Rényi Entropies

Rényi entropies are a one-parameter family of entropy measures built from powers of a density operator. For a density operator $\rho$ and a parameter $\alpha>0$ with $\alpha\ne1$,

$$S_\alpha(\rho) = \frac{1}{1-\alpha} \log_2 \operatorname{Tr}(\rho^\alpha).$$

The limit $\alpha\to1$ gives the von Neumann entropy,

$$\lim_{\alpha\to1}S_\alpha(\rho) = S(\rho) = -\operatorname{Tr}(\rho\log_2\rho).$$

For a pure bipartite state, applying $S_\alpha$ to a reduced state describes how its Schmidt weights are distributed:

$$S_\alpha(A) = S_\alpha(\rho_A), \qquad \rho_A=\operatorname{Tr}_B\lvert\Psi\rangle\langle\Psi\rvert.$$

Rényi entropies are especially useful because $S_2$ is determined by the purity $\operatorname{Tr}(\rho^2)$, integer orders connect to replica methods, and different values of $\alpha$ probe different parts of the eigenvalue spectrum.

Helpful background. Subsystem entropy introduces spectral entropy; Schmidt decomposition turns a pure-state entanglement question into an eigenvalue calculation.

Rényi entropy from a finite spectrum

Unless stated otherwise, the Hilbert space is finite dimensional and logarithms are base two. These assumptions justify differentiating finite sums and make the limits below unambiguous.

Let $\rho$ have spectral decomposition

$$\rho = \sum_k p_k \lvert k\rangle\langle k\rvert, \qquad p_k\ge0, \qquad \sum_k p_k=1.$$

Then

$$\operatorname{Tr}(\rho^\alpha) = \sum_k p_k^\alpha,$$

so

$$S_\alpha(\rho) = \frac{1}{1-\alpha} \log_2 \left( \sum_k p_k^\alpha \right).$$

The logarithm base fixes the units. Base $2$ gives bits; natural logarithms give nats. This page states finite-dimensional examples in bits when numerical values are given.

Rényi entropy depends only on the eigenvalues of $\rho$. It is invariant under unitary conjugation:

$$S_\alpha(U\rho U^\dagger) = S_\alpha(\rho).$$

For a pure state, all Rényi entropies vanish:

$$S_\alpha(\lvert\psi\rangle\langle\psi\rvert)=0.$$

For a maximally mixed state $\rho_\ast=I/d$,

$$S_\alpha(\rho_\ast) = \log_2 d$$

for every $\alpha>0$.

Pure Bipartite Entanglement

Let a normalized bipartite pure state have Schmidt decomposition

$$\lvert\Psi\rangle = \sum_r s_r \lvert u_r\rangle_A \lvert v_r\rangle_B.$$

Define Schmidt probabilities

$$p_r=s_r^2.$$

The reduced density operator on $A$ has nonzero eigenvalues $p_r$, so

$$S_\alpha(A) = \frac{1}{1-\alpha} \log_2 \left( \sum_r p_r^\alpha \right).$$

The same value is obtained from $\rho_B$:

$$S_\alpha(A)=S_\alpha(B)$$

for pure bipartite states. They vanish exactly for product pure states and are invariant under local unitaries. All positive orders are nonincreasing under deterministic pure-to-pure LOCC, but the stronger requirement of nonincrease on average over recorded outcomes holds generally for $0<\alpha\le1$, not for every $\alpha>1$ (Vidal). The last exercise gives an explicit counterexample at order two.

The case $\alpha\to1$ is the usual Entanglement Entropy:

$$S_1(A) = -\sum_r p_r\log_2 p_r.$$

The case $\alpha\to0$ gives the logarithm of the Schmidt Rank:

$$S_0(A) = \log_2 \operatorname{SR}(\Psi),$$

provided the support is finite-dimensional.

The von Neumann limit

Let $f(\alpha)=\sum_{k:p_k>0}p_k^\alpha$. Then $f(1)=1$ and

$$f'(\alpha)=\sum_{k:p_k>0}p_k^\alpha\ln p_k.$$

L’Hôpital’s rule, with the base-two derivative factor retained, gives

$$\lim_{\alpha\to1}\frac{\log_2 f(\alpha)}{1-\alpha} =-\frac{f'(1)}{\ln2}=-\sum_k p_k\log_2p_k=S(\rho).$$

Zero eigenvalues contribute zero and are excluded from the differentiated sum. In infinite dimensions, convergence of the power sums and the interchange of limits with infinite sums need additional assumptions; a finite von Neumann entropy alone does not justify every Rényi limit.

