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Can anyone help me get a handle on how to go about solving for the energies and wave function for a particle in an infinite box where there is a step function? That is, the potential is infinite for $x<0$ and $x>a$, $0$ in the range $0<x<\frac{a}{2}$, and $V_0$ in the range $\frac{a}{2}<x<a$. This is for self-study from an old problem set posted on MIT OpenCourseWare.

The problem-set asserts that the wave function in the zero potential region is $Asin(k_1*x)$, with k relating to the ordinary energy for a particle-in-a box ($\sqrt{2mE/\hbar^2})$. And it says that the wave function in the stepped up region is $B*sin(k_2(a-x))$, with k2 being ($\sqrt{2m(E-V_0)/\hbar^2}$).

The problem seems to revolve around ensuring that the wave function (and its derivative) actually meet up at the point $x=a/2$. But I'm really confused about how to implement this: I can solve for B in terms of A, $k_1$, and $k_2$ by equating the two wave functions at the point $a/2$, but the math quickly gets too confusing to follow for me. I end up relating $k_1$ to $k_2$ with some messy cotangent functions, but I don't see how that gets me to a single wave function, or the energy levels that I need. Can someone sketch the solution?

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  • $\begingroup$ In problems like this you usually cannot solve the equations for the coefficients and $k$'s in closed form. $\endgroup$
    – Javier
    Commented Nov 9, 2020 at 21:29

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Let's make the following substitutions:

$z = k_2a$, $z_0 = a\sqrt{2mV_0}/\hbar$

By taking $k_1^2 - k_2^2$ you will see that $k_1a = \sqrt{z_0^2+z^2}$. Then by equating the "cotangent mess" you speak of (equating the wave functions (1) and their derivatives (2) and dividing (2) by (1)) you should obtain the following equation:

$$ \frac{\sqrt{z_0^2+z^2}}{a}\mathrm{cot}\Bigg(\frac{\sqrt{z_0^2+z^2}}{2}\Bigg) = -\frac{z}{a}\mathrm{cot}\Bigg(\frac{z}{2}\Bigg)$$

This is very difficult to solve analytically, however you can easily find approximate solutions by Taylor expanding! The Taylor expansion of $\mathrm{cot}(x)$ is $ \frac{1}{x} - \frac{x}{3} + O(x^3)$. By Taylor expanding both sides of the above equation you should get a final answer of:

$$z \sim \sqrt{12 - \frac{z_0^2}{2}}$$

Then just plug in the values for what $z$ is and you should get a very good estimate of the energies. You can then plug those into the wavefunctions given in the problem.

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  • $\begingroup$ Thanks, but could I ask for one last clarification? I see that you're creating new variable z that multiplies the wavenumber by the length of the box. Where do I get my quantization from? I understand this in the generic PIB case, but I'm confused about which boundary condtions give me my quantization here, and what I would do with z, even once I compute it. $\endgroup$ Commented Nov 9, 2020 at 18:59
  • $\begingroup$ Based on the above, I can only find 2 energies, which are the positive and negative square roots of what I gave for z. I don't think there are more solutions than these 2 energies (I may be wrong but I don't see how you would get any other solutions for E). $\endgroup$
    – Framazu
    Commented Nov 9, 2020 at 20:17

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