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Ashcroft Mermin Eq. 2.60

  1. Aug 8, 2012 #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 dont 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?
     
  2. jcsd
  3. Aug 8, 2012 #2
    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
     
  4. Feb 12, 2013 #3

    mzh

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    @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]?
     
  5. Feb 12, 2013 #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
     
  6. Feb 12, 2013 #5

    mzh

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    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.
     
  7. Feb 12, 2013 #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: Feb 12, 2013
  8. Feb 12, 2013 #7

    mzh

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    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].
     
  9. Feb 12, 2013 #8
    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
     
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