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How to Calculate Probability using Density Operator?

  1. Oct 5, 2016 #1
    Hello, I'm trying to understand how to calculate de probability of finding a system in a specific eigenstate using the density operator. In the book of Balian, Haar, Gregg I've found a good definition of it being the expectation value of the projector Pr in the orientation of the eingenstate.
    P(a) = tr(D.Pra)
    The problem is, since I have then a product between de density matrix tr(D.Pra), Pra would have to be a matrix of the same rank of D, write ? To calculate then the tr. What is the right way to construct the projection ?
    Transform the state, a vector, into a matrix ? Beeing a matrix with just one element in the main diagonal for each of the a directions relative to each eingenstate ?

    I'm a new member, so I'm not used to ask questions online. Thank you.
     
  2. jcsd
  3. Oct 10, 2016 #2
    Hi. I'll start from the beginning. If u have a density operator, call it $$\hat{D},$$ and some eigenstate, call it $$|\psi>.$$ You can simply make a projector to that eigenstate: $$ \hat{P}= | \psi> <\psi|.$$Then the probability of finding your system in that eigenstate is defined as $$ Tr( \hat{D} \hat{P}). $$ If u have a problem of writing that projector operator, here's help. U must write that bra state in some basis, and just multiply bra and ket vectors, u get a matrix with the same dimensions as your density operator and that's it. If you have problems with that multiplication of bra and ket vectors, search an article on wikipedia and that will help you. You can then multiply density operator and that projector and take trace of that final matrix.
     
  4. Oct 10, 2016 #3

    A. Neumaier

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    If your density operator is ##\rho## and your pure state is ##\psi##, the projector is ##P=\psi\psi^*## and the probability is
    ##p## = Trace ##\rho P## = Trace ##\rho\psi\psi^*=\psi^*\rho\psi##
    since the trace of ##uv^*## is ##v^*u##.
     
  5. Oct 11, 2016 #4
    Thank you.
    But, the definition of the density operator is not
    ρ=|ψ><ψ| or ρ=∑|ψi><ψi| ?

    Then it'is like the projector and the density operator are the same thing, or almost the same thing.
     
  6. Oct 11, 2016 #5

    A. Neumaier

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    The density operator of a pure state is indeed its projector. But the projector has a second use as an observable for the response of a system in an arbitrary state to a measurement of the pure state.
     
  7. Oct 11, 2016 #6
    Thanks, that is enlightening
     
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