Fidelity, Chernoff Bounds, and State Overlap

Fidelity measures how closely two states overlap; the quantum Chernoff exponent gives the optimal exponential rate for symmetric iid discrimination. They are related, but they are not the same quantity and do not have the same domain. In QFT, a numerical overlap is meaningful only after the observable algebra, fidelity convention, mode or split realization, copy model, and high-energy control have been declared.

Required background. Start from Relative Entropy for QFT States. Helpful background. Hypothesis Testing and Asymptotic Distinguishability fixes the error model and copy resource behind asymptotic exponents.

The chapter’s task-comparison table, structural map, and validity map keep overlap claims separate from asymmetric relative entropy and constrained channel norms.

Algebraic transition probability and root fidelity

This page uses the root fidelity

$$F(\rho,\sigma) =\left\lVert\sqrt\rho\sqrt\sigma\right\rVert_1 =\operatorname{Tr} \sqrt{\sqrt\rho\,\sigma\sqrt\rho}, \qquad 0\leq F\leq1.$$

The corresponding Uhlmann transition probability is

$$P_{\rm U}(\rho,\sigma)=F(\rho,\sigma)^2.$$

Both quantities are often called “fidelity.” A bare number such as $F=0.8$ is therefore ambiguous until “root” or “squared” has been stated. Uhlmann’s intrinsic definition starts from states on one common $C^*$-algebra and maximizes the squared overlap of vector representatives over common representations; see Uhlmann 1976, Eq. (4), p. 274 and Eq. (23), p. 277. On a type-I algebra it reduces to $P_{\rm U}=F^2$ with the trace formula above.

This distinction matters in continuum QFT. Normal states on a type-III local algebra have an intrinsic transition probability, but no local density matrix and no ambient trace. A lattice cutoff, finite-mode algebra, or split inclusion may supply density operators for a controlled approximation; it does not become the intrinsic definition merely because its formula is familiar.

Two metric conventions used later are

$$d_{\rm B}(\rho,\sigma)=\sqrt{2(1-F)}, \qquad P(\rho,\sigma)=\sqrt{1-F^2},$$

the Bures distance and purified distance for normalized states. Under a quantum channel or restriction to a smaller algebra, fidelity cannot decrease. Losing observables can make states look more alike.

Trace distance and discrimination bounds

For density operators, define

$$T(\rho,\sigma)=\frac12\lVert\rho-\sigma\rVert_1.$$

With equal priors, the optimal one-copy error and success probabilities are

$$p_e^*=\frac12(1-T), \qquad p_{\rm succ}^*=\frac12(1+T).$$

The Fuchs–van de Graaf inequalities in the root-fidelity convention are

$$1-F\leq T\leq\sqrt{1-F^2}.$$

They turn an overlap into certified, but generally nonexact, discrimination bounds. For $n$ iid copies, multiplicativity $F(\rho^{\otimes n},\sigma^{\otimes n})=F(\rho,\sigma)^n$ gives

$$\frac12\left(1-\sqrt{1-F^{2n}}\right) \leq p_{e,n}^* \leq\frac12F^n.$$

The inequalities and their operational interpretation are Fuchs and van de Graaf 1999, Theorem 1 and Eq. (46), pp. 1222–1223. If both states are pure, the upper trace-distance inequality is saturated and the lower bound on $p_{e,n}^*$ is exact.

Fidelity remains finite without support inclusion, whereas relative entropy is directional and may diverge. For example,

$$\rho=|0\rangle\langle0|, \qquad \sigma=(1-\delta)|0\rangle\langle0| +\delta|1\rangle\langle1|$$

has $F=\sqrt{1-\delta}$ and $D(\rho\Vert\sigma)=-\log(1-\delta)$, but $D(\sigma\Vert\rho)=+\infty$. Symmetric overlap cannot replace the support-sensitive information in either relative-entropy direction.

Chernoff overlap and the iid theorem

For finite-dimensional density operators and $0<s<1$, set

$$Q_s(\rho,\sigma)=\operatorname{Tr}\rho^s\sigma^{1-s}, \qquad Q(\rho,\sigma)=\inf_{0<s<1}Q_s(\rho,\sigma),$$

and

$$\xi_{\rm QCB}(\rho,\sigma)=-\log Q(\rho,\sigma).$$

If endpoints are included, $A^0$ must be defined as the support projection of $A$ and singular endpoint limits must be handled explicitly. Writing $0<s<1$ avoids hiding that convention.

