Derive the N to the three-halves free energy

Before this lesson: Turn the matrix integral into a Fermi gas.

Why does the ABJM sphere free energy grow as $N^{3/2}$ ? The answer is not obtained by counting elementary matrix entries. In the Fermi-gas formulation, exclusion forces successive auxiliary particles into higher one-particle levels. The number of available levels below energy $E$ grows as $E^2$ . Combining these two facts produces both the power and its coefficient.

Keep the same compact three-dimensional $U(N)_k\times U(N)_{-k}$ theory, positive integer $k$ , rank $N$ per factor and no extra quotient. The undeformed round-sphere background, absolute normalization and phase prescription are fixed. The full generic theory has $\mathcal N=6$ , enhanced to $\mathcal N=8$ at $k=1,2$ ; localization selects a supercharge within the chosen manifest $\mathcal N=2$ subalgebra. We take $N\to\infty$ at fixed $k$ and use $F=-\log Z$ . This is a sphere observable, not a physical thermal free energy of ABJM on flat space.

Count states before filling them

The positive operator derived in the prerequisite is

\begin{aligned} \widehat\rho&=e^{-U(\widehat x)/2}e^{-T(\widehat p)}e^{-U(\widehat x)/2} =e^{-\widehat H},\qquad [\widehat x,\widehat p]=i\hbar,\quad \hbar=2\pi k,\\ U(x)&=\log[2\cosh(x/2)],\qquad T(p)=\log[2\cosh(p/2)]. \end{aligned} \tag{AB19.1}

Its eigenvalues are $e^{-E_j}$ . Define the counting function $n(E)=\#\{j:E_j<E\}$ . It is a staircase. Its derivative is a distribution, $g(E)=\sum_j\delta(E-E_j)$ , called the density of states. A smooth approximation to $g$ describes an energy average, not a continuous replacement for every exact level. We use $g$ rather than $\rho$ for this density to avoid confusing it with the operator.

At high energy, the leading count comes from the area inside the classical energy contour. The phase-space cell has area $2\pi\hbar$ : in a long position interval of length $L$ , periodic plane waves have momenta separated by $2\pi\hbar/L$ . A momentum window $\Delta p$ therefore contains approximately $L\Delta p/(2\pi\hbar)$ states. This fixes the measure, including its factor of $k$ :

n(E)\sim\frac{\mathcal A(E)}{2\pi\hbar},\qquad \mathcal A(E)=\int_{H_{\rm cl}(x,p)<E}dx\,dp, \qquad H_{\rm cl}=U(x)+T(p). \tag{AB19.2}

Here $H_{\rm cl}$ is the leading classical symbol. The noncommuting operator $\widehat H$ is not exactly $U(\widehat x)+T(\widehat p)$ . The use of (AB19.2) is the high-energy Weyl approximation for this spectral problem, as in FG (4.12)–(4.17). We will distinguish its classical geometry from the quantum justification below.

The energy contour becomes a diamond

The exact identity

U(x)=\frac{|x|}{2}+\log(1+e^{-|x|}) \tag{AB19.3}

shows that its correction to a straight line is nonnegative and bounded by $\log2$ . The same holds for $T$ . Thus $H_\diamond=(|x|+|p|)/2\leq H_{\rm cl}\leq H_\diamond+\log4$ . For $E\geq\log4$ , the smooth allowed region lies between two diamonds:

\{H_\diamond<E-\log4\}\subseteq\{H_{\rm cl}<E\} \subseteq\{H_\diamond<E\}. \tag{AB19.4}

Each quadrant of the outer diamond is a right triangle with legs $2E$ . Its area is $2E^2$ , giving

8(E-\log4)^2\leq\mathcal A(E)\leq8E^2, \qquad \frac{\mathcal A(E)}{E^2}\longrightarrow8. \tag{AB19.5}

This proves the leading classical coefficient without assuming that the approximation is uniform near the four corners. Dividing by $2\pi\hbar=4\pi^2k$ yields

