# Understanding a 1-dimensional QM bound state

My lecturer has prescribed some practice problems, including the following:

Consider the potential $$V(x)= \begin{cases} \ + \infty & x<0 \\ -V_0 & 0a \end{cases}.$$ What is the minimum value of $$V_0>0$$ such that there is a bound state?

This has confused me. In the one-dimensional case $$E\geq -V_0$$ (the absolute minimum) always; while in the case of a bound state $$E. This leads me to believe $$V_0=0$$ is the minimum such value, but $$V_0>0$$ from the premise of the question.

I usually take pure math courses so I'm not as well versed in the usual heuristics physicists employ. Is this just a case of playing fast and lose with strict/non-strict equalities (for example, in the definition of the potential itself)?

• Does that not clear up your question, though? I guess I am confused what you are asking. The question is: what is the minimum value of $|V_0|$ such that there exists a solution to the time-independent Schrodinger equation with energy $E$ satisfying $-|V_0| < E < 0$? Solving the equation and demanding continuity of the wavefunction and the first derivative will give you an equation for $E$ that may or may not have solutions in this range depending on $V_0$.
– d_b
Commented Dec 9, 2019 at 0:52
• Does this answer your question? 1D Finite potential well: solutions with $\sinh$ and $\cosh$? Commented Dec 9, 2019 at 1:08
• Commented Dec 9, 2019 at 1:10

This is a classic finite well problem. The ansatz solution \begin{align} \psi(x)&=\left\{\begin{array}{cl}A\sin(kx) & 0< x and continuity conditions on $$\psi$$ and its derivative at $$x=a$$ lead to a transcendental equation depending on $$E$$ and $$V_0$$ through $$k$$ and $$\kappa$$.

Zeroes of this transcendental equations correspond to bound states so you need to find the condition on $$V_0$$ to have at least one $$0$$. Note that this condition will also depend on the mass $$m$$ of the particle and the width $$a$$ of the well.

You can connect your solution to the finite well problem because the boundary condition make your problem equivalent to finding the odd-parity states of a finite well extending from $$-a$$ to $$a$$.

• Thank you! I've found the transcendental equation $-cot(z)=\sqrt{(\frac{z_0}{z})^2 -1}$, where $z=a\frac{\sqrt{(E+V_0)2m}}{\hbar}$, $z_0=\frac{a}{\hbar} \sqrt{2m V_0}$. I've concluded by looking at graphs that since these graphs always intersect $V_0$ can be as small as we want - is this correct? Commented Dec 9, 2019 at 22:17
• @jonathan There is NOT always a solution when the solution must be $0$ at the origin. Check the curves on the linked wiki page to see this (the colors are not perfect but the graphics are still clear.) Commented Dec 9, 2019 at 22:36

I think you may be helped if you tried to draw the potential. It sounds like you are confused on the concepts, not the math involved. On the left side of the graph, it's infinite. On the right, it's zero everywhere except between 0 and a. And in that region, the potential is $$-V_0$$ So for finite box length, you will have a bound state if $$E. At $$-\infty$$ $$V=\infty$$ so that's no problem. At $$+\infty, V=0$$ So in order for you wavefunction to be bound in this system, its energy must be negative.

If $$V_0=0$$, then you have no well, just a free particle confined to be from $$x=0\rightarrow\infty$$, and correspondingly no bound states. For larger $$V_0$$ is, you have a deeper your finite potential energy well between 0 and a. If $$V_0 = -\infty$$, then you effectively have an infinitely deep well there with the usual bound states of $$E_n = -\infty + \frac{\hbar^2\pi^2}{2ma^2}$$ & $$\psi_n=\sqrt{\frac{2}{a}}sin{\frac{n\pi x}{a}}$$

So for finite positive values of $$V_0$$, you are in the in between range and may or may not have bound states depending on the length of the box and the depth of $$V_0$$. To solve for the allowed energies, you'll need to set the appropriate boundary conditions on $$\psi$$ and $$\frac{d\psi}{dx}$$ at x=0 and x=a which were mentioned in the comments above. If you have questions on how to approach that math, feel free to ask, but I hope this helps you understand the problem. If it's not clearer, I'm happy to try again.

• This is misleading. You could equally well apply this argument to the symmetric finite square well in the shallow limit, and erroneously conclude that there are well depths at which there are no bound states. Commented Dec 9, 2019 at 7:45
• OP's potential (which has a hard wall on the left!) has no bound states for small-enough $V_0$. Commented Dec 9, 2019 at 15:15