# Semi-Infinite Potential Square Well: Negative Potential

I am having trouble with a problem involving a Semi-infinite potential square well:

I have written down some notes that I added to the post. I have problems understanding the physical situation. See the figure in the notes. When the potential is just shifted upwards, so that the $$V(x)=0$$ region is $$0, I have no trouble understanding the solution. However, I am just really confused about the negative potential.

As you will see in the notes, I have interpreted Region I as a bound state (e.g. an electron bound to a proton). Region II, I believe, is the free state (e.g. "bound" electron in energy level $$n=\infty$$), since $$V(x) = 0$$. Is this correct?

I face some troubles with the boundary conditions, as one of my wave-functions simply disappear!

Answering your quantum mechanics question seems to become a hobby of mine :). First let me say that there is nothing wrong with shifting the potential up so that $V_{0} = 0$. This choice would indeed simplify the problem as it gets rid of negative energy values and thus avoids the following point of confusion: when $E < 0$ the wave function of a particle in region $\text{II}$ is not a plane wave $\mathrm{e}^{\mathrm{i} k x}$ ($k \in \mathbb{R}$) like that of a free particle moving in vacuum but a decaying exponential $\mathrm{e}^{-\kappa x}$ (with $\kappa \in \mathbb{R}$). So, the choice $V_{0} = 0$ is advisable. Shifting the potential back later is always possible and might be easier.

Nevertheless, of course the problem is solvable irrespective of the choice of $V_{0}$. I think your problem in this case is that you enforced an incorrect boundary condition. To avoid confusion I will start from the beginning:

I will only consider the case $E < 0$ in the following, so $E$ is always a negative number between $-V_{0}$ and $0$. For region $\text{II}$ you have to solve the Schroedinger Equation:

\begin{align} - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= E \psi_{\text{II}}(x) \\ \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= - \frac{2mE}{\hbar^{2}} \psi_{\text{II}}(x) \ . \end{align}

Ok, here comes the first "problem" with the choice of $V_{0}$: In your solution you chose $\frac{2mE}{\hbar^{2}} = k^{2}$ as you would naturally do for a free particle. But since $E$ is negative $k$ will be a complex number (and not a real number as it would be for a free particle) and so this differential equation does not have an oscillatory solution $\psi_{\text{II}}(x) = A \sin(kx) + B \cos(kx)$ with $k \in \mathbb{R}$. Rather you get a decaying exponential function by doing the following: $- \frac{2mE}{\hbar^{2}}$ will be a positive number so you simply set $- \frac{2mE}{\hbar^{2}} = \kappa^{2}$ to get

\begin{align} \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= \kappa^{2} \psi_{\text{II}}(x) \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ . \end{align}

Now, this differential equation has a solution of the form

\begin{align} \psi_{\text{II}}(x) = A_{\text{II}} \mathrm{e}^{\kappa x} + B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ . \end{align}

At this point you can enforce the boundary condition that the wavefunction $\psi_{\text{II}}$ must be square integrable over the whole interval $a < x < \infty$ (Note that you don't have to consider the interval $-\infty < x < \infty$ but only the interval on which $\psi_{\text{II}}$ is defined). Since $\lim_{x \to \infty} \mathrm{e}^{\kappa x} = \infty$ this mean that the prefactor $A_{\text{II}}$ must be equal to zero, so

So much for region $\text{II}$. Let's move on to region $\text{I}$: Here you have to solve the Schroedinger Equation:

\begin{align} - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} - V_{0} \psi_{\text{I}}(x) &= E \psi_{\text{I}}(x) \\ \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &= - \frac{2m(E + V_{0})}{\hbar^{2}} \psi_{\text{I}}(x) \ . \end{align}

Again, care must be taken: Since $-V_{0} < E < 0$ the term $(E + V_{0})$ is positive and so $- \frac{2m(E + V_{0})}{\hbar^{2}}$ is a negative number and you can set $- \frac{2m(E + V_{0})}{\hbar^{2}} = -k^{2}$ which leads to

\begin{align} \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &= - k^{2} \psi_{\text{I}}(x) \qquad \quad \text{with} \qquad k \in \mathbb{R} \ . \end{align}

This differential equation has a solution of the form:

\begin{align} \psi_{\text{I}}(x) = A_{\text{I}} \mathrm{e}^{\mathrm{i} k x} + B_{\text{I}} \mathrm{e}^{-\mathrm{i} k x} \qquad \quad \text{with} \qquad k \in \mathbb{R} \ . \end{align}

(You could also formulate an equivalent solution in terms of $\sin$ and $\cos$ functions using Euler's formula if that is more to your liking. But I prefer plane waves ;)) This is the point where you made your main mistake: You discarded one of the terms in your $\psi_{\text{I}}$ wavefunction because it blew up for $x \to \infty$. But $\psi_{\text{I}}(x)$ is only defined on the interval $0 \leq x \leq a$ and within this interval both exponential terms are perfectly square-integrable and you can't discard any of them.

