fidelity of quantum states

The fidelity between two quantum states $\backslash rho$ and $\backslash sigma$, expressed as density matrices, is commonly defined as:R. Jozsa, Fidelity for Mixed Quantum States, J. Mod. Opt. 41, 2315--2323 (1994). DOI: http://doi.org/10.1080/09500349414552171{{cite book |last1=Nielsen |first1=Michael A. |url=http://www.michaelnielsen.org/qcqi/ |title=Quantum Computation and Quantum Information |last2=Chuang |first2=Isaac L. |publisher=Cambridge University Press |year=2000 |isbn=978-0521635035 |doi=10.1017/CBO9780511976667}}

:$F(\backslash rho,\; \backslash sigma)\; =\; \backslash left(\backslash operatorname\{tr\}\; \backslash sqrt\{\backslash sqrt\{\backslash rho\}\; \backslash sigma\; \backslash sqrt\{\backslash rho\}\}\backslash right)^2.$

The square roots in this expression are well-defined because both $\backslash rho$ and $\backslash sqrt\backslash rho\backslash sigma\backslash sqrt\backslash rho$ are positive semidefinite matrices, and the square root of a positive semidefinite matrix is defined via the spectral theorem. The Euclidean inner product from the classical definition is replaced by the Hilbert–Schmidt inner product.

As will be discussed in the following sections, this expression can be simplified in various cases of interest. In particular, for pure states, $\backslash rho=|\backslash psi\_\backslash rho\backslash rangle\backslash !\backslash langle\backslash psi\_\backslash rho|$ and $\backslash sigma=|\backslash psi\_\backslash sigma\backslash rangle\backslash !\backslash langle\backslash psi\_\backslash sigma|$, it equals:$$F(\backslash rho,\; \backslash sigma)=|\backslash langle\backslash psi\_\backslash rho|\backslash psi\_\backslash sigma\backslash rangle|^2.$$This tells us that the fidelity between pure states has a straightforward interpretation in terms of probability of finding the state $|\backslash psi\_\backslash rho\backslash rangle$ when measuring $|\backslash psi\_\backslash sigma\backslash rangle$ in a basis containing $|\backslash psi\_\backslash rho\backslash rangle$.

Some authors use an alternative definition $F\text{'}:=\backslash sqrt\{F\}$ and call this quantity fidelity. The definition of $F$ however is more common.{{cite book |last=Bengtsson |first=Ingemar |title=Geometry of Quantum States: An Introduction to Quantum Entanglement |title-link=Geometry of Quantum States |publisher=Cambridge University Press |year=2017 |isbn=978-1-107-02625-4 |location=Cambridge, United Kingdom New York, NY}}{{cite book |last1=Walls |first1=D. F. |title=Quantum Optics |last2=Milburn |first2=G. J. |publisher=Springer |year=2008 |isbn=978-3-540-28573-1 |location=Berlin}}{{cite book |last=Jaeger |first=Gregg |title=Quantum Information: An Overview |publisher=Springer |year=2007 |isbn=978-0-387-35725-6 |location=New York London}} To avoid confusion, $F\text{'}$ could be called "square root fidelity". In any case it is advisable to clarify the adopted definition whenever the fidelity is employed.

Given two random variables $X,Y$ with values $(1,\; ...,\; n)$ (categorical random variables) and probabilities $p\; =\; (p\_1,p\_2,\backslash ldots,p\_n)$ and $q\; =\; (q\_1,q\_2,\backslash ldots,q\_n)$, the fidelity of $X$ and $Y$ is defined to be the quantity

:$F(X,Y)\; =\; \backslash left(\backslash sum\; \_i\; \backslash sqrt\{p\_i\; q\_i\}\backslash right)^2$.

