# Trace distance

In quantum mechanics, and especially quantum information and the study of open quantum systems, the trace distance is a metric on the space of density matrices and gives a measure of the distinguishability between two states. It is the quantum generalization of the Kolmogorov distance for classical probability distributions.

## Definition

The trace distance is defined as half of the trace norm of the difference of the matrices:$T(\rho ,\sigma ):={\frac {1}{2}}\|\rho -\sigma \|_{1}={\frac {1}{2}}\mathrm {Tr} \left[{\sqrt {(\rho -\sigma )^{\dagger }(\rho -\sigma )}}\right],$where $\|A\|_{1}\equiv \operatorname {Tr} [{\sqrt {A^{\dagger }A}}]$ is the trace norm of $A$, and ${\sqrt {A}}$ is the unique positive semidefinite $B$ such that $B^{2}=A$ (which is always defined for positive semidefinite $A$). This can be thought of as the matrix obtained from $A$ taking the algebraic square roots of its eigenvalues. For the trace distance, we more specifically have an expression of the form $|C|\equiv {\sqrt {C^{\dagger }C}}={\sqrt {C^{2}}}$ where $C=\rho -\sigma$ is Hermitian. This quantity equals the sum of the singular values of $C$, which being $C$ Hermitian, equals the sum of the absolute values of its eigenvalues. More explicitly, $T(\rho ,\sigma )={\frac {1}{2}}\operatorname {Tr} |\rho -\sigma |={\frac {1}{2}}\sum _{i=1}^{r}|\lambda _{i}|,$ where $\lambda _{i}\in \mathbb {R}$ is the $i$-th eigenvalue of $\rho -\sigma$, and $r$ is its rank.
The factor of two ensures that the trace distance between normalized density matrices takes values in the range $[0,1]$.

## Connection with the total variation distance

The trace distance serves as a direct quantum generalization of the total variation distance between probability distributions. Given two probability distributions $P$ and $Q$, their total variation distance is defined as$\delta (P,Q)={\frac {1}{2}}\|P-Q\|_{1}={\frac {1}{2}}\sum _{k}|P_{k}-Q_{k}|.$When extending this concept to quantum states, one must account for the fact that for quantum states different measurement can produce different distributions. A natural approach is to consider the (classical) total variation distance between the measurement outcomes produced by two states for a fixed choice of measurement, and then maximize over all possible measurements. This procedure leads precisely to the trace distance between the quantum states. More explicitly, this is the quantity$\max _{\Pi }{\frac {1}{2}}\sum _{i}|\operatorname {Tr} (\Pi _{i}\rho )-\operatorname {Tr} (\Pi _{i}\sigma )|,$with the maximization performed with respect to all possible POVMs $\{\Pi _{i}\}_{i}$.
To understand why this maximum equals the trace distance between the states, note that there is a unique decomposition $\rho -\sigma =P-Q$ with $P,Q\geq 0$ positive semidefinite matrices with orthogonal support. With these operators we can write concisely $|\rho -\sigma |=P+Q$. Furthermore $\operatorname {Tr} (\Pi _{i}P),\operatorname {Tr} (\Pi _{i}Q)\geq 0$, and thus $|\operatorname {Tr} (\Pi _{i}P)-\operatorname {Tr} (\Pi _{i}Q))|\leq \operatorname {Tr} (\Pi _{i}P)+\operatorname {Tr} (\Pi _{i}Q)$. We thus have$\sum _{i}|\operatorname {Tr} (\Pi _{i}(\rho -\sigma ))|=\sum _{i}|\operatorname {Tr} (\Pi _{i}(P-Q))|\leq \sum _{i}\operatorname {Tr} (\Pi _{i}(P+Q))=\operatorname {Tr} |\rho -\sigma |.$This shows that$\max _{\Pi }\delta (P_{\Pi ,\rho },P_{\Pi ,\sigma })\leq T(\rho ,\sigma ),$where $P_{\Pi ,\rho }$ denotes the classical probability distribution resulting from measuring $\rho$ with the POVM $\Pi$, $(P_{\Pi ,\rho })_{i}\equiv \operatorname {Tr} (\Pi _{i}\rho )$, and the maximum is performed over all POVMs $\Pi \equiv \{\Pi _{i}\}_{i}$.
To conclude that the inequality is saturated by some POVM, we need only consider the projective measurement with elements corresponding to the eigenvectors of $\rho -\sigma$. With this choice,$\delta (P_{\Pi ,\rho },P_{\Pi ,\sigma })={\frac {1}{2}}\sum _{i}|\operatorname {Tr} (\Pi _{i}(\rho -\sigma ))|={\frac {1}{2}}\sum _{i}|\lambda _{i}|=T(\rho ,\sigma ),$where $\lambda _{i}$ are the eigenvalues of $\rho -\sigma$.

