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?

 

[paranormal]

To solve this, we can start by rearranging the equation to isolate G_q(\omega) on one side:

(\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)\}

Next, we can substitute in the expression for G_{f,f'}(\omega) from the given information:

(\hbar\omega-h)\frac{1}{N}\sum_{\vec{q}} e^{i\vec{q}(\vec{f}-\vec{f'})}G_{\vec{q}}(\omega)-i2\langle\hat{S}_f^z\rangle\delta_{f,f'}=\langle\hat{S}^z\rangle\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)\}

Now we can rearrange and factor out the G_q(\omega) term:

G_{\vec{q}}(\omega)[(\hbar\omega-h)-i2\langle\hat{S}_f^z\rangle\delta_{f,f'}+\langle\hat{S}^z\rangle\sum_gI(f-g)e^{i\vec{q}(\vec{f}-\vec{f'})}]=0

Since this equation must hold for all values of \vec{q}, we can set the bracketed term equal to zero:

(\hbar\omega-h)-i2\langle\hat{S}_f^z\rangle\delta_{f,f'}+\langle\hat{S}^z\rangle\sum_gI(f-g)e^{i\vec{q}(\vec{f}-\vec{f'})}=0

Now we can rearrange this to solve for G_q(\omega):

G_{q}(\omega)=\frac{i2\langle