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?

  • $\begingroup$ iirc that is roughly estimating the number of states in a volume of $\text d^3p$. $\endgroup$ – Phoenix87 Jan 19 '15 at 11:40
  • 1
    $\begingroup$ ... also to avoid a problem with units? $\endgroup$ – Emilio Pisanty Jan 19 '15 at 14:49

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 $L/h$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.