In the analysis of Harmonic Oscillator, it is claimed that $\langle\hat H\rangle$ cannot be zero, why is it so?

I mean $\hat H = \frac{ \hat p^2 }{2m } + \frac12 k \hat x^2$, and $$\left<x^2\right> = \int dx (x\psi(x))^\dagger (x\psi(x)) = 0$$ would imply that $x\psi(x) = 0 \quad \forall x.$ In particular, this is true when $x = 0$, so we have two options; either $\psi(x) = 0$ or $\psi(x) = \delta(x)$.

So, why can't $\psi(x) = \delta(x)$ in the case of Harmonic oscillator ?

Note: $\hat H = \frac{ \hat p^2 }{2m } + \frac12 k \hat x^2 $


The state $\psi(x) = \delta(x)$ is a perfectly valid state for the harmonic oscillator to occupy. (With caveats, though: it is not normalizable, so it's not a physically-accessible state. Still, it's a perfectly reasonable thing for the mathematical formalism to handle.) As you note, it has a position uncertainty equal to zero, as well as a vanishing expectation value $⟨x^2⟩=0$.

However, it does not have a vanishing momentum uncertainty, and in fact if you expand it as a superposition of plane waves, $$ \delta(x) = \frac{1}{2\pi\hbar} \int_{-\infty}^\infty e^{ipx/\hbar}\mathrm dp = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^\infty A(p) e^{ipx/\hbar}\mathrm dp, $$ you require an even weight $A(p) \equiv 1/\sqrt{2\pi\hbar}$ for all momenta, which means that the momentum-squared expectation value $$ ⟨p^2⟩ = \int_{-\infty}^\infty |A(p)|^2 p^2\mathrm dp = \frac{1}{2\pi\hbar}\int_{-\infty}^\infty p^2\mathrm dp = \infty $$ diverges to infinity. (This result is required by the uncertainty principle, but the derivation here does not rely on it - it's an independent proof of that fact. Still, you can see the consistency in that $\Delta x=0$ and $\Delta p \geq \hbar/2\Delta x$ can only be satisfied by having $\Delta p = \infty$.)

This then implies that the expectation value of the hamiltonian is also infinity: $$ ⟨H⟩ = \frac{1}{2m}⟨p^2⟩ + \frac12 k ⟨x^2⟩ = \infty. $$

As for this,

In the analysis of Harmonic Oscillator, it is claimed that $\langle\hat H\rangle$ cannot be zero, why is it so?

this is the zero-point energy of the oscillator, which has been explored multiple times on this site. If you want to ask why this is, you should ask separately, with a good showing of the previous questions here and what it is about them you do not understand.

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  • $\begingroup$ This post appears to challenge your first sentence. You might want to comment about this. $\endgroup$ – Ruslan Feb 2 '19 at 17:59
  • $\begingroup$ @Ruslan See my response there. $\endgroup$ – Emilio Pisanty Feb 3 '19 at 0:29
  • $\begingroup$ (As for why this answer does not go into those details - this is an explicit pedagogical choice, given the level at which the original question was pitched (as well as a now-deleted answer by OP). If people disagree with that choice, they're welcome to add further answers expanding on those details.) $\endgroup$ – Emilio Pisanty Feb 3 '19 at 0:30

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