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Minimum volume of phase space

  1. Sep 25, 2014 #1
    In several books I have seen the statement that due to Heisenbergs principle no particle can be localized into a region of phase space smaller than ##(2 \pi \hbar)^3##. However, Heisenbergs uncertainty principle states that ##dx dp \geq \hbar/2## -- so a direct translation of this should imply that the minimal volume must be ##(\hbar/2)^3## and not ##(2 \pi \hbar)^3##. So, what am I missing here?
     
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  3. Sep 25, 2014 #2

    vanhees71

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    You don't miss anything! It's just a wrong explanation for the "phase-space-cell factor" [itex]h^3=(2 \pi \hbar)^3[/itex] in statistical mechanics.

    The correct explanation is as follows. Suppose we have a gas of particles in a container of volume [itex]V[/itex]. It's important to have a container of finite size. The shape doesn't matter. To keep the argument simple, suppose it's a cube with length [itex]L[/itex]. Take this container as a typical box in a very large box containing the gas. Then it makes sense to assume that the container is as good as any other box within the big box, and we can impose periodic boundary conditions. Then the momentum operator for particles in the container is well defined, and the eigenvalues of the three momentum components are discrete. The normalized eigenvectors are
    [tex]u_{\vec{p}}(\vec{x})=\frac{1}{L^{3/2}} \exp \left (\frac{\mathrm{i} \vec{p} \cdot \vec{x}}{\hbar} \right )[/tex]
    with the boundary conditions enforcing that
    [tex]\vec{p} \in \frac{2 \pi \hbar}{L} \mathbb{Z}^3.[/tex]
    Now to get the thermodynamic potentials you have to sum over the states or count states. In this case you have to sum over all these momenta. Calculating such sums is much more tedious than doing integrals, and often you are not interested on finite-size effects at all, i.e., you make the container very large and take the "thermodynamic" limit. If you make the container very large, i.e., [itex]L[/itex] becomes very large, the grid of momenta becomes very fine, and it is appropriate to replace the sum by an integral. To get this limit right, we need a density of states in a given momentum volume [itex]\mathrm{d}^3 \vec{p}[/itex]. In the discretized version in such a momentum volume you have
    [tex]\mathrm{d} \Omega = \mathrm{d}^3 \vec{p} \frac{L^3}{(2 \pi \hbar)^3}=\mathrm{d}^3 \vec{p} \frac{V}{(2 \pi \hbar)^3}[/tex]
    momentum states for your single particle.

    So in the thermodynamic limit you can substitute the sum over momentum states by an integral with the just calculated integration measure:
    [tex]\sum_{\vec{p} \in \frac{2 \pi \hbar}{L} \mathbb{Z}^3} \rightarrow \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{V}{(2 \pi \hbar)^3}.[/tex]
    Note that here also the dimensions are correct, because both the sum symbol and the integral are formally of dimension 1.

    That's the true reason for the phase-space volume [itex](2 \pi \hbar)^3[/itex] in statistical mechanics. It has nothing to do with the uncertainty relation in a direct way, but with the number of states in a given momentum range for a particle in a container of volume [itex]V[/itex] in the thermodynamical limit!
     
  4. Oct 8, 2014 #3
    Hmm.. this is similar to the argument when considering sinusoidal modes of the electromagnetic waves in a box: then one argues that only standing waves for which ##\frac{n \Lambda}{2} = L## for ##n =1,2, \ldots## are allowed. However, when using the debrogeli relation
    ## p = 2\pi \hbar/\lambda## this would indicate that
    $$p = \frac{\pi \hbar}{L} n.$$

    Why do you get a factor which is off by two compared to this reasoning?
     
  5. Oct 9, 2014 #4
    Vanhees is right about why the ##(2 \pi \hbar)^3## appears in statistical mechanics. But the statement you mention in your OP is a qualitative one for which the exact constants aren't important. For example the seemingly precise statement ##\Delta x \Delta p \geq \hbar/2## actually requires you to choose some more or less arbitrary convention to define the uncertainties ##\Delta x## and ##\Delta p##. The typical choice is to use the standard deviation to define the uncertainty but this is a convention and we could just as well define ##\Delta x## to be twice the standard deviation in ##x##. This would change the numerical factor on the right-hand side, so the numerical factor is just conventional. However the factor of ##h## is important and is independent of your conventions for defining uncertainty.
     
  6. Oct 10, 2014 #5

    vanhees71

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    The [itex]\Delta x[/itex] and [itex]\Delta p[/itex] are not arbitrary quantities but the standard deviations and thus well-defined statistical quantities with a clear statistical meaning. The Heisenberg-Robertson uncertainty relation, however, is not the physical reason for the choice of the phase-space-volume scale in statistical mechanics but the density of states according to quantum theory. Note that there is no such well-defined phase-space-volume scale in classical mechanics and that makes classical statistical physics a somewhat complicated issue, while it is easy to solve all quibbles with the foundations of classical statistics by deriving it as the appropriate limit of quantum statistics.
     
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