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From what I have read, the evolution of a quantum state is determined by the Hamiltonian (Schrodinger equation). However, I'm trying to understand if the Hamiltonian itself can be fully derived from the quantum state, or if it needs to be defined externally. From my understanding, the Hamiltonian includes information about the potential energy (when particles are interacting, etc), and that the laws of physics are actually "embedded" in the Hamiltonian, and not in the actual state vector. Is this correct, or does the state vector itself contain information about all the laws of physics? I hope my question is clear... Thanks!

Qmechanic
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Dmitry Pugachev
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1 Answers1

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No. The state is determined by the preparation procedure, which is quite distinct and independent from the Hamiltonian.

As additional reading on this I recommend this excellent article by a master in foundational issues:

Peres, Asher. "What is a state vector?." American Journal of Physics 52.7 (1984): 644-650.

The abstract is by itself enlightening:

“ A state vector is not a property of a physical system (nor of an ensemble of systems). It does not evolve continuously between measurements, nor suddenly ‘‘collapse’’ into a new state vector whenever a measurement is performed. Rather, a state vector represents a procedure for preparing or testing one or more physical systems. No ‘‘quantum paradoxes’’ ever appear in this interpretation. The formulation of dynamical laws may involve path integrals and/or S‐matrix theory.”

Edit: I understand the “state” question as meaning the initial state $\vert\Psi(0)\rangle$.

ZeroTheHero
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  • However, could one determine (at least something about) the Hamiltonian if we could perform many experiments where we could track the change in the state over time? Certainly this doesn't determine the Hamiltonian, but could this be used to "find it"? This is what I thought when I read in the question "I'm trying to understand if the Hamiltonian itself can be fully derived from the quantum state..." (I didn't down vote). – BioPhysicist May 19 '20 at 20:15
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    @BioPhysicist I think the OP is asking if the Hamiltonian can somehow be read directly from the initial state. If you want to determine the Hamiltonian doing some experiments you can do it probably much more simply if you know how to measure energy (as long as you can find a tomographically complete set of observables). –  May 19 '20 at 20:46
  • Take the (infinitely many copies of) initial state and perform repetitive measurements of energy (which gives you an energy eigenstate) followed by a tomographically complete set of measurements (to determine what this eigenstate is, say, in position basis). This way, eventually, you can determine all energies and corresponding energy eigenstates which gives you the Hamiltonian using spectral decomposition. –  May 19 '20 at 20:47
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    @BioPhysicist If you only have one state to work with, the answer is no. To see why, suppose that the state you happen to have is a stationary state of the Hamiltonian. The only thing you can find out about the Hamiltonian in this case is that the state you have is one of its eigenstates; you learn nothing else about the structure. – probably_someone May 19 '20 at 20:50
  • @DvijD.C. Possibly. But they do start the post off talking about the time evolution of the state. And they never say "initial state". They just say "the quantum state", which is evolving according to the Hamiltonian – BioPhysicist May 19 '20 at 20:51
  • @probably_someone Oh of course. I am just trying to get inside the OP's head a bit :) – BioPhysicist May 19 '20 at 20:51
  • @DvijD.C. This assumes that the state's evolution is nontrivial. I think if you're stuck with an energy eigenstate, then you can't learn much by watching it change over time. – probably_someone May 19 '20 at 20:52
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    @BioPhysicist Yes, I agree. It's a bit unclear what OP means by "quantum state". –  May 19 '20 at 20:53
  • @probablysomeone Umm, no, I'm not using non-trivial time evolution at all. In fact, I'm using the fact that energy eigenstates evolve trivially with time. Say, I'm given (infinitely many copies of) an arbitrary quantum state. I first measure its energy and bring it to an eigenstate of Hamiltonian. Now, I use the fact that it evolves trivially to perform subsequent measurements by a tomographically complete set of observables to actually figure out what this state is explicitly (as in what $\phi(x)$). And repeat until I get all energy eigenstates. –  May 19 '20 at 20:58
  • @probablysomeone . I agree you can't learn much about a state by observing it's time evolution if it's trivial. But the point is that you'd need to perform a tomographically complete set of measurements at every time slice to even actually determine what time evolution looks like. You don't just see the time evolution. But then it's better to use them actively to determine all the energy eigenstates rather than passively to just observe the time evolution. –  May 19 '20 at 20:59
  • let me rephrase the original question: given $|\psi(t)\rangle$ for every $0 \le t \le T$ and we know that $|\psi(t) \rangle= e^{-\mathfrak j \hat {H} t/\hbar} |\psi(0)\rangle$ under what condition we can find $\hat {H}$? Is there a "logarithm" that would allow us to invert it? – hyportnex May 19 '20 at 21:35
  • @hyportnex but is this what the OP asks? My understanding of the question is that the OP is asking about $\vert\psi(0)\rangle$. – ZeroTheHero May 19 '20 at 22:05
  • OP wrote: "I'm trying to understand if the Hamiltonian itself can be fully derived from the quantum state" – hyportnex May 19 '20 at 22:11
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    @hyportnex so I guess it's unclear as to "which state" we are talking about. – ZeroTheHero May 19 '20 at 22:55
  • OK, then, do you think I should ask my question separately? – hyportnex May 19 '20 at 23:03
  • @hyportnex so if your question is "does the time-evolution of an arbitrary initial state allows one to determine this state if only energy can be measured" then yes! – ZeroTheHero May 19 '20 at 23:05
  • @hyportnex Of course there is a formal logarithm of the evolution operator, even of the propagator. Play with the free case. – Cosmas Zachos May 19 '20 at 23:22
  • @DvijD.C. wrote "I'm using the fact that energy eigenstates evolve trivially with time." - is that true? Yes, the energy eigenkets (state-vectors) have a time varying phase but the physical eigenstates (rays) aren't time dependent are they? – Alfred Centauri May 20 '20 at 01:11