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Question from Ashcroft/Mermin: electronic density eqn

  1. Apr 13, 2015 #1
    I'm taking a SS physics course and going through A&M and found something hard for me to understand. It is on page 44, starting with eqn 2.60. Another thread almost addressed my question, but I cannot post replies on the thread (click here).

    I think I understand what is going on in the 1st equals sign (they are transforming to spherical coords', so the integral over k-space becomes an integral from 0 to infinity like when you integrate a sphere from r=0 to r=R). But my question is what happens during the 2nd equals sign? And further, why do the bounds on the integral over ε go from -∞ to +∞ when the expression for ε=(hk)2/2m, hence, how is it possible to even consider an ε<0 when the k in that expression is squared?

    EDIT: I forgot to mention that I notice in eqn 2.61 for g(ε) that for ε<0, g(ε)=0, hence, effectively eqn 2.60 becomes what makes sense (the integrand amounts to 0 for any ε<0), however, my question still remains about why it is even possible/physical to consider bounds of ε that are ε<0? If ε cannot be less than 0, it just seems weird to me to make an equation that explicitly states that, rather than just change the bounds of the integral in 2.60 to be from ε=0 to ε=∞.
     
  2. jcsd
  3. Apr 13, 2015 #2

    DrDu

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    For non-free particles, E can become negative. Consider as an example the hydrogen atom where there is a set of discrete states with ## E=-Ry/n^2## besides the continuum of states with positive energy. Similarl bound states exist also in solids so it makes good sense to extend the integral to minus infinity.
     
  4. Apr 13, 2015 #3
    Thanks for your answer.
    I guess I was just confused because I'm still in the part of the book that is talking about the free electron gas, hence ε ∝ k^2 and cannot be negative. But you're saying that ε can take different forms than the expression I used, just depending on the potential the particle is in?

    But if this is the case, that ε can be negative, then why is it the case that the the function g(ε) = 0 whenever ε < 0 ? In this case, its the same as if ε cannot be negative.
     
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