Solving Ferromagnet Ques: Get G_q(\omega) from (hbarw-h)G_ff'(\omega)

[Petar Mali]

Homework Statement

How from

(\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)\}

get

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

where

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

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

Homework Equations

The Attempt at a Solution

I really don't have a clue what to do. h is constant.

If I use that spin don't depands of indices

\langle\hat{S}^z\rangle=\langle\hat{S}_g^z\rangle=S\sigma

and use

(\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)\}What now?

 

[nrqed]

Homework Statement

How from

(\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)\}

get

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

where

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

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

Homework Equations

The Attempt at a Solution

I really don't have a clue what to do. h is constant.

If I use that spin don't depands of indices

\langle\hat{S}^z\rangle=\langle\hat{S}_g^z\rangle=S\sigma

and use

(\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)\}


What now?

What is the definition of I(f-g) ? We need that to make any progress.

 

[Petar Mali]

I is exchange interraction!

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

\sum_g I(f-g)=\sum_{\vec{\lambda}}I(\vec{\lambda})=J(0)\equiv J_0

 

[Petar Mali]

Any idea?