# Particle in a box plus step (ground state)

I am trying to come up with a QM problem that:

1. Can be solved analytically
2. Contains a potential that is a sum of some analytically solvable potential and another contribution: $$V'=V_0 + V$$
3. Is then also easily accessible to perturbation theory by analytically solvable integrals of the form $$\int_\Omega\psi_0^* V\psi_j d\tau$$

After having checked several possibilties$$\dagger$$ I thought I attempt a particle in the box with an additional step:

$$V(x)=\begin{cases} \infty & x\lt -\frac{\pi}{2} \\ 0 & -\frac{\pi}{2} \leq x\lt 0 \\ \delta & 0 \leq x\lt \frac{\pi}{2} \\ \infty & \frac{\pi}{2} \le x \end{cases}$$

so that we can assume $$0<\delta in the SE: $$\frac{\partial^2 \psi}{\partial x^2} + \frac{8 \pi^2 m}{h^2}(E-V)\psi = 0$$ That means it should be a small step and in addition I am only interested in the the ground state solution (for the moment).

For that and according to textbook knowledge I make the following ansatz for the wavefunction:

$$\psi(x)=\begin{cases} 0 & x\lt -\frac{\pi}{2} \\ a \sin(k_1 x) + b \cos(k_1 x) & -\frac{\pi}{2} \leq x\lt 0 \\ c \sin(k_2 x) + d \cos(k_2 x) & 0 \leq x\lt \frac{\pi}{2} \\ 0 & \frac{\pi}{2} \le x \end{cases}$$

The real positive $$k_1$$ and $$k_2$$ follow directly from inserting $$\psi$$ in to the SE: $$k_{i} = \sqrt{\frac{8 \pi^2 m}{h^2}(E-\delta_{i2}\delta)}$$

Now I have to determine the coefficients $$a,b,c,d$$, for that the following conditions come to my mind:

1. $$\psi(-\frac{\pi}{2})=\psi(\frac{\pi}{2})=0$$
2. $$\psi(x)$$ is continuous at $$x=0$$
3. $$\psi(x)$$ is smooth at $$x=0$$
4. $$<\psi|\psi>=1.$$

Boundary condition (1) yields as we know the energy quantization, but I omit the higher values (they are even spaced as expected):

$$k_1 = \arctan{\left(\frac{b}{a}\right)}\frac{2}{\pi}$$ $$k_2 = \arctan{\left(\frac{d}{c}\right)}\frac{2}{\pi}$$

From the continuity (2.) I get: $$b=d$$

from smoothness (3.):

$$a \arctan{\left(\frac{b}{a}\right)} = - c \arctan{\left(\frac{d}{c}\right)}$$

and from the normalization (3.):

$$\frac{\pi\left(-cd + \arctan{\left(\frac{d}{c}\right)}(c^2 + d^2) \right)}{4\arctan{\left(\frac{d}{c}\right)}} = 1$$

Now problems start: I am not sure if the latter three equations can be enough to solve for $$a,b,c,d$$, nor do I see how to do that.

Can someone help?

Footnote:

$$\dagger$$ and a close hit with particle in the box and Tellers $$V = \frac{\alpha}{\cos^2 x}+\frac{\alpha}{\sin^2 x}$$ with $$x\in[-\frac{\pi}{2};\frac{\pi}{2}]$$ potential problem, that is solved here (Pöschl, G. & Teller, E. Z. Physik (1933) 83: 143. https://doi.org/10.1007/BF01331132) but it doesn't work out since the perturbation with respect to the box seems too large for all solvable $$\alpha$$ values.

• Not exceedingly happy with the "homework-and-exercise" tag, (wouldn't mind to get the solution straight away ..) but here you go ... – Rudi_Birnbaum Mar 7 at 17:22

## 1 Answer

It's slightly easier if you set the box to start at $$x=0$$, $$V(x)=\begin{cases} \infty & x\lt 0 \\ 0 & 0 \leq x\lt \frac{\pi}{2} \\ \delta & \frac{\pi}{2}\leq x\lt \pi \\ \infty & \pi \le x \end{cases}$$
which means that you can simplify your wavefunction by taking explicit account of the boundary conditions $$\psi(x)=\begin{cases} 0 & x\lt 0 \\ a \sin(k_1 x) & 0 \leq x\lt \frac{\pi}{2} \\ b \sin(k_2 (\pi-x)) & \frac{\pi}{2} \leq x\lt \pi \\ 0 & \pi \le x . \end{cases}$$ With that, you're reduced to the continuity of $$\psi$$ and $$\psi'$$ at the boundary: \begin{align} a\sin(k_1 \pi/2) & = b \sin(k_2\pi/2) \\ ak_1\cos(k_1\pi/2) & = -bk_2 \cos(k_2\pi/2), \end{align} or in other words $$\begin{pmatrix} \sin(k_1 \pi/2) & \sin(k_2\pi/2) \\ k_1\cos(k_1\pi/2) & -k_2 \cos(k_2\pi/2) \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.$$ Since you're looking for a nonzero solution, you require $$\det\mathopen{}\begin{pmatrix} \sin(k_1 \pi/2) & \sin(k_2\pi/2) \\ k_1\cos(k_1\pi/2) & -k_2 \cos(k_2\pi/2) \end{pmatrix} =0,$$ or in other words $$k_2 \sin(k_1 \pi/2)\cos(k_2\pi/2) + k_1\cos(k_1\pi/2)\sin(k_2\pi/2) =0.$$

Now, let's simplify this a bit. Since you've already set the box length to a useful dimensionless number, it doesn't make sense to keep dimensionful $$\hbar$$ and $$m$$, so this just sets $$k_1=\sqrt{2E}$$ and $$k_2=\sqrt{2(E-\delta)}$$, or in other words $$k_2=k$$ and $$k_1=\sqrt{k^2+2\delta}$$, so we have $$k \sin\mathopen{}\left(\sqrt{k^2+2\delta}\,\pi/2\right)\mathclose{}\cos(k\pi/2) + \sqrt{k^2+2\delta}\cos\mathopen{}\left(\sqrt{k^2+2\delta}\,\pi/2\right)\mathclose{}\sin(k\pi/2) =0.$$ Frankly, I think this is as far as you can get. This is easily solvable numerically (say, by Newton's method starting with $$k=1$$), but it looks to be out of the envelope of what you could solve analytically - which is not surprising, given that you've essentially set up a finite square well, which cannot be solved beyond a reduction to a transcendental equation.

Here are some mathematical considerations on the solubility and solutions of this equation.

• Great! "a finite square well, which cannot be solved beyond a reduction to a transcendental equation." why is that so? – Rudi_Birnbaum Mar 7 at 19:52
• @Rudi_Birnbaum I'm not sure that there is a concrete reason for that that one can point to. If there is, then I'd sure like to know. – Emilio Pisanty Mar 7 at 19:54
• For $\delta=0$ I get $E=1/2$ and $k=1$(that should be the particle in the box of length $\pi$ in atomic(?) units), fine. For $\delta=0.1$ I get about $k=0.97$. I would have expected the energy to increase if the box gets narrower, any idea? – Rudi_Birnbaum Mar 7 at 20:28
• That's because $k$ is set to $k_2$ for simplicity. The energy does rise. – Emilio Pisanty Mar 7 at 20:36