# Understanding bound states in quantum mechanics

Suppose I have this potential: $$\ V(x)= \left\{ \begin{array}{ll} +\infty & x < 0 \\ -V_0 & 0\leq x\leq a\\ 0 & x>a \\ \end{array} \right. \$$ for $$a>0$$ and $$V_0>0$$. My job is to prove that there are no bound states for some energy, $$E<0$$, such that $$V_0<-E$$.

One way to do this would be to look at Schrodinger Time independent equation:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi(x)=E\psi(x)$$

and simply associate the first part with the second derivative with the kinetic energy and as $$E<0$$ and $$V(x)=-V_0$$ that would imply a negative kinetic energy but that has no physical meaning, thus there are no bound states.

My problem is with the math. Even though that makes sense, if I try to solve the equation somehow I get to this:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi(x)=E\psi(x) \\ \frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}(|E|-V_0)\psi(x)$$

which has indeed a solution:

$$\psi(x<0)=0 \\ \psi(0a)=Ce^{-k_1x}$$

for some constants $$A$$,$$B$$,$$C$$ and $$k^2=\frac{2m}{\hbar^2}(|E|-V_0)$$ and $$k_1^2=\frac{2m}{\hbar^2}(|E|)$$. This is a weird solution but still a solution to the equation that decays exponentially at $$x\to \infty$$ and is already $$0$$ at $$x<0$$. What am I missing and how does this prove that there aren't any bound states?

EDIT:

After some implying continuity and smoothness I get:

$$\ \left\{ \begin{array}{ll} A+B=0 \\ Ae^{ka}+Be^{-ka}=Ce^{-k_1a} \\ Ake^{ka}-Bke^{-ka}=-Ck_1e^{-k_1a} \\ \end{array} \right. \$$

From which I can use only the two first equations and get:

$$\ \left\{ \begin{array}{ll} A=-B \\ 2A\sinh(ka)e^{k_1a}=C \\ \end{array} \right. \$$

Allowing me to write:

$$\psi(x<0)=0 \\ \psi(0a)=2A\sinh(ka)e^{-k_1(x-a)}$$

And finally with the normalization I get:

$$A=\bigg(\frac{1}{k}(\cosh(ak)-1)+\frac{2}{k_1}\sinh^2(ka)\bigg)^{-1/2}$$

which is a mere constant. What am I getting wrong?

As for the energy, I can divide the last two equations and get:

$$\frac{1}{k}\tanh (ka)=-\frac{1}{k_1} \\ \tanh (ka)=-\frac{k}{k_1}\\ \tanh (ka)=-\sqrt{(1-V_0/|E|)}\\ ka=\tanh^{-1}(-\sqrt{1-V_0/|E|}) \\ \frac{2m}{\hbar^2}(|E|-V_0)a^2=(\tanh^{-1}(-\sqrt{1-V_0/|E|}))^2 \\ |E|=V_0+\frac{\hbar^2}{2ma^2}(\tanh^{-1}(-\sqrt{1-V_0/|E|}))^2$$

• inside the well you need complex exponentials? – ZeroTheHero Jun 6 at 0:12
• No, only real exponentials are needed... – Bidon Jun 6 at 0:15
• You still have an equation left, so you should solve for E in it. – Adam Jun 6 at 12:01
• @Adam I did that. See my edit. Even with that, it only tells me that for a given $V_0$ there is only one $|E|$ alowed and there is a $V_{0_{min}}$ such that, for any $V_0>V_{0_{min}}$ would imply $|E|<0$ and that is impossible. – Bidon Jun 6 at 12:20
• You should check your calculation. The units are not correct. You should get an implicit equation for E that needs to be solved numerically. – Adam Jun 6 at 12:23

HINT: You have the equation: $$\frac{1}{ak}\tanh (ka)=-\frac{1}{a k_1}$$ where $$k_1$$ and $$k$$ are both positive. Try graphing the function $$\tanh x/x$$ and see where it could have a solution for $$\tan{x}/x < 0$$.

• oh... I should have seen it. That completely solves it. Thank you. This was driving me insane – Bidon Jun 6 at 13:13

Hint: Your wavefunction should be continuous at $$x=0,a$$ and also differentiable at $$x=a$$. Can you fulfill those requirements?

• I think so, it comes out a bit weird with some hyperbolic functions but I have 3 equations for 3 variables, it should be possible... – Bidon Jun 6 at 0:10
• Go for it. Write the equations in the question and show us what you get. – eranreches Jun 6 at 0:16
• Please see my edit. I am really blind here... – Bidon Jun 6 at 11:42

Bound states are characterized by the property $$\psi(x) \xrightarrow{\vert x \vert \rightarrow \infty} 0$$ (you can find this is in any standard quantum textbook, for example Shankar page 160). In addition, they must be smooth everywhere (consider your boundaries). With the smoothness constraint, you should see that your wave function with energies greater than 0 fail to satisfy the first condition above.

• Please see my edit. I can't really see what I am getting wrong – Bidon Jun 6 at 11:42