# On the probability of agreeing with different density matrices?

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.

• it's easy to define agreement in this case because both states are classical: diagonal in the same bases. For more general pairs of states, this kind of "agreement" will depend on the choice of measurements performed on both sides. And generalising further (because really, "same outcome" when measurements are performed on different sides just means correlated outcomes), you're then essentially asking about the correlations between the measurement outcomes, which is certainly a well-studied type of quantity (look up eg accessible mutual information)
– glS
Jul 13 at 7:44
• Since I'm taking trace it should be independent of the basis? Maybe a measure of agreement of outcome + how quantum the system is behaving? (Since it vanishes when the off-diagonal of either density Matrix is 0) Jul 13 at 8:18
• but as far as i can tell you're only taking the trace because in this case it's equal to the expression you got for the "agreement". That won't be true in general, so then what's the rationale behind considering the trace?
– glS
Jul 13 at 12:57
• Yes the time evolution of soley that quantity is not self contained :/ Jul 13 at 16:12

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.