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

Can we prepare mixed states?

  1. Mar 15, 2014 #1

    naima

    User Avatar
    Gold Member

    As mixed states and density matrices are the generalization of pure states, i wondered if it was possible to prepare a given mixed state. I know that decoherence give mixed states. Are there other ways to get them?
    measurements on pure states always give pure states (POVM also). So how?
     
  2. jcsd
  3. Mar 15, 2014 #2
    Sure. To get the mixed state density matrix ##\rho = a |\psi\rangle\langle\psi| + b |\phi\rangle\langle\phi|##, prepare ##|\psi\rangle## with probability ##a## and ##|\phi\rangle## with probability ##b##, but don't look to see which one you prepared.
     
  4. Mar 15, 2014 #3
    I doubt his question was on the form of a mixed state density matrix.
     
  5. Mar 15, 2014 #4

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    A mixed state is a mixture of classical probability (which reflects ignorance about the true state) and quantum probability (which is involved even when you know for certain what the state is). The interesting thing about mixed states is that there is no unambiguous way to tease apart the two sources of probability.

    As The_Duck says, you can interpret a mixed state, described by a density matrix [itex]\rho[/itex] as arising from these two kinds of probability:

    [itex]\rho = \sum_j P_j |\Psi_j \rangle \langle \Psi_j |[/itex]

    This is the mixed state from choosing the pure state [itex]\Psi_j[/itex] with probability [itex]P_j[/itex]. But given [itex]\rho[/itex], there is no unique way to find the [itex]P_j[/itex] and the [itex]\Psi_j[/itex]
     
  6. Mar 15, 2014 #5

    WannabeNewton

    User Avatar
    Science Advisor

    Doesn't matter. The_Duck's response highlighted the fact that preparation of a mixed state involves classical probabilities (ignorance about the state of the system) arising from convex combinations of pure states, which is the main idea behind such a preparation. If you dislike the abstract example then just apply it verbatim to a Stern-Gerlach experiment.
     
  7. Mar 15, 2014 #6

    atyy

    User Avatar
    Science Advisor

    In the sense that decoherence gives mixed states, preparing any pure state of a system consisting of subsystems will give a mixed state, provided you make a measurement on the subsystem. This is an improper mixed state.

    In contrast, the answer that The Duck gave you is on how to prepare a proper mixed state.
     
  8. Mar 15, 2014 #7
    What I meant was that the OP probably understood the theory provided.
     
  9. Mar 15, 2014 #8

    naima

    User Avatar
    Gold Member

    I am looking for a setup that will give mixed states. Without a computer with random choice.
     
  10. Mar 15, 2014 #9

    atyy

    User Avatar
    Science Advisor

    Make a measurement on a known pure state, but don't look at the answer.
     
  11. Mar 15, 2014 #10

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Proper mixed states are reflections of ignorance about the true state of the system. So you don't have to do anything special to get a proper mixed state. The mixed state is the way you describe the situation; it's not inherent in the situation.

    If you create an electron and have no idea what its spin is, then you would describe it as the mixed state that is mixture of spin-up and spin-down (along any axis you like--you get the same density matrix, regardless of which axis you choose).

    Improper mixed states are produced by entanglement, where the system of interest is coupled to another system (the environment, or a measuring device). When you trace over the degrees of freedom of the system you don't care about, you get a density matrix for the system you do care about that is a mixed state.
     
  12. Mar 15, 2014 #11

    naima

    User Avatar
    Gold Member

    ????????????????????
    A measurement on a pure state will give a pure state.
     
  13. Mar 15, 2014 #12

    WannabeNewton

    User Avatar
    Science Advisor

    There can be a device which prepares a particular pure state from a prescribed collection of pure states, the latter of which you are aware of. However you do not inquire about which particular pure state was in the end prepared by the apparatus. This is an instance of a mixed state preparation procedure. It's artificial but valid nonetheless. All you need is a classical ensemble of some kind.
     
  14. Mar 15, 2014 #13

    naima

    User Avatar
    Gold Member

    we are so accustomed to pure states that we forgive that pure states are also tools that are used to manage the fact that we ignore what will be the outcomes of setups. We use beam splitters for them, logic gates too. I do not think that for mixed states the only machinery is "dont look at the result".
     
  15. Mar 15, 2014 #14

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    As I said, whether a state is mixed or not is a reflection of your ignorance about the state. If the state is either spin-up, or spin-down, and you don't know which, then it's in a mixed state.
     
  16. Mar 15, 2014 #15

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    For (proper) mixed states, there is nothing more needed than ignorance. So "don't look at the result" is perfectly valid as a way to produce a mixed state.

    You might be right, that pure states may themselves be subjective, in some sense, but it's more complicated to understand pure states that way.
     
  17. Mar 15, 2014 #16

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How about this then? Send a beam of silver atoms through a Stern-Gerlach device that splits the beam into a left beam and a right beam. If you put up a screen that blocks the right beam but lets the left beam through a hole, you will have prepared a pure spin state. Let's call it "up". What if you move the screen so that it's very close to the magnet, and the hole is only very slightly to the left of the path that the atoms would have taken if there had been no magnetic field? Wouldn't you have to describe the spin state of the atoms that go through the hole as mixed?

    What we're doing here is essentially to do a Stern-Gerlach experiment badly (on purpose). In this scenario, we're not just "not looking at the result", because there's no reliable result to look at.

    I agree with this.
     
  18. Mar 15, 2014 #17

    atyy

    User Avatar
    Science Advisor

    Yes, a measurement will give you a definite result, and you can take that to be a pure state. But by the Born rule, each time you do a measurement on the same initial pure state, you will get a different result (assuming the initial pure state was not an eigenstate of the measurement operator). So by accumulating the ensemble of pure states that result from multiple identical measurements on the same initial pure state, you will get a mixed state according to the Born rule. This mixed state is a proper mixed state.

    The other way to get a mixed state is to simply prepare a pure state of a large system, in which the subsystems are entangled. The reduced density matrix of a subsystem will be an improper mixed state.
     
  19. Mar 15, 2014 #18

    naima

    User Avatar
    Gold Member

    I agree . A perfect which way information will give a mixed outcome.
    If i have imperfect which-way information this will change the "//www.atomwave.org/rmparticle/ao%20refs/aifm%20refs%20sorted%20by%20topic/decoherence%20refs/englert%20visibility.pdf" [Broken].
    thank you for the idea.
     
    Last edited by a moderator: May 6, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook