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Determinant of density matrix

  1. Nov 2, 2009 #1

    Fredrik

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    Why is the determinant of a mixed state density matrix always positive?

    In the specific case of a 2-dimensional Hilbert space, the density matrix (as well as any other hermitian matrix) can be expressed as

    [tex]\rho=\frac 1 2 (I+\vec r\cdot\vec \sigma)[/tex]

    so its determinant is

    [tex]\det\rho=\frac 1 4(1-\vec r^2)[/itex]

    We have [itex]|\vec r|=1[/itex] if and only if we're dealing with a pure state, so we seem to need the condition [itex]\det\rho\geq 0[/itex] to see that the set of mixed states is the interior of the sphere rather than the exterior.
     
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  3. Nov 3, 2009 #2
    I think that if you have mixed states statistical operator you have

    [tex]{\hat{\rho}}^2 < \hat{\rho}[/tex]

    and of course for statistical operator [tex]Tr\hat{\rho}=1[/tex] and [tex]\hat{\rho}^{\dagger}=\hat{\rho}[/tex]

    I think that from that three conditions you have [tex]det\hat{\rho}>0[/tex]
     
    Last edited: Nov 3, 2009
  4. Nov 3, 2009 #3

    Fredrik

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    Thanks, but now I have two problems instead of one. I don't see why [itex]\rho^2<\rho[/itex] for mixed states. (Do you mean that all expectation values of [itex]\rho-\rho^2[/itex] are positive?) I also don't see why these results imply that [itex]\det\rho>0[/itex].
     
  5. Nov 3, 2009 #4

    strangerep

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    The "state operators" in the original post don't necessarily qualify as such...

    Summarizing from Ballentine section 2.3 (pp 50-51)...

    An acceptable state operator (a) has unit trace, (b) is self-adjoint, and
    (c) satisfies [itex]\langle u | \rho | u \rangle \ge 0[/itex], for all [itex]| u \rangle[/itex].

    Property (b) implies that a state operator's eigenvalues [itex]\rho_n[/itex] are real.

    Property (c) implies that a state operator's eigenvalues must satisfy [itex]\rho_n \ge 0[/itex].

    Hence the determinant is real and non-negative, being the product of the eigenvalues.

    [EDIT: Removed unnecessary silly mistake herein noted by Fredrik in post #5 below.]
     
    Last edited: Nov 3, 2009
  6. Nov 3, 2009 #5

    Fredrik

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    Thank you strangerep. That was surprisingly simple (at least when I only consider finite-dimensional Hilbert spaces, but that's the case I'm interested in for the moment). [Edit: Removed comment about silly mistake in #4 :smile:]
     
    Last edited: Nov 4, 2009
  7. Nov 4, 2009 #6
    Just to show c)

    [tex]\hat{\rho}=\sum^K_{k=1}w_k|\psi_k> <\psi_k|=\sum^K_{k=1}w_k\hat{P}_k[/tex]

    and


    [tex]\left\langle \psi_n\left|\hat{P}_k\right|\psi_n\right\rangle=||\hat{P}_k\psi_n||^2[/tex]
     
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