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Homework Help: Density of states in QM

  1. Jan 15, 2010 #1
    Hi. I'm studying the transition rates between a state a and a state b in the continuos level.
    In the book "Physics of atoms and molecules" by Bransden and Joachain it is said:
    We have to calculate the density of final states. To do this let the volume V be a cube of side L. We can impose periodic boundary conditions on the wave function, that is:
    [tex]k_x=\frac{2\pi}{L}n_x[/tex]
    [tex]k_y=\frac{2\pi}{L}n_y[/tex]
    [tex]k_z=\frac{2\pi}{L}n_z[/tex]
    where nx, ny and nz are positive or negative integers, or zero. Since L is very large we can treat nx, ny and nz as continuous variables, and the number of states in the range d[tex]\vec{k}=dk_xdk_ydk_z[/tex] is:
    [tex]dn_xdn_ydn_z=\left(\frac{L}{2\pi}\right)^3dk_xdk_ydk_z=\left(\frac{L}{2\pi}\right)^3k^2dkd\Omega[/tex]

    I can't understand the last equality, [tex]\Omega[/tex] is the solid angle, but how do I relate it to [tex]dk_xdk_ydk_z[/tex]?
     
  2. jcsd
  3. Jan 15, 2010 #2

    nicksauce

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    It is nothing tricky, it is just spherical coordinates.

    [tex]d^3k = k^2dkd\Omega = k^2\sin{\theta}dkd\theta d\phi.[/tex]

    If nothing depends on angle, then the solid angle can be integrated out to give a factor of 4pi.

    (Edit: Actually you also need to divide by 8. All the n's are >0, whereas the factor of 4pi assumes that k can be negative and positive).
     
    Last edited: Jan 15, 2010
  4. Jan 15, 2010 #3
    Thank you!:smile:
     
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