On the probability of agreeing with different density matrices?

Question

Let's say I have a density matrix and I (person $1$) suspect it to be of the form:

$$ \rho_1 = p_1 |\psi \rangle \langle \psi | + p_2 |\phi \rangle \langle \phi |$$

$|\psi \rangle$ and $| \phi \rangle$ are orthogonal wavefunctions and $p_i$ are the probabilities.

But for the same system someone (person $2$) else assigns different probabilities and thus have a different density matrix.

$$ \rho_2 = \tilde p_1 |\psi \rangle \langle \psi | + \tilde p_2 |\phi \rangle \langle \phi |$$

Now, imagine someone asks $1$ and $2$ before the measurement to write what he thinks the outcome of the measurement will be. Then, there are four possible scenarios:

$$P(1 \text{ says } | \psi \rangle + 2 \text{ says } | \psi \rangle) = p_1 \tilde p_1 $$

$$P(1 \text{ says } | \phi \rangle + 2 \text{ says } | \phi \rangle) = p_2 \tilde p_2 $$

$$P(1 \text{ says } | \psi \rangle + 2 \text{ says } | \phi \rangle ) = p_1 \tilde p_2 $$ $$P(1 \text{ says } | \phi \rangle + 2 \text{ says } | \psi \rangle ) = p_2 \tilde p_1 $$

Hence the chance of them saying the same thing is essentially:

$$P(\text{Agree}) = p_1 \tilde p_1 + p_2 \tilde p_2$$

But this is nothing more than:

$$P(\text{Agree}) = \text{Tr} (\rho_1 \rho_2) $$

In fact this quantity, evolves unitarily and self contained since:

$$ i \hbar \frac{\partial \rho_1 \rho_2}{\partial t} = [H,\rho_1 \rho_2 ] $$

Thus, we can think of this as a measure of agreement (be careful for $\rho_1 = \rho_2$ then $\text{Tr} (\rho_1 \rho_2) \leq 1$).

Now, I don't want to reinvent the wheel and I'm sure someone has barked up this tree or it's completely wrong? Possible use cases is the difference of descriptions of density matrices in Wigner's Friend or even perhaps seeing how this quantity changes when going from a classical distribution to quantum probabilities.

Where can I read more about this?

Answer 1

This is a good question but it is entirely classical, because all of the density matrices are diagonal in the $|\psi\rangle,|\phi\rangle$-basis and the measurement is assumed to be in the same basis. Let's make it more general.

Let's stick to measurements in the $|\psi\rangle,|\phi\rangle$-basis but let each person have their own state $\rho_1$ and $\rho_2$. The "chance of agreeing" implicitly defined here is $$P(\mathrm{agree})=\langle \psi|\rho_1|\psi\rangle \langle \psi|\rho_2|\psi\rangle+\langle \phi|\rho_1|\phi\rangle \langle \phi|\rho_2|\phi\rangle.$$ The trace quantity, however, is $$\mathrm{Tr}(\rho_1\rho_2)=\langle \psi|\rho_1|\psi\rangle \langle \psi|\rho_2|\psi\rangle+\langle \phi|\rho_1|\phi\rangle \langle \phi|\rho_2|\phi\rangle+ \langle \psi|\rho_1|\phi\rangle \langle \phi|\rho_2|\psi\rangle+\langle \phi|\rho_1|\psi\rangle \langle \psi|\rho_2|\phi\rangle.$$ These two are only equal when $\rho_1$ and/or $\rho_2$ have no off-diagonal components in the measurement basis.