Finally, you just have to remember that $\psi_{\text{III}}(x) = 0 \ \forall \ x$ and enforce the boundary conditions given by the system:

\begin{align} \psi_{\text{I}}(0) &= \psi_{\text{III}}(0) \quad &&\Rightarrow \quad A_{\text{I}} \mathrm{e}^{0} + B_{\text{I}} \mathrm{e}^{0} = 0 \quad &&\Rightarrow \quad B_{\text{I}} = -A_{\text{I}} \\ \psi_{\text{I}}(a) &= \psi_{\text{II}}(a) \quad &&\Rightarrow \quad A_{\text{I}} \mathrm{e}^{\mathrm{i} k a} - A_{\text{I}} \mathrm{e}^{-\mathrm{i} k a} = B_{\text{II}} \mathrm{e}^{-\kappa a} && \\ & &&\hphantom{\Rightarrow} \ \quad A_{\text{I}} \underbrace{\bigl(\mathrm{e}^{\mathrm{i} k a} - \mathrm{e}^{-\mathrm{i} k a} \bigr)}_{= \, 2 \mathrm{i} \sin(k a)} = B_{\text{II}} \mathrm{e}^{-\kappa a} \quad &&\Rightarrow \quad B_{\text{II}} = 2 \mathrm{i} \sin(k a) \mathrm{e}^{\kappa a} A_{\text{I}} \end{align}

Edit:

Concerning your comment: You can indeed enforce another boundary condition. The probability flux is required to be continuous across any surface, since otherwise the surface would contain sources or sinks. This entails the requirement that both $\psi$ and $\nabla \psi$ must be continuous. The continuity condition $\psi_{\text{I}}(a) = \psi_{\text{II}}(a)$ has already been enforced but the one for the gradient is still left:

\begin{align} \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \ . \end{align}

From the above discussion we know that:

\begin{align} \psi_{\text{I}}(x) &= A_{\text{I}} \underbrace{\left(\mathrm{e}^{\mathrm{i} k x} - \mathrm{e}^{-\mathrm{i} k x} \right)}_{= \, 2 \mathrm{i} \sin(k x)} = 2 \mathrm{i} A_\text{I} \sin (kx) = C_\text{I} \sin(kx) \end{align}

and

\begin{align} \psi_{\text{II}}(x) &= \underbrace{2 \mathrm{i} A_{\text{I}} \sin(k a)}_{= \, \psi_{\text{I}}(x=a)} \, \mathrm{e}^{\kappa a} \mathrm{e}^{- \kappa x} = \psi_{\text{I}}(x \! = \! a) \, \mathrm{e}^{- \kappa (x - a)} \end{align}

with $C_\text{I} = 2 \mathrm{i} A_{\text{I}}$, so

\begin{align} \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \\ k C_\text{I} \cos(ka) &= - \kappa C_\text{I} \sin(ka) \\ \kappa &= - \frac{k}{\tan(k a)} \ . \end{align}

While this equation looks innocent enough it is not easily solved for $E$ and you will need the assistance of something like Mathematica for that.

• Yes, it seems that way :) I totally follow your discussion on making sure $k$ or $\kappa$ is a real value, and generating the general solution thereafter; that makes sense! I got the this for the wave-function ($with 2Ai = A_I$): $$\Psi (x) = \begin{cases} A_I\sin (kx) & 0\leq x \leq a \\ A_I e^{\kappa a} \sin (ka) e^{- \kappa x} & x>a \\ 0 & x<0 \end{cases}$$ But does not continuity at $x=a$ need me to set: $$\psi_I' (a) = \psi_{II}' (a)$$ Based on the equations above, this just gave me $-k = \kappa$ , which is incorrect. How would I go about determining the allowed anergy values? – Yoda May 21 '14 at 8:23
• @AndersMB I think you might have made a small mistake while differentiating the sin function in $\psi_{\text{I}}$. I've added a new paragraph to my answer which shows the way to the (hopefully) correct solution. – Philipp May 21 '14 at 10:30
• Thanks! Sometimes, when I differentiate sine and cosine functions, I forget to switch the trig functions! Just for fun, I found that $E = \frac{V_0}{\tan^{-2}(ka)-1}$. The Problem asked me to determine the allowed energies, but I am not sure how detailed I should be. If, as you say, I need software to do so, then I probably do not need to go very deep into it. Thanks again! You are a lifesaver. I must watch out for these little mistakes at the exam! – Yoda May 21 '14 at 11:00
• @AndersMB Yeah, it's easy to forget something like that. And there is still a problem with this equation: the argument of the tangens contains $E$ (disguised as $k$) so this is still an implicit equation for $E$ that needs to be solved computationally. – Philipp May 21 '14 at 11:07
• Right - I'll just stop there, then, and move on :) – Yoda May 21 '14 at 11:34

I believe there are two seperate mistakes, one in region I and one in region II.

For region I, it looks like you made a sign error, and that you should have $-V_0 - E$ instead of $V_0 - E$.

While this may seen trival at first, the sign is important to the form of the solution to the differential equation (wavefunction).

For region II, the way the diagram of the question is drawn, E is negative. Assure that this fact is properly accounted for in the form of the solution.