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The paper Quantum fidelity measures for mixed states considers various differently-normalized variants of the Hilbert-Schmidt inner product $\mathrm{Tr}(A^\dagger B)$ on linear operators as candidate measures of the fidelity $\mathcal{F}$ between two density operators $\rho$ and $\sigma$ - that is, $$\mathcal{F} = \frac{\mathrm{tr}(\rho \sigma)}{f \left(\mathrm{tr}(\rho^2), \mathrm{tr}(\sigma^2) \right)}$$ for various choices of normalization function $f(x,y)$. For various choices of $f$, they say which of the Jozsa axioms are and are not respected by that choice:

J1a. $\mathcal{F}(\rho, \sigma) \in [0, 1]$

J1b. $\mathcal{F}(\rho, \sigma) = 1 \iff \rho = \sigma$

J1c. $\mathcal{F}(\rho, \sigma) = 0 \iff \rho \sigma = 0$

J2. $\mathcal{F}(\rho, \sigma) = \mathcal{F}(\sigma, \rho)$

J3. $\mathcal{F}(\rho, \sigma) = \mathrm{tr}(\rho \sigma)$ if either $\rho$ or $\sigma$ is a pure state

J4. $\mathcal{F}(U \rho U^\dagger, U \sigma U^\dagger) = \mathcal{F}(\rho, \sigma)$ for any unitary operator $U$.

But oddly enough, they never discuss which of these axioms are respected by the simplest choice of normalization of all: $f \equiv 1$, which gives the Hilbert-Schmidt inner product itself as the candidate fidelity.

Which of the Jozsa axioms does the Hilbert-Schmidt inner product respect? It's easy to see that it satisfies axioms J2-J4, but what about J1a-J1c?

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You could probably reach the same conclusions by identifying that $tr(\rho \sigma)$ is just the expectation value of $\rho$ under the mixed state $\sigma$, but let's do it explicitly:

for $\rho = \sum_i p_i |\psi_i\rangle \langle \psi_i|$ and $\sigma = \sum_i q_i |\phi_i\rangle \langle \phi_i|$, we have:

$tr(\rho \sigma)$ = $\sum_{ij} p_i q_j tr(|\psi_i\rangle \langle \psi_i|\phi_j\rangle \langle \phi_j|)$ = $\sum_{ij} p_i q_j |\langle \psi_i|\phi_j\rangle |^2$ by the trace cyclic property and the fact that the trace of a scalar is the scalar.

Then:

J1a is true because $p_i$, $q_j$ and $|\langle \psi_i|\phi_j\rangle |$ are all larger or equal zero and smaller or equal one so: $0 \leq \sum_{ij} p_i q_j |\langle \psi_i|\phi_j\rangle |^2 \leq \sum_{ij} p_i q_j = (\sum_i p_i)(\sum_j q_j) = 1$

J1b is false because $tr(\rho \sigma) = tr(\rho^2) < 1$ for $\rho = \sigma$ a non-pure state

J1c is true because right to left direction is trivial. For left to right direction suppose $tr(\rho \sigma) = \sum_{ij} p_i q_j |\langle \psi_i|\phi_j\rangle |^2$ = 0, so $\langle \psi_i|\phi_j\rangle = 0$ for all $i$, $j$ in the mixed states (i.e. $p_i, q_j \neq 0$). Then $\rho \sigma = (\sum_i p_i |\psi_i\rangle \langle \psi_i|)(\sum_j q_j |\phi_j\rangle \langle \phi_j|) = \sum_{ij} p_i q_j |\psi_i\rangle \langle \psi_i|\phi_j\rangle \langle \phi_j| = 0$ by the assumption.

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  • $\begingroup$ How is your last example incompatible with J1c? $\endgroup$ – tparker Feb 18 at 13:24
  • $\begingroup$ $F(\rho, \sigma) = tr(\rho \sigma) = 0$ while $\rho \neq 0$ and $\sigma \neq 0$. This is incompatible with the iff statement of J1c $\endgroup$ – Yehuda Naveh Feb 18 at 14:14
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    $\begingroup$ I believe that you have misread the statement of J1c. The RHS doesn't say "$\rho = 0$ or $\sigma = 0$"; it says "$\rho \sigma = 0$". The former interpretation doesn't even make sense, given the trace condition on density matrices. $\endgroup$ – tparker Feb 18 at 14:36
  • $\begingroup$ You are correct. I will edit my answer a bit later, it in fact reverses the conclusion for J1c $\endgroup$ – Yehuda Naveh Feb 18 at 15:34
  • $\begingroup$ J1c is now fixed $\endgroup$ – Yehuda Naveh Feb 18 at 16:36

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