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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?
     
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  3. Feb 20, 2009 #2

    Avodyne

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    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

    [tex]\rho(E)=\int{d^dp\,d^dx\over(2\pi\hbar)^d}\,\delta
    \bigl(E-H(p,x)\bigr)\bigl[1+O(\hbar)\big][/tex]

    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

    Avodyne

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    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
    \bigl(E-H(p,x)\bigr)[/tex]

    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

    Avodyne

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    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

    [tex]N(E)=\int{d^dp\,d^dx\over(2\pi\hbar)^d}\,\theta
    \bigl(E-H(p,x)\bigr)[/tex]

    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.
     
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