Answer
In all cases $\psi=0$, which is not a physically acceptable solution as it cannot be normalized.
Work Step by Step
The time-independent Schrodinger equation for a particle of mass m inside the well,where potential V=0,is given by-
$(-\hbar^{2}/2m)\frac{d^{2}\psi}{dx^{2}}=E\psi$
where
$\hbar$=Reduced Plank's constant
E=Energy of the particle
For E=0,the above equation becomes
$(-\hbar^{2}/2m)\frac{d^{2}\psi}{dx^{2}}=0$
$\frac{d^{2}\psi}{dx^{2}}=0$
The general solution is
$\psi(x)=A+Bx$, where A and B are two arbitrary constants.
At x=0, $\psi(x=0)=A$
Now the wave function becomes-
$\psi(x)=Bx$
From boundary condition we have $\psi(a)=0$,i.e,
$\psi(x=a)=Ba=0\implies B=0$
$\therefore\psi=0$
Now for E$\lt0$ we can write
$\frac{d^{2}\psi}{dx^{2}}=k^{2}\psi$
where $k^{2}=-2mE/\hbar^{2}$, k=Real positive constant
So, $\psi(x)=Ae^{kx}+Be^{-kx}$
$\therefore\psi(0)=A+B=0\implies A=-B$
$\psi(x)=A(e^{kx}-e^{-kx})$
Again,the boundary condition $\psi(a)=0$ yields,
$\psi(a)=A(e^{ka}-e^{-ka})=0$
So, either A=0,which implies $\psi=0$
or $(e^{ka}-e^{-ka})=0$
$e^{ka}=e^{-ka}$
$e^{2ka}=1$
$2ka=ln(1)=0$
$k=0$
k=0 implies that $\psi=0$
So in all cases we find that $\psi=0$ which is not an physically acceptable solution as it cannot be normalized.
