3
$\begingroup$

So the explanation in my text book is this: we have a phase space volume $\Delta x\Delta y\Delta z\Delta p_x\Delta p_y\Delta p_z$. Now they state that Heisenberg's uncertainty principle states $\Delta p_x\Delta x\geq h$ so our minimal phase space volume is $h^3$.

However, shouldn't Heisenberg's uncertainty principle be $\Delta p_x\Delta x \geq \frac{h}{4\pi}$?

$\endgroup$
3
$\begingroup$

If you take a particle in a box (3D infinite square well), then the density of momentum states per unit volume is given by $$ g(p)\ dp = g_s\ \frac{4\pi p^2}{h^3}\ dp,$$ where $g_s$ is a spin degeneracy factor (e.g. $g_s=2$ for electrons, protons etc. because they can have 2 spin states per momentum state). This means that the volume of phase space occupied by one momentum state must be $h^3$.

The treatment you give in your question is a rough attempt to justify this result in terms of the (usually) more familiar Heisenberg uncertainty principle. Such a treatment cannot be exact since for a particle in a 1D infinite potential well, the product of uncertainty in momentum and position is actually given by $$ \Delta p \Delta x = \frac{h}{4\pi} \left(\frac{n^2 \pi^2}{3} - 2\right)^{1/2}$$ (see this link), where $n \geq 1$ is the principle quantum number and thus the product of momentum and position uncertainty is always bigger than $h/4\pi$ and would then need to be averaged in some way over all the momentum states.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much! I've seen a more thorough derivation of the density of states which indeed makes more sense. Good to know that the heisenberg derivation is not exact. $\endgroup$ – harrymuana May 27 '15 at 12:13

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.