Page's Random-State Theorem

Page’s theorem quantifies a kinematic fact: a Haar-random pure state on a bipartite Hilbert space gives its smaller factor nearly maximal entropy. When the dimension ratio $m/n$ tends to zero, the same result also implies closeness to the maximally mixed state in trace norm. Each factor dimension enters the exact mean. The theorem is a powerful benchmark for unitary evaporation, but it is not a Hamiltonian, a locality statement, or a proof that black-hole dynamics samples the Haar ensemble.

Required background. Quantum Information supplies Haar measure, reduced density operators, Schmidt spectra, and von Neumann entropy used in the theorem.

Helpful background. Semiclassical Gravity supplies the entropy scale used to estimate effective dimensions; Quantum Fields in Curved Spacetime supplies the contrasting Hawking state; Black-Hole Information supplies the unitarity question for which this is a benchmark.

Haar-random bipartite entropy

Theorem. Let $\lvert\psi\rangle$ be distributed with normalized Haar measure on

$$\mathcal H_A\otimes\mathcal H_B\simeq\mathbb C^m\otimes\mathbb C^n, \qquad 1\leq m\leq n.$$

For $\rho_A=\operatorname{Tr}_B\lvert\psi\rangle\langle\psi\rvert$ and natural logarithms,

$$\mathbb E_{\mathrm{Haar}}S(\rho_A) =H_{mn}-H_n-\frac{m-1}{2n}, \qquad H_N=\sum_{k=1}^{N}\frac1k.$$

Page gave this exact expression as a conjecture Page 1993, Eqs. (1)–(11); Sen supplied the proof Sen 1996, pp. 1–3 of the published article. The hypotheses include finite dimensions, no energy or charge constraint, no locality structure, and no preferred basis. Violating any of them changes the ensemble and can change the answer.

Induced eigenvalue measure

A Haar vector can be generated by drawing an $m\times n$ complex Gaussian matrix $G$ with independent entries and setting

$$\rho_A=\frac{GG^\dagger}{\operatorname{Tr}(GG^\dagger)}.$$

Unitary invariance makes the eigenvectors irrelevant to entropy. The normalized joint density of eigenvalues $\lambda_i$ on the simplex is

$$p(\lambda_1,\ldots,\lambda_m) =C_{m,n}\, \delta\!\left(1-\sum_{i=1}^m\lambda_i\right) \prod_{i=1}^m\lambda_i^{n-m} \prod_{i\lt j}(\lambda_i-\lambda_j)^2,$$

where $C_{m,n}$ is fixed by unit integral. The power $n-m$ comes from the rectangular Wishart matrix; the squared Vandermonde factor is eigenvalue repulsion for complex matrices.

Introduce the moment function

$$F(q)=\mathbb E\operatorname{Tr}\rho_A^q =\int\!p(\boldsymbol\lambda)\sum_{i=1}^{m}\lambda_i^q\,d^m\lambda.$$

Because $F(1)=1$ and the integral is differentiable near $q=1$ in finite dimensions,

$$\mathbb E S(\rho_A)=-F'(1).$$

Sen evaluates the simplex integral by converting it to a Laguerre-polynomial integral for the unnormalized Wishart eigenvalues. Differentiating the resulting gamma-function factors produces digamma functions $\psi(k+1)=H_k-\gamma$; the Euler constants cancel, leaving $H_{mn}-H_n-(m-1)/(2n)$. This is the proof architecture: induced measure, moment integral, orthogonal-polynomial evaluation, and differentiation at one. Harlow gives the associated Haar and black-hole presentation Harlow 2016, § 5.3 and Appendix D.

Large-dimension interpretation

Using

$$H_N=\ln N+\gamma+\frac{1}{2N}-\frac{1}{12N^2}+O(N^{-4}),$$

one obtains, for fixed $m$ and large $n$,

$$\mathbb E S(\rho_A) =\ln m-\frac{m^2-1}{2mn} +\frac{1-m^{-2}}{12n^2}+O(n^{-4}).$$

Thus the deficit from the maximum $\ln m$ is approximately $m/(2n)$ when both dimensions are large and $m\ll n$. At equal dimensions $m=n=d$,

$$\mathbb E S(\rho_A)=\ln d-\frac12+O(d^{-2}).$$

When $m/n\to0$, the additive entropy deficit tends to zero. At equal dimensions the deficit instead approaches one half nat: $S(\rho_A)/\ln d\to1$, but this relative entropy statement does not imply that $\rho_A$ converges to $I_A/d$ in trace norm. Concentration of measure makes a typical state’s entropy close to the relevant mean at large total dimension. The mean formula and the chosen notion of “nearly maximally mixed” must therefore be kept distinct.

Mean and typicality are logically separate statements. The exact harmonic-number formula computes the first moment of entropy. A concentration bound then uses the high real dimension of the unit sphere and the continuity of entropy away from very small eigenvalues to suppress large deviations. One should therefore report the mean formula as exact and any tail probability with its own norm and constants. For small $mn$, the distribution can remain broad even though the mean is known exactly; for large $mn$, a rare low-entanglement state is possible but occupies a small Haar-measure set.

