# Exchange interaction in solid state physics

1. Aug 5, 2009

### Petar Mali

$$I(\vec{n}-\vec{m})=\frac{1}{N}\sum_{\vec{k}}I(\vec{k})e^{i\vec{k}(\vec{n}-\vec{m})}$$

$$I(\vec{n}-\vec{m})=\frac{1}{\sqrt{N}}\sum_{\vec{k}}I(\vec{k})e^{i\vec{k}(\vec{n}-\vec{m})}$$

In which cases is better to use first and in which second relation? $$N$$ is number of knottes of lattice.

And when I can say $$I(\vec{n}-\vec{m})=I(|\vec{n}-\vec{m}|)$$?

And is it a way to show

$$\sum_{\vec{k}}I(\vec{k})=\sum_{-\vec{k}}I(\vec{k})$$?

2. Aug 8, 2009

### A.Rida

please i am anew member can you explain the meanings of the symbols

3. Aug 8, 2009

### genneth

This has nothing to do with exchange interactions, and is a problem of Fourier transforms. These are elementary questions, so I suggest a textbook, which will explain far better and in greater detail than anyone here can.

4. Aug 13, 2009

### Petar Mali

$$I$$ is exchange interaction.

$$I(\vec{n}-\vec{m})=I(|\vec{n}-\vec{m}|)$$

Sometimes $$I$$ depends only of absolyte value of argument. This is physical reasons and do not have any relationship with Fourier transform in general. $$I$$ is sometimes tensor $$I^{\beta}_{\alpha}$$.

In physics is not always the same think to work with

$$I(\vec{n}-\vec{m})=\frac{1}{N}\sum_{\vec{k}}I(\vec{k})e^{i\v ec{k}(\vec{n}-\vec{m})}$$

or with

$$I(\vec{n}-\vec{m})=\frac{1}{\sqrt{N}}\sum_{\vec{k}}I(\vec{k} )e^{i\vec{k}(\vec{n}-\vec{m})}$$

I just whant to know when is better to do with first or with second relation?

Thanks

5. Aug 15, 2009