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Norton's dome demonstrate Classical non-determinism.

I am wondering if there is "quantum version" of Norton's dome? In other words, is there any Hamiltonian where Schrodinger's equation is non-deterministic?

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  • $\begingroup$ Check out spontaneous symmetry breaking. $\endgroup$ – Feynman's Cat May 13 '20 at 20:58
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    $\begingroup$ I would clarify that Norton's dome does not "demonstrate" (i.e. show experimentally, or establish the truth of) non-determinism. Rather, since uncaused action contradicts the very tenets of Newtonian laws (and contradicts all science), he effectively poses the question to the scientist to find the fault in the reasoning - and it appears that at least one fault in his reasoning, is the admittance of the possibility of uncaused action, within a Newtonian system that axiomatically excludes the possibility. $\endgroup$ – Steve May 13 '20 at 21:47
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    $\begingroup$ Related physicsoverflow question: physicsoverflow.org/40217 $\endgroup$ – Wolpertinger May 14 '20 at 8:10
  • $\begingroup$ @Feynman'sCat I admit that this gets into philosophical waters, but I would argue that this question is getting at something a little different than SSB. (I would argue that) SSB is something that occurs because of decoherence due to environmental interactions, whereas this question is getting at the possibility of non-deterministic pure idealized Schrodinger time evolution in a hypothetical perfectly isolated system (where SSB arguably would not occur). $\endgroup$ – tparker May 16 '20 at 2:10
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I would say that in a precise sense a quantum version of Norton's dome is impossible. Let me explain.

Classical mechanics is governed by second order ordinary differential equations (ODE) (Newton's law) which can be cast as first order ODE (Hamilton equations). Now a first order ODE has a unique solution, with given initial condition, if the function is Lipschitz. Norton's dome lacks uniqueness by constructing a Hamiltonian such that the relevant function is not Lipschitz (and the ODE indeed has multiple solutions corresponding to the same initial condition). Mathematically this is perfectly valid.

Let us now turn to quantum mechanics. Evolution here is described by Schroedinger's equation. Now by the axioms of quantum mechanics (the rules of the game) the Hamiltonian should be self adjoint. At this point Stone's theorem asserts (among other things) that the solution of Schroedinger's equation with a given initial condition will be unique. It appears to me that classically, instead, there is no conceptual obstruction to constructing a potential leading to a non Lipschitz ODE.

With this I don't mean to say that any situations that can be considered "analogous" to Norton's dome is impossible quantum mechanically, but translating the principle in its most pure and obvious way doesn't work. This seems to be also the OP's point of view.

In shorts, by Stone's theorem, there is no admissible (i.e. self adjoint) Hamiltonian such that Schroedinger's equation admits multiple solutions for a given initial condition.

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  • $\begingroup$ Thanks for your answer! is this the Stone's theorem you mentioned? $\endgroup$ – Shing May 15 '20 at 11:20
  • $\begingroup$ @Shing yes exactly. $\endgroup$ – lcv May 15 '20 at 17:13
  • $\begingroup$ The theorem is important (in my view) because for unbounded Hamiltonians the Schroedinger's equation is not Lipschitz. But being linear there's not much that can go wrong. $\endgroup$ – lcv May 15 '20 at 17:21
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Two-part answer incoming, a short first part and then some more details for people who are interested.

Although there are some differences, the phenomenon of optical bistability has similar features to Norton's dome. In simple terms, optical bistability is that certain coupled light-matter quantum systems feature two stable states, leading to a non-equilibrium phase transition and dynamical switching between the states. The latter results for example in hysteresis for certain observables.

The quantum features of optical bistability have been investigated for example in Zens2019 (arxiv, journal). The two stable states correspond to two solutions of the semi-classical equations. On the quantum level, the system does not jump between the solutions any more, but there is a switching time. However, in the thermodynamic limit (that is for many particles), this switching time diverges, such that the semi-classical behaviour is recovered and one essentially has a true quantum system with something like Norton's dome.