The fidelity deals with the marginal distribution of the random variables. It says nothing about the joint distribution of those variables. In other words, the fidelity $F(X,Y)$ is the square of the inner product of $(\backslash sqrt\{p\_1\},\; \backslash ldots\; ,\backslash sqrt\{p\_n\})$ and $(\backslash sqrt\{q\_1\},\; \backslash ldots\; ,\backslash sqrt\{q\_n\})$ viewed as vectors in Euclidean space. Notice that $F(X,Y)\; =\; 1$ if and only if $p\; =\; q$. In general, $0\; \backslash leq\; F(X,Y)\; \backslash leq\; 1$. The measure $\backslash sum\; \_i\; \backslash sqrt\{p\_i\; q\_i\}$ is known as the Bhattacharyya coefficient.

Given a classical measure of the distinguishability of two probability distributions, one can motivate a measure of distinguishability of two quantum states as follows: if an experimenter is attempting to determine whether a quantum state is either of two possibilities $\backslash rho$ or $\backslash sigma$, the most general possible measurement they can make on the state is a POVM, which is described by a set of Hermitian positive semidefinite operators $\backslash \{F\_i\backslash \}$. When measuring a state $\backslash rho$ with this POVM, $i$-th outcome is found with probability $p\_i\; =\; \backslash operatorname\{tr\}(\; \backslash rho\; F\_i\; )$, and likewise with probability $q\_i\; =\; \backslash operatorname\{tr\}(\; \backslash sigma\; F\_i\; )$ for $\backslash sigma$. The ability to distinguish between $\backslash rho$ and $\backslash sigma$ is then equivalent to their ability to distinguish between the classical probability distributions $p$ and $q$. A natural question is then to ask what is the POVM the makes the two distributions as distinguishable as possible, which in this context means to minimize the Bhattacharyya coefficient over the possible choices of POVM. Formally, we are thus led to define the fidelity between quantum states as:

:$F(\backslash rho,\backslash sigma)\; =\; \backslash min\_\{\backslash \{F\_i\backslash \}\}\; F(X,Y)\; =\; \backslash min\_\{\backslash \{F\_i\backslash \}\}\; \backslash left(\backslash sum\; \_i\; \backslash sqrt\{\backslash operatorname\{tr\}(\; \backslash rho\; F\_i\; )\; \backslash operatorname\{tr\}(\; \backslash sigma\; F\_i\; )\}\backslash right)^\{2\}.$

It was shown by Fuchs and CavesC. A. Fuchs, C. M. Caves: [http://prl.aps.org/abstract/PRL/v73/i23/p3047_1 "Ensemble-Dependent Bounds for Accessible Information in Quantum Mechanics"], Physical Review Letters 73, 3047(1994) that the minimization in this expression can be computed explicitly, with solution the projective POVM corresponding to measuring in the eigenbasis of $\backslash sigma^\{-1/2\}|\backslash sqrt\backslash sigma\backslash sqrt\backslash rho|\backslash sigma^\{-1/2\}$, and results in the common explicit expression for the fidelity as$$F(\backslash rho,\; \backslash sigma)\; =\; \backslash left(\backslash operatorname\{tr\}\; \backslash sqrt\{\backslash sqrt\{\backslash rho\}\; \backslash sigma\; \backslash sqrt\{\backslash rho\}\}\backslash right)^2.$$

An equivalent expression for the fidelity between arbitrary states via the trace norm is:

:$F(\backslash rho,\; \backslash sigma)=\; \backslash lVert\; \backslash sqrt\{\backslash rho\}\; \backslash sqrt\{\backslash sigma\}\; \backslash rVert\_\backslash operatorname\{tr\}^2\; =\; \backslash left(\backslash operatorname\{tr\}|\backslash sqrt\backslash rho\backslash sqrt\backslash sigma|\backslash right)^2,$

where the absolute value of an operator is here defined as $|A|\backslash equiv\; \backslash sqrt\{A^\backslash dagger\; A\}$.

