Exchange interaction in solid state physics

[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

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

 

[A.Rida]

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

 

[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.

 

[Petar Mali]

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.

$$I$$ 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

 

[saaskis]

It does not matter which convention you use, as long as you use the appropriate inverse transformation. If you include the square root, you should also include it in the inverse transformation.