & x\leq -d \\
       -V_0 \delta(x) & x > -d 
   \end{cases}
$$
with d > 0
I would like to compute the condition that has to be verified in order to have at least one bound state.
From $\hat{H} \psi(x) = E_n \psi(x)$ and assuming we have a bound state (E < 0), I get as general solution for $\psi(x):$
$$
\begin{cases} 
      0 & x < -d \\
      A(-e^{-2kd} e^{kx} + e^{-k x}) & -d < x < 0 \\
      Ce^{-kx} & x > 0
   \end{cases}
$$
where $k = \sqrt{\frac{2m|E|}{\hbar^2}} $
By imposing the boundary conditions in $x = 0$: $\Delta \frac{d \psi(x)}{dx} = -\frac{2 m V_o}{\hbar^2} \psi(0), \hspace{0.5 cm} \psi(0^-) = \psi(0^+)$ $\hspace{0.5 cm}$  and $\hspace{0.5 cm}$ $\psi(-d) =0$,
I conclude that the quantization of the energy of the system is given  by:
$$
\begin{equation} 
      \frac{1+e^{-2kd}}{1-e^{-2kd}} = 1 - \frac{2 m V_0}{\hbar^2 k}
\end{equation}
$$
Which doesn't make any sense due to the fact that $kd > 0$ and the left member of this equation is bigger than 1 in this region.
Could someone explain me what is wrong?
 A: Be careful with those deltas
So in the first part of your question you correctly seem to determine a form
$$\psi(x)=A(e^{-k|x|}-e^{-k(x+2d)}),$$
having the basic property that on every interval it has an $Ae^{kx}+Be^{-kx}$ dependence and it must go to zero at $x=-d$ and there is a continuity in the value but discontinuity in the derivative at $x=0.$
Then you have the desire to set $k$ to make the “kink” in the wavefunction match the Delta function potential via the Schrödinger equation,$$E\psi(x) = -\frac{\hbar^2}{2m} \psi''(x)-V_0\delta(x)\psi(x)$$
(One word of caution, much like the Dirac Delta function potential, the units of $V_0$ are Joule-meters.) So we integrate both sides in a small interval $\int_{-\epsilon}^{\epsilon}\mathrm dx$ and get your relation
$$
E\cdot\psi(0)\cdot 2\epsilon= -\frac{\hbar^2}{2m} [\psi'(\epsilon)-\psi'(-\epsilon)] - V_0\psi(0).$$Of course in the limit the left hand side goes to zero and we get your “boundary condition,” but maybe indulge me a bit and instead of interpreting this as a boundary condition, let's interpret it as an experiment to determine $V_0$. So someone gives us this wavefunction (say via tomography) and asks us how deep the delta potential must have been to create this shape.
Now the second term of $\psi(x)$ has a continuous first derivative so it doesn't change between $-\epsilon$ and $+\epsilon$ and so it just subtracts itself away. Instead we're just looking at $\exp(-k|x|)$ which transitions abruptly from slope $+k$ to slope $-k$ at x=0,
$$
0= -\frac{\hbar^2}{2m} [{-k A}-kA] - V_0~A(1-e^{-2kd}),\\
V_0=\frac{\hbar^2 k}{m(1-e^{-2kd})}.$$
So that looks fine and easy to calculate.
This is where I think you might have gone wrong? Don't get me wrong, I like your numerator of $1-e^{2kd}$ more, it looks like I could turn it into a hyperbolic tangent, maybe derive it easier by translating $x\to x+d$ or so, so that on the interval $0<x<d$ then $\phi(x)=A\sinh(kx),\phi'(x)=Ak\cosh(kx),$ wham, bam, done in time for tea. But I think you just did $\phi'(0^-)/\phi(0)$ or so and it just doesn't work that way? Not sure.
excuse me as I nerd-snipe myself
Now we also have an interesting problem which is that we apparently are depending on this function $f(x)=x/(1-e^{-x})$ to be invertible if we want to go the direction that you wish to go, as $$k=\frac{1}{2d} ~f^{-1}\left({2dmV_0
\over\hbar^2
}\right),$$
And then you can get the kinetic energy with $\hbar^2k^2/2m$ as per usual. Now, we know that $f$ is not going to have an inverse in terms of any sorts of elementary functions but the real danger is if it doesn't have an inverse at all. If $f$ fails to have an inverse then given some $V_0$ there is no such $k$.
One easy way to show that something is invertible is to show that it is monotonically increasing (sufficient, but not necessary), and I think we can do that here? The derivative seems to have a positive squared denominator and a numerator factorable as $e^x(e^x-x-1)$ and the Taylor expansion of the second $e^x$ makes this also a product of two positive numbers... So if the derivative is always positive then it's monotonically increasing and hence invertible.
Actually Wolfram Alpha is even nicer and tells us that the inverse is just $f^{-1}(x) = x+W(-xe^{-x}),$ where $W$ is the “product log function”.
So I think you might still need a condition of adequacy for the physics, probably $k>0$ so that $\psi(x)$ is normalizable off to positive infinity. Thankfully that's just $f(0)$ which we can easily calculate and thankfully the only condition we need from that is $V_0>0$ which I am kind of surprised and grateful for.
A: The position delta distribution $\delta(x)$ has units of inverse length, so that $\int \mathrm dx\ \delta(x) = 1$ is dimensionally consistent. Therefore, dimensional consistency suggests your potential should be
$$
V(x) = \begin{cases}
\infty & x < -d \\
-V_0 a \delta(x) & x > -d
\end{cases}
$$
where $a$ is a new parameter with units of length.
In the limit $d\to\infty$ you would expect to recover the single bound state of the isolated delta-well potential; the additional parameter suggests you should recover the bound state in the limit $d \gg a$.
In your quantization condition, note that $\frac{2mV_0}{\hbar^2 k}$ has the same units as $k$, but you’re using it in a dimensionless context.
However, changing $V_0\to V_0 a$ doesn’t fix the problem you’ve identified in your quantization condition, that one side is definitely larger than one while the other side is definitely smaller.  The problem is still present even if you take the $d\to\infty$ limit and write
$$
\lim_{d\to\infty} \frac{1 + e^{-2kd}}{1 - e^{-2kd}} = 1 \overset{?!}{=} 1 - \frac{2mV_0 a}{\hbar^2 k}
$$
There are two possible explanations for this scenario:

