# How does Ehrenfest's theorem apply to the quantum harmonic oscillator?

Ehrenfest's theorem, to my level of understanding, says that expectation values for quantum mechanical observables obey their Newtonian mechanics counterparts, which means that we can use newton's laws on expectation values. However, in the case of the quantum harmonic oscillator, this clearly does not look newtonian because the expectation value of the position does not oscillate like the newtonian $$\sin\omega t$$.

These states are of the form $$\psi=K(n, \xi)e^{-\xi^2/2}$$. Why do they not obey Ehrenfest's theorem? They don't give a harmonic oscillator, imo. • Maybe this question of mine can answer your question: physics.stackexchange.com/questions/267835/… Mar 10, 2019 at 9:46
• Though it does not address the question asked, I used to have students plot the position probability distribution for the quantum and classical cases over a range of energies corresponding to, say, $n = 1,2,3,4,10$ and a "big" number like $20$. I'll post a figure in chat: chat.stackexchange.com/transcript/message/49407141#49407141. Mar 10, 2019 at 23:19

It is actually true, in an almost trivial way. The Ehrenfest theorem states that, $$\begin{equation} \frac{d}{dt}\langle x\rangle=\langle p\rangle,\quad \frac{d}{dt}\langle p\rangle =- \langle V'(x)\rangle \end{equation}$$ However for all eigenfunctions for the Harmonic oscillator $$\langle x\rangle=0$$ (and therefore $$\langle V'(x)\rangle=0$$) and $$\langle p\rangle=0$$. So the Ehrenfest theorem on the eigenstates reduces to $$0=0$$.

You can see that the general version of the Ehrenfest theorem works trivially for all eigenstates. It states that for the arbitrary observable $$A$$ its expectation value satisfies the equation, $$\begin{equation} \frac{d}{dt} \langle A\rangle=\frac{1}{i\hbar}\langle [A,H]\rangle+\langle \frac{\partial A}{\partial t}\rangle \end{equation}$$ However on the eigenstates, $$\begin{equation} \langle\psi_n| [A,H]|\psi_n\rangle=\langle\psi_n|AH-HA|\psi_n\rangle=E_n\langle\psi_n|A-A|\psi_n\rangle=0 \end{equation}$$ So the expectation value of the observable that doesn't explicitly depend on time, doesn't evolve on the eigenstates which is what you would expect.

So where does the Ehrenfest theorem lead to the classical dynamics? You need to consider the localized wavepackets. The simplest example would be the coherent state of the Harmonic oscillator that is the Gaussian wavepacket that follows the classical trajectory For the Harmonic oscillator the Ehrenfest theorem is always "classical" if only in a trivial way (as in case of the eigenstates). However in general the Ehrenfest theorem reduces to the classical equation of motion only on such localized wavepackets that concentrate near the classical trajectory as $$\hbar$$ goes to zero. The key point happens to be the interchange $$\langle V'(x)\rangle \mapsto V'(\langle x\rangle)$$ that on general states can't be done. So if you want to recover some classical dynamics from the quantum theory look on the localized wavepackets.

1. Your version of $$\psi$$ is (I'm sure you know) derived from the Time Independent Schrodinger equation, $$\hat{H}\psi=E\psi$$. To find time-dependent solutions, we solve $$i\frac{\partial}{\partial t}\psi=\hat{H}\psi.$$ You were trying to solve for stationary states, and the entire point of those is that $$|\psi(x)|^2$$ does not change over time. Still, for these stationary solutions solutions,$$\frac{\mathrm{d}}{\mathrm{d}t}\left=\left=0; \frac{\mathrm{d}}{\mathrm{d}t}\left=-\left<\frac{\mathrm{d}}{\mathrm{d}x}V (x)\right>=0,$$ which is in accordance with the Ehrenfest theorem (albeit uninformatively).
2. Be careful about the time dependence of your reported $$\psi$$: you're actually dealing with $$\Psi(x, t)=\psi(x)e^{-itE/\hbar}$$ for those stationary states. Of course, this doesn't relate to the Ehrenfest theorem, but it's something worth mentioning: the complex and real parts are oscillating, as shown by the pink and blue lines in this diagram (from the wikipedia page on QHO): Do not make the mistake of suggesting that all derivatives with respect to time are necessarily automatically equal to zero because the wavefunction is time-independent. Contrary to what the shorthand notation suggests, we do have a (separable) time-dependent bit.

3. Referring to the same diagram, observe parts G and H: these represent coherent states, which can be understood using Ehrenfest's theorem because $$|\psi|^2$$ looks like a Gaussian which follows a classical $$\sin$$ or $$\cos$$ function.

See

Kanasugi, H., and H. Okada. “Systematic Treatment of General Time-Dependent Harmonic Oscillator in Classical and Quantum Mechanics.” Progress of Theoretical Physics, vol. 93, no. 5, 1995, pp. 949–960., doi:10.1143/ptp/93.5.949.

What you wrote down is just a complete set of solutions of the time independent Schrödinger equation. \begin{align} [- \frac{\hbar²}{2m} \Delta + V(x) ] \Psi(x) = E \Psi(x) \end{align} Of course this solutions don't carry any time dependence, because the time-independent Schrödinger equation (in this representation) only makes statements about functions that depend on the spatial coordinates only (in this case, this is only x).

How to get to the time-dependent solutions? A particular property of the solutions $$\Psi_{E}$$ of the time-independent Schrödinger Equation is that $$\Psi_{E}(x) e^{-i \frac{E}{\hbar}t}$$ is a solution to the time-dependent Schrödinger equation.

If you would apply this to your suggested set of solutions of the harmonic oscillator, you would arrive at time dependent solutions. THOSE are the ones that the Ehrenfest-Theorem makes a statement about. You would then calculate that the expectation values $$$$ and $$

$$

do not change. But you would as well calculate that both of this values are 0. This is in perfect agreement with the Theorems of Ehrenfest, and the classical analogon would be a particle resting at the deepest spot of the harmonic potential.