Quantum harmonic oscillator meaning

Question

Imagine we want to solve the equations $$ i \hbar \frac{\partial}{\partial t} \left| \Psi \right> = \hat{H}\left| \Psi \right> $$ where $$\hat{H} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \frac{1}{2} m \omega^2 x^2.$$

Denote $\left| \Psi_n \right>$ each eigenstate of the equation. Imagine initial condition $$ \left| \Psi^0 \right> = \frac{\pi}{\sqrt{6}}\sum_{n=0}^\infty \frac{1}{n+1} \left| \Psi_n \right> $$ Then, $\hat{H}\left| \Psi^0 \right>$ does not converge. What is, in a mathematically rigorous manner, $\hat{H}$?

Answer 1

$\hat{H}$ is an unbounded operator. It is a fact from Functional Analysis that Hermitian unbounded operators cannot be defined on the entire Hilbert space, only in subspaces of it: this result is known as the Hellinger–Toeplitz theorem. Hence, it is not surprising that there are vectors for which $\hat{H} |\psi\rangle$ is not defined.

$\hat{H}$ is not alone in this sense. Position and momentum are unbounded too (and thus it is not surprising that the sum of their squares is unbounded).

Answer 2

The state $$ \lvert \Psi^0 \rangle = \frac{\sqrt{6}}{\pi}\sum_{n=0}^\infty \frac{1}{n+1} \lvert \Psi_n \rangle $$ is in the Hilbert space, as it is square integrable. However, it is not a (standard) solution to the time-dependent Schrodinger equation, exactly because, as OP notices, $\hat{H}\lvert \Psi^0 \rangle$ is not square-integrable and hence isn't in the Hilbert space.

However, if we proceed naively and just assume that we can time-evolve this state in the standard way, which is to say $$ \lvert \Psi^0(t) \rangle = \frac{\sqrt{6}}{\pi}\sum_{n=0}^\infty \frac{1}{n+1} e^{-i\hbar\omega(n+1/2)t}\lvert \Psi_n \rangle\,, $$ then we get a (seemingly) perfectly-well defined (time-dependent) quantum state, because this state is square-integrable and it evolves in the "standard" way.

This is what is called (I believe!) a weak solution to the partial differential equation (or it is at least related to the same idea). The idea is that while the derivatives of the function might not exist, we can transform the differential equation into an (almost-equivalent) integral equation, and it turns out that more functions satisfy the integral equation than satisfy the differential equation.

In fact, these weak solutions can be non-differentiable. The same sort of thing shows up in the context of the wave equation. One can solve for the normal modes of a string with fixed ends, but it turns out that you can expand functions that don't satisfy the boundary conditions as a linear combination of these normal modes, and the expansion works. The function then satisfies the wave equation in only the weak sense.

So, physically speaking, this might be the way of resolving things. Look for weak solutions to the time-dependent Schrodinger equation rather than "strong" solutions, and perhaps we can call these perfectly well-defined quantum states. Whether this is relevant physically is another matter.

---

See also Sobolev space.

Answer 3

An aspect complementary to the already existing answer: the problem with the initial state $$|\psi(0)\rangle = \frac{\sqrt{6}}{\pi} \sum\limits_{n=0}^\infty \frac{1}{n+1} |\phi_n\rangle, \quad \langle \phi_m |\phi_n\rangle= \delta_{mn}, \quad H |\phi_n\rangle = \hbar \omega (n+1/2) |\phi_n\rangle,$$ (note the correct normalization ensuring $\langle \psi(0) |\psi(0)\rangle=1$) is not its time evolution. $|\psi(t)\rangle$ is perfectly well defined by the unitary transformation $$|\psi(t)\rangle = e^{-iHt /\hbar} |\psi(0)\rangle = \frac{\sqrt{6}}{\pi} \sum\limits_{n=0}^\infty \frac{e^{-i\omega(n+1/2)t}}{n+1}| \phi_n \rangle,$$ in spite of the fact that $|\psi(0)\rangle $ (and, more generally, $|\psi(t)\rangle$) is not in the domain of $H$.

The real problem with the initial state $|\psi(0)\rangle$ is the fact that it is not even in the domain of $\sqrt{H}$, such that the expectation value of $H$ in this state does not exist (diverges). In other words, $|\psi(0)\rangle$ is sick because it corresponds to a state with infinite energy.