Trace distance

The trace distance is half of the trace norm of the difference of the matrices:

${\displaystyle T(\rho ,\sigma ):={\frac {1}{2}}||\rho -\sigma ||_{1}={\frac {1}{2}}\mathrm {Tr} \left[{\sqrt {(\rho -\sigma )^{\dagger }(\rho -\sigma )}}\right].}$

(The trace norm is the Schatten norm for p=1.) The purpose of the factor of two is to restrict the trace distance between two normalized density matrices to the range [0, 1] and to simplify formulas in which the trace distance appears.

Since density matrices are Hermitian,

${\displaystyle T(\rho ,\sigma )={\frac {1}{2}}\mathrm {Tr} \left[{\sqrt {(\rho -\sigma )^{2}}}\right]={\frac {1}{2}}\sum _{i}|\lambda _{i}|,}$

where the ${\displaystyle \lambda _{i}}$ are eigenvalues of the Hermitian, but not necessarily positive, matrix ${\displaystyle (\rho -\sigma )}$.

By using the Hölder duality for Schatten norms, the trace distance can be written in variational form as [1]

${\displaystyle T(\rho ,\sigma )={\frac {1}{2}}\sup _{-\mathbb {I} \leq U\leq \mathbb {I} }\mathrm {Tr} [U(\rho -\sigma )]=\sup _{0\leq P\leq \mathbb {I} }\mathrm {Tr} [P(\rho -\sigma )].}$

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 ${\displaystyle \rho }$ or ${\displaystyle \sigma }$, each with probability ${\displaystyle {\frac {1}{2}}}$ and sends it to Bob who has to discriminate between the two states using a binary measurement. Let Bob assign the measurement outcome ${\displaystyle 0}$ and a POVM element ${\displaystyle P_{0}}$ such as the outcome ${\displaystyle 1}$ and a POVM element ${\displaystyle P_{1}=1-P_{0}}$ to identify the state ${\displaystyle \rho }$ or ${\displaystyle \sigma }$, respectively. His expected probability of correctly identifying the incoming state is then given by

${\displaystyle p_{\text{guess}}={\frac {1}{2}}p(0|\rho )+{\frac {1}{2}}p(1|\sigma )={\frac {1}{2}}\mathrm {Tr} (P_{0}\rho )+{\frac {1}{2}}\mathrm {Tr} (P_{1}\sigma )={\frac {1}{2}}\left(1+\mathrm {Tr} \left(P_{0}(\rho -\sigma )\right)\right).}$

Therefore, when applying an optimal measurement, Bob has the maximal probability

${\displaystyle p_{\text{guess}}^{\text{max}}=\sup _{P_{0}}{\frac {1}{2}}\left(1+\mathrm {Tr} \left(P_{0}(\rho -\sigma )\right)\right)={\frac {1}{2}}(1+T(\rho ,\sigma ))}$

of correctly identifying in which state Alice prepared the system.[2]

The trace distance has the following properties[1]

• It is a metric on the space of density matrices, i.e. it is non-negative, symmetric, and satisfies the triangle inequality, and ${\displaystyle T(\rho ,\sigma )=0\Leftrightarrow \rho =\sigma }$
• ${\displaystyle 0\leq T(\rho ,\sigma )\leq 1}$ and ${\displaystyle T(\rho ,\sigma )=1}$ if and only if ${\displaystyle \rho }$ and ${\displaystyle \sigma }$ have orthogonal supports
• It is preserved under unitary transformations: ${\displaystyle T(U\rho U^{\dagger },U\sigma U^{\dagger })=T(\rho ,\sigma )}$
• It is contractive under trace-preserving CP maps, i.e. if ${\displaystyle \Phi }$ is a CPT map, then ${\displaystyle T(\Phi (\rho ),\Phi (\sigma ))\leq T(\rho ,\sigma )}$
• It is convex in each of its inputs. E.g. ${\displaystyle T(\sum _{i}p_{i}\rho _{i},\sigma )\leq \sum _{i}p_{i}T(\rho _{i},\sigma )}$

For qubits, the trace distance is equal to half the Euclidean distance in the Bloch representation.