Since the trace of a matrix is equal to the sum of its eigenvalues

:$F(\backslash rho,\; \backslash sigma)=\; \backslash sum\_j\backslash sqrt\{\backslash lambda\_j\},$

where the $\backslash lambda\_j$ are the eigenvalues of $\backslash sqrt\{\backslash rho\}\; \backslash sigma\; \backslash sqrt\{\backslash rho\}$, which is positive semidefinite by construction and so the square roots of the eigenvalues are well defined. Because the characteristic polynomial of a product of two matrices is independent of the order, the spectrum of a matrix product is invariant under cyclic permutation, and so these eigenvalues can instead be calculated from $\backslash rho\backslash sigma$.{{cite journal |last1=Baldwin |first1=Andrew J. |last2=Jones |first2=Jonathan A. |title=Efficiently computing the Uhlmann fidelity for density matrices |journal=Physical Review A |date=2023 |volume=107 |page=012427 |doi=10.1103/PhysRevA.107.012427|arxiv=2211.02623 }} Reversing the trace property leads to

:$F(\backslash rho,\; \backslash sigma)=\; \backslash left(\backslash operatorname\{tr\}\backslash sqrt\{\backslash rho\backslash sigma\}\backslash right)^2$.

If (at least) one of the two states is pure, for example $\backslash rho=|\backslash psi\_\backslash rho\backslash rangle\backslash !\backslash langle\backslash psi\_\backslash rho|$, the fidelity simplifies to$$F(\backslash rho,\backslash sigma)=\backslash operatorname\{tr\}(\backslash sigma\backslash rho)=\backslash langle\; \backslash psi\_\backslash rho|\backslash sigma|\backslash psi\_\backslash rho\backslash rangle.$$This follows observing that if $\backslash rho$ is pure then $\backslash sqrt\backslash rho=\backslash rho$, and thus$$$$

F(\rho, \sigma) = \left(\operatorname{tr} \sqrt{ | \psi_\rho \rangle \langle \psi_\rho | \sigma | \psi_\rho \rangle \langle \psi_\rho |} \right)^2

= \langle \psi_\rho | \sigma | \psi_\rho \rangle \left(\operatorname{tr} \sqrt{ | \psi_\rho \rangle \langle \psi_\rho |} \right)^2

= \langle \psi_\rho | \sigma | \psi_\rho \rangle.

If both states are pure, $\backslash rho=|\backslash psi\_\backslash rho\backslash rangle\backslash !\backslash langle\backslash psi\_\backslash rho|$ and $\backslash sigma=|\backslash psi\_\backslash sigma\backslash rangle\backslash !\backslash langle\backslash psi\_\backslash sigma|$, then we get the even simpler expression:$$F(\backslash rho,\; \backslash sigma)=|\backslash langle\backslash psi\_\backslash rho|\backslash psi\_\backslash sigma\backslash rangle|^2.$$

Some of the important properties of the quantum state fidelity are:

• Symmetry. $F(\backslash rho,\backslash sigma)=F(\backslash sigma,\backslash rho)$.
• Bounded values. For any $\backslash rho$ and $\backslash sigma$, $0\backslash le\; F(\backslash rho,\backslash sigma)\; \backslash le\; 1$, and $F(\backslash rho,\backslash rho)=1$.
• Consistency with fidelity between probability distributions. If $\backslash rho$ and $\backslash sigma$ commute, the definition simplifies to $$F(\backslash rho,\backslash sigma)\; =$$

\left[\operatorname{tr}\sqrt{\rho\sigma}\right]^2 =

\left(\sum_k \sqrt{p_k q_k} \right)^2 =

F(\boldsymbol p, \boldsymbol q),where $p\_k,\; q\_k$ are the eigenvalues of $\backslash rho,\backslash sigma$, respectively. To see this, remember that if $[\backslash rho,\backslash sigma]=0$ then they can be diagonalized in the same basis: $$\backslash rho\; =\; \backslash sum\_i\; p\_i\; |\; i\; \backslash rangle\; \backslash langle\; i\; |$$

\text{ and }

\sigma = \sum_i q_i | i \rangle \langle i |,so that $\backslash operatorname\{tr\}\backslash sqrt\{\backslash rho\backslash sigma\}\; =$

\operatorname{tr}\left(\sum_k \sqrt{p_k q_k} |k\rangle\!\langle k|\right) =

\sum_k \sqrt{p_k q_k}.

• Explicit expression for qubits.

