How Is the Volume of an Element in K-Space Determined?

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Discussion Overview

The discussion centers on determining the volume of an element in k-space, particularly in the context of a free electron gas within a crystal lattice. Participants explore the relationship between k-space volume and the density of states, as well as the implications of different boundary conditions on the derivation of k values.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion regarding the volume of an element in k-space, referencing the uncertainty principle and questioning the assertion that V equals (2π)³.
  • Another participant provides a derivation based on state counting in a three-dimensional cube with periodic boundary conditions, leading to the conclusion that the measure in k-space is V/(2π)³ d³k.
  • A subsequent participant questions the k value used in the derivation, suggesting it should be k = π/L based on a different boundary condition (hard wall) and asks for clarification on the factor of 2 in the original formula.
  • A later reply clarifies the difference in boundary conditions, noting that the use of periodic boundary conditions results in a density of states that incorporates all octants of k-space, thus accounting for the factor of 2³ = 8.

Areas of Agreement / Disagreement

Participants demonstrate differing views on the appropriate boundary conditions and their implications for k values and density of states. There is no consensus on the correct approach, as multiple models are presented and debated.

Contextual Notes

The discussion highlights the dependence on boundary conditions and the resulting variations in k-space volume calculations. There are unresolved aspects regarding the assumptions made in different derivations.

Master J
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I've been trying to figure out the volume of an element in k-space (free electron gas in a crystal lattice).

In deriving the density of states, dn/dE = dn/dk . dk/dE. My dn = 4pi(k^2).dk / V, where V is the volume of an element in k-space.
I came across the spurious, Dk.Dx ~ 2pi (an uncertainty principle relation), where V was said to be (2pi)^3, but that UP doesn't make sense to me.


Could someone shed some light on this?

The ever so grateful MJ...
 
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A relatively simple derivation follows from state counting. Consider a three dimensional cube with periodic boundary conditions and sides of length L. The single particle energy eigenstates are of the form exp(i k x) where k is given by [tex]k = \frac{2\pi}{L}(n_x, n_y,n_z)[/tex] with arbitrary integers n.

Summing over all these states can be written as [tex]\sum_n = \sum_{n_x,n_y,n_z} \Delta n_x \Delta n_y \Delta n_z = \sum_{n_x,n_y,n_z} \Delta k_x \Delta k_y \Delta k_z \left(\frac{L}{2 \pi}\right)^3 \rightarrow \frac{V}{(2\pi)^3} \int d^3 k[/tex].

Thus we may approximate sums over discrete single particles states in a big box via an integral over three dimensional k space with the measure [tex]\frac{V}{(2\pi)^3} d^3 k[/tex].

Is this what you were looking for?
 
Ah, indeed.

The k value slightly confuses me tho. Should it not be k = pi / L (sorry I'm rubbish at LaTeX)??

If we let the electron wavefunction say be sin(kx) in the cube, with the condition taht it is zero at the edges, that gives k = n pi / L. Where does the 2 come from?
 
I used periodic boundary conditions while you are using hard wall boundary conditions. It is true that in your case all the allowed values of k make are [tex]\pi n/L[/tex] but with n positive. Thus compared to my analysis you have a density of states of [tex]( L^3/\pi^3) d^3 k[/tex] but you're also allowed to only integrate over the totally positive k octant. In my formula I integrate over all 8 octants thus making up the missing factor of [tex]2^3 = 8[/tex].

Hope this helps.
 
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