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About mixed state and pure state

  1. Apr 10, 2008 #1
    I don't have any problems dealing with mechanical calculations(I think), but yet I have some conceptual problems with mixed state(= statistical mixture?) and pure state in QM.

    - pure state : |Φ> = 1/sqrt2 ( |↑> + |↓> )

    - mixed state : ρ = 1/2 ( |↑><↑| + |↑><↑| )

    What is the difference besides the coefficients?

    And, some denote that |↑><↑| is an operator but others a state.

    I am really confused. Somebody give me clarification.
     
  2. jcsd
  3. Apr 10, 2008 #2

    Fredrik

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    |↑><↑| is an operator, but since there's exactly one operator |u><u| for each state vector |u>, it still makes some sense to call it a "state".

    The "pure state" above represents a system that's in a superposition of "spin up" and "spin down". The "mixed state" represents a system that's either in the the "spin up" state, or in the "spin down" state.

    It's also worth noting that that particular density matrix (or density operator if your prefer) is actually equal to [itex]1/2(|\leftarrow\rangle|\leftarrow\rangle+|\rightarrow\rangle|\rightarrow\rangle)[/itex], so it also represents a system that's either in the "spin right" state or the "spin left" state. I think you can actually let "?" represent an arbitrary direction and interpret that density matrix as representing a system that's either in the "spin ?" state or "spin opposite of ?" state.
     
  4. Apr 10, 2008 #3
    Thx for the reply. appreciate that.

    I am still kinda confused. We call the "pure state" above as a superposition of spin up and down, and when we measure it, we can then know that the system is either spin up or down state. But for "mixed state" system, we just say the system is either in spin up or down state which is pre-determined. Am I right?
     
  5. Apr 10, 2008 #4

    nrqed

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    Yes, you are right.

    A mixed state is a mixture of different state in th eclassical sense: say 50% of the particles are actually spin up and 50% are actually spin down. Each particle has a well-defined spin state. If we measure the spin of one particle we may get up or down with a 50% probability but that has nothing to do with quantum superposition.
     
  6. Apr 10, 2008 #5
    It is better to look at this from the point of view of statistical mechanics. There you have an ensemble of identical systems. The density matrix tells you what fraction of those systems are in state [tex]\left|\uparrow\right\rangle[/tex] and what fraction are in the state [tex]\left|\downarrow\right\rangle[/tex]. So if you want to know expectation values of operators, you must take into account both the quantum expectation value, and the fact that you have a certain fraction of systems in one state, with one expectation value, and a certain fraction in another state.

    Now, if you believe that quantum mechanics is correct, then because of its linearity properties, you as an experimenter can only interact with one universe (your own), with one pure state [tex]\left|\Psi\right\rangle[/tex]. You are free to postulate an ensemble of universes, but your expectation values will only depend on [tex]\left|\Psi\right\rangle[/tex] (again, because of linearity). So why do we care about mixed states? Well, the universe is a closed system, but most of the time we deal with open systems, meaning there is a part of our system that we cannot measure.

    So lets start with a density matrix of a pure state. You'll agree that this is equivalent to a pure state, so just a change of notation. Now, if we can only make measurements of one part of the system, then in order to take expectation values using the (pure) density matrix, we need to "trace out" the variables of the system we cannot measure. This is called a "reduced density matrix". The reduced density matrix will generally "look like" a mixed state, i.e., an ensemble.

    One example is that if your system is in contact with an infinitely large heat reservoir, then the entire system (system + reservoir) is in a pure state, but if you don't know about the reservoir then your system will be in a mixed state with thermal distribution. The same thing happens in classical statistical mechanics where the entire system is described by a microcanonical ensemble, but your system is best described by a canonical ensemble.

    I have a write-up about it here: http://www.physics.thetangentbundle.net/wiki/Quantum_mechanics/density_matrix#Reduced_density_matrix if you are interested.
     
  7. May 9, 2010 #6
    A pure quantum state is a state which can be described by a single ket vector and
    A mixed state cannot be described as a ket vector. Instead, it is described by its associated density matrix (or density operator), usually denoted ρ
     
  8. May 10, 2010 #7
    Code (Text):
    I don't have any problems dealing with mechanical calculations(I think), but yet I have some conceptual problems with mixed state(= statistical mixture?) and pure state in QM.

    - pure state : |Φ> = 1/sqrt2 ( |↑> + |↓> )

    - mixed state : ρ = 1/2 ( |↑><↑| + |↑><↑| )

    What is the difference besides the coefficients?

    And, some denote that |↑><↑| is an operator but others a state.

    I am really confused. Somebody give me clarification.    
    IN PURE STATE THERE ARE SOME ANOTHER STATES ( ρ=|Φ><Φ| = ( |↑><↑| + |↓><↑| +|↓><↓| +|↑> <↓|) BUT IN MIXED STATE THERE ARE NOT ANY NONCLASSICAL STATES .IN COMPLETLEY MIXED NONDIAGONAL ELEMENTS ARE ZERO
     
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