Purity and Second Rényi Entropy

The second Rényi entropy is

$$S_2(\rho) = -\log_2 \operatorname{Tr}(\rho^2).$$

The quantity

$$\operatorname{Tr}(\rho^2)$$

is the purity. It equals $1$ for pure states and is less than $1$ for mixed states.

For a bipartite pure state, $S_2(\rho_A)$ is a convenient spectral diagnostic of entanglement; it is not generally an average LOCC monotone. In terms of Schmidt probabilities,

$$S_2(A) = -\log_2 \left( \sum_r p_r^2 \right).$$

For a Bell state, $p_1=p_2=1/2$, so

$$S_2(A) = -\log_2 \left( \frac14+\frac14 \right) = 1$$

bit.

For

$$\lvert\Psi_\theta\rangle = \cos\theta\,\lvert00\rangle + \sin\theta\,\lvert11\rangle,$$

the second Rényi entropy in bits is

$$S_2(A) = -\log_2 \bigl( \cos^4\theta+\sin^4\theta \bigr).$$

It vanishes at product endpoints and reaches one bit at $\theta=\pi/4$.

Other Useful Orders

The order $\alpha=0$ is the rank entropy:

$$S_0(\rho) = \log_2 \operatorname{rank}\rho,$$

where the rank is the number of nonzero eigenvalues. It is sensitive to arbitrarily small nonzero eigenvalues, so it is mathematically clean but not robust under small perturbations.

The limit $\alpha\to\infty$ gives the min-entropy:

$$S_\infty(\rho) = -\log_2 p_{\max},$$

where $p_{\max}$ is the largest eigenvalue of $\rho$.

For a fixed density operator, Rényi entropies are nonincreasing as $\alpha$ increases:

$$\alpha<\beta \quad \Longrightarrow \quad S_\alpha(\rho)\ge S_\beta(\rho).$$

Small $\alpha$ gives more weight to the number of populated eigenvalues. Large $\alpha$ emphasizes the largest eigenvalues.

The same two Schmidt probabilities give different spectral entropies. All three vanish for a product state and equal one bit at $p=1/2$; between those limits, larger Rényi order emphasizes the larger probability. The curves describe pure joint states when interpreted as entanglement.

Why purity can be measured with two copies

For the SWAP operator $F$ acting on two identical copies of a subsystem,

$$F|i\rangle|j\rangle=|j\rangle|i\rangle, \qquad \operatorname{Tr}[F(\rho\otimes\rho)]=\operatorname{Tr}(\rho^2).$$

To see the identity, insert a product basis: the trace is $\sum_{ij}\rho_{ij}\rho_{ji}$. Thus purity is a two-copy expectation value even though it is nonlinear in one copy of the state. Estimating $S_2$ requires estimating this expectation and then taking $-\log_2$; experimental uncertainty must be propagated through that nonlinear step. If the two copies differ, the same protocol measures $\operatorname{Tr}(\rho\sigma)$ instead.

Purity can also be inferred using suitable randomized-measurement protocols, each with its own preparation and sampling assumptions. Numerically, it avoids diagonalization when $\rho$ or an appropriate contraction is already available, but constructing the reduced state can remain expensive. For spatial scaling and tensor-network uses, see many-body entanglement entropy.

Replica Trick Preview

For an integer $n>1$,

$$S_n(\rho_A) = \frac{1}{1-n} \log_2 \operatorname{Tr}(\rho_A^n).$$

The quantity $\operatorname{Tr}(\rho_A^n)$ can sometimes be computed by considering $n$ copies of the system. In statistical mechanics and QFT, this idea leads to replica methods: compute for integer $n$, identify an appropriate continuation, and take the limit $n\to1$ to obtain von Neumann entropy. Integer values alone do not specify an arbitrary analytic continuation uniquely; the derivation must control the continuation and its regulator limits.

This is only a preview. In QFT, entanglement entropies of spatial regions can depend on regulators, boundary geometry, and continuum limits. The finite-dimensional Rényi formulas here are the conceptual entry point; Entanglement in QFT Preview explains the wider field-theory setting.

