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Density operator

  1. Jul 25, 2008 #1
    so I've been reading about the density operator formulation of quantum mechanics and I have some questions

    what is the density operator analog of the schrodinger equation that determines the time evolution?

    and how do you perform a projection measurement on a quantum system in the density operator formulation?

    how would you use it to describe a system like the infinite square well?
     
  2. jcsd
  3. Jul 28, 2008 #2

    Avodyne

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    [itex]i\hbar\dot\rho=[H,\rho][/itex], where [itex]\rho[/itex] is the density operator.
    Replace [itex]\rho[/itex] with [itex]P\rho P[/itex]. where [itex]P[/itex] is the projection operator.
    You use the above formulae for time evolution and measurement.
     
  4. Jul 30, 2008 #3
    is it possible to solve for the density operator and possible outcomes of a measurement without having to use the standard wave mechanics?

    sorry I'm not very used to seeing an equation which appears to only contain operators.

    also do you mean the partial derivative with respect to time when you write p dot?
     
  5. Jul 30, 2008 #4

    CarlB

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    Yes. In some cases, density operators allow for simpler calculations. Example, find the state corresponding to spin-1/2 in the direction of the unit vector [tex]\vec{u} = (u_x,u_y,u_z)[/tex].

    The operator for spin in that direction is [tex]\sigma_u = \vec{\sigma}\cdot u[/tex] where [tex]\vec{\sigma}[/tex] is the 3-vector of the Pauli spin matrices. This squares to 1 so an eigenstate for spin in the u direction is [tex]1+\sigma_u[/tex]. These are normalized so they are idempotent so the normalized state is [tex](1+\sigma_u)/2[/tex], which you can verify is idempotent.

    Then to convert this into a state vector (wave function) solution to the eigenvector problem for spin in the u direction, simply taken any nonzero column vector in the matrix. Since there are two columns in the matrix, there are two choices. Ignoring the division by 2, and not bothering to normalize the vector, the two choices are:
    [tex]\left(\begin{array}{c}1+u_z\\u_x+iu_y\end{array}\right),
    \left(\begin{array}{c}u_x-iu_y\\1-u_z\end{array}\right)[/tex]
    For the Dirac case general solution, see:
    http://en.wikipedia.org/wiki/Dirac_spinor
    where they currently call the 4x4 matrix the "projection operator", another, more general, name for the density matrix quantum state.

    By the way, the wave function analogy to "taking any nonzero column" is to choose any point where the density operator function is nonzero in the second entry. That is, if [tex]\rho(x,x')[/tex] is a density matrix solution to Schroedinger's equation, and b is a point where [tex]\rho(x,b)[/tex] is not identically zero over x, then [tex]\psi(x) = \rho(x,b)[/tex] is a solution to Schroedinger's equation.


    Yes. Of course it's actually rho. To write LaTeX for this, enter the following, without the spaces: [ t e x ] \ d o t { \ r h o } [ / t e x].
     
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