If $\backslash rho$ and $\backslash sigma$ are both qubit states, the fidelity can be computed as

M. Hübner, Explicit Computation of the Bures Distance for Density Matrices, Phys. Lett. A 163, 239--242 (1992). DOI: https://doi.org/10.1016/0375-9601%2892%2991004-B

:$F(\backslash rho,\; \backslash sigma)\; =\; \backslash operatorname\{tr\}(\backslash rho\backslash sigma)+2\backslash sqrt\{\backslash det(\backslash rho)\backslash det(\backslash sigma)\}.$

Qubit state means that $\backslash rho$ and $\backslash sigma$ are represented by two-dimensional matrices. This result follows noticing that $M=\backslash sqrt\{\backslash rho\}\backslash sigma\backslash sqrt\{\backslash rho\}$ is a positive semidefinite operator, hence $\backslash operatorname\{tr\}\backslash sqrt\{M\}=\backslash sqrt\{\backslash lambda\_1\}+\backslash sqrt\{\backslash lambda\_2\}$, where $\backslash lambda\_1$ and $\backslash lambda\_2$ are the (nonnegative) eigenvalues of $M$. If $\backslash rho$ (or $\backslash sigma$) is pure, this result is simplified further to $F(\backslash rho,\backslash sigma)\; =\; \backslash operatorname\{tr\}(\backslash rho\backslash sigma)$ since $\backslash mathrm\{Det\}(\backslash rho)\; =\; 0$ for pure states.

Direct calculation shows that the fidelity is preserved by unitary evolution, i.e.

:$\backslash ;\; F(\backslash rho,\; \backslash sigma)\; =\; F(U\; \backslash rho\; \backslash ;\; U^*,\; U\; \backslash sigma\; U^*)$

for any unitary operator $U$.

Let $\backslash \{E\_k\backslash \}\_k$ be an arbitrary positive operator-valued measure (POVM); that is, a set of positive semidefinite operators $E\_k$ satisfying $\backslash sum\_k\; E\_k=I$. Then, for any pair of states $\backslash rho$ and $\backslash sigma$, we have

$$$$

\sqrt{F(\rho,\sigma)} \le \sum_k \sqrt{\operatorname{tr}(E_k\rho)}\sqrt{\operatorname{tr}(E_k\sigma)} \equiv \sum_k \sqrt{p_k q_k},

where in the last step we denoted with $p\_k\; \backslash equiv\; \backslash operatorname\{tr\}(E\_k\; \backslash rho)$ and $q\_k\; \backslash equiv\; \backslash operatorname\{tr\}(E\_k\; \backslash sigma)$ the probability distributions obtained by measuring $\backslash rho,\backslash \; \backslash sigma$ with the POVM $\backslash \{E\_k\backslash \}\_k$.

This shows that the square root of the fidelity between two quantum states is upper bounded by the Bhattacharyya coefficient between the corresponding probability distributions in any possible POVM. Indeed, it is more generally true that $$F(\backslash rho,\backslash sigma)=\backslash min\_\{\backslash \{E\_k\backslash \}\}\; F(\backslash boldsymbol\; p,\backslash boldsymbol\; q),$$ where $F(\backslash boldsymbol\; p,\; \backslash boldsymbol\; q)\backslash equiv\backslash left(\backslash sum\_k\backslash sqrt\{p\_k\; q\_k\}\backslash right)^2$, and the minimum is taken over all possible POVMs. More specifically, one can prove that the minimum is achieved by the projective POVM corresponding to measuring in the eigenbasis of the operator $\backslash sigma^\{-1/2\}|\backslash sqrt\backslash sigma\backslash sqrt\backslash rho|\backslash sigma^\{-1/2\}$.{{Cite book |last=Watrous |first=John |url=http://dx.doi.org/10.1017/9781316848142 |title=The Theory of Quantum Information |date=2018-04-26 |publisher=Cambridge University Press |doi=10.1017/9781316848142 |isbn=978-1-316-84814-2}}

