If a pendulum of mass $m$ and length $l$ is considered quantum mechanically, what will be the approximate ground state energy? I have seen that the solution of the quantum pendulum is obtained by solving the Mathieu's equation form of the Schrodinger equation and it also depends on a parameter 'q' called the energy barrier.
But can we approximate the energy levels of the quantum pendulum to that of the quantum harmonic oscillator? Under what limit can we do this?
 A: I could not find the reference I mentioned where they perform perturbation theory starting from both the quantum harmonic oscillator limit and the quantum rotor limit and connect their solutions and energies with the exact solutions of the quantum pendulum. Here I will just briefly scheme the basics of this procedure.
The Hamiltonian for a quantum pendulum can be written in terms of the angle with respect to the vertical $\phi$ with $-\pi \leq \phi < \pi$ producing the Schrödinger equation
\begin{equation}
0 = \left[ - \frac{\hbar^2}{2 I} \frac{\partial^2}{\partial \phi^2} + I \omega^2 \left( 1 - \cos \phi \right) - E \right] \psi\left( \phi \right) \, 
\end{equation}
with $I = m R^2$ and $\omega = \sqrt{\frac{g}{R}}$. With a transformation like
\begin{equation}
z = \frac{\pi}{2} + \frac{\phi}{2} \, ,
\end{equation}
you can put it the form of Mathieu equation,
\begin{equation}
0 = \frac{\partial^2 y}{\partial z^2} + \left[ \lambda - 2 q \cos(2z) \right] y \,
\end{equation}
with
\begin{equation}
q = \frac{R^4}{L^4_{HO}}
\end{equation}
where
\begin{equation}
L_{HO} = \sqrt{\frac{\hbar}{2 m \omega}}
\end{equation}
is the standard factor of the harmonic oscillator position operator when written in terms of creation annihilation operators, $ x = \sqrt{\frac{\hbar}{2 m \omega}} \left( a + a^\dagger \right)$. Notice that the energy is in $\lambda$,
\begin{equation}
\lambda(q) = - 2 q + 4 \sqrt{q} \frac{E(q)}{\hbar \omega} \, ,
\end{equation}
or, in other words,
\begin{equation} \tag{E}\label{E}
\frac{E(q)}{\hbar \omega} = \frac{\sqrt{q}}{2} + \frac{1}{4 \sqrt{q}} \lambda(q) \, .
\end{equation}
Several properties are well known of the periodic solutions with period $\pi$ of the Mathieu equations, which are the ones of interest to us in this case. In particular, the value of $\lambda$ for these particular solutions is often expanded around two limits: $q \rightarrow \infty$ and $q \rightarrow 0$. Both are useful in their own way.
You can slowly reproduce the terms in the expansion of $\lambda$ around $q \rightarrow \infty$ if you use perturbation theory breaking your Schrödinger equation in the form
\begin{equation}
0 = \left[ - \frac{\hbar^2}{2 I} \frac{\partial^2}{\partial \phi^2} + \frac{1}{2}I \omega^2 \phi^2 - E \right] \psi\left( \phi \right) + \Delta H \psi\left( \phi \right) \, 
\end{equation}
where we see the order-$0$ Hamiltonian is of the harmonic oscillator kind, and $\Delta H$ contains all the $\mathcal{O}(\phi^k)$, $k \geq 4$ terms of the expansion of the potential term around $\phi = 0$. The energies of the order-$0$ Hamiltonian are the well known
\begin{equation}
E^{(HO)}_k = \frac{1}{2} \left( 2k + 1 \right) \hbar \omega \, ,
\end{equation}
with $k \in \{0\} \cup \mathbf{Z^+}$. When putting the subsequent terms in the expansion, in turns out the term proportional to $\sqrt{q}$ in (\ref{E}) is ultimately cancelled, and the $E^{(HO)}_k$ contribution is followed by a $\mathcal{O}\left(q^{-1/2}\right)$ correction.
On the other hand, you can slowly reproduce the terms in the expansion of $\lambda$ around $q \rightarrow 0$ if you use perturbation theory breaking your Schrödinger equation in the form
\begin{equation}
0 = \left[ - \frac{\hbar^2}{2 I} \frac{\partial^2}{\partial \phi^2} - E \right] \psi\left( \phi \right) + \Delta \tilde{H} \psi\left( \phi \right) \, 
\end{equation}
where we see the order-$0$ Hamiltonian is of the free rotor kind, and $\Delta \tilde{H}$ contains all the original potential term. The energies of the order-$0$ Hamiltonian are
\begin{equation}
E^{(R)}_n = \frac{n^2}{\sqrt{q}} \hbar \omega \, , 
\end{equation}
with $n \in \{0\} \cup \mathbf{Z^+}$. This unperturbed spectrum is also degenerate for $n \geq 1$. I have not fully checked this case, but I believe this order-$0$ energy is universally followed by the $\frac{\sqrt{q}}{2}$ term in (\ref{E}), followed then by higher $q$-power contributions which at some point should break the unperturbed degeneracy.
Turns out Mathematica has functions for the values of $\lambda$ that we need. Using the notation of Mathieu functions, their characteristic values for periodic solutions of period $\pi$ are labeled $a_{2l}(q)$ and $b_{2l+2}(q)$, for even and odd solutions respectively. The result for the exact values of the energy can be written
\begin{eqnarray}
\frac{E_r}{\hbar \omega} &=& \frac{\sqrt{q}}{2} + \frac{1}{4 \sqrt{q}} a_{r}(q) \, \, \, , \, \, \,\mathrm{for} \, \,  r \, \,  \mathrm{even} \, ,
\\
\frac{E_r}{\hbar \omega} &=& \frac{\sqrt{q}}{2} + \frac{1}{4 \sqrt{q}} b_{r+1}(q) \, \, \, , \, \, \, \mathrm{for} \, \,  r \, \, \mathrm{odd} \, .
\end{eqnarray}
Plotting this allows us to visualize very nicely the two limits we have discussed: from the degenerate free rotor at small $q$ to the uniformly spaced spectrum of the harmonic oscillator for large $q$. We also see how, for the ground state, the energy changes continuously from $0$ to $\frac{\hbar \omega}{2}$:

A: My best guess would be $\dfrac{h\nu}{2}$. Note that it requires quantum gravity to check my answer .
