How is the density of final states calculated in quantum mechanics?

[eoghan]

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:
$$k_x=\frac{2\pi}{L}n_x$$
$$k_y=\frac{2\pi}{L}n_y$$
$$k_z=\frac{2\pi}{L}n_z$$
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$$\vec{k}=dk_xdk_ydk_z$$ is:
$$dn_xdn_ydn_z=\left(\frac{L}{2\pi}\right)^3dk_xdk_ydk_z=\left(\frac{L}{2\pi}\right)^3k^2dkd\Omega$$

I can't understand the last equality, $$\Omega$$ is the solid angle, but how do I relate it to $$dk_xdk_ydk_z$$?

 

[nicksauce]

It is nothing tricky, it is just spherical coordinates.

$$d^3k = k^2dkd\Omega = k^2\sin{\theta}dkd\theta d\phi.$$

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).

 

[eoghan]

Thank you!