*

*You’ve made some error in setting up your boundary conditions. When you fix that error, you’ll uncover the known solution $E=-{m(V_0a)^2}/{2\hbar^2}$ for the $d\to\infty$ limit, and a correction of order $a/d$ for a distant infinite barrier.


*The bound state of the delta potential is “delicate” enough that an infinite potential barrier, at any distance, destroys it.  In that case you might look at a finite barrier
$$
V(x) = \begin{cases}
V_1 & x < -d \\
-V_0 a \delta(x) & x > -d
\end{cases}
$$
and expect the existence of the bound state to have a condition like $V_1 d \lesssim V_0 a$.
I’m leaning towards #1, another error, but I wouldn’t be too surprised either way.

There is a sign error in your middle definition of $\psi$:
$$
\psi|_{x=-d} = A\cdot(-e^{-3kd} + e^{-kd}) \neq 0
$$
Using instead
$$
\psi|_{-d<x<0} = A\cdot(-e^{+2kd}e^{+kx} + e^{-kx} ) 
$$
gives the quantization condition
$$
k = \frac{mV_0 a}{\hbar^2} ( 1 - e^{-2kd} )
$$
For $d\to\infty$ this does in fact reduce to the condition for the unperturbed attractive delta-well, and we recover the single solution $k_\infty = \frac{mV_0 a}{\hbar^2}$.  For finite $d$, there is a solution with $k=0$ which corresponds (after normalization) to the zero-everywhere wavefunction. There is a nontrivial solution only if the right-hand-side is initially steeper in $k$ than the left-hand-side; that is, if
$$
%\frac{\mathrm d}{\mathrm dk} k_\infty(1-e^{-2kd}) = k_\infty \cdot (+2 d) > 1
2d > \frac1{k_\infty}
$$
The value of the solution will involve the Lambert W function (a normal person would find it numerically). Satisfyingly, it is the case that a "deeper" or "wider" attractive well, with larger $V_0$ or $a$, is more likely to retain its bound state at a given $d$. My intuition about splitting the "strength" of the well into two factors with interpretable units turned out to be not-helpful in this case.
A: My guess would be to say that you are right which would mean that there are no bound states for this potential.
I think this is the case since for a simple delta-function attractive well there is only one bound state, so if we add another condition (in this case the infinit step) it could remove that unique allowed state, which would result in no allowed bound state.
EDIT: If your exercise states that there is a bound state, my guess must be wrong but I still don't see a mistake in your calculations...