EDIT - To relate to lcv's nice answer: the example here avoids the formal restrictions that he outlines in a limiting sense (the thermodynamic/many-particle limit).


I will try to illustrate and sketch some of the mathematics behind this effect, which should also clarify the similarity to Norton's dome. The OP asked for a Hamiltonian, so here we have a Hamiltonian for a spin coupled to a single cavity mode with external driving (simplified version of the one in the cited paper)

$$H = \Delta_C a^\dagger a + g \sum_{j=1}^N (a^\dagger \sigma^-_j + a \sigma^+_j) + i\eta(t) (a^\dagger - a) \,,$$

where $a$ is the cavity modes, $\sigma_i^{z/\pm}$ are the Pauli operators for spin $i$ and $\eta(t)$ is the time dependent driving strength. $g$ is the coupling constant (independent of $i$ for simplicity) and $\Delta_C$, $\Delta$ are detunings. In Zens2019 there are also incoherent decay terms, which I will ignore here.

The semi-classical equations for this Hamiltonian are (again from the cited paper) $$\frac{d}{dt}\langle a \rangle = -i\Delta_c a - iNg\langle\sigma_i^-\rangle + \eta \,,$$ $$\frac{d}{dt}\langle \sigma^-_i \rangle = ig\langle\sigma_i^z\rangle\langle a\rangle \,,$$ $$\frac{d}{dt}\langle \sigma^z_i \rangle = 2ig\langle\sigma_i^-\rangle\langle a^\dagger\rangle + 2ig\langle\sigma_i^+\rangle\langle a\rangle\,.$$

Let us check what happens if we start with the spins all in the excited state ($\langle\sigma^z_i\rangle=1$, $\langle\sigma^+_i\rangle=\langle\sigma^-_i\rangle=0$) and no photons in the cavity mode ($\langle a\rangle=0$) and no driving field ($\eta=0$). We then have $$\frac{d}{dt}\langle \sigma^z_i \rangle = \frac{d}{dt}\langle \sigma^-_i \rangle = \frac{d}{dt}\langle a \rangle = 0 \,.$$

Hm, so semi-classically our spin in the excited state does not do anything even though it is coupled to a cavity mode? Well, it can be shown that these types of coupled non-linear differential equations have a second solution, a so-called singular solution. The spins in the excited state is an unstable equilibrium initial state that allows for both of these solutions. This is where we can see the similarity to Norton's dome.

These are the semi-classical equations, but as described in the initial paragraph of the answer, this phenomenon has an impact on the quantum behaviour in the thermodynamic limit, as investigated in the referenced paper. Optical bistability itself is usually considered a steady-state phenomenon, that is with a driving field, but the effect is related to these features of the differential equations.

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    $\begingroup$ This is a nice effect, thanks for sharing. But I'm not sure if this is similar to Norton's dome. The semiclassical equations are Lipschitz. And instability exists also classically (the top of a pendulum) but does not violate determinsm. $\endgroup$ – lcv May 14 '20 at 9:34
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    $\begingroup$ @lcv Thanks! To respond: I agree that the example is not precisely like Norton's dome. But I think that the analogy is much closer than you state. As far as I understand, the semi-classical equations are not Lipschitz. If you reduce them to a single ODE, you get $\dot{\alpha}^2 = J^2 - (M - \alpha^2)^2$ as shown in one of the references. I am not sure how to check explicitly if this is Lipschitz, but it has to not be, because it seems to avoid Picard–Lindelöf by the existence of the singular solution. This seems very closely related to Norton's dome to me. $\endgroup$ – Wolpertinger May 14 '20 at 10:29
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    $\begingroup$ Ops, very interesting. I must admit I've been a bit cavalier and did not actually check if the semiclassical equations were Lipschitz. Great! Now I understand your answer better. I agree, than you have a Norton's dome situation semiclassically. $\endgroup$ – lcv May 14 '20 at 11:04
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    $\begingroup$ @lcv Thansk for the comments either way, I think this helped clarify! For myself too ;) $\endgroup$ – Wolpertinger May 14 '20 at 12:16

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