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Is this approximation OK?

  1. Oct 6, 2009 #1
    For [tex]k_D<<k_F[/tex]

    [tex]|\frac{\hbar^2k^2_F}{2m}-\frac{\hbar^2k^2}{2m}|\approx \frac{\hbar^2k_F}{m}|k_F-k|[/tex]

    Where [tex]k[/tex] goes from [tex]k-k_D[/tex] to [tex]k+k_D[/tex]

    [tex]k_F[/tex] - Fermi wave vector
    [tex]k_D[/tex] - Debay wave vector
  2. jcsd
  3. Oct 6, 2009 #2
    I suppose you mean [itex]k[/itex] goes between [itex]k_F+k_D[/itex] and [itex]k_F-k_D[/itex]?

    Yes it is valid. Let's denote [itex]k=k_F+\delta k[/itex], then

    |\frac{\hbar^2k_F^2}{2m}-\frac{\hbar^2k^2}{2m}|=|\frac{\hbar^2k_F\delta k}{m}+\frac{\hbar^2 \delta k^2}{2m}|

    since [itex]\delta k<k_D\ll k_F[/itex] we can neglect the last term (quadratic in [itex]\delta k[/itex] and get

    |\frac{\hbar^2k_F\delta k}{m}|=\frac{\hbar^2k_F}{m}|k_F-k|
  4. Oct 7, 2009 #3
    Thanks a lot! :) Yes from [tex]k_F-k_D[/tex] to [tex]k_F+k_D[/tex].

    I think that you have just a little mistake

    You must write like


    |\frac{\hbar^2k_F^2}{2m}-\frac{\hbar^2k^2}{2m}|=|-\frac{\hbar^2k_F\delta k}{m}-\frac{\hbar^2 \delta k^2}{2m}|


    To get [tex]C|k_F-k|[/tex] or in case you wrote you will get


    You helped me a lot!
  5. Oct 7, 2009 #4
    I assumed that [itex] |\ldots |[/itex] meant taking the absolute value. If this is so then overall signs do not matter.

    Anyway, You're welcome.
  6. Oct 7, 2009 #5
    Yes! My mistake!
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