As was previously shown, the square root of the fidelity can be written as $\backslash sqrt\{F(\backslash rho,\backslash sigma)\}=\backslash operatorname\{tr\}|\backslash sqrt\backslash rho\backslash sqrt\backslash sigma|,$which is equivalent to the existence of a unitary operator $U$ such that

$$\backslash sqrt\{F(\backslash rho,\backslash sigma)\}=\backslash operatorname\{tr\}(\backslash sqrt\backslash rho\backslash sqrt\backslash sigma\; U).$$Remembering that $\backslash sum\_k\; E\_k=I$ holds true for any POVM, we can then write$$\backslash sqrt\{F(\backslash rho,\backslash sigma)\}=\backslash operatorname\{tr\}(\backslash sqrt\backslash rho\backslash sqrt\backslash sigma\; U)=$$

\sum_k\operatorname{tr}(\sqrt\rho E_k \sqrt\sigma U)=\sum_k\operatorname{tr}(\sqrt\rho \sqrt{E_k} \sqrt{E_k}\sqrt\sigma U) \le

\sum_k\sqrt{\operatorname{tr}(E_k\rho)\operatorname{tr}(E_k \sigma)},where in the last step we used Cauchy-Schwarz inequality as in $|\backslash operatorname\{tr\}(A^\backslash dagger\; B)|^2\backslash le\backslash operatorname\{tr\}(A^\backslash dagger\; A)\backslash operatorname\{tr\}(B^\backslash dagger\; B)$.

The fidelity between two states can be shown to never decrease when a non-selective quantum operation $\backslash mathcal\; E$ is applied to the states:{{Cite arXiv|last=Nielsen|first=M. A. |author-link=Michael Nielsen|date=1996-06-13|title=The entanglement fidelity and quantum error correction|eprint=quant-ph/9606012}}$$F(\backslash mathcal\; E(\backslash rho),\backslash mathcal\; E(\backslash sigma))\; \backslash ge$$

F(\rho,\sigma), for any trace-preserving completely positive map $\backslash mathcal\; E$.

We can define the trace distance between two matrices A and B in terms of the trace norm by

:$$

D(A,B) = \frac{1}{2}\| A-B\|_{\rm tr} \, .

When A and B are both density operators, this is a quantum generalization of the statistical distance. This is relevant because the trace distance provides upper and lower bounds on the fidelity as quantified by the Fuchs–van de Graaf inequalities,C. A. Fuchs and J. van de Graaf, "Cryptographic Distinguishability Measures for Quantum Mechanical States", IEEE Trans. Inf. Theory 45, 1216 (1999). arXiv:quant-ph/9712042

:$$

1-\sqrt{F(\rho,\sigma)} \le D(\rho,\sigma) \le\sqrt{1-F(\rho,\sigma)} \, .

Often the trace distance is easier to calculate or bound than the fidelity, so these relationships are quite useful. In the case that at least one of the states is a pure state Ψ, the lower bound can be tightened.

:$$

1-F(\psi,\rho) \le D(\psi,\rho) \, .

We saw that for two pure states, their fidelity coincides with the overlap. Uhlmann's theorem{{cite journal |last1=Uhlmann |first1=A. |year=1976 |title=The "transition probability" in the state space of a ∗-algebra |url=http://www.physik.uni-leipzig.de/~uhlmann/PDF/Uh76a.pdf |journal=Reports on Mathematical Physics |volume=9 |issue=2 |pages=273–279 |bibcode=1976RpMP....9..273U |doi=10.1016/0034-4877(76)90060-4 |issn=0034-4877}} generalizes this statement to mixed states, in terms of their purifications:

Theorem Let ρ and σ be density matrices acting on Cⁿ. Let ρ^{{frac|1|2}} be the unique positive square root of ρ and

$$$$

| \psi _{\rho} \rangle = \sum_{i=1}^n (\rho^{{1}/{2}} | e_i \rangle) \otimes | e_i \rangle \in \mathbb{C}^n \otimes \mathbb{C}^n

be a purification of ρ (therefore $\backslash textstyle\; \backslash \{|e\_i\backslash rangle\backslash \}$ is an orthonormal basis), then the following equality holds:

:$F(\backslash rho,\; \backslash sigma)\; =\; \backslash max\_\{|\backslash psi\_\{\backslash sigma\}\; \backslash rangle\}\; |\; \backslash langle\; \backslash psi\; \_\{\backslash rho\}|\; \backslash psi\; \_\{\backslash sigma\}\; \backslash rangle\; |^2$

where $|\; \backslash psi\; \_\{\backslash sigma\}\; \backslash rangle$ is a purification of σ. Therefore, in general, the fidelity is the maximum overlap between purifications.

