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A Weinberg: Quantum Mechanics Without State Vectors

  1. Mar 29, 2016 #21
    Why is the rhs of your equation positive? In your link it is negative.
     
  2. Mar 29, 2016 #22

    ShayanJ

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    ## [\rho,H]=-[H,\rho]##
     
  3. Mar 29, 2016 #23

    vanhees71

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    The von Neumann equation for the density operator in the Schrödinger picture is
    $$\frac{1}{\mathrm{i}\hbar} [\hat{\rho},\hat{H}]+(\partial_t \hat{\rho})_{\text{expl.}}=0.$$
     
  4. Mar 29, 2016 #24
    So by reformulating quantum theory the weirdness of the double slit experiment and the violation of Bell's inequality (= the negation of local realism) disappear?
    That's pretty weird.
     
  5. Mar 29, 2016 #25

    marcus

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    Maybe someone else would like to comment. That's how I read what is said in the abstract
    "With this definition of a physical state, even in entangled states nothing that is done in one isolated system can instantaneously effect the physical state of a distant isolated system. This change in the description of physical states opens up a large variety of new ..."
     
  6. Mar 29, 2016 #26
    I think you're reading too much into that quoted passage. Run of the mill QM doesn't say that "one isolated system can instantaneously effect the physical state of a distant isolated system." That is just a bit of quantum flapdoodle [Gell-Mann]. QM merely says the results of Alice and Bob's measurements are correlated and those correlations will not disappear with a reformulated theory.
    How those correlations take place is indeed a mystery, but no more so than how masses make gravitational attraction.
     
  7. Apr 3, 2016 #27
    It would be great to better understand Weinberg's proposal. But I'm not used to density matrix formulation. I only have basic understanding of the state vector formalism he is criticizing. What is the physical significance of the extra terms in his Lindblad equation? First we have ##L_n ρ(t) L_m^\dagger##. To me this looks similar to ##O_{ij} = <A_i | \hat{O} | A_j>## which just defines the matrix elements for operator ##\hat{O}## in a basis ##|A_1>,...|A_N>##. So is ##L_n ρ(t) L_m^\dagger## doing nothing more than casting the density matrix of the system in the L basis? And then I wonder what the physical significance of subtracting ##½(ρ(t) L_m^\dagger L_n + L_m^\dagger L_n ρ(t))## is. Is the part that's added to the quantum Liouville equation doing nothing more than constructing a matrix out of the unspecified ##h_{n,m}## elements in the ##L## basis? Thanks!
     
  8. Apr 3, 2016 #28

    A. Neumaier

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    pp.21-24 of my slides http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf might be of help in interpreting the Lindblad equation in a concrete situation. In a microscopic derivation, all three terms arise together; it therefore does not make sense to interpret the three terms separately. The combination is the generator of a completely positive map - that's what counts.
     
    Last edited: Apr 5, 2016
  9. Apr 5, 2016 #29
    Thanks for this. A concrete illustration of the Lindblad equation is certainly helpful and your slides are clearly written. But I'm still puzzled.

    One prima facie difference between your equation and the general Lindblad equation is that the sum (the sigma) is not present in yours. For an N-dimensional system there is a sum of ##N^2-1## terms. Why is that missing in yours? Why isn't there such a sigma in front of each of your three additional terms (corresponding to a, b, and c)?

    Your equation also appears to be missing the ##h_{n,m}## term, i.e. the term for the hermitian matrix elements - the constants that determine the dynamics. What you instead have outside each bracket is just one constant. For example outside the first set of brackets you have 0.002 x 138 - the cavity loss rate. Shouldn't the ##h_{n,m}## term be specified so that it's clear that the matrix elements are continuous functions of the relevant constant (such as cavity loss rate)?

    In trying to make sense of your claim that the terms make no sense separately, but together function to guarantee "complete positivity", I stumbled across this. It helped to define ##L_n ρ(t) L_m^\dagger## as a test for whether ρ(t) is positive definite: ρ(t) is positive definite relative to operator ##L## if ##L_n ρ(t) L_m^\dagger## yields a positive number. Is the idea that (##L_n ρ(t) L_m^\dagger - ½(ρ(t) L_m^\dagger L_n + L_m^\dagger L_n ρ(t))## just functions to ensure all eigenvalues are positive?
     
  10. Apr 5, 2016 #30

    A. Neumaier

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    https://en.wikipedia.org/wiki/Lindblad_equation gives the most general form. The number ##N## of terms could be arbitrary but one can show that this many are enough. My example just has three nonzero diagonal terms and no nonzero off-diagonal term.

    In the initial wikipedia form, only the complete expression generates a completely positive map; thus even the individual expressions whose coefficients are the ##h_{n,m}## don't mean anything. But there is an equivalent diagonal form, derived in the subsection https://en.wikipedia.org/wiki/Lindblad_equation#Diagonalization
    which is of the same type as mine, just with more terms, and which can be interpreted physically.

    ρ(t) is positive definite iff ##x^*ρ(t)x>0## for all nonzero ##x##. But this is unrelated to complete positivity.

    Actually my original statement was slightly incorrect - I had meant ''generates a completely positive map'' not ''is a completely positive map''. (I corrected my statement.)
     
  11. Jan 11, 2017 #31

    A. Neumaier

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    More information about this is in the discussion in a related thread, starting here.
     
  12. Apr 16, 2017 #32
    This question hasn't been answered yet, and it puzzles me as well. In the second paper, Weinberg says
    "Whether in open systems in ordinary quantum mechanics or in closed systems in some modified version of quantum mechanics, in order to avoid instantaneous communication at a distance in entangled states, it is important to require that the density matrix at one time depends on the density matrix at any earlier time, but not otherwise on the state vector at the earlier time."

    Since the density operators of both subsystems in a Bell experiment are completely mixed, how can local measurements (be it his suggested Lindblad evolution or any other kind) be correlated?
     
  13. Apr 18, 2017 #33

    kith

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    The reduced density matrix contains all information which is relevant for experiments which are performed on the subsystem alone.

    Weinberg stresses that if the outcomes of such a measurement on subsystem A depended on the reduced density matrix of subsystem B, FTL communication would be possible by altering the reduced density matrix of B.

    What Weinberg's statement is silent about is whether the reduced density matrices contain all information about the combined system of A and B. And indeed, they don't contain information about how the outcomes of individual measurements on the subsystems are correlated.

    The information about the correlations is only contained in the density matrix of the combined system and not in the reduced density matrices of the subsystems.
     
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