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Must ensembles be homogeneous?

  1. Nov 9, 2014 #1
    An ensemble is a collection of systems, all prepared in the same way. Does this mean that all the systems are in the same state? I have seen some authors create ensembles where 30% of the systems are in a state, s, and 70% of the systems are in a state, t . As far as measurements go, this ensemble represents the mixed state 0.3 s + 0.7 t . Is this accepted as a valid ensemble for this mixed state?
     
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  3. Nov 9, 2014 #2

    dextercioby

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    Yes, the same pure state per individual system. It's one of the underlying (i.e. not explicit) assumptions in QM: the ability to prepare a system in a certain state.
     
  4. Nov 9, 2014 #3
    What about the mixed (non-pure) state example that is in the original post?
     
  5. Nov 9, 2014 #4

    dextercioby

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    I honestly don't know how to physically prepare a statistical mixture. A pure one can be done through a Stern-Gerlach type of experimental setup. So, sorry for not helping more, I'm rather into the mathematical delopments of QM than to the their connection with real-life experiments.
     
    Last edited: Nov 9, 2014
  6. Nov 9, 2014 #5
    Thanks for the info. Perhaps, someone else will know about ensembles of mixed states.
     
  7. Nov 9, 2014 #6

    atyy

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    Yes, the ensemble can be mixed - it can be any ensemble that is represented by a density matrix. One way to prepare a mixed state is to perform a measurement, but don't sort according to the outcome. The ensemble will be a statistical mixture of all possible measurement outcomes. The mixed density matrix prepared by ignorance of a measurement outcome is called a "proper mixture".

    A mixed state can also be prepared by preparing a system that is in a pure state. If the subsystems are entangled, then the reduced density matrix of a subsystem will be mixed. As an example of a mixed state preparation, take a Bell state of a pair of spins. The Bell state is a pure state. If you make observations on only one spin in the entangled pair, the reduced density matrix describing the state of that spin is mixed. The mixed density matrix prepared by ignorance of some degrees of freedom in a total pure state is called an "improper mixture".

    There is a relationship between the two ways of preparing a mixed state. Take the Bell state again. If we always measure on spin A first, we will cause the wave function to collapse. However, the person measuring spin B, if performing his measurement locally and not sorting according to the outcome of the measurement on spin A, will basically be doing a measurement on a mixed state prepared by measurement followed by not sorting. This is an example of local measurements being completely unable to distinguish between proper and improper mixtures. Nonlocal measurements can distinguish between them.
     
    Last edited: Nov 9, 2014
  8. Nov 9, 2014 #7
    Thank you for the detailed response. From the question in the original post, the mixed state 0.3 s + 0.7 t is represented by the ensemble having 30% of the systems in state s and 70% of the systems in state t, but this is not a proper mixture unless the states s and t happen to be the two eigen states of a measurement operator, with eigenvalues 0.3 and 0.7, respectively. Do you see any problem with using this ensemble, which is neither proper nor improper, to represent the mixed state?
     
  9. Nov 10, 2014 #8

    bhobba

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    You understand about proper and improper mixtures :D:D:D:D:D:D:D.

    Improper ones happen all the time due to decoherence.

    Proper ones are easily done simply by randomly mixing up pure ensembles. Say for axample you want 1/6th in a certain pure state and 5/6th in another pure state simply take a dice, throw it, and every time you get a 1 select it from one state otherwise from another. Do that a large number of times.

    I will leave you to think about the implications of the fact it can conceptually be made as large as you like, but not infinite. Its the same issue you get in stochastic models used in applied math all the time.

    Thanks
    Bill
     
  10. Nov 10, 2014 #9

    atyy

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    The state you describe is a proper mixture for any s and t that are pure states. A proper mixture is just a statistically weighted mixture of any pure states, so as long as the relative weights add up to 1, it's ok. http://en.wikipedia.org/wiki/Density_matrix
     
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