Ashcroft Mermin Eq. 2.60

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In summary, Eq. 2.60 of Ashcroft, Mermin states that to integrate out the spherical coordinates and to integrate Phi and Theta, we first introduce the state density g(E) and then use the Heaviside theta to fix the integration boundaries.
  • #1
mzh
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Dear Physics Forums users
Eq. 2.60 of Ashcroft, Mermin is
[itex] \int \frac{d\vec{k}}{4\pi^3} F(E(\vec{k}) = \int_0^\infty \frac{k^2dk}{\pi^2}F(E(\vec{k})) = \int_{-\infty}^{\infty} dE g(E) F(E) [/itex]

I understand the first transformation is done by introducing spherical coordinates (as written in the text) and to integrate out [itex]\phi[/itex] and [itex]\theta[/itex].
I also get the second transformation, where we insert the expression for the energy [itex]E=\frac{\hbar^2 k^2}{2m} [/itex], but what I don't understand is the new boundaries. How do we arrive at the boundaries from minus to plus infinity? Does E range from -infinity to plus infinity?

Thanks for any hints?
 
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  • #2
mzh said:
Dear Physics Forums users
Eq. 2.60 of Ashcroft, Mermin is
[itex] \int \frac{d\vec{k}}{4\pi^3} F(E(\vec{k}) = \int_0^\infty \frac{k^2dk}{\pi^2}F(E(\vec{k})) = \int_{-\infty}^{\infty} dE g(E) F(E) [/itex]

I understand the first transformation is done by introducing spherical coordinates (as written in the text) and to integrate out [itex]\phi[/itex] and [itex]\theta[/itex].
I also get the second transformation, where we insert the expression for the energy [itex]E=\frac{\hbar^2 k^2}{2m} [/itex], but what I don't understand is the new boundaries. How do we arrive at the boundaries from minus to plus infinity? Does E range from -infinity to plus infinity?

Thanks for any hints?

Hello,

I think your problem will be gone if you note that in the last expression there is a function [itex]g(E)[/itex] that could be equal to zero for the values of energy that do not belong to the spectra of the system considered.

edit
Note also that you do not need to assume [itex]E=\frac{\hslash ^2 k^2}{2m}[/itex] because a different expression for [itex]E(k)[/itex] would only cause a change in the explicit form of [itex]g(E)[/itex].

Ilm
 
  • #3
@Ilmrak. Thanks for the reply... ok, so what's the motivation for introducing [itex]g(E)[/itex]? And, how to come up with the definition of [itex]g(E)[/itex]?
 
  • #4
Hello,

I's been a while since the last time I read that book and I don't have one here to read.

That said, if I remember well what that formula was about, the motivation to introduce [itex] g(E)[/itex] is that the last expression obtained in Eq. 2.60 is much more general then the one you started with. Different systems will have different functions [itex] g(E)[/itex].

The function [itex] g(E)[/itex] is defined as the density of the states with energy [itex]E[/itex], i.e. [itex] g(E) \mathrm{d}E[/itex] is equal to the numer of states with energy between [itex]E[/itex] and [itex]E+\mathrm{d}E[/itex]. This number obviously depends on the specific system considered.

I hope this helps a bit,

Ilm
 
  • #5
Hello. That's right, [itex]g(E)[/itex] is the state density. But what I still don't see is how the substitution [itex]k\rightarrow E[/itex] is being made.
 
  • #6
If I understand your problem you don't get the last step of that equation.
Then the solution is very simple.
Try with this expression for the states density

[itex]
g(E) = \frac{1}{\pi^2 \hslash ^3} \theta(E) \sqrt{2m^3 E} \; ,
[/itex]

You will see that, being [itex] E= \frac{\hslash ^2}{2m} k^2[/itex], then

[itex]\frac{k^2}{\pi^2} \mathrm{d}k = g(E) \mathrm{d}E[/itex].

Ilm

Edit: the Heaviside theta is obviously to fix the integration boundaries.
 
Last edited:
  • #7
Ilmrak said:
Edit: the Heaviside theta is obviously to fix the integration boundaries.

ahm, yeah heard of it...

yes, it's the last step of the equation that I'm struggling with.

I guess my conceptual problem is that to me the term substitution of variables when calculating integrals involves some sort of [itex]u(x) = 2x[/itex] mechanism when calculating the integral [itex]\int \sin 2x dx[/itex].
 
  • #8
mzh said:
ahm, yeah heard of it...

yes, it's the last step of the equation that I'm struggling with.

I guess my conceptual problem is that to me the term substitution of variables when calculating integrals involves some sort of [itex]u(x) = 2x[/itex] mechanism when calculating the integral [itex]\int \sin 2x dx[/itex].

In particular here we have [itex] E(k)= \frac{\hslash ^2}{2m} k^2[/itex], and then [itex]\mathrm{d}E=\frac{\hslash ^2}{m} k \, \mathrm{d}k[/itex].

If you are not familiar with the Heaviside theta don't worry, it's very simple!

[itex]
\theta(t)= 0 \quad \mathrm{if} \quad t<0,
[/itex]
[itex]
\theta(t)= 1 \quad \mathrm{if} \quad t>0.
[/itex]

Then:

[itex] \int_{-a}^b \mathrm{d}t \; f(t) \, \theta (t) = \int_{0}^b \mathrm{d}t \; f(t) \quad \forall \,a,b\in ℝ^+ \; , \, \forall f \in C^0(ℝ).
[/itex]

Ilm
 

What is Eq. 2.60 in Ashcroft Mermin?

Eq. 2.60 in Ashcroft Mermin refers to an equation in the textbook "Solid State Physics" by Neil W. Ashcroft and N. David Mermin. It is an equation that describes the density of states in a metal, which is a measure of the number of electron states per unit energy in the material.

Why is Eq. 2.60 important in solid state physics?

Eq. 2.60 is important in solid state physics because it is a fundamental equation that helps us understand the behavior and properties of metals. It is also used in various calculations related to electronic structure, conductivity, and thermodynamic properties of materials.

How is Eq. 2.60 derived?

Eq. 2.60 is derived using quantum mechanics principles and the free electron model. It takes into account the energy levels and momentum of electrons in a metal and uses statistical mechanics to calculate the density of states.

What are the assumptions made in Eq. 2.60?

The main assumptions made in Eq. 2.60 are that the electrons in the metal are free and non-interacting, and that the metal is in a state of thermal equilibrium. These assumptions simplify the calculations and allow us to better understand the behavior of real materials.

Are there any limitations to Eq. 2.60?

Yes, there are limitations to Eq. 2.60. It is based on the free electron model, which does not take into account the effects of electron-electron interactions and the crystal lattice structure of the material. Therefore, it may not accurately describe the properties of all metals, especially at low temperatures or high pressures.

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