# From sum to integral

1. Sep 5, 2009

### Petar Mali

In solid state we often have case

$$\sum_{\vec{k}}F(\vec{k})=\frac{V}{h^3}\int_{I bz} F(\vec{p})d^3\vec{p}$$

Integral goes into first Briolen zone.

We can always say that

$$\frac{V}{h^3}\int_{I bz} F(\vec{p})d^3\vec{p}=4\pi \frac{V}{h^3}\int^{\infty}_{0}F(p)p^2dp$$

In 2D we will have integral

$$\frac{S}{h^2}\int_{I bz} F(\vec{p})d^2\vec{p}$$

where $$d^2\vec{p}=2\pi pdp$$

Am I right?

Can you tell me what I will have in 1D? Thanks!

2. Sep 5, 2009

### Bob_for_short

Only if F(p) is isotropic, i.e., it does not depend on angles.

In 2D we will have integral
Again, only if F(p) is isotropic, i.e., it does not depend on angles.
L/h*...*dp

3. Sep 5, 2009

### olgranpappy

Two other tiny comments: 1) It's spelled "Brilluoin" not "Briolen"; 2) I might not use the letters "I bz" to mean "first Brillouin zone" since it might be confused with "irreducible Brillouin zone". Cheers.

4. Sep 6, 2009

### Petar Mali

When I don't have isotropy in crystal lattice? Some example!

So you say

$$\sum_kF(k)=\frac{L}{h}\int^{\infty}_0dpp$$

?

I don't have some $$\pi$$ or something?

5. Sep 6, 2009

### Petar Mali

Thanks! Yes I meant first Brilluoin zone! I will have that in mind. And what is irreducible Brilluoin zone?

6. Sep 6, 2009

### olgranpappy

In any real crystal the symmetry is at most cubic not completely isotropic. For example, in simple cubic polonium there will not be a spherical fermi surface--the energy is not a quadratic function of the momentum but rather will have cubic terms. Thus, the expression for the DOS in polonium would be an example of a sum in momentum space in which the integrand is not isotropic.

7. Sep 6, 2009

### Bob_for_short

No, in 1D case there is no pi. 2*pi and 4*pi arise from integration over angles. In case of 3D space the total solid angle is 4*pi. In 2D space the total angle is 2*pi. They follow from definition of dp.