# Ehrenfest theorem initial conditions

The Ehrenfest theorem states that the expectation value of position $$x$$ obeys the following equation: $$m\frac{\mathrm{d^2}}{\mathrm{d}t^2}\langle x\rangle=\langle F(x)\rangle$$ But there I only need 2 initial conditions and for the wavefunction I need infinitely many initial conditions. So from where come all the initial conditions? I think that they come from the higher moments of the position operator but for the harmonic oscillator there are also infinitely many initial conditions and the centered moments vanish.

• why do you need infinitely many conditions for the wavefunction? You know the wavefunction here else you would be able to evaluate averages... Apr 24, 2022 at 20:03
• @ZeroTheHero yes but the expectation value of position of the harmonic oscillator behaves classical according to the Ehrenfest theorem. So if I know the initial position and momentum of the expectation value I know how the system behaves. My question is now why I need more initial conditions in quantum mechanics and where they come from. Apr 24, 2022 at 20:15
• But... a constraint of an expectation value, $d^2\langle x\rangle /dt ^2= -\omega^2 \langle x\rangle$ is not all there is to QM!?!! Indeed, there is an infinity of constraints of x -moments, which depend on the particular wavefunction involved.... Apr 24, 2022 at 20:19
• well... this is an average so there are infinitely many states that will produce the same average. Although this is not a time-dependent example, any normalizable state of the form $\sum_{n=0}^{\infty} \alpha_{2n} \vert 2n\rangle$ will have $\langle x\rangle=0$ so clearly the average value is not enough to specify the state. Apr 25, 2022 at 0:16

So if I know the initial position and momentum of the expectation value I know how the system behaves. My question is now why I need more initial conditions in quantum mechanics and where they come from.

That's just it, you really don't know how the system behaves! While, for the quantum oscillator, the first moment has the magically classical property $$\frac{𝑑^2⟨𝑥⟩}{𝑑𝑡^2}=−𝜔^2⟨𝑥⟩,$$ specified by two initial conditions, you can check that for the great majority of quantum oscillator states (barring coherent states) this dramatically fails for higher moments $$\langle x^n\rangle$$.

That is to say that many-many states described by the oscillator TDSE do "interesting things.

This is actually easiest to "see" in deformation quantization, where the coordinate-space probability density profile is gotten by integrating out the momentum dependence of the rigidly rotating Wigner function.

That is, whereas the x-distribution's mean oscillates classically, the rest of it morphs and wiggles in a dramatic way, around that semiclassical mean.

For instance, look at the simplest quantum flip-flop, that is, integrate over the p-axis.

There is much-much more in ICs you need to specify in QM, and they all come from the astounding $$\hbar$$-dependence of the Schroedinger equation. Choosing to focus on the semi-classical first moment is almost throwing out the baby with the ... ahh, stop that runaway metaphor!

• Thank you! Do you have any reference where I can read about the $\hbar$-dependence of the SE? Apr 25, 2022 at 16:42
• Sorry, not really.... the classical limit sacrifices the ℏ-dependent information of QM... Note the SE has one more independent variable, x , and a different dependent variable, ψ, than Newton's equation of motion in the same dimension... What was x(0) , now has morphed to ψ(x,0)... Apr 25, 2022 at 16:54