Fidelity of quantum states

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In quantum information theory, fidelity is a measure of the "closeness" of two quantum states. It is not a metric on the space of density matrices.

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In probability theory, given two random variables p = (p1...pn) and q = (q1...qn) on the probability space X = {1,2...n}. The fidelity of p and q is defined to be the quantity

F(p,q) = \sum _i \sqrt{p_i q_i}.

In other words, the fidelity F(p,q) is the inner product of (\sqrt{p_1}, \cdots ,\sqrt{p_n}) and (\sqrt{q_1}, \cdots ,\sqrt{q_n}) viewed as vectors in Euclidean space. Notice that when p = q, F(p,q) = 1. In general, 0 \leq F(p,q) \leq 1.

Making the appropriate modification for the matricial notion of square root and mimicking the above definition give the fidelity of two quantum state.

Given two density matrices ρ and σ, the fidelity is defined by

F(\rho, \sigma) = \operatorname{Tr} [\rho^{\frac{1}{2}} \sigma \rho^{\frac{1}{2}}]^{\frac{1}{2}}.

By M½ of a positive semidefinite matrix M, we mean its unique positive square root given by the spectral theorem. The Euclidean inner product from the classical definition is replaced by the Hilbert-Schmidt inner product. When the states are classical, i.e. when ρ and σ commute, the definition coincides with that for probability distributions.

Notice by definition F is non-negative, and F(ρ,ρ) = 1. In the following section it will be shown that it can be no larger than 1.

Consider pure states \rho = | \phi \rangle \langle \phi | and \sigma = | \psi \rangle \langle \psi |. Their fidelity is

F(\rho, \sigma) = \operatorname{Tr} [\rho \; \sigma \rho]^{\frac{1}{2}} = \operatorname{Tr} \; [ | \phi \rangle \langle \phi | \psi \rangle \langle \psi |\phi \rangle \langle \phi | ]^{\frac{1}{2}} = | \langle \phi | \psi \rangle |.

This is sometimes called the overlap between two states. If, say, |\phi\rangle is an eigenstate of an observable, and the system is prepared in | \psi \rangle, then F(ρ, σ)2 is the probability of the system being in state |\phi\rangle after the measurement.

Let ρ and σ be two density matrices that commute. Therefore they can be simultaneously diagonalized by unitary matrices, and we can write

\rho = \sum_i p_i | i \rangle \langle i | and \sigma = \sum_i q_i | i \rangle \langle i |

for some orthonormal basis \{ | i \rangle \}. Direct calculation shows the fidelity is

F(\rho, \sigma) = \sum_i \sqrt{p_i q_i}.

This shows that, heuristically, fidelity of quantum states is a genuine extension of the notion from probability theory.

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

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

for any unitary operator U.

We saw that for two pure states, their fidelity coincides with the overlap. Uhlmann's theorem generalizes this statement to mixed states, in terms of their purifications:

Theorem Let ρ and σ be density matrices acting on Cn. Let ρ½ be the unique positive square root of ρ and

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

be a purfication of ρ (therefore {|ei >} is an orthonormal basis), then the following equality holds:

F(\rho, \sigma) = \max_{|\psi_{\sigma} \rangle} | \langle \psi _{\rho}| \psi _{\sigma} \rangle |

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


Proof: A simple proof can be sketched as follows. Let |Ω > denote the vector

| \Omega \rangle= \sum | e_i \rangle \otimes | e_i \rangle

and σ½ 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

| \psi_{\sigma} \rangle = ( \sigma^{\frac{1}{2}} V_1 \otimes V_2 ) | \Omega \rangle

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

| \langle \psi _{\rho}| \psi _{\sigma} \rangle | = \langle \Omega | ( \rho^{\frac{1}{2}} \otimes I) ( \sigma^{\frac{1}{2}} V_1 \otimes V_2 ) | \Omega \rangle = \operatorname{Tr} ( \rho^{\frac{1}{2}} \sigma^{\frac{1}{2}} V_1 V_2 ).

But in general, for any square matrix A and unitary U, it is true that |Tr(AU)| ≤ Tr (A*A)½. Furthermore, equality is achieved if U* is the unitary operator in the polar decomposition of A. From this follows directly Uhlmann's theorem.

Some immediate consequences of Uhlmann's theorem are

  • Fidelity is symmetric in its arguments, i.e. F (ρ,σ) = F (σ,ρ). Notice this is not obvious from the definition.
  • F (ρ,σ) lies in [0,1], by the Cauchy-Schwarz inequality.
  • F (ρ,σ) = 1 if and only if ρ = σ, since Ψρ = Ψσ implies ρ = σ.
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