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Eigenstate-to-eigenstate evolution with no intermediate superposition?

  1. Sep 16, 2013 #1
    Eigenstates of some observable O are represented by orthonormal vectors in complex Hilbert space.

    Is it true that the only possible way that the state vector can evolve from one eigenstate of O to the next, is to rotate between the two eigenvectors so that intermediate state vectors are superpositions of the eigenstates of O?

    That is, are properties that (i) can vary (i.e. not constant) but (ii) cannot superpose, simply impossible to represent in quantum mechanics?

    My suspicion is that you might be able to rotate the state vector through the complex plane in such a way that you go from one eigenstate to the next but avoiding superpositions of those eigenstates. But not sure.

    Interested in your thoughts!
     
  2. jcsd
  3. Sep 17, 2013 #2
    So what is it in general? I mean lets say in the middle of the rotation through the complex plane that you suggested.

    In my humble opinion I think the answer to the above question is yes in some sense, we must also consider that we are making measurements in the eigen-space of the operator!

    Then what will you have to say about the state in some other non-commuting space? what if we measure that state?

    I am just thinking loudly!! Hope I made some sense!
     
  4. Sep 18, 2013 #3

    Nugatory

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    What would be an example of a property that cannot superpose? As you say, these are orthogonal vectors in a vector space; so their linear combinations must also be vectors in that vector space.
     
  5. Sep 18, 2013 #4
    Yes but it is not mandatory to treat their linear sums as representing anything in the world. Textbooks often say that if a system is associated with some vector space then any Hermitian operator on that space designates a property - as if that followed simply from being a Hermitian operator. But this is misleading. For if the dynamics is stipulated in such a way that the state vector cannot rotate into such a sum, then for that reason alone there is no reason to assume that such a vector is an eigenvector of some measureable property.

    To answer your question, maybe the property of being conscious cannot superpose, as Wigner thought. Or as Penrose once thought (I think) the property of being a spacetime region cannot superpose. Then the question is whether such properties can be represented quantum mechanically (one might want to do this if one thought either of those properties collapses wave functions). This raises the question: if values for such properties are represented by orthonormal vectors, but superpositions are ruled out, can we make sense of time-evolution of their state vectors, or are such properties destined to never change...
     
  6. Sep 18, 2013 #5

    strangerep

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    Well, there's charge. Also total angular momentum (e.g., you can't have a superposition of spin-1/2 and spin-1, iirc.)

    I.e., any property for which superselection rules apply. :wink:

    Of course, lots of reasonable Hamiltonians tend to commute with such operators, so the conundrum doesn't arise.
     
  7. Sep 18, 2013 #6

    strangerep

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    Let's take a step back...

    What matters when constructing a quantum model of a system is the dynamical group. I.e., the largest group of transformations which maps solutions of the equations of motion into other solutions. Typically, the group has a continuous (identity-connected) subgroup, but there might also be discrete elements corresponding to parity, charge, and various other things.

    Then one finds a maximal mutually-commuting subset of the generators of the group (usually containing the Hamiltonian), then we find the spectrum (by demanding that these operators be representable as self-adjoint operators on an abstract Hilbert space). The spectrum of the Casimir operators for the group then determines one set of superselection rules, i.e., different values for the Casimirs correspond to distinct concrete Hilbert spaces, and so on. The procedure for quantizing ordinary angular momentum is a simple example of this. (Are you familiar with the latter, as presented in, say, Ballentine's textbook?)
     
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