What is semi-classical level counting?

[jostpuur]

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 $E$ could be counted by computing the area enclosed by the contour $H(x,p)=E$ 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?

 

[Avodyne]

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

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

where $\delta$ is the Dirac delta function.

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

$$\rho(E)=\sum_\alpha \delta(E-E_\alpha)$$

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

$$\rho(E)=\sum_\alpha \langle\alpha|\delta(E-\hat H)|\alpha\rangle$$

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

$$\rho(E)=\mathop{\rm Tr}\delta(E-\hat H)$$

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

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

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

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

(continued next post)

 

[Avodyne]

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

If the hamiltonian was of the form $\hat H ={1\over 2m}\hat p^2 + V(\hat x)$, and we were interested in a "short" time $\tau$, 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]<br /> <br /> Then we could insert a complete set of momentum eigenstates to get<br /> <br /> [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]<br /> <br /> Now I can remove the hats from the operators $\hat p$ and $\hat x$, because they are acting on their eigenstates, and pull these factors out front, since they are now just numbers. So we now have<br /> <br /> [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]<br /> <br /> Now I use $\langle x|p\rangle=\langle p|x\rangle^*=e^{ipx/\hbar}/(2\pi\hbar)^{d/2}$, and we have<br /> <br /> $$\langle x|e^{-i\hat H\tau/\hbar}|x\rangle \approx \int {d^dp\over\,(2\pi\hbar)^d}\,e^{-iH(p,x)\tau/\hbar}$$<br /> <br /> Plugging this into our last formula for the density of states, we get<br /> <br /> $$\rho(E) \approx \int_{-\infty}^{+\infty}{d\tau\over2\pi\hbar}\int{d^dp\,d^dx\over(2\pi\hbar)^d}\,e^{iE\tau/\hbar}\,e^{-iH(p,x)\tau/\hbar}$$<br /> <br /> Now carry out the integral over $\tau$ to get<br /> <br /> $$\rho(E) \approx \int{d^dp\,d^dx\over(2\pi\hbar)^d}\,\delta<br /> \bigl(E-H(p,x)\bigr)$$<br /> <br /> Ta da!<br /> <br /> Of course, I cheated, because I used a small-$\tau$ approximation, then integrated over all $\tau$. Look up the "Gutzwiller trace formula" to see how corrections to this result (which is sometimes called the "Weyl formula") are computed.[/tex][/tex][/tex]

 

[Avodyne]

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

$$N(E)=\int_0^E dE'\,\rho(E')$$

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

$$N(E)=\int{d^dp\,d^dx\over(2\pi\hbar)^d}\,\theta<br /> \bigl(E-H(p,x)\bigr)$$

where $\theta$ is the step function. In one dimension, this is just the area of the $x$-$p$ plane that is inside the contour specified by $H(p,x)=E$.

 

[jostpuur]

ok. Looks great.

 

[jostpuur]

Can anyone recommend a book about this topic?

It seems that most books about quantum theory don't include this kind of topics.