So the way I understand this, the way measurement is taught is that

you have a wave function $\Psi(t)$. It's evolution over time is : $$i \hbar \frac{d}{d t}\vert\Psi(t)\rangle = \hat H(t)\vert\Psi(t)\rangle$$

If you have an observable $\hat{O}$ it defines a basis of the Hilbert space s.t. $$ \Psi(t) = \sum e_k O_k(t)$$ where $O_k$ are the eigenstates, $e_k$ the eigenvalues. If you measure $\hat{O}$ in the real world, you will read the results $O_k$ with probability $\frac{e_k^2}{\sum_j e_j^2}$.

The question is, can we craft $\hat{H}$ such that the evolution of $\psi$ converges towards $O_k$ as the $O_k$ become attractors of the solutions of Schrödinger's equation. If we craft $\hat H(t)$ correctly we can have the size of the regions of attractor $O_k$ be $\frac{e_k^2}{\sum_j e_j^2}$ to respect the "discrete" view of measurement.

Is such a Hamiltonian findable ? Have I missed something glaringly obvious ?

Ps: I'm not a physicist.

  • $\begingroup$ Are you imposing additional conditions to make the $O_k$ attractors? $\endgroup$ – probably_someone Jun 3 at 11:25
  • $\begingroup$ Do we need to ? To the best of my knowledge I don't think there is. I imagine there would be sort of a time constant in order to reach $O_k$ and that would have to be small I guess (but because we handcraft $\hat H$ I guess that's fine). $\endgroup$ – Narsilou Jun 3 at 11:36
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    $\begingroup$ $e_k$ are the eigenvalues?? $\endgroup$ – Qmechanic Jun 3 at 12:01
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    $\begingroup$ If I'm not mistaken, yes. $\endgroup$ – Narsilou Jun 3 at 12:23
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    $\begingroup$ Why should a generic state be a linear combination of eigenstates of an operator with the coefficients being the eigenvalues? It can't be--it is only a specific state. Also, since you are working in the Schrödinger picture, the coefficients would depend on time, not the eigenbasis. So, a generic state would be $\sum_k c_k(t) | O_k\rangle$ not $\sum_k e_k | O_k(t)\rangle$. Your question is not answered or made moot via this technical correction but I think it is useful to clarify it. The answer to your question is negative as explained by Chiral Anomaly. $\endgroup$ – Feynmans Out for Grumpy Cat Jun 4 at 7:07

The short answer is no, the attractor idea doesn't work, because time evolution is unitary.

In more detail, the equation $$ i\hbar\frac{d}{dt}|\Psi(t)\rangle =\hat H(t)|\Psi(t)\rangle \tag{1} $$ implies that the inner product between any two given solutions is independent of time: $$ \frac{d}{dt}\langle\Psi_1(t)|\Psi_2(t)\rangle = 0. \tag{2} $$ This is called unitarity, and it rules out the attractor idea.

Intuitively, unitarity says that time evolution can't make an initially uniform distribution of state-vectors in the Hilbert space cluster toward the basis vectors in any given orthonormal basis, not even slightly. Unitarity says that an initially uniform distribution remains uniform forever.

To put a little more detail behind this intuition, consider one of the eigenstates $O_k$ of the measured observable, and consider some small neighborhood of this eigenstate in the Hilbert space. By small neighborhood, I mean a neighborhood in which every state-vector $\psi$ is approximately proportional to $O_k$, in the sense that the component of $\psi$ orthogonal to $O_k$ has a much smaller norm than the component of $\psi$ that is proportional to $O_k$. Since unitarity preserves inner products, if we run this neighborhood backward in time arbitrarily far, all of those state-vectors will still be in a small neighborhood of (a past version of) the eigenstate $O_k$. But the overwhelming majority of the state-vectors in the Hilbert space are not in a small neighborhood of any of the basis vectors in any given orthogonal basis, so the overwhelming majority of the state-vectors in the Hilbert space are not in any basin of attraction.

Here's an analogy: A unitary transformation in Hilbert space is essentially the complex-valued version of a rigid rotation in (many-dimensional) Euclidean space. The fact that a rigid rotation preserves angles is not compatible with the attractor idea.

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    $\begingroup$ Aren't hermitian hamiltonian just a convenient convention that make time evolution unitary ? Why do we require time evolution to be unitary ? Clearly measurement is not so why can't we adapt Schrödinger's equation to allow non unitary time evolution ? $\endgroup$ – Narsilou Jun 4 at 7:52
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    $\begingroup$ @Narsilou Good question. This is simpler in the Heisenberg picture, where time parameterizes observables and not states. Then the question is simply: can we find an observable (at any time) with the property that almost all state-vectors are in a small neighborhood of one of its eigenstates? The answer is no, not unless the observable is trivial (proportional to the identity operator), because observables are linear operators in quantum theory. People have tried to invent non-linear theories to implement the attractor idea, but (by definition) they're not called quantum theory anymore. $\endgroup$ – Chiral Anomaly Jun 4 at 13:47
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    $\begingroup$ @Narsilou In the Schrodinger picture, you can see the fact that the time evolution operator has to be unitary as long as you grant two basic postulates: time-translational symmetry (there is no preferred $t=0$) and that quantum mechanics is a linear theory. This is the famous Wigner theorem and it is beautifully proven in an appendix of Chapter $2$ of Weinberg's Quantum Theory of Fields, Vol. 1. $\endgroup$ – Feynmans Out for Grumpy Cat Jun 4 at 15:09
  • $\begingroup$ Thanks for the answers I'll try to look into it @ChiralAnomaly Do you have references or keywords to those non-linear theories for attractor idea ? Google never led me to anything satisfactory to me. $\endgroup$ – Narsilou Jun 5 at 7:29
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    $\begingroup$ @Narsilou You can try starting with the "GRW theory": en.wikipedia.org/wiki/…. $\endgroup$ – Chiral Anomaly Jun 6 at 0:07

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