hilbert space - Hamiltonian of quantum harmonic oscillator with $psi(x)=delta(x)$: comparison to classical mechanics

I was just reading the question Why can't $\psi(x)=\delta(x)$ in the case of a harmonic oscillator? The accepted answer says that $\psi(x)=\delta(x)$ is a mathematically valid state, though it's not physically possible because delta distributions aren't normalizable. If it's possible, I'd like to ignore the fact that it's un-physical, and instead try to fill in the holes in my understanding of the mathematical model.

The answer concludes that the expectation value of the Hamiltonian of the system is infinity, which makes sense to me because it follows from the proposition that $\left$ is infinity.

However, I see this as a contradiction of my impression that quantum mechanical systems at high energies are easily approximated to classical behavior. For instance, with a particle in a square well, the position probability distribution at high energies is approximately constant across the whole allowed region, which is exactly what classical physics predicts for the same energy.

Going back to the case of harmonic oscillators, I think that a classical model with zero energy would have a position distribution of $\delta (x)$, because the particle's localized in one exact place and it isn't oscillating or moving in any way. So this tells me that a quantum harmonic oscillator with an extremely high energy behaves like a classical zero-energy system, which just generally sounds wrong.

Is it wrong/meaningless to make such comparisons of a zero-energy classical particle to a quantum harmonic oscillator with $\psi(x)=\delta(x)$? Alternatively, is this apparent contradiction actually logically justifiable?

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I think that there's something fishy in my comparison when I say that a classical system shows a position distribution of $\delta(x)$, since the momentum distribution is also a delta function (with a spike at $p=0$). However, the QM handling of momentum would predict a completely different state. But I can't build more out of that reasoning to see if it's relevant.

Furthermore, the lowest energy state of a QM oscillator is nonzero, so there is no valid QM counterpart of a classical oscillator with $\psi(x)=\delta(x)$. However, this holds true for the particle in an infinite well too, so I would expect a symmetry whereby in both cases, high-energy QM descriptions are similar to classical mechanics.

Answer

The most classical-like states of harmonic oscillator are Coherent states of harmonic oscillator, not $\delta(x)$. The reason is that coherent state has balanced uncertainty of coordinate and momentum: $$\Delta x =\sqrt{\frac{\hbar}{2m\omega}}; \Delta p = \sqrt{\frac{m\hbar\omega}{2}},$$ and in the classical limit ($\hbar\rightarrow 0$) $\Delta x \approx 0$, $\Delta p \approx 0$.

In a state with $\phi(x)=\delta(x)$ the momentum uncertainty $\Delta p=\infty$, it does not tend to zero.

UPDATE:

It is not correct to assume that a particle with high energy can always be described by classical mechanics. Consider for example two states $|\psi_1\rangle$ and $|\psi_2\rangle$, in which particle has high energy. Any superposition of them is a valid state in quantum mechanics but does not make sense in classical mechanics.