Graduate Doubts in a lattice translation example

[Lebnm]

I have two question about a exemple given in the Sakurai's quantum mechanics book, section 4.3. Let's consider an electron in a periodic potential $V(x + a) = V(x)$, that has the form of a wave. We will take the potential to go to infinity between two latteces sites, such that its form change to something like this : ...U U U U U U ... In this case, the book says that one possible candidate to the ground state of the system is a state $|n \rangle$ where the electron is completely localized in the $n$th site, and so we have $\hat{H}|n\rangle = E_{0} |n \rangle$. But why? I can't see why this state have to be the lowest energy state.
The second question is: Let's go back to the first case, where the potential is finite. In this case, the states $|n \rangle$ are not more hamiltonian eigenstates, that is, $H$ is not diagonal in this basis. I also don't understend why this happen. This seems intuitive, but I can't find a good reason for this.

 

[DrClaude]

When the potential goes to infinity, the wave function goes to zero. Therefore, a localized solution, for which the wave function must go to zero outside one potential well (otherwise, it is not localized), is compatible with the full Hamiltonian. This is no longer the case when the potential barrier between wells is finite.

 

[hilbert2]

For example, in a system where the potential is a rectangular wave of infinite amplitude:

$V(x) = 0\hspace{20pt}$ when $\hspace{20pt}\sin\left(\frac{2\pi x}{a}\right)<0$, and

$V(x) = \infty\hspace{20pt}$ when $\hspace{20pt}\sin\left(\frac{2\pi x}{a}\right)\geq 0$,

any solution where the $\psi (x)$ is one of the particle-in-box eigenstates on one of the intervals where $V(x)=0$ and $\psi (x)$ is zero everywhere else, is an eigenstate of the full hamiltonian. If $\psi (x)\neq 0$ on more than one interval, it has to be the same particle-in-box eigenstate on all of them.

 

[hilbert2]

Actually, there's something quite special about this kind of systems... Suppose I have a double infinite square well, where the width of the left "box" is $L_1$ and that of the right box is $L_2$. The ground state wavefunction is such that all probability density is in the box of longer width, and is there the same as for the ground state of a normal particle-in-a-box system.

Now, if $L_1 \neq L_2$ and I put equal amounts of probability in both boxes, with the wavefunctions coinciding with the ground states of ordinary particle in box systems with those widths, then the probability density $|\psi (x,t)|^2$ will be time-independent but the state is not an eigenstate of $H$ because the time dependent phase factor $e^{-iEt/\hbar}$ is not the same in the left and right compartments.

I wonder if there are any other systems where $\frac{\partial |\psi (x,t)|^2}{\partial t} = 0$ does not imply that $\psi (x,t)$ is an energy eigenstate?

 

[Lebnm]

thank you!