\boxed{n(E)\sim C E^2,\qquad C=\frac{2}{\pi^2k}.} \tag{AB19.6}

The diamond calculation is FG (4.20)–(4.26). The replacement of the exact quantum count by this leading area requires more than the inequalities in (AB19.5). Fixed integer $k$ does not send $\hbar$ to zero. The relevant energies instead become large compared with its fixed scale. FG’s paragraph after (4.27) and its quantum analysis in section 5.3 argue, using an all-orders WKB analysis, that the leading $E^2$ coefficient survives: quantum corrections change lower-order terms. We use that leading spectral input, while deriving the area and thermodynamic consequences here. This is not a nonperturbative remainder theorem for the exact staircase. The full expansion, including exponentially small sectors, is not proved by a diamond drawing.

Fill the levels and recover the free energy

First imagine filling the lowest $N$ levels sharply. The top occupied energy, called the Fermi energy, obeys

N\sim C E_F^2,\qquad E_F\sim\sqrt{N/C}. \tag{AB19.7}

Replacing the level sum by its leading density gives the total occupied energy

\sum_{j<N}E_j\sim\int_0^{E_F}E\,(2CE)\,dE =\frac{2C}{3}E_F^3 =\frac{2}{3\sqrt C}N^{3/2}. \tag{AB19.8}

The low-energy lower endpoint is immaterial to this leading term; the true spectrum does not begin at zero. This explains the power: the energy of a typical occupied level grows as $\sqrt N$ , and there are $N$ occupied levels. Fermions are essential. Putting every particle into a single lowest level would give a different result.

The actual auxiliary gas has inverse temperature one, so its occupation is rounded rather than a sharp step. Its exact generating function gives

J(\mu)=\sum_j\log(1+e^{\mu-E_j}),\qquad \overline N(\mu)=J'(\mu)=\sum_j\frac1{1+e^{E_j-\mu}}. \tag{AB19.9}

The bar denotes a grand-canonical mean, not a change in the physical integer rank. The occupation falls from nearly one to nearly zero in an energy interval of order one around $\mu$ . At large $\mu$ that is narrow compared with the occupied range. Integrating by parts gives $J=\int_0^\infty n(E)/(1+e^{E-\mu})\,dE$ . Using the leading count and replacing the rounded edge by its leading step therefore gives

J(\mu)\sim\frac C3\mu^3,\qquad \overline N(\mu)\sim C\mu^2. \tag{AB19.10}

For example, extending just the model density $2CE$ down to zero produces only an order-one correction to $\overline N$ from the rounding; Exercise 3 computes it. It does not change the coefficient of $\mu^2$ .

Choose $\mu_*$ so that $\overline N=N$ . Coefficient extraction of $e^{N\mu}$ from the generating function gives the leading Legendre expression $F=N\mu_*-J(\mu_*)$ . Its sign follows from $Z\sim e^{J-N\mu_*}$ , not from the sign of $J$ alone. Consequently

\boxed{F_{S^3}(N,k)\sim\frac{2}{3\sqrt C}N^{3/2} =\frac{\pi\sqrt{2k}}3N^{3/2}, \qquad N\longrightarrow\infty,\quad k\in\mathbb Z_{>0}\ \text{fixed}.} \tag{AB19.11}

The earlier saddle calculation supplies the coefficient-contour argument; we have now derived its cubic input from the one-particle count. The leading number variance is $J''\sim2C\mu$ , so its standard deviation divided by its mean tends to zero as $\mu^{-3/2}$ . This is consistent with the equivalence of the two ensembles at leading order. It does not equate their finite-rank partition functions.

FG (4.27) writes the opposite sign because its free energy is $\log Z$ . The independently obtained planar strong-coupling answer in DMP (5.22), (5.34)–(5.37), after its $g_s=2\pi i/k$ and sign translations, has the same leading coefficient in the overlapping strong-coupling regime. It is separate evidence, not permission to exchange the fixed- $k$ and fixed- $N/k$ limits.

