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!

Homework Help: Feromagnet question

  1. Aug 6, 2010 #1
    1. The problem statement, all variables and given/known data
    How from

    [tex](\hbar\omega-h)G_{f,f'}(\omega)=i2\langle\hat{S}_f^z\rangle\delta_{f,f'}+\langle\hat{S}^z\rangle\sum_gI(f-g)\{G_{f,f'}(\omega)-G_{g,f'}(\omega)\}[/tex]

    get

    [tex]G_{q}(\omega)=\frac{i\hbar}{2\pi}\frac{2\langle\hat{S}^z\rangle}{\hbar\omega-h-\epsilon(q)}[/tex]

    where

    [tex]G_{f,f'}(\omega)=\frac{1}{N}\sum_{\vec{q}} e^{i\vec{q}(\vec{f}-\vec{f'})}G_{\vec{q}}(\omega)[/tex]

    [tex]\delta_{f,f'}=\frac{1}{N}\sum_{q}e^{i\vec{q}(\vec{f}-\vec{f'})}[/tex]


    2. Relevant equations





    3. The attempt at a solution

    I really don't have a clue what to do. [tex]h[/tex] is constant.

    If I use that spin don't depands of indices

    [tex]\langle\hat{S}^z\rangle=\langle\hat{S}_g^z\rangle=S\sigma[/tex]

    and use

    [tex](\hbar\omega-h)\frac{1}{N}\sum_{\vec{q}} e^{i\vec{q}(\vec{f}-\vec{f'})}G_{\vec{q}}(\omega)=i2S\sigma\frac{1}{N}\sum_{q}e^{i\vec{q}(\vec{f}-\vec{f'})}+S\sigma\sum_gI(f-g)\{\frac{1}{N}\sum_{\vec{q}} e^{i\vec{q}(\vec{f}-\vec{f'})}G_{\vec{q}}(\omega)-\frac{1}{N}\sum_{\vec{q}} e^{i\vec{q}(\vec{g}-\vec{f'})}G_{\vec{q}}(\omega)\}[/tex]


    What now?
     
    Last edited: Aug 6, 2010
  2. jcsd
  3. Aug 6, 2010 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member


    What is the definition of I(f-g) ? We need that to make any progress.
     
  4. Aug 6, 2010 #3
    [tex]I[/tex] is exchange interraction!

    [tex]\sum_f I(f)e^{-i\vec{g}\cdot\vec{f}}=J(\vec{q})[/tex]

    [tex]\sum_g I(f-g)=\sum_{\vec{\lambda}}I(\vec{\lambda})=J(0)\equiv J_0[/tex]
     
  5. Aug 8, 2010 #4
    Any idea?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook