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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 16, 2013 #2
    Well, I would say that you're describing one way to view how quantum states change, but there is another. Your view is saying that if the state vector is in an eigenstate of some operator at one moment, then it might evolve according to the schrodinger equation into a superposition of eigenstates. For example, if we had a double dirac delta potential (one where there are two wells centered at x1 and x2 respectively, all completely time-independent), then the bound states could be represented in the position eigenbasis with basis vectors |x1>, |x2>, and a state prepared initially in |x1> will evolve under the schrodinger equation into a superposition of both states. (The particle will get a nonzero probability of tunnelling into the other dirac well.) This is like the state, initially on an axis, rotating into an off-axis point, with the axes stationary.

    The second way a quantum state can change is if the hamiltonian changes. For example, a hydrogen atom in its first excited state is an energy eigenstate (an eigenstate of the hamiltonian), so it should be stable forever under schrodinger evolution. (It will always have a 100% probability of being in that excited state.) In other words, if the hamiltonian does not change and the state vector is sitting on an axis in the energy eigenbasis, then the state vector will sit exactly on that axis forever. However, if an electromagnetic wave passes by the hydrogen atom, the EM wave can cause a stimulated decay of the atom. (It develops a nonzero probability of being in a decayed state.) This transition can occur because the electromagnetic wave causes a quickly time-varying term in the hamiltonian, and the schrodinger evolution of the state vector cannot "keep up" with quickly-varying energy eigenstates.

    In general, you could look at this second way for the quantum state changing it in this way: The state |ψ> might be an eigenstate of the hamiltonian at one instant, but if a sudden change in the hamiltonian occurs, |ψ> will not have had time to change much (since schrodinger evolution is continuous), but the eigenstates have all changed suddenly. So the state |ψ> is now a superposition of the new eigenstates, and has a nonzero probability of being measured in multiple of the new eigenstates. So this picture is more like the axes rotating and the state staying stationary (as opposed to the other way around, as in the double dirac-delta potential scenario).

    First of all, the state vector doesn't live in the complex plane. It lives in a Hilbert space, which is a multidimensional complex space (usually a function space) which is basically euclidean. This is an important distinction.

    Anyway, one example of what you seem to be describing is called the Adiabatic theorem. This theorem states that if the hamiltonian changes over time, but it changes very slowly, then the nth eigenstate of the hamiltonian at one instant will evolve into the nth eigenstate of the hamiltonian at an instant later with 100% certainty. In other words, if the Hamiltonian is changing, then the energy eigenstates will be changing too, but since the changes in the eigenstates are slow enough, schrodinger evolution won't cause "transitions", it will simply keep an eigenstate as an eigenstate without forming superpositions. This picture would be like the state sitting on an axis, and the axes slowly rotating, with the state being stuck to one rotating axis.
     
    Last edited: Sep 17, 2013
  4. Sep 16, 2013 #3

    Office_Shredder

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    You won't be able to go from one specific eigenstate to another without intermediate state vectors - one dimensional subspaces of a Hilbert space are always closed (as are any finite dimensional subspaces), so our starting state is v, and our ending state is w, and u(t) is a continuous curve with u(0) = v and u(1) = w, then there is a largest value of t for which u(t) lies in the span of v (i.e. our state is v), and a smallest t for which it lies in the span of w, and these are obviously not equal, and any value of t in between is going to be an intermediate state.

    If you had an infinite dimensional eigenspace which was not closed (I'm not sure if this is possible - it's clear that it requires your linear operator to not be continuous, but my intuition for that happening is that you can get arbitrarily large values near the origin, so I'm unclear about whether you can have non-closed eigenspaces in the first place) then it might be possible to jump from one eigenspace to the next, which from my rudimentary understanding of quantum mechanics would mean you are jumping discontinuously between measurable values

    As for rotating through the complex plane - remember that your base field is the complex numbers, so multiplying by i is the same as multiplying by any other scalar. It's like trying to get from the vector (1,0) to (0,1) by multiplying by a real number
     
  5. Sep 17, 2013 #4
    Thanks Jolb! Just to be clear, you think the process that is closest to the one I'm looking for, is one in which the quantum state of the system remains stuck in the same state?
    At one point you say "the energy eigenstates will be changing" but presumably when you say "the nth eigenstate of the hamiltonian at one instant will evolve into the nth eigenstate of the hamiltonian at an instant later" it doesn't sound like there is any evolution or change at all given that we appear to stay in the nth state. Have I missed something?
     
  6. Sep 17, 2013 #5
    Sorry can you please explain the significance of the space being nonclosed? Why do you think that would enable eigenstate jumps?
     
  7. Sep 17, 2013 #6
    Well, under adiabatic changes in the hamiltonian, a particle prepared in the nth energy eigenstate stays stuck there, but that eigenstate is changing! For example, imagine we had a particle in a 1-d box of length L. If the particle is in the 1st excited state (an energy eigenstate), and then the box very slowly grows from length L to length 2L, the particle will end up in the 1st excited state of the box of length 2L with 100% certainty. Clearly the state has changed--the region where it has nonzero probability of being observed has doubled in size! But it stayed in the 1st excited state.

    Contrast this with a box that suddenly (e.g. instantaneously) grows from L to 2L. The particle initially in the 1st excited state will in general evolve into a state with nonzero probability of being observed in either the ground state, or the 1st excited state, or the 2nd excited state, ...
     
    Last edited: Sep 17, 2013
  8. Sep 17, 2013 #7
    It's true that the state has changed, but if the change is only that the region where it has nonzero probability of being observed as doubled, then its position wave function has spread meaning its evolved into a superposition of many more positions than before? But then this doesn't illustrate eigenstate-to-eigenstate evolution, but superposition spread?

    I get the impression that there's a real gap in my understanding here. For I can understand the above example given the description I gave, in this example...

    My description would entail that there is no difference, in both cases excited state stays fixed, position wavefunction spreads.


    One example that I thought would illustrate my point would be the so-called "total-of-nothing boxes" that change the states of systems by multiplying their state vectors by -1.
    So if you put a particle in state |A> through the box it will "change" the state to -|A>; but since |A>=-|A> there is not physical change.
    But if |A> = c|B> + c|B>
    And if |A'> = c|B> - c|B>
    And using a two-path experiment you put the +c|B> component through the total-of-nothing box instead, and recombine the paths, you will find the state to be |A'> rather than |A>.
    Isn't that a transition from eigenstate |A> to orthogonal eigenstate |A'> that doesn't go through superpositions of |A> and |A'>?
     
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