The ordering $m\leq n$ is only a naming convention for the smaller factor. If the physical subsystem of interest has the larger dimension, purity gives the same nonzero Schmidt spectrum, so its entropy equals that of the smaller complement and the theorem is applied after exchanging $m$ and $n$.

The entropy deficit also controls a useful notion of closeness. Relative entropy from the maximally mixed state satisfies

$$D\!\left(\rho_A\middle\Vert\frac{I_A}{m}\right)=\ln m-S(\rho_A).$$

Quantum Pinsker’s inequality then gives

$$\left\lVert\rho_A-\frac{I_A}{m}\right\rVert_1 \leq\sqrt{2\bigl(\ln m-S(\rho_A)\bigr)}.$$

Thus a small entropy deficit for an individual state guarantees small trace distance. For the Haar ensemble, Jensen’s inequality gives the valid mean bound

$$\mathbb E\left\lVert\rho_A-\frac{I_A}{m}\right\rVert_1 \leq \sqrt{2\left(\ln m-\mathbb E S(\rho_A)\right)}.$$

This tends to zero when $m/n\to0$, turning “nearly maximally mixed” into a statement about every bounded measurement on $A$. At $m=n$, the right-hand side remains order one, so entropy typicality alone supplies no trace-norm convergence. None of these statements says that the correlations are easy to decode from $B$, because decoding complexity is not fixed by the spectrum of $\rho_A$.

Another normalization check follows from swapping the subsystem names. The larger subsystem has additional zero eigenvalues but the same nonzero Schmidt values, so its entropy is exactly the smaller subsystem entropy for each pure state, not merely after ensemble averaging. Any numerical sample that violates this equality has formed the partial trace or logarithm incorrectly.

Independent checks and counterensemble

The first exact check is $m=1$. The formula gives $H_n-H_n=0$, as required for a one-dimensional subsystem. For $m=n=2$ it gives

$$H_4-H_2-\frac14=\frac13$$

nats, below the maximum $\ln2$. One can check this case directly without using the harmonic-number formula. For $m=n=2$, the eigenvalues are $\lambda$ and $1-\lambda$, and the induced measure reduces to $p(\lambda)=3(2\lambda-1)^2$ on $0\leq\lambda\leq1$. It is normalized, and symmetry gives

$$\begin{aligned} \mathbb E S &=-6\int_0^1(2\lambda-1)^2\lambda\ln\lambda\,d\lambda\\ &=-6\left(-\frac14+\frac49-\frac14\right) =\frac13. \end{aligned}$$

Here $\int_0^1x^a\ln x\,dx=-1/(a+1)^2$ supplies an elementary independent check. A second Haar integral gives the average purity

$$\mathbb E\operatorname{Tr}\rho_A^2=\frac{m+n}{mn+1}.$$

For $n\gg m$, this approaches $1/m$, confirming near-maximal mixing without differentiating the entropy integral.

A decisive counterexample is the structured ensemble of product states $\lvert a\rangle\otimes\lvert b\rangle$, with each factor independently Haar random. It has the same Hilbert-space dimensions and unitary invariance within each factor, yet $S(A)=0$ for every draw. Energy-shell, charge-constrained, local-circuit, and integrable ensembles can likewise differ from global Haar measure. Dimension alone does not enforce Page behavior unless typicality in the relevant ensemble is justified.

Black-hole use and limitation

In an idealized evaporation model, one sets $m=\min(d_R,d_B)$ and $n=\max(d_R,d_B)$. As the smaller factor switches from radiation to remaining black hole, the theorem motivates a turnover. It does not derive $d_R(t)$ or $d_B(t)$, prove rapid mixing, impose energy conservation, or specify the radiation algebra in gravity. Generic Haar theory belongs to Quantum Information, the Hawking state to Quantum Fields in Curved Spacetime, and explicit microscopic string or holographic dynamics elsewhere.

Exercise

Evaluate the theorem for $(m,n)=(2,3)$ and compare the mean entropy with both its maximum and the purity benchmark.

Solution

The entropy is $H_6-H_3-1/6=(49/20)-(11/6)-1/6=9/20$ nats. The maximum is $\ln2\approx0.693$, so the mean has a finite-size deficit of about $0.243$. The mean purity is $(2+3)/(6+1)=5/7$, larger than the maximally mixed value $1/2$, consistently indicating residual spectral unevenness. Purity does not determine entropy by itself, but both checks point in the same direction.

References

• Harlow, Daniel. 2016. “Jerusalem Lectures on Black Holes and Quantum Information.” Reviews of Modern Physics 88: 015002. Open text.
• Page, Don N. 1993. “Average Entropy of a Subsystem.” Physical Review Letters 71: 1291–1294. Open text.
• Sen, Siddhartha. 1996. “Average Entropy of a Quantum Subsystem.” Physical Review Letters 77: 1–3. DOI. Open PDF, preprint titled “Average Entropy of a Subsystem”.