Previously, we learned how to measure the closeness between two quantum operators using the trace distance. However, this metric relies on diagonalization or optimization methods, which can turn particularly costly for high-dimensional systems. Is there an easier way to compute the similarity between quantum states? An alternative to trace distance is the _fidelity_. Given two states $\rho$ and $\phi$, it is defined as $$ F(\rho,\sigma) = \text{Tr} \sqrt{\rho^{1/2}\sigma\rho^{1/2}}. $$ The fidelity turns out to be a quantum generalization of the Bhattacharyya coefficient. For instance, when the states $\rho$ and $\phi$ commute, they can be diagonalized in the same basis. Therefore, we can write $$ \rho = \sum_i r_i \vert i \rangle \langle i \vert, \qquad \sigma = \sum_i s_i \vert i \rangle \langle i \vert. $$ We observe that the fidelity is reduced to the classical Bhattacharyya coefficient between the vectors formed by the eigenvalues of $\rho$ and $\phi$ respectively: $$ \begin{split} F(\rho,\sigma) & = \text{Tr}\sqrt{\sum_i r_i s_i \vert i \rangle \langle i \vert} \\ & = \text{Tr}\sum_i \sqrt{r_i s_i} \vert i \rangle \langle i \vert \\ & = \sum_i \sqrt{r_i s_i} \\ & = F(r_i s_i). \end{split} $$ Using fidelity as a measure of similarity is very useful and convenient when dealing with at least one pure state. Suppose that we prepare a pure state $\vert \psi \rangle$ and send it using a quantum information protocol. Ideally, the receiver should get the same state, but what if they get something else? If the channel is noisy, the state at the receiving end might be mixed, though we sent a pure state If we send the pure state $\rho = \vert \psi \rangle\! \langle \psi \vert$, and the final—and possibly mixed—state is described via a density matrix $\sigma$, then the fidelity can be computed via $$ \begin{equation} F(\rho,\sigma) = \sqrt{\langle \psi \vert \sigma \vert \psi \rangle}. \end{equation} \tag{1} $$ If the state $\sigma$ is also pure, let's say $\sigma = \vert \varphi \rangle \! \langle \varphi \vert$, the fidelity is equivalent to the absolute value of their overlap: $$ \begin{equation} F( \psi , \varphi) = \vert \langle \varphi \vert \psi \rangle \vert . \end{equation} \tag{2} $$ In such situations, we can see that the fidelity is closely related to the probability of the state $\sigma$ will pass a yes/no test associated with the pure state $\vert \psi \rangle$. Note that computing equations 1 and 2 does not require any diagonalization in principle. That provides a really simplified way to calculate the similarity between quantum states, both analytically and numerically! This is important for tasks such as measuring the quality of an experiment by keeping track of how close is the produced state to the ideal one. We always need to quantify the uncertainty introduced by noise. In the next exercise, let's demonstrate equations 1 and 2 and see how the fidelity is related to the Bathacharyya coefficient when we have two pure states! --- ***Exercise D.4.1.(a)*** Suppose we are given pure state $\rho = \vert \psi \rangle\!\langle\psi \vert$ and an arbitrary (potentially mixed) state $\sigma$. Show that $$ \text{Tr}\sqrt{(\vert \psi \rangle\!\langle\psi \vert)^{1/2}\sigma (\vert \psi \rangle\!\langle\psi \vert)^{1/2}} = \sqrt{\langle\psi \vert\sigma \vert \psi \rangle } $$

Hint

Note that the square of any pure state $\rho$ is equal to itself: $$ \rho^2 = \rho. $$

Solution

Since $\rho = \rho^2$, we have $$ (\rho)^{1/2} = (\rho^2)^{1/2} = \rho = \vert \psi \rangle\!\langle\psi \vert. $$ Thus, $$ \begin{split} \text{Tr}\sqrt{(\vert \psi \rangle\!\langle\psi \vert)^{1/2}\sigma (\vert \psi \rangle\!\langle\psi \vert)^{1/2}} & = \text{Tr}\sqrt{(\vert \psi \rangle\!\langle\psi \vert\sigma \vert \psi \rangle\!\langle\psi \vert} \\ & = \text{Tr}\sqrt{\langle\psi \vert\sigma \vert \psi \rangle \ \vert \psi \rangle\!\langle\psi \vert} \\ & = \sqrt{\langle\psi \vert\sigma \vert \psi \rangle} \ \text{Tr} \vert \psi \rangle\!\langle\psi \vert \\ & = \sqrt{\langle\psi \vert\sigma \vert \psi \rangle} . \end{split} $$ Note that if $\sigma$ is also a pure state, for instance $\sigma = \vert \varphi \rangle\!\langle\varphi \vert,$ we have $$ \sqrt{\langle\psi \vert\sigma \vert \psi \rangle} = \sqrt{\langle\psi \vert \varphi \rangle\!\langle\varphi \vert \psi \rangle} = \vert \langle\psi \vert \varphi \rangle \vert. $$

***Exercise D.4.1.(b)*** Consider the pure states $$ \vert \psi \rangle= \sum_{i=1}^n \sqrt{p_i} \vert i \rangle \quad \text{and} \quad \vert \varphi \rangle= \sum_{i=1}^n \sqrt{q_i} \vert i \rangle, $$ where the vectors $\vert i \rangle$ form an orthonormal basis and $p_i$ and $q_i$ are elements of the probability distributions $p = [p_1,\cdots,p_n]$ and $q = [q_1,\cdots,q_n]$ respectively. Show that the fidelity $F( \psi , \varphi)$ is equivalent to the Bhattacharyya coefficient between the vectors $p$ and $q$, $$ F(\psi , \varphi) = F(p,q) = \sum_i \sqrt{p_i q_i}. $$

Solution

Using Equation 1, we have $$ \begin{split} F(\psi,\varphi) &= \left \vert \sum_{i,j} \sqrt{p_i q_j} \ \langle i \vert j \rangle\ \right \vert \\ &= \left \vert \sum_{i,j} \sqrt{p_i q_j} \ \delta_{i,j} \right \vert \\ &= \sum_{i} \sqrt{p_i q_j} \\ &= \sqrt{p} \cdot \sqrt{q} \\ &= F(p, q) . \end{split} $$

---

Codercise D.4.1 - Fidelity between pure states

Codercise D.4.2 - Trace distance from the fidelity

Codercise D.4.3 - Fidelity in PennyLane

Defining fidelity

Properties of the fidelity

Fidelity and trace distance

Uhlmann's theorem