Worked checks with the smooth contour and the quantum gas

The smooth classical contour can be integrated without replacing it by straight lines. Its positive turning point is $x_*=2\operatorname{arcosh}(e^E/4)$ , and solving for $p$ gives

\mathcal A(E)=8\int_0^{x_*} \operatorname{arcosh}\!\left(\frac{e^E}{4\cosh(x/2)}\right)dx, \qquad E\geq\log4. \tag{AB19.12}

This is a classical area, not an exact list of quantum energies. Direct integration gives $\mathcal A(4)/(8\cdot4^2)=0.89689770$ and $\mathcal A(16)/(8\cdot16^2)=0.99357448$ : the ratio approaches one even though the lowest contour begins at $E=\log4$ . A second normalization check integrates a smooth function of the same Hamiltonian. For $t>0$ , substitute $u=e^x$ to obtain

\begin{aligned} K_{\rm cl}(t)&=\int\frac{dx\,dp}{4\pi^2k}\,e^{-tH_{\rm cl}} =\frac1{4\pi^2k}\left[\frac{\Gamma(t/2)^2}{\Gamma(t)}\right]^2,\\ \int_{-\infty}^{\infty}\frac{dx}{[2\cosh(x/2)]^t}& =\int_0^\infty\frac{u^{t/2-1}}{(1+u)^t}\,du,\\ K_{\rm cl}(t)&\sim\frac{4}{\pi^2k\,t^2}\qquad(t\to0^+). \end{aligned} \tag{AB19.13}

The beta integral on the second line fixes the gamma-function ratio. Since the Laplace transform of $2CE$ is $2C/t^2$ , the last line independently reproduces (AB19.6). This is FG (4.28)–(4.30), with our measure displayed. The equality is exact for the classical integral; $K_{\rm cl}(t)$ is not generally the exact quantum $\operatorname{Tr}e^{-t\widehat H}$ . Agreement at $t=1$ with $\operatorname{Tr}\rho=1/(4k)$ does not establish equality at other $t$ .

For an actual finite-rank comparison one can instead diagonalize the positive kernel of the preceding lesson, then build $Z_N$ from its elementary symmetric coefficients. If $c_n$ are the coefficients already accumulated, adding one eigenvalue $\lambda$ updates them by $c_n\mapsto c_n+\lambda c_{n-1}$ , using the old coefficient list on the right. This includes exclusion exactly for the discretized operator. At $k=1$ it gives:

$N$	Numerical $F(N,1)$	$F/[\pi\sqrt2\,N^{3/2}/3]$
$8$	$32.40038$	$0.966878$
$16$	$92.86529$	$0.979783$
$32$	$264.95614$	$0.988338$

These values use the kernel itself, with no Airy approximation as input. Symmetric weighted Gauss–Legendre matrices on $[-L,L]$ , $L=64,80$ , were refined from 768 to 1024 nodes. The last refinement changed these free energies by less than $1.3\times10^{-11}$ and the two windows differed by less than $10^{-9}$ . These are numerical stability tests, not rigorous error bars for the infinite operator. The displayed precision is deliberately lower. The several-percent discrepancy from the leading formula is an asymptotic correction, much larger than that observed numerical variation.

The coefficient in (AB19.11) concerns $N\to\infty$ . It does not determine a finite-rank answer, the linear grand-potential term, or the additive constant. The next lesson studies those corrections and the sectors a perturbative resummation omits.

Exercises

1. Change the two slopes

Consider a different auxiliary model with high-energy Hamiltonian $H=\alpha|x|+\beta|p|$ , $\alpha,\beta>0$ , at the same fixed $\hbar$ and with one state per cell. Derive its counting coefficient and leading canonical free energy. Recover the ABJM coefficient when $\alpha=\beta=1/2$ and $\hbar=2\pi k$ . Is changing a slope equivalent to changing the measure by hand?

Hint 1

The first-quadrant triangle has intercepts $E/\alpha$ and $E/\beta$ .

Hint 2

Divide the area of all four triangles by $2\pi\hbar$ , then repeat the filling integral.

