Which of the Jozsa axioms does the Hilbert-Schmidt inner product violate?

Question

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?

Yehuda Naveh · Accepted Answer

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 $rho = sum_i p_i |psi_irangle langle psi_i|$ and $sigma = sum_i q_i |phi_irangle langle phi_i|$, we have:

$tr(rho sigma)$ = $sum_{ij} p_i q_j tr(|psi_irangle langle psi_i|phi_jrangle langle phi_j|)$ = $sum_{ij} p_i q_j |langle psi_i|phi_jrangle |^2$ 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 $|langle psi_i|phi_jrangle |$ are all larger or equal zero and smaller or equal one so:
$0 leq sum_{ij} p_i q_j |langle psi_i|phi_jrangle |^2 leq sum_{ij} p_i q_j = (sum_i p_i)(sum_j q_j) = 1$

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 $tr(rho sigma) = sum_{ij} p_i q_j |langle psi_i|phi_jrangle |^2$ = 0, so $langle psi_i|phi_jrangle  = 0$ for all $i$, $j$ in the mixed states (i.e. $p_i, q_j neq 0$). Then $rho sigma = (sum_i p_i |psi_irangle langle psi_i|)(sum_j q_j |phi_jrangle langle phi_j|) = sum_{ij} p_i q_j |psi_irangle langle psi_i|phi_jrangle langle phi_j| = 0$ by the assumption.

Which of the Jozsa axioms does the Hilbert-Schmidt inner product violate?

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