Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is semi-classical level counting?

  1. Feb 19, 2009 #1
    Wikipedia article Hilbert-Polya conjecture has a link to an article H=xp and the Riemann zeros by Berry & Keating. They mention that the number of energy levels below given [itex]E[/itex] could be counted by computing the area enclosed by the contour [itex]H(x,p)=E[/itex] in the phase space. What is that all about? Does there exist some theorem concerning this? Can it be justified easily? What information sources are there about this?
  2. jcsd
  3. Feb 20, 2009 #2


    User Avatar
    Science Advisor

    Yes, there is a general theorem. Consider a quantum system with [itex]d[/itex] degrees of freedom that is obtained by canonical quantization of a classical hamiltonian [itex]H(p,x)[/itex]. Suppose that the classical motion is bounded in phase space for energy [itex]E[/itex], so that the quantum states have discrete energy eigenvalues. Let [itex]\rho(E)dE[/itex] be the number of energy eigenstates with energy between [itex]E[/itex] and [itex]E+dE[/itex] (where [itex]dE[/itex] should be much larger than the mean spacing between energy eigenvalues, but small compared to any classically relevant energy scale). Then


    where [itex]\delta[/itex] is the Dirac delta function.

    To derive this, first write the exact density of states as

    [tex]\rho(E)=\sum_\alpha \delta(E-E_\alpha)[/tex]

    where the energy eigenstates are indexed by [itex]\alpha[/itex], and [itex]E_\alpha[/itex] is an energy eigenvalue. This can be written as

    [tex]\rho(E)=\sum_\alpha \langle\alpha|\delta(E-\hat H)|\alpha\rangle[/tex]

    where [itex]\hat H[/itex] is the hamiltonian operator. Then we note that [itex]\sum_\alpha\langle\alpha|\ldots|\alpha\rangle[/itex] is equivalent to a trace over the Hilbert space, so we have

    [tex]\rho(E)=\mathop{\rm Tr}\delta(E-\hat H)[/tex]

    Then use [itex]\delta(E)=\int_{-\infty}^{+\infty}{d\tau\over2\pi\hbar}\,e^{iE\tau/\hbar}[/itex] to get

    [tex]\rho(E)=\mathop{\rm Tr}\int_{-\infty}^{+\infty}{d\tau\over2\pi\hbar}\,e^{i(E-\hat H)\tau/\hbar}=\int_{-\infty}^{+\infty}{d\tau\over2\pi\hbar}\,e^{iE\tau/\hbar}\mathop{\rm Tr}e^{-i\hat H\tau/\hbar}[/tex]

    Now we have the trace of the time evolution operator, which can be written as a sum over position eigenstates,

    [tex]{}\mathop{\rm Tr}e^{-i\hat H\tau/\hbar}=\int d^dx\,\langle x|e^{-i\hat H\tau/\hbar}|x\rangle[/tex]

    (continued next post)
    Last edited: Feb 20, 2009
  4. Feb 20, 2009 #3


    User Avatar
    Science Advisor

    Now I'm going to do some hand waving. Rigorous justification is possible but lengthy.

    If the hamiltonian was of the form [itex]\hat H ={1\over 2m}\hat p^2 + V(\hat x)[/itex], and we were interested in a "short" time [itex]\tau[/itex], we could use the Campbell-Baker-Hausdorf formula to write

    [tex]e^{-i\hat H\tau/\hbar} = e^{-i\hat p^2\tau/2m\hbar} \,e^{-iV(\hat x)\tau/\hbar}\,e^{O(\tau^2)}[/itex]

    Then we could insert a complete set of momentum eigenstates to get

    [tex]\langle x|e^{-i\hat H\tau/\hbar}|x\rangle \approx \int d^dp\,\langle x|e^{-i\hat p^2\tau/2m\hbar}|p\rangle\langle p|e^{-iV(\hat x)\tau/\hbar}|x\rangle[/itex]

    Now I can remove the hats from the operators [itex]\hat p[/itex] and [itex]\hat x[/itex], because they are acting on their eigenstates, and pull these factors out front, since they are now just numbers. So we now have

    [tex]\langle x|e^{-i\hat H\tau/\hbar}|x\rangle \approx \int d^dp\,e^{-i(p^2/2m+V(x))\tau/\hbar}\langle x|p\rangle\langle p|x\rangle[/itex]

    Now I use [itex]\langle x|p\rangle=\langle p|x\rangle^*=e^{ipx/\hbar}/(2\pi\hbar)^{d/2}[/itex], and we have

    [tex]\langle x|e^{-i\hat H\tau/\hbar}|x\rangle \approx \int {d^dp\over\,(2\pi\hbar)^d}\,e^{-iH(p,x)\tau/\hbar}[/tex]

    Plugging this into our last formula for the density of states, we get

    [tex]\rho(E) \approx \int_{-\infty}^{+\infty}{d\tau\over2\pi\hbar}\int{d^dp\,d^dx\over(2\pi\hbar)^d}\,e^{iE\tau/\hbar}\,e^{-iH(p,x)\tau/\hbar}[/tex]

    Now carry out the integral over [itex]\tau[/itex] to get

    [tex]\rho(E) \approx \int{d^dp\,d^dx\over(2\pi\hbar)^d}\,\delta

    Ta da!

    Of course, I cheated, because I used a small-[itex]\tau[/itex] approximation, then integrated over all [itex]\tau[/itex]. Look up the "Gutzwiller trace formula" to see how corrections to this result (which is sometimes called the "Weyl formula") are computed.
  5. Feb 20, 2009 #4


    User Avatar
    Science Advisor

    Oh, I should have mentioned: the total number of states with energy less than [itex]E[/itex] is given by

    [tex]N(E)=\int_0^E dE'\,\rho(E')[/tex]

    Since the integral of a delta function is a step function, we have


    where [itex]\theta[/itex] is the step function. In one dimension, this is just the area of the [itex]x[/itex]-[itex]p[/itex] plane that is inside the contour specified by [itex]H(p,x)=E[/itex].
  6. Feb 21, 2009 #5
    ok. Looks great. :cool:
  7. Feb 23, 2010 #6
    Can anyone recommend a book about this topic?

    It seems that most books about quantum theory don't include this kind of topics.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook