2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ How is your last example incompatible with J1c? $\endgroup$
    – tparker
    Feb 18 '20 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 $\endgroup$ Feb 18 '20 at 14:14
  • 2
    $\begingroup$ 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 '20 at 14:36
  • $\begingroup$ You are correct. I will edit my answer a bit later, it in fact reverses the conclusion for J1c $\endgroup$ Feb 18 '20 at 15:34
  • $\begingroup$ J1c is now fixed $\endgroup$ Feb 18 '20 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.