Solution

The area is $2E^2/(\alpha\beta)$ , hence

C_{\alpha\beta}=\frac1{\pi\hbar\alpha\beta},\qquad F\sim\frac23\sqrt{\pi\hbar\alpha\beta}\,N^{3/2}.

The ABJM slopes give $C=2/(\pi^2k)$ and the coefficient in (AB19.11). A changed slope changes the auxiliary Hamiltonian and its phase-space region. The measure instead follows from the commutator and cannot be altered independently while claiming to describe the same system. This exercise defines a toy spectral change; it does not identify an arbitrary ABJM mass deformation with a slope change.

2. Replace the diamond by an oscillator

Take independent fermions with one-particle energies $E_j=\omega(j+1/2)$ , $j=0,1,\ldots$ , at fixed $\omega>0$ and inverse temperature one. Find the large- $N$ power of $F$ . Then obtain the exact canonical partition function by writing occupied integers as $0\leq n_1<\cdots<n_N$ . Explain why exclusion alone does not guarantee the power $3/2$ .

Hint 1

Here $n(E)\sim E/\omega$ , rather than $CE^2$ . The filled energy is an arithmetic sum.

Hint 2

Write $n_j=(j-1)+\sum_{a=1}^{j}g_a$ with independent nonnegative integers $g_a$ . Each gap contributes to $N-a+1$ occupied energies.

Solution

The lowest filled energy is $\omega\sum_{j=1}^N(j-1/2)=\omega N^2/2$ . Summing every gap as a geometric series gives

Z_N=\frac{e^{-\omega N^2/2}}{\prod_{m=1}^{N}(1-e^{-m\omega})},\qquad F_N=\frac\omega2N^2+\sum_{m=1}^N\log(1-e^{-m\omega}).

At fixed positive $\omega$ the sum converges to a finite negative constant: its large- $m$ summands are of order $-e^{-m\omega}$ . The leading growth is therefore $N^2$ . The exclusion rule is identical to the ABJM auxiliary gas, but the count grows linearly with energy. Both the statistics and the one-particle growth law are needed to determine the exponent. A limit that also sends $\omega$ to zero would require a new analysis.

3. Does thermal rounding determine the rank shift?

Use only the toy smooth density $g_0(E)=2CE$ for all $E\geq0$ . Show that its mean particle number obeys

N_0(\mu)=C\mu^2+\frac{C\pi^2}{3} -2C\int_0^\infty\frac{v\,dv}{1+e^{v+\mu}},\qquad \mu>0.

Bound the last term. Does its constant correction determine the exact linear term in the ABJM grand potential?

Hint 1

Split the occupation integral at $E=\mu$ and reflect the part below it using $1/(1+e^{-v})=1-1/(1+e^v)$ .

Hint 2

Use $\int_0^\infty v/(1+e^v)\,dv=\pi^2/12$ , obtained by integrating the alternating exponential series. For a bound, replace the denominator $1+e^{v+\mu}$ by $e^{v+\mu}$ .

Solution

After reflection, the sharp filled part gives $C\mu^2$ . Extending both edge integrals to the whole positive $v$ axis gives $4C\int_0^\infty v/(1+e^v)\,dv=C\pi^2/3$ . The overextended lower edge must be subtracted and is exactly the integral displayed in the question. Thus

0< C\mu^2+\frac{C\pi^2}{3}-N_0(\mu)<2Ce^{-\mu}.

This predicts a linear term $C\pi^2\mu/3$ in the toy $J_0$ . With ABJM’s leading $C$ its coefficient would be $2/(3k)$ . But the leading high-energy count did not specify the lower-order classical count, quantum shifts or the low-energy completion. Those also affect the linear term. Thermal rounding of the leading density alone therefore cannot determine the physical rank shift. It correctly leaves the cubic coefficient, and hence (AB19.11), unchanged.

Continue with the Airy function and the corrections it omits. The leading state count supplies one coefficient; a normalized rank-by-rank prediction needs more information.

The source companion identifies the precise WKB input and separates it from the exact spectral trace expansion.