Let the priors $\pi_0,\pi_1$ be fixed and nonzero, and let $p_{e,n}^*$ be the minimum Bayes error for discriminating $\rho^{\otimes n}$ from $\sigma^{\otimes n}$ using unrestricted collective measurements. The theorem is the logarithmic limit

$$\lim_{n\to\infty} -\frac1n\log p_{e,n}^* =\xi_{\rm QCB}(\rho,\sigma).$$

It does not assert a constant-factor equivalence $p_{e,n}^*\asymp e^{-n\xi_{\rm QCB}}$. The achievability inequality is due to Audenaert et al. 2007, Eqs. (1)–(4) and Theorem 1; the matching converse is Nussbaum and Szkoła 2009, Theorem 2.2 and Eq. (9), pp. 1046–1047.

The midpoint overlap is not generally fidelity:

$$Q_{1/2} =\operatorname{Tr}\sqrt\rho\sqrt\sigma \leq \left\lVert\sqrt\rho\sqrt\sigma\right\rVert_1 =F.$$

Equality holds for commuting states, but not for arbitrary mixed states, and the minimizing $s$ need not equal $1/2$. If both states are pure, then $Q_s=F^2$ throughout the open interval. If only $\rho=|\psi\rangle\langle\psi|$ is pure, then

$$Q_s=\langle\psi|\sigma^{1-s}|\psi\rangle,$$

which generally depends on $s$, but decreases to $F^2=\langle\psi|\sigma|\psi\rangle$ as $s\downarrow0$; the analogous endpoint statement holds when only $\sigma$ is pure. Thus whenever at least one state is pure, the infimum is $Q=F^2$ and

$$\xi_{\rm QCB}=-\log F^2=-2\log F.$$

This is precisely where confusing $F$ with $F^2$ produces a factor-of-two error.

A displaced thermal wavepacket mode

Use the same regulated wavepacket mode as on the hypothesis-testing page. Let

$$\tau_{\beta,\omega} =(1-e^{-\beta\omega})e^{-\beta\omega a^\dagger a}, \qquad \bar n=\frac1{e^{\beta\omega}-1},$$

and compare equal-covariance Gaussian states

$$\rho_j=D(\alpha_j)\tau_{\beta,\omega}D(\alpha_j)^\dagger, \qquad \Delta\alpha=\alpha_1-\alpha_0.$$

The following formulas define the trace-class overlaps of the oscillator states exactly. The finite-dimensional Chernoff theorem above applies after also freezing an occupation cutoff $M$ and normalizing the projected states. One must then certify convergence as $M$ grows; identifying the untruncated oscillator’s operational exponent requires an applicable infinite-dimensional theorem or a controlled order of limits.

With $[q,p]=i$ and vacuum covariance $I/2$, the root fidelity is

$$F(\rho_1,\rho_0) =\exp\left[ -\frac{|\Delta\alpha|^2}{2(2\bar n+1)} \right].$$

This follows directly from the displacement term in the general Gaussian fidelity formula Banchi, Braunstein, and Pirandola 2015, Eqs. (8)–(9), pp. 260501-2–260501-3.

The entire Chernoff family can also be evaluated. Write $r=e^{-\beta\omega}=\bar n/(1+\bar n)$. A power of the thermal state is proportional to another thermal state:

$$\tau_{\beta,\omega}^{\,s} =\frac{(1-r)^s}{1-r^s}\, \tau_{s\beta,\omega}.$$

Taking the Hilbert–Schmidt overlap of the two displaced thermal factors gives

$$Q_s =\exp\left[ -|\Delta\alpha|^2 \frac{(1-r^s)(1-r^{1-s})}{1-r} \right].$$

The exponent is symmetric under $s\leftrightarrow1-s$ and is largest at $s=1/2$, so $Q_s$ is minimized there:

$$\begin{aligned} Q &=\exp\left[ -|\Delta\alpha|^2\tanh\frac{\beta\omega}{4} \right],\\ \xi_{\rm QCB} &=|\Delta\alpha|^2\tanh\frac{\beta\omega}{4}. \end{aligned}$$

The $s$-overlap calculation, including general Gaussian states, is developed in Calsamiglia et al. 2008, § VII.A, Eqs. (88)–(94), pp. 032311-12–032311-13.