A simple proof can be sketched as follows. Let $\backslash textstyle\; |\backslash Omega\backslash rangle$ denote the vector

:$|\; \backslash Omega\; \backslash rangle=\; \backslash sum\_\{i=1\}^n\; |\; e\_i\; \backslash rangle\; \backslash otimes\; |\; e\_i\; \backslash rangle$

and σ^{{frac|1|2}} be the unique positive square root of σ. We see that, due to the unitary freedom in square root factorizations and choosing orthonormal bases, an arbitrary purification of σ is of the form

:$|\; \backslash psi\_\{\backslash sigma\}\; \backslash rangle\; =\; (\; \backslash sigma^\{\{1\}/\{2\}\}\; V\_1\; \backslash otimes\; V\_2\; )\; |\; \backslash Omega\; \backslash rangle$

where V_i's are unitary operators. Now we directly calculate

:$$

| \langle \psi _{\rho}| \psi _{\sigma} \rangle |^2

= | \langle \Omega | ( \rho^{{1}/{2}} \otimes I) ( \sigma^{{1}/{2}} V_1 \otimes V_2 ) | \Omega \rangle |^2

= | \operatorname{tr} ( \rho^{{1}/{2}} \sigma^{{1}/{2}} V_1 V_2^T )|^2.

But in general, for any square matrix A and unitary U, it is true that |tr(AU)| ≤ tr((A^*A)^{{frac|1|2}}). Furthermore, equality is achieved if U^* is the unitary operator in the polar decomposition of A. From this follows directly Uhlmann's theorem.

We will here provide an alternative, explicit way to prove Uhlmann's theorem.

Let $|\backslash psi\_\backslash rho\backslash rangle$ and $|\backslash psi\_\backslash sigma\backslash rangle$ be purifications of $\backslash rho$ and $\backslash sigma$, respectively. To start, let us show that $|\backslash langle\backslash psi\_\backslash rho|\backslash psi\_\backslash sigma\backslash rangle|\backslash le\backslash operatorname\{tr\}|\backslash sqrt\backslash rho\backslash sqrt\backslash sigma|$.

The general form of the purifications of the states is:$$\backslash begin\{align\}$$

|\psi_\rho\rangle &=\sum_k\sqrt{\lambda_k}|\lambda_k\rangle\otimes|u_k\rangle, \\

|\psi_\sigma\rangle &=\sum_k\sqrt{\mu_k}|\mu_k\rangle\otimes|v_k\rangle,

\end{align}were $|\backslash lambda\_k\backslash rangle,\; |\backslash mu\_k\backslash rangle$ are the eigenvectors of $\backslash rho,\backslash \; \backslash sigma$, and $\backslash \{u\_k\backslash \}\_k,\; \backslash \{v\_k\backslash \}\_k$ are arbitrary orthonormal bases. The overlap between the purifications is$$\backslash langle\backslash psi\_\backslash rho|\backslash psi\_\backslash sigma\backslash rangle\; =$$

\sum_{jk}\sqrt{\lambda_j\mu_k} \langle\lambda_j|\mu_k\rangle\,\langle u_j|v_k\rangle =

