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

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?

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.

• How is your last example incompatible with J1c? – tparker Feb 18 at 13:24
• $F(\rho, \sigma) = tr(\rho \sigma) = 0$ while $\rho \neq 0$ and $\sigma \neq 0$. This is incompatible with the iff statement of J1c – Yehuda Naveh Feb 18 at 14:14
• 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. – tparker Feb 18 at 14:36
• You are correct. I will edit my answer a bit later, it in fact reverses the conclusion for J1c – Yehuda Naveh Feb 18 at 15:34
• J1c is now fixed – Yehuda Naveh Feb 18 at 16:36