For the shared benchmark, $\bar n=1/2$, $|\Delta\alpha|^2=1$, and $\beta\omega=\log3$. Therefore

$$\begin{aligned} F&=e^{-1/4}\simeq0.778800783071,\\ P_{\rm U}=F^2&=e^{-1/2}\simeq0.606530659713,\\ Q&=e^{-(2-\sqrt3)} \simeq0.764946645195,\\ \xi_{\rm QCB}&=2-\sqrt3 \simeq0.267949192431. \end{aligned}$$

The strict ordering $Q<F$ is expected: this mixed pair does not identify the Chernoff coefficient with fidelity. The machine-readable benchmark records the exact formulas, both fidelity conventions, and numerical values.

Exact finite-copy control at zero temperature

The limit $\bar n\to0$ turns the same pair into the vacuum $|0\rangle$ and coherent state $|\Delta\alpha\rangle$. For $|\Delta\alpha|^2=1$,

$$F=e^{-1/2}, \qquad Q=F^2=e^{-1}, \qquad \xi_{\rm QCB}=1.$$

Both hypotheses are pure, so the $n$-copy Helstrom error is exact:

$$p_{e,n}^* =\frac12\left(1-\sqrt{1-e^{-n}}\right).$$

Copies $n$	$F^n$	$T_n=\sqrt{1-F^{2n}}$	Exact $p_{e,n}^*$
1	0.606530660	0.795060098	0.102469951
2	0.367879441	0.929873495	0.0350632525
5	0.082084999	0.996625332	0.00168733385
10	0.006737947	0.999977300	$1.13501113\times10^{-5}$

For the mixed $\bar n=1/2$ pair, $F$ gives rigorous finite-copy bounds and $Q$ is the exact trace-class Chernoff coefficient. At each fixed, certified occupation cutoff, the corresponding $Q_M$ gives the optimal iid exponent. Neither quantity should be mislabeled as the exact one-copy Helstrom trace norm, and passing from the cutoff exponents to an untruncated oscillator exponent requires a separate theorem or a controlled double limit.

Removing a mode regulator

For a product of regulated normal modes with common temperatures and displacement differences $\Delta\alpha_k$,

$$-\log F_K =\frac12\sum_{k\leq K} \frac{|\Delta\alpha_k|^2}{2\bar n_k+1},$$

while the exact finite-mode coefficient obeys

$$-\log Q_K =\sum_{k\leq K} |\Delta\alpha_k|^2 \tanh\frac{\beta\omega_k}{4}.$$

The sums make the ultraviolet question explicit. A nonzero continuum fidelity needs convergence of the first series. A finite limiting Chernoff coefficient needs convergence of the second. If either sum diverges, the regulated overlap can tend to zero even though every fixed-$K$ calculation is finite. Turning $-\log Q_K$ into an operational many-copy exponent also requires a justified occupation-cutoff or infinite-dimensional theorem before taking $K\to\infty$. Correlated modes, inequivalent representations, or a nonproduct state require a new analysis rather than these factorized formulas.

Local and global overlaps can also differ maximally. The orthogonal Bell states $|\Phi^\pm\rangle=(|00\rangle\pm|11\rangle)/\sqrt2$ have global fidelity zero, but both restrict to $I/2$ on either qubit and therefore have local fidelity one. In QFT, changing from a global algebra to a detector algebra is equally consequential.

For operator-algebraic overlap and noncommutative divergence results, continue to the rigorous mathematical treatment. For energy-constrained channel inputs rather than state pairs, continue to Operational Distinguishability and Continuity Bounds.

Common pitfalls

Leaving fidelity unsquared or squared unnamed. Translate every source to one convention before inserting it into a metric, error bound, or exponent.

Calling $Q_{1/2}$ the fidelity. The trace and trace norm coincide for commuting states, not in general. The Chernoff optimizer can also lie away from $1/2$.

Reading the exponent as an exact error. $\xi_{\rm QCB}$ controls the logarithmic iid limit under collective measurements. It does not by itself determine the prefactor or the one-copy Helstrom error.

Turning spatial separation into iid copies. Correlated field restrictions do not obey $Q_s(\rho^{\otimes n},\sigma^{\otimes n})=Q_s(\rho,\sigma)^n$. The tensor-product preparation must be physical and explicit.

Exercises

1. Translate the four overlap conventions

For pure states $|\psi\rangle$ and $|\phi\rangle$, express $F$, $P_{\rm U}$, $d_{\rm B}$, and purified distance $P$ in terms of $c=|\langle\psi|\phi\rangle|$. Which convention enters the pure-state Chernoff exponent?

