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[Quantum Mechanics] Quantum Fisher Information for a Pure State

  1. Mar 8, 2012 #1
    Hi everyone.

    1. The problem statement, all variables and given/known data
    We are given N spins 1/2. A rotation is defined as
    [itex]\rho_\theta=e^{-i\theta J_n}\rho_\theta e^{i\theta J_n}[/itex]
    on an Hilbert Space H, with
    [itex]J_n=n_xJ_x+n_yJ_y+n_zJ_z\:,\quad n_x^2+n_y^2+n_z^2=1[/itex],
    and [itex]\theta[/itex] isn't related to any observable.
    Given a quantum state [itex]\rho=\sum_ir_i|r_i\rangle\langle r_i|[/itex],
    the Formula for the Quantum Fisher Information i've come to is
    [itex]F[\rho,J]=2\sum_{i,j}\frac{(r_i-r_j)^2}{r_i+r_j}|\langle r_i|J|r_j\rangle|^2[/itex] (which is indeed right).
    Problem is that I have to calculate the Quantum Fisher Information for a Pure state [itex]\rho=|\psi\rangle\langle\psi|[/itex].
    The solution should be [itex]F[\rho,J]=4\Delta_\psi^2J[/itex],
    where [itex]\Delta_\psi^2J=\langle\psi|J^2|\psi\rangle-(\langle\psi|J|\psi\rangle)^2[/itex] is the variance of J, but I can't come to it


    2. Relevant equations
    I have to use the given equation for Fisher Information with the fact that [itex]\rho[/itex] is pure.


    3. The attempt at a solution
    I have difficulties in how to procede. In pure states all the coefficient [itex]r_i[/itex] should be 0, except for one of the, which should be 1.
    Any idea?
    Many thanks, this is quite urgent :(
     
  2. jcsd
  3. Mar 8, 2012 #2
    With ρ = | ri >< ri | I get the result... (split J2 and insert Ʃj | rj >< rj |)
     
  4. Mar 8, 2012 #3
    Problem is that I don't get how [itex]r_i[/itex] and [itex]r_j[/itex] behave with this particular
    [itex]\rho[/itex]
    What do you mean by splitting [itex]J^2[/itex]?
     
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