From sum to integral in solid state

[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!

 

[Bob_for_short]

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

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

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

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

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

L/h*...*dp

 

[olgranpappy]

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.

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.

 

[Petar Mali]

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

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?

 

[Petar Mali]

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.

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

 

[olgranpappy]

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

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.

 

[Bob_for_short]

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

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.