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Density matrix

  1. Mar 7, 2014 #1
    Is it not possible to deduce quantum states from a density matrix?
     
  2. jcsd
  3. Mar 7, 2014 #2

    ShayanJ

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    Its like classical statistical mechanics. You don't know the properties of every individual particle but you know the distribution of speeds and energies. When you have a (mixed)density matrix of a system, it means you don't know the state of each particle but just the distribution of states.(or something like that!).
     
  4. Mar 7, 2014 #3
    (1 0)'*(1 0) = (1 0; 0 0), ok
    but if (a1 a2)'*(a1 a2) = (1 0; 0 0) and in general case there is no solution to find a1 and a2, is this correct?

    thanks
     
  5. Mar 7, 2014 #4

    naima

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    Is (1 0; 0 0) a notation for the density matrix in a v1 v2 basis?
    if yes this density matrix is |v1> < v1|
     
  6. Mar 8, 2014 #5
    hi,

    this was just an example, the main question is whether states can be derived from density matrix

    thanks
     
  7. Mar 8, 2014 #6

    naima

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    a density matrix is more general that a pure state.
    a pure state |v> in a 2 dimentional system has |v><v| for density matrix (1 0;0 0) but all density matrix are not like that.
    take (1/3 0; 0 2/3) it is the density matrix 1/3 |v1><v1| + 2/3 |v2><v2| = 1/3 (1 0;0 0) + 2/3 (0 0; 0 1).
    It is to be used when somebody gives you apure state v1 (or v2) with 1/3 (2/3) probability.
     
  8. Mar 8, 2014 #7

    bhobba

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    I am not sure what you mean.

    But a quantum state, by definition, is a positive operator of unit trace which can represented as a matrix called the density matrix.

    You can however find more detail such as what mixed and pure states are in Ballentine - Quantum Mechanics - A Modern Development - Chapter 2.

    Its use is the Born Rule which says given an observable O these exists a positive operator of unit trace such that the expected outcome of O, E(O) = Trace (PO). By definition P is called the state of the system. Ballentine develops QM from just two axioms - the Born Rule is the second - the first is associated with any observation is a Hermitian operator whose eigenvalues are the possible outcomes.

    To some extent Born's Rule is implied by the first axiom via Gleason's Theorem:
    http://kof.physto.se/cond_mat_page/theses/helena-master.pdf [Broken]

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  9. Mar 10, 2014 #8
    In general, a state vector cannot be deduced from a density matrix. The reason was already mentioned, it is that the density matrix is more general than the state vector. One example is mixed states, which can be described in the density matrix notation, but not as a state vector.
     
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