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Consider Ehrenfest's theorem: \begin{align} m\frac{d\langle x\rangle}{dt}=\langle p\rangle \\ \frac{d\langle p\rangle}{dt}=-\langle V'(x)\rangle. \end{align} Suppose $V(x)=x^2+x^{n+1}$ where $n>1$. Then $\langle V'(x)\rangle\neq V'(\langle x \rangle)$. Therefore the evolution of $\langle x\rangle$ and $\langle p\rangle$ is different to the classical prediction: recall that classically \begin{align} m\dot{x} &= p \\ \dot{p} &=- V'(x). \end{align} Now, we are told that if $\hbar\to 0 $ quantum mechanics reproduces classical physics. Therefore I would expect that $\langle x^n \rangle-\langle x \rangle^n$ is $O(\hbar)$ (because it is this term which is responsible for the deviation of the evolution of $\langle x\rangle$ and $\langle p\rangle$ from the classical prediction). However, from the quantum mechanical formulation, I don't see why this term is actually $O(\hbar)$.

So my question is: is $\langle x^n \rangle-\langle x \rangle^n$ actually $O(\hbar)$? If it isn't then how do I explain why $\hbar\to 0$ reproduces classical predictions for the evolution of $\langle x\rangle$ and $\langle p\rangle$?

[Notes: For one, the smallness of $\langle x^n \rangle-\langle x \rangle^n$ depends on the details of the state: if $\psi(x)=\delta(x-x_0)$ for example then $\langle x^n \rangle-\langle x \rangle^n=0$. But the classical limit should not depend on the details of the state, there must be some general arguments for why $\langle x^n \rangle-\langle x \rangle^n$ is small -- my own feeling is that it is controlled by $\hbar$, but, as my question says, I can't see why.]

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    $\begingroup$ There is nothing wrong with that quantity depending on the particularities of the state involved, as long as it has the suitable dimensional-analysis dependence on ℏ. Bog-swill or not, you might reassure the reader you have mastered the Gaussian state of arbitrary width, and confirmed the general result for $\langle x^{2n}\rangle$. $\endgroup$
    – Cosmas Zachos
    Commented Oct 14, 2023 at 18:45
  • $\begingroup$ @CosmasZachos Very good, thank you $\endgroup$
    – dennis
    Commented Oct 14, 2023 at 19:42
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    $\begingroup$ Linked. $\endgroup$
    – Cosmas Zachos
    Commented Oct 14, 2023 at 23:08
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    $\begingroup$ Linked. $\endgroup$
    – Cosmas Zachos
    Commented Oct 17, 2023 at 15:14

1 Answer 1

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The key point is that ℏ is dimensionful, with units of action (or angular momentum), and the classical limit is that in which the characteristic action quantities of the system are huge multiples of this ℏ. So, as you might find out in the quantum oscillator, and/or phase-space quantization, it makes sense to use natural units; in this case, instead of units (L,M,T), use units of action, S, energy, E, and P (why not?), and then absorb the units of E and P into the physical variables (nondimensionalize) so the only remaining units are those of action, S, namely ℏ.

In these units, both x and p are balanced to have units of $\sqrt \hbar$, and your quantity is automatically of $O(\hbar^{n/2})$. (And the uncertainty principle is memorable.)

Specifically, $$ [S]= \frac{ML^2}{T}, \qquad [E]=\frac{ML^2}{T^2}, \qquad [P]= \frac{ML}{T}, $$ so that $$ [M]= \frac{P^2}{E}, \qquad [T]=\frac{S}{E}, \qquad [L]= \frac{S}{P}, $$ where you observe the operators on the Schrödinger equation have units of E, (and the 1-D wavefunction has units of $\sqrt{P/S}$).

Again, if, in these units, the relevant actions are much larger than $\hbar$, the wavefunctions of the states are spiked, and you manifestly have your result; otherwise not, and you are stuck in an irredeemably quantum state.

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  • $\begingroup$ I could, however, arbitrarily introduce a length-scale $\ell$. In that case $\langle x^n \rangle\propto \ell^n$ (if I take the state $e^{-x^2/\ell^2}$ for example). $\endgroup$
    – dennis
    Commented Oct 14, 2023 at 21:29
  • $\begingroup$ In fact, I can easily construct an initial state with large spread in $x$, so that initially there is large deviation from classical evolution. The best you can hope for is that at late times the spread becomes small, and classical evolution is followed, but there is no good reason for that to actually happen. (And notice, none of what I said had to do with the size of $\hbar$! Actually, $\hbar$ enters in the time-evolution operator but it's size is not going to dictate whether the state becomes more or less spread.) $\endgroup$
    – dennis
    Commented Oct 14, 2023 at 21:39
  • $\begingroup$ Indeed, you may introduce states which will not have a reasonable classical limit, to the extent that there is no absolutely no action restriction on states. You may also describe classical mechanics in the KvN formulation. However the TDSE does have its scale set by ℏ, which controls the type of of state one is ultimately discussing. Classical states are the ones whose characteristic actions are much larger than ℏ, otherwise they are quantum. These limits are most visible in deformation quantization. Bog swill redux. $\endgroup$
    – Cosmas Zachos
    Commented Oct 14, 2023 at 22:29
  • $\begingroup$ Actually, if we use Euclidean time evolution, we're in better shape: take an arbitrary initial state $\psi(x)$. Now use the Euclidean time evolution operator $e^{-\hbar^{-1}H\tau}$ where $H$ is the Hamiltonian: $H=-\frac{\hbar^2}{2}\partial_x^2+V(x)$. When $\hbar \to 0$, $V(x)$ is the dominant term, so the wavefunction just gets multiplied by $e^{-\hbar^{-1}V(x)\tau}$. Now $V(x)$ has a global minimum at some point $x_0$. At late times, $e^{-\hbar^{-1}V(x)\tau}$ is damping, and it localises the wavefunction around $x_0$. The spread is low and classical evolution ensues. $\endgroup$
    – dennis
    Commented Oct 14, 2023 at 23:21

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