Solution

For rank-one states,

$$F=c, \qquad P_{\rm U}=c^2, \qquad d_{\rm B}=\sqrt{2(1-c)}, \qquad P=\sqrt{1-c^2}.$$

The Chernoff coefficient is $Q=c^2=P_{\rm U}$, so $\xi_{\rm QCB}=-\log c^2=-2\log F$. Using the root fidelity directly inside the logarithm would miss the factor of two.

2. Minimize the thermal Chernoff overlap

Let $0<r<1$. Show that

$$g(s)=(1-r^s)(1-r^{1-s})$$

is maximized at $s=1/2$, and derive $\xi_{\rm QCB}=|\Delta\alpha|^2(1-\sqrt r)/(1+\sqrt r)$.

Solution

Expanding,

$$g(s)=1+r-r^s-r^{1-s}.$$

The arithmetic–geometric mean inequality gives $r^s+r^{1-s}\geq2\sqrt r$, with equality exactly at $s=1/2$. Hence $g(s)\leq(1-\sqrt r)^2$. Dividing by $1-r=(1-\sqrt r)(1+\sqrt r)$ gives

$$\frac{g(1/2)}{1-r} =\frac{1-\sqrt r}{1+\sqrt r}.$$

Substitution into $-\log Q_{1/2}$ yields the stated exponent.

3. Check the finite-copy coherent benchmark

For the vacuum and a coherent state with $|\Delta\alpha|^2=1$, derive the trace distance and exact equal-prior error for $n$ copies. Verify the $n=2$ table row.

Solution

The one-copy root fidelity is $|\langle0|\Delta\alpha\rangle|=e^{-1/2}$. Tensor products give $F_n=e^{-n/2}$. The trace distance between two pure states saturates the upper Fuchs–van de Graaf inequality:

$$T_n=\sqrt{1-F_n^2}=\sqrt{1-e^{-n}}.$$

Thus

$$p_{e,n}^*=\frac12(1-T_n).$$

At $n=2$, $F_2=e^{-1}\simeq0.367879441$, $T_2=\sqrt{1-e^{-2}}\simeq0.929873495$, and $p_{e,2}^*\simeq0.0350632525$.

4. Restriction can erase global distinguishability

Compute the fidelity of $|\Phi^+\rangle$ and $|\Phi^-\rangle$ globally and after tracing out either qubit. Explain the direction of fidelity monotonicity.

Solution

The two Bell vectors are orthogonal, so their global root fidelity is zero. Tracing out either qubit gives $I/2$ for both states, whose fidelity is one. Partial trace is a quantum channel, and fidelity is monotone upward under channels:

$$F(\mathcal N(\rho),\mathcal N(\sigma)) \geq F(\rho,\sigma).$$

Discarding observables cannot reveal a distinction that the full algebra failed to see; it can only preserve or erase distinctions.

References

• Audenaert, Koenraad M. R., et al. “Discriminating States: The Quantum Chernoff Bound.” Physical Review Letters 98 (2007): 160501. DOI. Open preprint.
• Banchi, Leonardo, Samuel L. Braunstein, and Stefano Pirandola. “Quantum Fidelity for Arbitrary Gaussian States.” Physical Review Letters 115 (2015): 260501. DOI. Open preprint.
• Calsamiglia, John, Ramon Muñoz-Tapia, Lluís Masanes, Antonio Acín, and Emilio Bagan. “Quantum Chernoff Bound as a Measure of Distinguishability between Density Matrices: Application to Qubit and Gaussian States.” Physical Review A 77 (2008): 032311. DOI. Open preprint.
• Fuchs, Christopher A., and Jeroen van de Graaf. “Cryptographic Distinguishability Measures for Quantum-Mechanical States.” IEEE Transactions on Information Theory 45, no. 4 (1999): 1216–1227. DOI. Open preprint.
• Nussbaum, Michael, and Arleta Szkoła. “The Chernoff Lower Bound for Symmetric Quantum Hypothesis Testing.” The Annals of Statistics 37, no. 2 (2009): 1040–1057. DOI. Open PDF.
• Uhlmann, Armin. “The ‘Transition Probability’ in the State Space of a *-Algebra.” Reports on Mathematical Physics 9, no. 2 (1976): 273–279. DOI. Open PDF.