Mixed-State Caution

For a mixed joint state $\rho_{AB}$, the Rényi entropy of $\rho_A$ measures local mixedness, not entanglement by itself. The same warning that applies to von Neumann subsystem entropy applies to every $S_\alpha(\rho_A)$.

For example,

$$\rho_{\mathrm{cc}} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert$$

has $\rho_A=I/2$, so every Rényi entropy of $A$ is one bit. The state is separable. Its local entropy reflects classical correlation with $B$, not pure-state entanglement.

By contrast, for a Bell pure state, the same one-qubit reduced state $I/2$ arises from entanglement with the other qubit. The joint-state context determines the interpretation.

Comparing entropy orders

For Schmidt probabilities $(0.9,0.1)$,

$$S_0=1,\quad S_1\simeq0.469,\quad S_2=-\log_2(0.82)\simeq0.286,\quad S_\infty=-\log_2(0.9)\simeq0.152.$$

All values describe the same spectrum. Rank entropy counts both occupied modes equally, while larger orders emphasize the dominant weight. $S_2$ means the second Rényi entropy, not the square of von Neumann entropy.

References

• P. Calabrese and J. Cardy, “Entanglement entropy and quantum field theory”, Journal of Statistical Mechanics: Theory and Experiment P06002 (2004), DOI: 10.1088/1742-5468/2004/06/P06002.
• A. Rényi, “On measures of entropy and information,” in Proceedings of the Fourth Berkeley Symposium on Mathematical Statistics and Probability, Vol. 1, University of California Press (1961), pp. 547–561.
• G. Vidal, “Entanglement monotones”, Journal of Modern Optics 47, 355–376 (2000).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018).
• A. Wehrl, “General properties of entropy”, Reviews of Modern Physics 50, 221–260 (1978).

Exercises

1. Compute $S_\alpha$ for a pure state.

Solution

A pure state has eigenvalues $1,0,0,\ldots$. Therefore

$$\operatorname{Tr}(\rho^\alpha) = 1^\alpha = 1,$$

so

$$S_\alpha(\rho) = \frac{1}{1-\alpha}\log_21 = 0.$$

2. Compute $S_\alpha$ for the maximally mixed state $I/d$.

Solution

The eigenvalues are all $1/d$. Hence

$$\operatorname{Tr}(\rho^\alpha) = d\left(\frac1d\right)^\alpha = d^{1-\alpha}.$$

Thus

$$S_\alpha(\rho) = \frac{1}{1-\alpha}\log_2 d^{1-\alpha} = \log_2 d.$$

3. Show that SWAP measured on two different diagonal states $\rho=\operatorname{diag}(p,1-p)$ and $\sigma=\operatorname{diag}(q,1-q)$ yields $pq+(1-p)(1-q)$. When does this equal the purity of $\rho$?

Solution

The SWAP expectation is $\operatorname{Tr}(\rho\sigma)=pq+(1-p)(1-q)$. Subtracting $\operatorname{Tr}\rho^2$ gives $(q-p)(2p-1)$, so equality holds for $q=p$ or $p=1/2$. Agreement in a special case does not verify identical preparation.

4. A pure two-qutrit state has Schmidt probabilities $(0.9,0.05,0.05)$. Alice applies Kraus operators $K_0=\operatorname{diag}(\sqrt5/3,0,0)$ and $K_1=\operatorname{diag}(2/3,1,1)$. Verify completeness and show that the average second Rényi entanglement increases slightly.

Solution

$K_0^\dagger K_0+K_1^\dagger K_1=I$. Both branches have probability $1/2$. Their normalized Schmidt probabilities are $(1,0,0)$ and $(0.8,0.1,0.1)$. Consequently

$$S_2^{\rm in}=-\log_2(0.815)\simeq0.295128, \qquad \sum_m q_m S_2^{(m)}=-\tfrac12\log_2(0.66)\simeq0.299731.$$

The increase shows why the order-two entropy is not a general average entanglement monotone. There is no contradiction with von Neumann entropy monotonicity or with deterministic pure-to-pure LOCC restrictions: this protocol has distinct recorded branches.

5. For Schmidt probabilities $p,1-p$, write $S_2$ in bits.

Solution

The second Rényi entropy is

$$S_2 = -\log_2 \bigl( p^2+(1-p)^2 \bigr).$$

It is zero at $p=0$ and $p=1$, and equals one bit at $p=1/2$.