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Understanding Mixed States, Pure States

  1. Sep 12, 2010 #1
    This question is from an example taken from Zurek 1991 (Decoherence and the Transition from Quantum to Classical).

    Start with a spin state of an electron, [tex]\Psi[/tex]= a|+> + b|->

    Question 1: Why is this considered a pure state? I figured it would not be called "pure" since it is a superposition. I guess it is pure because it is not entangled, right?

    Now measure the electron with some quantum device, d. The states |d+> and |d-> span the Hilbert space of the device.

    The resulting correlated system can be described by a density matrix rho = a^2|+><+|d+><d+| + ab*|+><-|d+><d-| + a*b|-><+|d-><d+| + b^2|-><-|d-><d-|

    Question 2: Is this also a pure state? Why?

    Question 3: If I then write down the reduced density matrix, by setting the off-diagonal terms to zero, does this become a mixed state? Why?
     
  2. jcsd
  3. Sep 12, 2010 #2
    Assuming [tex]\rho[/tex] is normalized so that [tex]\mbox{tr }\rho=1,[/tex] the state described by the density matrix is pure if [tex]\rho^2=\rho[/tex]. In this case [tex]\rho[/tex] is an orthogonal projection operator on a vector state (pure).

    [tex]\rho=|\psi\rangle\langle\psi|[/tex], [tex]\langle\psi|\psi\rangle =1.[/tex]

    Vector [tex]\psi[/tex] is determined by [tex]\rho[/tex] up to a complex phase.

    If [tex]\rho^2\neq \rho[/tex] then [tex]\rho[/tex] describes a mixed state, that is it has several different mutually orthogonal eigenvectors.
     
  4. Sep 12, 2010 #3
    Ahh, I see. Very helpful, thanks.

    Is it true that a mixed state cannot be factored into a bra times a ket? Or in other words, it does not project onto a state, so it can't be written in the form |psi><psi|?
     
  5. Sep 12, 2010 #4
    Right. That can be a definition of the truly mixed state.
     
  6. Sep 14, 2010 #5

    Just to clarify, a pure state can typically be rewritten, using a change of basis, in diagonal form, right? But the new basis states will be superposition states. In that case, wouldn't the pure state thus written also have several mutually orthogonal eigenvectors (i.e. the new basis states)?
     
  7. Sep 14, 2010 #6
    I am not sure I understand your question, but

    Say |1>,|2> are two basis vectors. Let

    [tex]|x>=\frac{1}{\sqrt{2}}(|1>+|2>)[/tex]

    Then the density matrix associated with |x> is

    [tex]\rho = |x><x|[/tex]

    From the very construction it follows that [tex]\rho^2=\rho[/tex] and [tex]tr(\rho)=1[/tex]. Therefor [tex]\rho[/tex] has eigenvalues 0 and 1. You can amuse yourself finding the two eigenvectors.
     
  8. Sep 14, 2010 #7
    Right, ok. I was just thinking that you were saying that only a density matrix that represents a mixed state would have a set of mutually orthogonal eigenvectors. But here, you have confirmed that a pure state will (trivially) have orthogonal eigenvectors as well.

    Of course, I may still be saying it wrong, but the help is much appreciated!
     
  9. Sep 15, 2010 #8
    Just remeber this rule: in a n-dimensional space a density matrix associated to a pure state has one eigenvector with eigenvalue 1 (unique up to a phase) and n-1 with eigenvalues zero (if n>2 then there is no way of choosing them in a unique way - unless you specify some basis that you like for some other reasons).
     
  10. Sep 15, 2010 #9
    I hope I don't bring more confusion here but...

    a mixed state can always be written as a sum over (projection operators of) pure states. I.e. the notation:

    [tex]\rho = \sum_i p_i |\psi_i\rangle\langle \psi_i|[/tex]

    refers to a sum over pure state projectors. Each state [tex]|\psi_i\rangle[/tex] is a pure state. Suppose you have some mixture of particles described by this density matrix. Then the probability that a particle is in the pure state [tex]|\psi_i\rangle[/tex] is given by [tex]p_i[/tex].

    If you have a system which is in a pure state then you can always choose a basis such that the density matrix is still written as the projection operator [tex]|\psi\rangle\langle\psi|[/tex], i.e. the density matrix of a pure state has a single eigenvector (like arkajad mentioned).

    Now the one thing I wanted to emphasize is that if you have a density matrix of a mixed state, then the corresponding pure states [tex]|\psi_i\rangle[/tex] need not be orthogonal with respect to each other. They can be, but they don't have to be. You can have [tex]|\psi_1\rangle[/tex] = spin up in z direction and [tex]|\psi_2\rangle[/tex] = spin up in the x direction. These states are not orthogonal to each other.
     
    Last edited: Sep 15, 2010
  11. Sep 15, 2010 #10
    Right. Density matrix must be non-negative of trace 1. Given any number m of 1-dimensional projection operators you cane form

    [tex]\rho=(P_1+\ldots +P_m)/m[/tex]

    and you have a non-negative operator of trace 1. This fact does not depend on whether [tex]P_iP_j=0,\; i\neq j[/tex] or not.
     
  12. Sep 16, 2010 #11
    Thanks for bringing me up to speed. I guess I hadn't been clear up until now on the distinction between the chosen basis, and the state of the system. Sounds like the state of the system is an objective quality of the system, and the basis states are arbitrarily chosen by me when describing the system.

    So tell me if I have this correct:

    Point 1) |K1> = a|+> + b|-> is a pure state, written in the basis |+> and |->. Right?

    Point 2) |K2> = a|+> - b|-> is another (orthogonal) pure state, written in the same basis.

    Point 3) However, these two form another orthonormal basis of the space. Ahhh...That is where I am getting myself confused...between states and bases. The Ket |K2> is often used to form an orthonormal basis with |K1>. But it's not relevant here as a state of the system.

    So if we stick with our original basis |+> and |->, then the state |K1> is a pure state.
    Question: Can this be turned into a density matrix, just this one pure state? Rho = |K1><K1|. The answer is yes, but it will introduce non-diagonal elements, since we have written the density matrix in terms of an arbitrary basis. Correct?
     
  13. Sep 16, 2010 #12
    Just a few corrections: |K1>,|K2> is an orthogonal system but not orthonormal - unless you normalize dividing by sqrt(2).

    When some autors talk about non-diagonal elements - they have in mind a "preferred basis". So, if your preferred basis is |+>,|-> then your Rho has non-diagonal elements in this basis. But is diagonal in |K1>,|K2> basis - which may be a preferred basis of your friend.
     
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