From Yoshioka (eq. 10.5), the grand partition is given by

$$ \Xi = \sum_{N=0}^{\infty}\ \int_0^\infty \exp \left[ \frac{[E-\mu N-S_I(E,N)T]}{k_B T} \right] \ .$$

It goes on that the value of the free energy $J=k_B T \ln \Xi(T,\mu)$ can be formally evaluated by expanding the integrand around the maximum of at ($E^*,N^ *$) up to second order, neglecting terms terms of order one with respect to terms of order N.

This should then result in (eq. 10.7) $$ J = E^*-S(E^*,N^*)T-\mu N^* \ .$$

I don't underdstand what is meant by expanding up to second order, in this context. How is use supposed to take a Taylor expansion here?


When integrating a sharply peaked function, you can approximate the integrand as a Gaussian in the neighborhood of the peak and evaluate from there.

Consider the integral $$I = \int \exp[f(x)] dx$$

where $f$ is some everywhere-differentiable function with a global maximum at $x=x^*$. One would then assume that the integrand would be sharply peaked at $x^*$, so the integral would be dominated by the contribution from a small neighborhood around that point.

We can Taylor expand $f$ around $x^*$ as follows:

$$f(x) \approx f(x^*) + f'(x^*)(x-x^*) + \frac{1}{2}f''(x^*)(x-x^*)^2 + O\big((x-x^*)^3\big)$$

The second term on the right vanishes by virtue of the fact that $x^*$ is a stationary point of $f$. Also, since $x^*$ is a maximum, $f''(x^*)<0$. Therefore, letting $\big|f''(x^*)\big|\equiv a^2$ for convenience, we have that to second order in $(x-x^*)$,

$$f(x) = f(x^*) - \frac{1}{2}a^2 (x-x^*)^2$$ and

$$I = \int \exp\big[f(x^*) - \frac{1}{2}a^2(x-x^*)^2\big] dx = \exp[f(x^*)] \int\exp\big[-\frac{1}{2}a^2(x-x^*)^2\big]dx$$ $$ = \exp[f(x^*)]\cdot \frac{\sqrt{2\pi}}{a} = \exp\big[f(x^*)\big] \cdot \sqrt{\frac{2\pi}{-f''(x^*)}}$$

where we've used the fact that $\int_{-\infty}^\infty \exp(-\alpha x^2)dx = \sqrt{\frac{\pi}{\alpha}}$ for $\alpha>0$, as well as the fact that since the dominant contribution to the integral comes from the small neighborhood around $x^*$, we may as well extend the Gaussian integral out to $\pm \infty$ (as the Gaussian is strongly suppressed outside of this neighborhood anyway).

Applying precisely the same procedure to a function $F$ of two variables $(E,N)$ is a fairly straightforward extension of this idea. You should find that the result is proportional to $\exp\big[F(E^*,N^*)\big]$, yielding the result you're looking for.

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