Particle in a box plus step (ground state) I am trying to come up with a QM problem that: 


*

*Can be solved analytically

*Contains a potential that is a sum of some analytically solvable potential and another contribution: $V'=V_0 + V$

*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 $0<\delta<E$ 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:


*

*$\psi(-\frac{\pi}{2})=\psi(\frac{\pi}{2})=0$ 

*$\psi(x)$ is continuous
at $x=0$ 

*$\psi(x)$ is smooth at $x=0$ 

*$<\psi|\psi>=1.$
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. 
 A: 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.
