Semi-infinite + finite potential well QM

[Kyuubi]

Homework Statement

A particle of mass m is subject to the potential given below (relevant equations) with V0 > 0.

Relevant Equations

\begin{align*}
V(x) = \left\{ {\begin{array}{*{20}{l}}\infty& x < 0\\-V_0&0<x<a\\0&x>a\end{array}} \right.,\\
\end{align*}

I want to verify some inspection I'm making at this problem. Because of the infinite barrier at $x=0$, we expect the wave function to take the value 0 there to preserve continuity. As such, we can make the conclusion that the wave function will just be a sine term in the [0,a] region.

But looking at the discussion of the finite well in Griffiths' QM, we are basically just taking the odd solution of the finite well, and instead of analyzing the $x>0$ half and saying that the $x<0$ region is replicated with $-\psi(-x)$, we are just saying that the left half is 0. This is also taking into consideration the fact that the even part of the solution is also not included.

So the solution to this problem should simply just be the odd solution of the centered finite well.
Is this a correct assessment?

Note: I am only interested in the bound states here as of now.

 

[kuruman]

So the solution to this problem should simply just be the odd solution of the centered finite well.
Is this a correct assessment?

I am concerned about the meaning of "just be the odd solution". As far as sketching the wavefunction, yes. But you would have to recalculate the energies of the bound states. This situation is a hybrid of the infinite potential well and a square well with bound states. I would solve the problem formally by writing the solutions in terms of constants to be determined by matching boundary conditions etc. instead of "just writing the odd solution." Don't forget to normalize in the end.

 

[vanhees71]

I'd just solve the boundary problem. You have $\psi(x)=0$ for $x<0$, which means that you need the boundary condition $\psi(0)=0$. Further at $x=a$ both $\psi$ and $\psi'$ should be continuous. Then for the possible energy eigenvalues you can solve for $-V_0 \leq E \leq 0$ and for $E>0$, separately. It should be clear then, which of these values refer to bound states ("discrete" energy eigenspectrum) and scattering states ("continuous" energy spectrum).