Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum mechanics without unitary evolution

  1. Feb 25, 2006 #1


    User Avatar

    A philosophy that underpins many approaches to understanding quantum mechanics (the many worlds interpretation in particular, but collapse models and other related ideas also) is that continuous Schroedinger evolution is somehow `nicer', `preferred', or `more fundamental' than the "damned quantum jumps".

    A measurement in QM can be described by a set of Kraus operators [itex]\left\{
    K_{i}\right\} [/itex] which satisfy
    \sum_{j}K_{j}^{\dagger }K_{j}=\mathbb{I}.
    For a system initially in some state [itex]\rho,[/itex] the final (collapsed) state after a measurement which yields outcome [itex]j[/itex] with probability [itex]p_{j}=Tr(\rho K_{j}^{\dagger }K_{j})[/itex] is
    \rho \rightarrow K_{j}\rho K_{j}^{\dagger }/p_j
    In standard quantum mechanics the continuous (Schroedinger) evolution takes the form
    \rho \rightarrow U\rho U^{\dagger }.
    where [itex]U[/itex] is a unitary operator.
    If the Hamiltonian governing this evolution has spectral decomposition
    H=\sum_{j=1}^{d}\lambda _{j}|j\rangle \langle j|
    then this unitary is given by
    U=\sum_{j=1}^{d}e^{\lambda _{j}t}|j\rangle \langle j|.
    (Note I'm just doing the finite dimensional case here for simplicity).
    Alternatively we can imagine that the system is actually undergoing a large number of very frequent measurements as follows. Define the Kraus operators
    K_{j}=\frac{1}{\sqrt{d}}\left[ \mathbb{I+}\left( e^{id\lambda _{j}\tau
    }-1\right) |j\rangle \langle j|\right]
    where [itex]\tau [/itex] is a very small time increment, and we presume a measurement occurs approximately every [itex]\tau[/itex] seconds. Since one can readily verify that [itex]K_{j}^{\dagger}K_{j}=\mathbb{I}/d[/itex] we see that regardless of the initial state [itex]\rho [/itex] the outcomes are all equally likely. Thus in a time [itex]t\gg \tau [/itex] roughly [itex]\tfrac{t}{d\tau}[/itex] of each specific outcome will be obtained, and it is easy to see then that the final state will be very close to the one which unitary evolution would have generated. One may think that [itex]\tau[/itex] needs to be very small (say Planck scale), but thinking about it I cannot see that we have experimental evidence of smooth evolution beyond the scale of optical vacumm fluctations ([itex]1/\omega^3[/itex]) with [itex]\omega[/itex] roughly an optical frequency.

    Thus we see that the "less fundamental" form of quantum evolution can actually subsume the supposedly more fundamental one. Perhaps our attachment to unitary evolution is simply an historical artifact better dispensed with!
  2. jcsd
  3. Feb 25, 2006 #2
    I see the point of the post in regards to another way to look at quantum evolution, but not about unitarity.

    I haven't ever been taught that unitarity is a necessary requirement of quantum mechanics. The time reversal operator is, in fact, anti-unitary. Unitarity simply provides some easily calculable results. In addition there is a nice little theorem that any operator is similar to a unitary (or anti-unitary) operator. So we lose nothing by using the unitary operators. (There's a detail about the operators I'm not recalling off the top of my head, but that's the gist of it. Perhaps they need to be observables?)

  4. Feb 25, 2006 #3


    User Avatar
    Science Advisor

    Unitarity in QM is nothing more, nothing less than basic conservation of probability, (antiunitary operators conserve probabilty; a phase is but a phase. It is, in fact, a vitally necessary condition for any, repeat any dynamical theory based on probability -- quantum or classical. One of the very clever aspects of QM is that it gives a very usable approach to probability when particle number is not conserved.

    Among other things, as d-->0 the Krause operator blow up. I've not worked through your computations, but this K sort'a looks like the transition amplitude for a finite time. If the state at the start is an eigenstate of H, why somewhat later would all states have equal probability? (Nature does not always work that way.)

    How do you deal with the dynamical evolution of, say, today's Slalom
    race at the Olympics? Do you use your Kraus approach, say, for a random walk problem?

    Your approach would be far better stated if you would do a whole problem -- say basic radioactive decay, or scattering from a standard 1-D potential well.
    Reilly Atkinson
    Last edited: Feb 25, 2006
  5. Feb 26, 2006 #4


    User Avatar

    "d" is the dimension of the Hilbert space, so d>=2.

    The Kraus map I gave is trace preserving, this is all that is required for proabability conservation. (See Nielsen and Chuang's "Quantum Computation", chapter 8 or any other similar textbook for an explanation).

    I'm not interested in anti-unitary operators, since they have nothing to do with evolution per se. Unitary evoltion is simply that resulting from the Schroedinger equation, and that is what is very well approximated by this sort of operation...
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Quantum mechanics without unitary evolution