Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Equation in a paper about Dicke states

  1. Apr 30, 2018 #1
    Can anyone with basic knowledge of Dicke States assist with explaining how we arrive at equation (4) in the paper 'Entanglement detection in the vicinity of arbitrary Dicke states': <Moderator's note: link fixed>

    $$\langle J^2_{x} \rangle_{\mu} = \sum_{i_1,i_2} \langle J_{xi_{_1}} \rangle_{\mu} \langle J_{xi_{_2}} \rangle_{\mu} + \sum_{i}\langle ( \Delta J_{xi})^2 \rangle_{\mu}~~~~~~~~~~~~~(4)$$

    Any assistance is appreciated.
     
    Last edited by a moderator: May 1, 2018
  2. jcsd
  3. May 1, 2018 #2

    DrClaude

    User Avatar

    Staff: Mentor

    I am not very knowledgeable about Dicke states, but isn't that equation simply a rewriting of
    $$
    \sigma^2 = \langle (\Delta x)^2 \rangle = \langle x^2 \rangle - \langle x \rangle^2
    $$
     
  4. May 1, 2018 #3
    @DrClaude Yes I think you are correct. Since ##J_{x}= \sum_{i=1}^{k}J_{xi}## it follows that $$\langle J_{x} \rangle^{2}_{\mu} = \langle \sum_{i=1}^{k}J_{xi} \rangle_{\mu}^{2} = \sum_{i_1, i_2} \langle J_{xi_1}\rangle_{\mu}\langle J_{xi_2} \rangle_{\mu}$$ and $$\langle ( \Delta J_x )^2 \rangle = \sum_{i} \langle ( \Delta J_{xi})^2 \rangle_{\mu}$$ hence $$\sum_{i}\langle ( \Delta J_{xi})^2 \rangle_{\mu} = \langle J^2_{x} \rangle_{\mu} - \sum_{i_1,i_2}\langle J_{xi_{1}}\rangle_{\mu}\langle J_{xi_2}\rangle_{\mu}$$
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Loading...