The energy of a harmonic oscillator with amplitude $A$, frequency $\omega$, and mass $m$ is
$$E=\frac 12 m \omega^2A^2 \, .$$
It is intuitive to think that the energy depends on the amplitude because more the amplitude means that the oscillator has more energy, and similarly if the angular frequency is high even then the energy will be more.
Now let's consider a quantum harmonic oscillator (QHO).
The energy is
$$E=\left( n+ \frac 12 \right ) h\nu \, .$$
No amplitude term is there! This is odd because, even if you argue that we are dealing in microscopic domain, we all can agree to the fact that, in general for any mass oscillating under some force, if we have more energy then the oscillator will move farther from its mean position and therefore will have more amplitude.
The relation of energy of QHO can't be wrong, where else the above conception of energy for an oscillator also doesn't seems to be wrong.
Best Answer
You can calculate the variance of the position coordinate, $\sigma_x^2$, for a general eigenstate of the energy $\psi_n$ to be $$\sigma_x^2=\frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right) \, .$$ We can replace the $n$ dependance by energy dependence using the relation $$E_n = \hbar\omega\left( n+\frac{1}{2} \right)$$ and we get $$\sigma_x^2=\frac{\hbar}{m\omega}\frac{E_n}{\hbar\omega} \, .$$ Rearranging we get $$E_n=m\omega^2\sigma_x^2 \, .$$
Remembering that for the classical case $$\sigma_x^2=\frac{1}{2}A^2$$ we retrieve the original relation in the quantum case as well.
Check the derivation at this link