I know that in the Kronig Penney model there are values of the energy $E$ for which solutions to the Schrodinger equation don't exist.
I understand that these forbidden values of $E$ form the band gaps.
How, intuitively, does the crystal structure lead to the existence of forbidden bands?
[Physics] Origin of band Gap
solid-state-physics
Related Solutions
Yes, you are correct, and you don't need to assume factorizability--- the situation for a general square-lattice periodic potential is just the same, the bangaps occur at the edge of the k-space domain defined by lattice periodicity.
The reason is all symmetry: the translation group is broken to lattice translations. This means momentum is not conserved, but violations can only come by adding multiples of the k-space reciprocal lattice. At high electronic energies (assuming a smooth V), you don't notice the potential and the bandgaps go to zero, but where V is important, there are bandgaps only when either the x and y momentum are an integer multiple of ($2\pi$ times) the reciprocal lattice size.
Momentum conservation
On the lattice, a momentum state $|k\rangle$ can mix off the potential with momentum states $|k'\rangle$ only when $|k'\rangle$ has the same eigenvalue for the operators $T_x$ and $T_y$, where
$$T_x= e^{iaP_x}$$ $$T_y = e^{iaP_y}$$
Since $T_x$ and $T_y$ commute with the full Hamiltonian. This means that a momentum state only mixes with other states which acquire the same phase for x and y translations by a, and this justifies Bloch's form for the eigenstates--- they are periodic in x and y up to a phase.
The mixing in general will split the energy of the states, which gives rise to the bands. The bandgaps occur at the exact point where the phase acquired when you translate one of x,y by a becomes equal to (an integer multiple of) $2\pi$.
To see why, it is easiest to start with a tight-binding model. Consider a potential which is a deep square well with very high walls, but periodic in x and y. The bound states are only mixed by tunneling, so that the bands are very narrow, their energy is nearly the same as the bound state energy, and these states are mixed with a phase which varies in a linear way from well to well.
If you give a periodic boundary condition on x which says that $\psi(x+a,y)=e^{i\phi}\psi(x)$ where $\phi$ is close to $2\pi$, but slightly less, then $\psi$ in each well is very close to an energy eigenstate, say the ground state, but with a little twist in the tunneling region. The moment this phase $\phi$ equals $2\pi$, however, there is the requirement that the result must be orthogonal to $0$ phase shift, and this condition was absent just before.
It is impossible to satisfy this condition when $\phi=2\pi$ by using a state which is close to the ground state. You must use the first excited state to continue past this value of k. This gives rise to the gap discontinuity.
The discontinuity appears immediately when you add a periodic potential. It always occurs at the first point where you move from the solution being the lowest energy state with the given lattice phase condition to the next lowest energy state with this phase condition. This happens exactly at the boundary of the k-space of the lattice.
Separable potentials
In this case, there is a zero-counting theorem for arranging the eigenstates. The n-th eigenstate has n zeros. In the Bloch case, the parameter "n" is the number of $2\pi$ twists of the phase as you scan from left to right. The first band has no twists, the second band has 1 twist, etc.
The additivity of the energy means that the 2-d bands are sums of the bands of each motion separately. I don't know what more there is to say.
General theorem
it should be possible to prove that the energy as a function of k is continuous except at the boundary of the lattice k-space region It is obvious physically, but I didn't give a proof.
Consider your equation
$$\frac{P}{Ka}\sin(Ka) + \cos(Ka) = \cos(ka) \, .$$
The right hand side can only ever attain values in the range $[-1,1]$. Therefore, if there is an energy $E$ which causes the left hand side to take a value outside the range $[-1,1]$, then that $E$ can never be realized for any value of $k$. In other words, that $E$ is forbidden and we say it sits in a "band gap".
We can see this graphically. Here I plot the energy $E$ versus the value of the left hand side of the equation for the case
$$a=1, \quad b=10^{-4}, \quad U_0=10^4 \,.$$
The vertical blue line indicates the point at which the left hand side of the equation is $-1$. Any points where we go to the left of that line are forbidden. The red band indicates the range of $E$ over which the left hand side of the equation is outside the range $[-1,1]$. As you can see, there is a band of values of $E$ where no possible $k$ can satisfy the equation. This is a band gap.
Best Answer
Diffraction of electron waves inside a periodic crystal structure, i guess? Superposition of wave functions. That is the intuitive way. One more thing, you can get this with perturbation theory too.Or with tight binding approximation. Anyways for electrons that satisfy special condition there will be constructive interference and they will interact strongly with the lattice. This happens exactly on the edge of brillouin zone...and you get from this linear combinations of incoming and reflected waves which then give you possible forms of wave functions. In one dimension you get two forms id est two combinations. When you write down the probability distribution and plot it against the lattice, you get that for the first linear combination electrons are piling up between ions and for the second near the atoms, so you get two distinct energies because of interaction. Where does the wave vector come in play? Through the Bragg condition where we see that this happens for special values of k-vector...and this is definition of Brillouin zone.