\operatorname{tr}\left(\sqrt\rho\sqrt\sigma U\right),where the unitary matrix $U$ is defined as$$U=\backslash left(\backslash sum\_k\; |\backslash mu\_k\backslash rangle\backslash !\backslash langle\; u\_k|\; \backslash right)\backslash ,\backslash left(\backslash sum\_j\; |v\_j\backslash rangle\backslash !\backslash langle\; \backslash lambda\_j|\backslash right).$$The conclusion is now reached via using the inequality $|\backslash operatorname\{tr\}(AU)|\backslash le\; \backslash operatorname\{tr\}(\backslash sqrt\{A^\backslash dagger\; A\})\backslash equiv\backslash operatorname\{tr\}|A|$: $$|\backslash langle\backslash psi\_\backslash rho|\backslash psi\_\backslash sigma\backslash rangle|=$$

|\operatorname{tr}(\sqrt\rho\sqrt\sigma U)| \le

\operatorname{tr}|\sqrt\rho\sqrt\sigma|.Note that this inequality is the triangle inequality applied to the singular values of the matrix. Indeed, for a generic matrix $A\backslash equiv\; \backslash sum\_j\; s\_j(A)|a\_j\backslash rangle\backslash !\backslash langle\; b\_j|$and unitary $U=\backslash sum\_j\; |b\_j\backslash rangle\backslash !\backslash langle\; w\_j|$, we have$$\backslash begin\{align\}$$

|\operatorname{tr}(AU)| &=

\left|\operatorname{tr}\left(\sum_j s_j(A)|a_j\rangle\!\langle b_j| \,\,\sum_k |b_k\rangle\!\langle w_k| \right)\right| \\ &=

\left|\sum_j s_j(A)\langle

w_j|a_j\rangle\right|\\ &\le

\sum_j s_j(A) \,|\langle w_j|a_j\rangle| \\

&\le

\sum_j s_j(A) \\

&= \operatorname{tr}|A|,

\end{align}where $s\_j(A)\backslash ge\; 0$ are the (always real and non-negative) singular values of $A$, as in the singular value decomposition. The inequality is saturated and becomes an equality when $\backslash langle\; w\_j|a\_j\backslash rangle=1$, that is, when $U=\backslash sum\_k\; |b\_k\backslash rangle\backslash !\backslash langle\; a\_k|,$ and thus $AU=\backslash sqrt\{AA^\backslash dagger\}\backslash equiv\; |A|$. The above shows that $|\backslash langle\backslash psi\_\backslash rho|\backslash psi\_\backslash sigma\backslash rangle|=$

\operatorname{tr}|\sqrt\rho\sqrt\sigma| when the purifications $|\backslash psi\_\backslash rho\backslash rangle$ and $|\backslash psi\_\backslash sigma\backslash rangle$ are such that $\backslash sqrt\backslash rho\backslash sqrt\backslash sigma\; U=|\backslash sqrt\backslash rho\backslash sqrt\backslash sigma|$. Because this choice is possible regardless of the states, we can finally conclude that$$\backslash operatorname\{tr\}|\backslash sqrt\backslash rho\backslash sqrt\backslash sigma|=\backslash max|\backslash langle\backslash psi\_\backslash rho|\backslash psi\_\backslash sigma\backslash rangle|.$$

Some immediate consequences of Uhlmann's theorem are

• Fidelity is symmetric in its arguments, i.e. F (ρ,σ) = F (σ,ρ). Note that this is not obvious from the original definition.
• F (ρ,σ) lies in [0,1], by the Cauchy–Schwarz inequality.
• F (ρ,σ) = 1 if and only if ρ = σ, since Ψ_ρ = Ψ_σ implies ρ = σ.

So we can see that fidelity behaves almost like a metric. This can be formalized and made useful by defining

:$\backslash cos^2\; \backslash theta\_\{\backslash rho\backslash sigma\}\; =\; F(\backslash rho,\backslash sigma)\; \backslash ,$

As the angle between the states $\backslash rho$ and $\backslash sigma$. It follows from the above properties that $\backslash theta\_\{\backslash rho\backslash sigma\}$ is non-negative, symmetric in its inputs, and is equal to zero if and only if $\backslash rho\; =\; \backslash sigma$. Furthermore, it can be proved that it obeys the triangle inequality, so this angle is a metric on the state space: the Fubini–Study metric.K. Życzkowski, I. Bengtsson, Geometry of Quantum States, Cambridge University Press, 2008, 131

• Quantiki: [https://quantiki.org/wiki/fidelity Fidelity]

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Category:Quantum information science