# QHO in Microcanonical Ensemble: Problem with alternate derivation

I am working through Franz Schwabl's book on Statistical Mechanics, and he has a number of derivations of thermodynamic quantities that are different than those I have seen before. I am also having difficulty finding them repeated elsewhere.

In particular, he has a method for calculating $$\Omega(E)$$, the number of states with a given energy $$E$$, of a series of $$N$$ independent Quantum Harmonic Oscillators ($$\mathcal{H} = \sum_{j=1}^N\hbar\omega(n_j+\frac{1}{2})$$) that I hadn't seen before. Proceeding from the result

$$\Omega(E) = \mathrm{Tr}\,\delta(\mathcal{H}-E)=\sum_{n_1=0}^{\infty}\cdots\sum_{n_N=0}^{\infty}\delta\left(E - \hbar\omega\sum_{j=1}^N\left(n_j+\frac{1}{2}\right)\right),$$

my strategy would be combinatoric: the delta-function turns the unrestricted sums over $$n_j$$ to a constraint on the total number of quanta. Calculating the number of ways you can partition $$n=\sum_{j=1}^Nn_j$$ quanta of energy among $$N$$ oscillators gives you $$\Omega(E)$$. This is the way we did it in undergraduate stat mech.

Schwabl's approach proceeds differently: by taking the Fourier Transform of the delta function, one obtains

$$\Omega(E) = \int \frac{dk}{2\pi}e^{ikE}\prod_{j=1}^N\left(e^{-ik\hbar\omega/2}\sum_{n_j=0}^{\infty}e^{-ik\hbar\omega n_j}\right)=\int\frac{dk}{2\pi}e^{ikE}\left(\frac{e^{-ik\hbar\omega/2}}{1-e^{-ik\hbar\omega}}\right)^N,$$

where this last step involves summing a divergent geometric series, declaring $$\sum_{\ell=0}^{\infty}e^{-i\alpha\ell} = \frac{1}{1-e^{-i\alpha}}$$ and ignoring the fact that this series doesn't converge in a conventional sense.

This simplifies to $$\Omega(E) = \int\frac{dk}{2\pi}e^{N(ik(E/N) - \log(2i\sin(k\hbar\omega/2)))}$$

which is solved using the saddle-point approximation. The maximum of the argument of the exponential occurs at a value

$$k_0 = \frac{1}{\hbar\omega i}\log\frac{\frac{E}{N}+\frac{\hbar\omega}{2}}{\frac{E}{N}-\frac{\hbar\omega}{2}}$$

Which is clearly imaginary, despite the fact that in a Fourier Transform $$k$$ is supposed to be a real number!

In spite of all this, if you evaluate the integral using the saddle point approximation at $$k=k_0$$, you get the same form for $$\Omega(E)$$ that one derives through the garden-variety combinatorial argument!

My question(s) are

Why does this work? Specifically:

• Why does it make sense to write the convergence of a divergent geometric series in the form given here (is this relying on some sense of convergence other than the typical one, and if so, what? And what does that imply about convergence in stat mech?), and

• Why can you use the saddle point approximation when the maximum value does not occur in the space over which you are integrating?

Answers to this question might rely on appeals to other situations in which this math occurs and has been rationalized, physically if not mathematically.

• It's worth noting that the microcanonical ensemble is a bad approach to deriving thermodynamic quantities, if that's your end goal. Not only is it unrealistic (when is energy known exactly?), ambiguous (volume entropy or surface entropy?), and corresponding poorly to thermodynamics (the "temperature" from MCE is not thermodynamic since thermal equilibrium between systems is not possible), but it also creates mathematical difficulties like you mention. – Nanite Mar 1 '14 at 14:36
• @Nanite I acknowledge this difficulty. I'm mostly just going through this book and trying to understand the leaps the author makes. He uses this technique (Fourier Transform of the delta function -> saddle point approximation) many times in the book, so it seems like it might be an important technique to understand. Furthermore, demonstrating ensemble equivalence can be a useful exercise in developing facility with the tools of stat mech. – A. Kennard Mar 1 '14 at 16:38
• "microcanonical ensemble is a bad approach to deriving thermodynamic quantities" Nanite, this seems like a personal opinion devoid of solid ground. Energy does not need to be known exactly (it never is), the entropy can be taken as log of surface or volume - they are practically the same for macro-systems. Temperature for microcanonical ensemble is a derived quantity based on thermodynamic relation $dS/dU = 1/T$. – Ján Lalinský Mar 1 '14 at 23:14

I) If we expect $\Omega(E)$ to depend analytically on the variable $\hbar\omega>0$ extended to (parts of) the complex plane, then we may regularize by introducing an $i\epsilon$ prescription, and substitute

$$\tag{1} \hbar\omega ~\longrightarrow ~ \hbar\omega (1-i\epsilon).$$

The variable

$$\tag{2} q~:=~ e^{-i\hbar\omega k}~\longrightarrow ~ e^{-(i+\epsilon)\hbar\omega k}$$

in the geometric series

$$\tag{3} \sqrt{q}\sum_{n=0}^{\infty} q^n$$

will then have

$$\tag{4} |q|~<~1,$$

so that the geometric series (3) is convergent. Then all steps in Schwabl's derivation of $\Omega(E)$ are mathematically well-defined. At the end of the calculation of $\Omega(E)$, we may put $\epsilon=0$.

II) Concerning a complex (as opposed to real) stationary solution in the method of steepest descent/ stationary phase method/saddle-point method, this is just part of the method. For a rigorous argument, one would have to consult the proof of the method. Heuristically, it is because when one evaluates the Gaussian integral over 'quantum fluctuations'

$$\tag{5} \int_{\mathbb{R}} \! dx~ e^{-\frac{a}{2}x^2+bx} ~=~\sqrt{\frac{2\pi}{a}}\exp\left(\frac{b^2}{2a}\right),$$

for two complex constants $a,b\in\mathbb{C}$, one only needs the condition

$$\tag{6} {\rm Re}(a)~>~0$$

to ensure convergence of the integral (5). There is no need to also assume that the stationary solution $\frac{b}{a}$ is real. Equation (5) follows from the fact that

$$\tag{7} \alpha\int_{\mathbb{R}} \! dx~e^{-\frac{1}{2}(\alpha x+\beta)^2} ~=~\int_{\gamma} \! d(\alpha x+\beta)~ e^{-\frac{1}{2}(\alpha x+\beta)^2} ~=~ \int_{\mathbb{R}} \! dx~ e^{-\frac{1}{2}x^2}~=~\sqrt{2\pi}$$

for any straight line in the complex plane

$$\tag{8} \gamma(x)~=~\alpha x+\beta, \qquad \alpha,\beta~\in~\mathbb{C},\qquad x~\in~\mathbb{R},$$

with slope

$$\tag{9} |\arg(\alpha)|<\frac{\pi}{4},$$

because in that case, it is possible to close the contour along exponentially suppressed arcs.

• Thanks! I'm not familiar with regularization, but it seems if I were to look into a book or internet material on QFT I'd find some explanations of it. I'll also have to read up on the saddle point method. Just to make things explicit, my understanding of how the saddle-point method can accommodate imaginary stationary solns is by turning the integral over $R$ into a contour by connecting the real line and the steepest descent path through the stationary soln. Assuming there are no poles inside the contour, then the real integral will be equal to the integral over the steepest descent path? – A. Kennard Mar 2 '14 at 1:30
• @A. Kennard: I agree. – Qmechanic Mar 2 '14 at 14:41