1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Density Matrix

  1. Jan 23, 2011 #1
    I have a question regarding the slide:

    http://theory.physics.helsinki.fi/~kvanttilaskenta/Lecture3.pdf

    On page 18-21 it gives the proof of the theorem that [tex]| \psi_i^{~} \rangle[/tex] and [tex]|\phi_{i}^{~}\rangle[/tex] generate the same density matrix iff [tex]|\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle[/tex] assuming that [tex]| \psi_i^{~}\rangle [/tex] is not necessarily normalized.

    What if [tex]| \psi_i^{~}\rangle [/tex] is normalized and [tex]| \phi_i^{~}\rangle[/tex] not independent?

    Would the necessary condition for which [tex] p = | \psi_i \rangle \langle \psi_i |= q = | \phi_j \rangle \langle \phi_j | [/tex] require that you have [tex]|\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle[/tex] ?

    We know for normalized states psi and phi that [tex] p = | \psi_i \rangle \langle \psi_i |= q = | \phi_j \rangle \langle \phi_j | [/tex] iff [tex] \sqrt{p_{i}} | \psi_i \rangle = \sum_j u_{ij} \sqrt{q_j} | \phi_j \rangle [/tex]
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Density Matrix
  1. Density matrix (Replies: 0)

  2. Density matrix (Replies: 6)

  3. Density matrix problem (Replies: 4)

  4. Density matrix (Replies: 4)

Loading...