The minima occur at $x=(2j+1)\pi$, where $V=-2V_0$. Writing $x=(2j+1)\pi+\xi$ gives
$$
V(x)
=V_0(\cos x-\cos2x)
=-2V_0+\frac52V_0\xi^2+O(V_0\xi^4).
$$
The <harmonic approximation near a potential minimum> therefore has frequency
$$
\omega_0=\sqrt{\frac{5V_0}{m}},
$$
because $\tfrac12m\omega_0^2\xi^2=\tfrac52V_0\xi^2$. The two lowest localized energies are
$$
\boxed{
E_0^{\rm loc}\simeq-2V_0+\frac12\hbar\omega_0,
\qquad
E_1^{\rm loc}\simeq-2V_0+\frac32\hbar\omega_0
}.
$$

In the <tight-binding model>, tunnelling between neighbouring wells broadens each localized level into a band,
$$
E_n(k)\simeq E_n^{\rm loc}-2J_n\cos(2\pi k).
$$
For large $V_0$, the hopping amplitudes $J_n$ are exponentially small compared with the oscillator spacing. Hence the gap between the first two bands is
$$
\boxed{
\Delta_{\rm strong}
\simeq E_1^{\rm loc}-E_0^{\rm loc}
=\hbar\sqrt{\frac{5V_0}{m}}
}
$$
up to exponentially small tight-binding bandwidth corrections.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 35C / b / ii / Solution

Near a nondegenerate minimum $x_0$, a smooth potential satisfies
$$
V(x)=V(x_0)+\frac12V''(x_0)(x-x_0)^2+\cdots.
$$
The localized levels are therefore approximately those of a harmonic oscillator of frequency $\sqrt{V''(x_0)/m}$; tunnelling broadens each level into a narrow band.


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 Incoming links: Harmonic approximation near a potential minimum