## Physical interpretation

By using the Hölder duality for Schatten norms, the trace distance can be written in variational form as
    $T(\rho ,\sigma )={\frac {1}{2}}\sup _{-\mathbb {I} \leq U\leq \mathbb {I} }\mathrm {Tr} [U(\rho -\sigma )]=\sup _{0\leq P\leq \mathbb {I} }\mathrm {Tr} [P(\rho -\sigma )].$
As for its classical counterpart, the trace distance can be related to the maximum probability of distinguishing between two quantum states:
For example, suppose Alice prepares a system in either the state $\rho$ or $\sigma$, each with probability ${\frac {1}{2}}$ and sends it to Bob who has to discriminate between the two states using a binary measurement. Let Bob assign the measurement outcome $0$ and a POVM element $P_{0}$ such as the outcome $1$ and a POVM element $P_{1}=1-P_{0}$ to identify the state $\rho$ or $\sigma$, respectively. His expected probability of correctly identifying the incoming state is then given by
    $p_{\text{guess}}={\frac {1}{2}}p(0|\rho )+{\frac {1}{2}}p(1|\sigma )={\frac {1}{2}}\mathrm {Tr} (P_{0}\rho )+{\frac {1}{2}}\mathrm {Tr} (P_{1}\sigma )={\frac {1}{2}}\left(1+\mathrm {Tr} \left(P_{0}(\rho -\sigma )\right)\right).$
Therefore, when applying an optimal measurement, Bob has the maximal probability
    $p_{\text{guess}}^{\text{max}}=\sup _{P_{0}}{\frac {1}{2}}\left(1+\mathrm {Tr} \left(P_{0}(\rho -\sigma )\right)\right)={\frac {1}{2}}(1+T(\rho ,\sigma ))$
of correctly identifying in which state Alice prepared the system.

## Properties

The trace distance has the following properties
* It is a metric on the space of density matrices, i.e. it is non-negative, symmetric, and satisfies the triangle inequality, and $T(\rho ,\sigma )=0\Leftrightarrow \rho =\sigma$
* $0\leq T(\rho ,\sigma )\leq 1$ and $T(\rho ,\sigma )=1$ if and only if $\rho$ and $\sigma$ have orthogonal supports
* It is preserved under unitary transformations: $T(U\rho U^{\dagger },U\sigma U^{\dagger })=T(\rho ,\sigma )$
* It is contractive under trace-preserving CP maps, i.e. if $\Phi$ is a CPTP map, then $T(\Phi (\rho ),\Phi (\sigma ))\leq T(\rho ,\sigma )$
* It is convex in each of its inputs. E.g. $T(\sum _{i}p_{i}\rho _{i},\sigma )\leq \sum _{i}p_{i}T(\rho _{i},\sigma )$
* On pure states, it can be expressed uniquely in term of the inner product of the states: $T(|\psi \rangle \langle \psi |,|\phi \rangle \langle \phi |)={\sqrt {1-|\langle \psi |\phi \rangle |^{2}}}$
For qubits, the trace distance is equal to half the Euclidean distance in the Bloch representation.

### Relationship to other distance measures
#### Fidelity

The fidelity of two quantum states $F(\rho ,\sigma )$ is related to the trace distance $T(\rho ,\sigma )$ by the inequalities
    $1-{\sqrt {F(\rho ,\sigma )}}\leq T(\rho ,\sigma )\leq {\sqrt {1-F(\rho ,\sigma )}}\,.$
The upper bound inequality becomes an equality when $\rho$ and $\sigma$ are pure states. [Note that the definition for Fidelity used here is the square of that used in Nielsen and Chuang]

#### Total variation distance

The trace distance is a generalization of the total variation distance, and for two commuting density matrices, has the same value as the total variation distance of the two corresponding probability distributions.