Let's take the very simple problem of what happens if I drop a 1 kg ball from a height of 1 meter. Classically, $F = mg$ and $g \approx 10 \frac{\mathrm{m}}{\mathrm{s}^2}$, so the ball feels a force of $10 \, \mathrm{N}$. From $S = ut + 1/2 gt^2$, we get that the ball hits the ground in about $\sqrt{1/5} \, \mathrm{s}$. No problem.

In principle, quantum mechanics should also be able to do this. So we start with the 3-dimensional time-independent Schrodinger equation with a potential of $V = -GM/r$: $$ -\frac{\hbar}{2m}\nabla^2 \psi - \frac{GM}{r}\psi = E\psi \,.$$

This is effectively the same equation as for solving the Hydrogen atom: a $1/r$ potential but with different coefficients. So the answer should take the same form, and we get all the energy levels, the n/l/m quantum numbers, etc. But qualitatively, this is clearly different from the classical result. It means the ball has many quasi-stable orbitals for example, and it doesn't give a precise prediction that the ball will hit the ground in $\sqrt{1/5} \, \mathrm{s}$. In the classical problem it's important that the ball was dropped from rest, otherwise the time taken to hit the ground will vary. But the quantum solution doesn't involve the initial conditions at all.

What has gone wrong? My first thought was that I need the time-dependent Schrodinger equation instead of the time-independent one, but that doesn't lead anywhere - it just means the ball oscillates between solutions. Am I missing some way to get the classical result from the Schrodinger equation?

Related but not quite the same: Is it possible to recover Classical Mechanics from Schrödinger's equation?

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    $\begingroup$ Have you tried calculating the spacing between these energy levels? $\endgroup$ Aug 1, 2018 at 5:16
  • $\begingroup$ @probably_someone the spacing would be minute, but I don't see how this smaller spacing could resolve the other fundamental inconsistencies (such as how initial conditions don't seem to matter in QM). $\endgroup$
    – Allure
    Aug 1, 2018 at 5:18
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    $\begingroup$ Ok, so we've basically recovered the continuum limit in energy. Initial conditions definitely do matter in QM. The Hamiltonian acts on the initial state to cause time evolution; different initial states evolve differently. So now, what's the expectation value of the position operator and the momentum operator as a function of time for a given initial state? $\endgroup$ Aug 1, 2018 at 5:25
  • $\begingroup$ Another possible duplicate here. It is itself a duplicate of this. $\endgroup$
    – knzhou
    Aug 1, 2018 at 9:15

2 Answers 2


This question has been asked many times before. I last answered it here and there's an excellent answer here that explicitly constructs a solution that obeys Newtonian gravity.

You're of course correct that if the ball were in a single stationary state, then it wouldn't do anything remotely like falling. It would occupy some standing wave about the Earth and not evolve in time. The reason we never see macroscopic objects in such states is because they are unstable in the same sense as Schrodinger's cat.

Your proposed stationary states would delocalize the ball over the entire Earth, over a scale of thousands of miles. But if you just look at the ball, you can measure where it is, collapsing this superposition. Or, if you don't like the collapse phrasing, the interaction of any stray photon with the ball will entangle the ball's position with the photon's state. This decoheres the enormous position superposition you've constructed and gives the ball a somewhat definite position to the perspective of any external observer.

That is to say, after a generic interaction with the environment the ball will quickly end up with its position peaked about a narrow value. (Not necessarily so narrow that the uncertainty principle comes into play, but effectively zero for all macroscopic purposes.) Such states exist in the Hilbert space, as complicated linear combinations of the energy eigenstates. To see that the peak in the wavefunction obeys Newton's laws, you can appeal to Ehrenfest's theorem, $$m \frac{d^2 \langle x \rangle}{dt} = - \left\langle \frac{dV}{dx} \right\rangle$$ which immediately gives that result.

You may still be troubled, because in classical mechanics we need to specify an initial position and initial velocity, while in quantum mechanics it seems we only need to specify the analogue of position. But you can't forget the wavefunction is complex. The "velocity" of a particle is encoded by how fast the phase winds around in position. For example, for the free particle, a constant wavefunction would give a totally stationary particle, while $e^{ikx}$ would give a moving particle. Since the ball is heavy, this property is stable under interaction with the environment, like the position.

