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In our lecture we used quite a couple of times that the sum over momentum states can be approximated by an integral over them.

But instead of substituting $\sum_p \rightarrow \int d^3p$, we replaced $\sum_p \rightarrow \frac{V}{h^3} \int d^3p.$ Now I think that the motivation for this $h$ is to avoid a problem of units, but I don't see where this $V$ comes from?

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  • $\begingroup$ iirc that is roughly estimating the number of states in a volume of $\text d^3p$. $\endgroup$ – Phoenix87 Jan 19 '15 at 11:40
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    $\begingroup$ ... also to avoid a problem with units? $\endgroup$ – Emilio Pisanty Jan 19 '15 at 14:49
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Let's consider the free Fermi gas confined in a $1D$ box of length $L$. The (box normalized) wavefunction is $\psi(x)=\frac{1}{\sqrt{L}}e^{ipx/\hbar}$. If we impose the periodical boundary condition, $\psi(0)=\psi(L)$, we get $p=nh/L$, where $n$ is an integer.

Now we want to count the number of states whose momentum is less than $p_F$. i.e: $\sum_{p<p_F}1$ , we get the number $N=Lp_F/h$.

If we want to use the integral instead of sum to do the job, that is:$N=\mathrm{const}\int_0^{p_F}1$, then you find that constant should be $L/h$.

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  • $\begingroup$ but the equation is still correct for arbitrary geometries, right? $\endgroup$ – Xin Wang Jan 26 '15 at 20:42
  • $\begingroup$ @XinWang Yes, I think so $\endgroup$ – an offer can't refuse Jan 27 '15 at 5:08

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