1. Limited time only! Sign up for a free 30min personal 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!

Many Body bogoliubov transformation

  1. Dec 6, 2016 #1

    Hl3

    User Avatar

    1. The problem statement, all variables and given/known data
    The occupation of each single-particle state with wave vector k =/= 0 in the ground state is given by nk = <0|bkbk|0>
    where b and b† are bogoliubov transformaition. Find an expression for nk.

    bk = cosh(θ)ak - sinh(θ) a-k
    bk = cosh(θ)ak - sinh(θ)a-k
    2. Relevant equations


    3. The attempt at a solution
    I don't fully understand the notaition with zeros. I belive nk would equal to 0, however question asks for the expression for nk. thnaks in advance
     
  2. jcsd
  3. Dec 6, 2016 #2
    No, it is not. The state ##|0>## is the vacuum of the annihilation operators ##a_k##. That is,
    $$a_k|0>=0$$ for all ##k##, but
    $$b_k|0>\neq 0$$ (unless ##\theta =0##).
     
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: Many Body bogoliubov transformation
  1. Many body system (Replies: 4)

Loading...