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I'm trying to figure out how the radius of a Fermi sphere $$p_F = \hbar (3 \pi^2 \frac{N}{V})^{1/3}$$ is derived from the formula $$dN_{spatial}=\frac{V \ d^3p}{\hbar^3}.$$ The solution states that I should arrive at the following $$\frac{2V}{(2 \pi \hbar)^3} \cdot \frac{4\pi}{3}p_F^3.$$ However I am not quite sure how I should go about after the following $$2\int dN_{spatial} = \frac{2V}{\hbar^3}\int d^3p= \ ?$$ I would love some help to solve this, and if you have any web resources I'd gladly accept that too (I've tried googling my way for a while now).

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I think it should be $h$ in the denominator of your second equation,

$$ {\rm d}N = \frac{V}{h^3}{\rm d}^3{\bf p} $$

Integrating at both sides, and taking into account the degeneracy

$$ N = \frac{2 V}{h^3} \int{\rm d}^3 {\bf p} = \frac{2 V}{h^3} \int {\rm d}\Omega\int_0^{p_F}{\rm d}p ~p^2 = \frac{2V}{h^3} \frac{4\pi}{3} p_F^3 $$

Using $2\pi\hbar = h$ you get

$$ N = \frac{2V}{(2\pi \hbar)^3} \frac{4\pi}{3}p_F^3 $$

and from here is just a matter of getting $p_F$

$$ p_F = \hbar \left(3\pi^2 \frac{N}{V} \right)^{1/3} = \hbar k_F $$

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  • $\begingroup$ You are correct, seems like a typo in the exercise. However could you explain the part of $\int d\Omega$ with degeneracy? $\endgroup$
    – Gjert
    Nov 30, 2018 at 15:10
  • $\begingroup$ That's the integral of the solid angle $d \Omega = \sin(\theta) d\theta d\phi$ $\endgroup$ Nov 30, 2018 at 15:13
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    $\begingroup$ @PhyCoMath It is just a short version of the integration over the angle . $\int {\rm d}\Omega = \int_0^{2\pi}{\rm d}\phi \int_0^{\pi}{\rm d}\theta\sin \theta$ $\endgroup$
    – caverac
    Nov 30, 2018 at 15:13
  • $\begingroup$ @caverac Oh, never seen that notation before. Thanks for enlightening me :) $\endgroup$
    – Gjert
    Nov 30, 2018 at 15:19
  • $\begingroup$ @PhyCoMath :) Happy to help $\endgroup$
    – caverac
    Nov 30, 2018 at 15:19

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