  • $\begingroup$ Can you elaborate on the last paragraph? I'm not quite seeing how "a constant wavefunction would give a totally stationary particle" - for a free particle the solutions are plane waves with $e^{ikx - i\omega t}$, so constancy would imply $\omega = 0$ -> wavefunction still varies with x -> particle can't be stationary (?) $\endgroup$
    – Allure
    Aug 17, 2018 at 4:22
  • $\begingroup$ @Allure By constant there I just meant "not varying with $x$". However, the two are equivalent, because $\omega = k^2/2m$. If $\omega = 0$, then $k = 0$. $\endgroup$
    – knzhou
    Aug 17, 2018 at 9:46
  • $\begingroup$ Thanks, had a brain freeze moment there. However I'm still uncertain: in that case, the wavefunction is constant everywhere, aka. the particle is equally likely to be found everywhere. How does that translate to "totally stationary particle"? Heisenberg's principle would mean that we can, in principle, know the momentum of the particle with utmost precision; however this doesn't mean the momentum is necessarily zero. Similarly I don't understand how $e^{ikx}$ would give a moving particle. Is $e^{ikx}$ the wavefunction? $\endgroup$
    – Allure
    Aug 17, 2018 at 13:11
  • $\begingroup$ @Allure A wavefunction of the form $e^{ikx}$ has definite momentum $k$ and hence totally indefinite position by the uncertainty principle. However it is a general principle that you can smear such wavefunctions over a small range of $k$ to form finite wavepackets which have the same qualitative features but a somewhat localized position. I glossed over this step in the interest of length. $\endgroup$
    – knzhou
    Aug 18, 2018 at 0:15
  • $\begingroup$ @Allure If you smear near zero $k$ you will get a bump that doesn’t move left or right, though it may slowly spread out symmetrically. If you smear near nonzero $k$ you get a traveling particle, doing the phase corkscrewing as it goes. $\endgroup$
    – knzhou
    Aug 18, 2018 at 0:16

The exact classical result is recovered only in the limit $\hslash\to 0$. If one does consider $\hslash$ with its real value, one would get corrections to the classical result, in term of powers of $\hslash$ (such corrections for an object of mass 1kg are extremely small).

The point is that, quantum mechanically, the initial conditions cannot be a position $x_0$ and momentum $\xi_0$ at a fixed time. In classical (statistical) mechanics, the initial condition is a probability distribution in the phase space (in the case we are considering, it is a delta distribution centered in the initial condition $(x_0,\xi_0)$). In quantum mechanics, the initial condition is a noncommutative probability, or quantum state, that usually is written as a density matrix acting on a Hilbert space "of wavefunctions". And due to the noncommutativity, it is never possible to interpret such density matrix as a delta distribution on the phase space.

Nonetheless, let us take $H_{\hslash}$ to be the quantum Hamiltonian of the system (the one you wrote for example), and $\varrho_\hslash(x_0,\xi_0)$ to be the density matrix associated to a so-called squeezed coherent state, a state having minimal quantum uncertainty (this is done to be in a case that corresponds to a classical point of the phase space $(x_0,\xi_0)$ (delta distribution); one could take a different quantum state, but then one should also take into account a different classical description of the system, corresponding to an initial classical probability distribution in phase space that may not be a delta). The evolved state of the quantum system is $\varrho_{\hslash}(t,x_0,\xi_0)=e^{-\frac{it}{\hslash}H_{\hslash}}\varrho_{\hslash}(x_0,\xi_0)\;e^{\frac{it}{\hslash}H_{\hslash}}$. The average position of the particle is then given by $$\mathrm{Tr} \,\varrho_{\hslash}(t,x_0,\xi)\,\hat{x}_{\hslash}\; ,$$ where $\hat{x}_{\hslash}$ is the position operator (I have put the $\hslash$-dependence on the position operator because in general quantum operators depend on $\hslash$, however in the standard QM representation of the canonical commutation relations the position operator is independent of $\hslash$, and all the dependence is on the momentum operator; one could change this by means of a unitary transformation).

The function $\mathrm{Tr} \,\varrho_{\hslash}(t,x_0,\xi_0)\hat{x}_{\hslash}$ is a function of time $t$, of position $x_0$, momentum $\xi_0$ through the initial quantum condition $\varrho_{\hslash}(x_0,\xi_0)$, and of $\hslash$. Now, it is possible to expand such function in powers of $\hslash$. And it turns out that $$\mathrm{Tr} \,\varrho_{\hslash}(t,x_0,\xi_0)\hat{x}_{\hslash}= S(t,x_0,\xi_0) + \mathrm{O}(\hslash)\; ,$$ where $S(t,x_0,\xi_0)$ is the classical trajectory of the particle at time $t$, corresponding to initial condition $(x_0,\xi_0)$. The corrections at a given power of $\hslash$ can be explicitly computed, or at least numerically bounded.

In general, to every (physically reasonable, let me not give the full details here) quantum density matrix $\varrho_{\hslash}$ there corresponds a classical probability measure $\mu$ on the phase space $\Omega$. The above result in the general case would then read: $$\mathrm{Tr} \,\varrho_{\hslash}(t)\hat{x}_{\hslash}= \int_{\Omega}\mathrm{d}\mu(x,\xi) \, S(t,x,\xi) + \mathrm{O}(\hslash)\; ,$$ where again $S(t,x,\xi)$ is the classical trajectory at time $t$, corresponding to initial condition $(x,\xi)$. The special case of a squeezed coherent state is included since in that case the measure is a delta distribution, as already remarked above.


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