# Why is $\omega = \sqrt{K/m}$ valid for a quantum oscillator?

I'm working in the 3rd edition of Modern Physics by Serway, Moses, and Moyer. In 6.6, it talks about a quantum oscillator. I don't fully understand how the definition of frequency works.

Now, we assume that a particle feels a force $F = -Kx$ like the classical version, and we define $K$ as the curvature of $U(x)$ at the point of stable equilibrium ($x = 0$). That is,

$$K = \left.\frac{d^2 U}{dx^2}\right|_a$$

where $x = a$ is the general point of stable equilibrium. It then uses the formula $\omega = \sqrt{\frac{K}{m}}$, though this is a classical equation. I don't understand why this is technically valid. I understand that that formula comes from the classical description $-Kx = ma = m \frac{d^2x}{dt^2}$ where $x = A\sin(\omega t + \phi_0)$. However, with quantum mechanics, why can we say that it oscillates by that description? We're not even dealing with a particle; it has wave and particle properties described by the wavefunction, so why can we immediately say that $U(x) = \frac{1}{2}Kx^2 = \frac{1}{2}m\omega^2x^2$? When we solve the Schrodinger equation, we find that the particle isn't limited to the interval $[-A, A]$, so why is it valid to use a formula that is derived from assuming that?

Any help is much appreciated!

• A quantum system in general does not have a frequency, but it has a spectrum. The energy differences between its eigenvalues are the relevant quantities. In case of the quantum harmonic oscillator these energy levels are equidistant and they correspond to the oscillation frequency of the equivalent classical oscillator. This is a very special system property and worth remembering for many applications, because a harmonic quantum oscillator can be excited up and down with one frequency (whereas, for instance, atoms need different excitation frequencies to change electronic state). – CuriousOne Aug 12 '14 at 20:27

We are still dealing with a particle, but the state of the particle at a given moment in time is no longer described by a pair $(x(t),p(t))$ of position and momentum in a classical phase space, it is instead a state $|\psi(t)\rangle$ in a Hilbert space. Because of this new description, we can no longer talk about the "motion" of a particle in a conventional sense, but rather are forced to talk about the "time evolution" of the state $|\psi(t)\rangle$.
Now as you point out, in quantum mechanics the parameter $\omega = \sqrt{k/m}$ cannot be physically interpreted in precisely the same way as in classical mechanics because the underlying model is different, but there are still vestiges of the classical harmonic oscillator that remain in the quantum harmonic oscillator that allow us to interpret $\omega$ in a similar way.
One of the most compelling of such classical remnants is as follows. The expectation value (average value) of any observable quantity $\hat O$ in quantum mechanics as a function of time is given by $\langle \hat O\rangle_\psi(t) = \langle \psi(t)|\hat O|\psi(t)\rangle$. One can show that for the quantum harmonic oscillator, the expectation value of the position operator in any state $|\psi(t)\rangle$ satisfies \begin{align} \frac{d^2\langle \hat x\rangle}{dt^2} = - \omega^2 \langle\hat x\rangle \end{align} where I have abbreviated $\langle \hat x\rangle_\psi(t) = \langle \hat x\rangle$. But this is simply the classical harmonic oscillator equation! So although it doesn't make sense to talk about an oscillating position of the quantum particle, it's still true that its average position oscillates in precisely the same way as the classical position!