# Probability of measuring two observables in a mixed state

Lets say i have density Matrix on the usual base

$$\rho = \left( \begin{array}{cccc} \frac{3}{14} & \frac{3}{14} & 0 & 0 \\ \frac{3}{14} & \frac{3}{14} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{4}{7} \\ \end{array} \right)$$

from this two states

$$|v_1\rangle=\frac{1}{\sqrt{2}}\left( \begin{array}{c} 1\\ 1 \\ 0 \\ 0 \\ \end{array} \right);|v_2\rangle=\left( \begin{array}{c} 0\\ 0 \\ 0 \\ 1 \\ \end{array} \right)$$ with weights $\frac{3}{7}$ and $\frac{4}{7}$ respectively

And two observables A and B

$$A=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right)$$

$$B=\left( \begin{array}{cccc} 3 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4 \\ \end{array} \right)$$

If you measure $A$ and $B$ at the same time, what you are doing is measuring $A$ with some state and measuring $B$ with not necessarily the same state but it may be the same.

So if I am trying to measure probability of measuring 2 and 4, that is $p(2)\times p(4)$ ?

Thats $p(2) = \frac{4}{7}$ and $p(4) = \frac{11}{14}$ then obtaining the two at the same time is $\frac{4}{7}\times\frac{11}{14} = \frac{22}{49}$

What confuses me is, why is lower than $\frac{4}{7}$, since $|v_2\rangle$ has that chance of being measured, the other state just adds chances of measuring 4 with B.

Whats really going on here?

• hint: Do $A$ and $B$ commute? – user1504 Apr 26 '14 at 2:15
• Yes they do, i dont see what you are trying to say. – spectator Apr 26 '14 at 2:17
• I have an answer, but I think I'm missing something here. When you say the "probability of measuring $2$ and $4$" do you mean the probability of getting $2$ for the measurement $A$ and $4$ for the measurement $B$? If so, I don't understand how you come up with $p(4) = \frac{11}{14}$...it seems to me the only way to get that is to be in state $\left|v_2\right\rangle$. – Jared Apr 26 '14 at 7:23
• What i mean by that is, The probability of measuring A and B at the same time gives you 2 and 4. $p(4)=Tr(\rho \Lambda)$ where $\Lambda$ is the projector for eigenvalue 4 – spectator Apr 26 '14 at 13:59

If you are measuring both observables at the same time (which is possible, since the two observables commute), then you have to do one measurement which measures both quantities. Therefore the measurement must result in one of the common eigenstates of $A$ and $B$. Now it turns out that $A$ and $B$ have an unique common set of eigenstates, which are just the basis states. Therefore the probabilities are just the diagonal elements of the density matrix, that is (naming $a$ the measurement result of $A$ and $b$ the measurement result of $B$): \begin{aligned} p(a=1 \land b=3) &= \frac{3}{14}\\ p(a=1 \land b=4) &= \frac{3}{14}\\ p(a=2 \land b=3) &= 0\\ p(a=2 \land b=4) &= \frac{4}{7} \end{aligned} Note that $p(a=2 \land b=4) \ne p(a=2)\cdot p(b=4)$ since the events are not independent of each other. But that's not specifically quantum; if you know $a=2$, you already know that $b$ must be $4$, while from learning $a=1$ you don't learn anything about the measurement result $b$. Already classical probability theory tells you that in that case, the product formula for the probabilities doesn't hold.
Indeed, given that there is a single combined measurement, you could split the measurement into two steps; first doing the measurement (which produces the values for $a$ and $b$), and then, reading the measurement results (which is a completely classical process, described by ordinary probability theory).
• I don't understand how you get $b = 3$. I agree that $\left\langle v_2 \right| B \left|v_2\right\rangle = 4$ (and similarly that $\left\langle v_2 \right| A \left|v_2\right\rangle = 2$), but doesn't $\left\langle v_1 \right| B \left|v_1\right\rangle = \frac{7}{2}$, not $3$? – Jared Apr 26 '14 at 10:17
• You're not measuring $|v_1\rangle$ or $|v_2\rangle$, you're measuring $A$ and $B$. Especially, by the postulates of quantum mechanics, every measurement of $B$ gives an eigenvalue of $B$ as result. The only eigenvalues of $B$ are $2$ and $4$, so those are the values you can get as result. $\langle v_2|B|v_2\rangle$ is the average measurement value of $B$ when doing repeated measurements, provided the original state is $|v_2\rangle$ (but in your case, the state is not $|v_2\rangle$, but the mixed state you mention. There the average measurement value of $B$ is given by … – celtschk Apr 26 '14 at 10:45
• … $\operatorname{tr}(B\rho)=53/14$ which, not coincidentally equals $3 p(b=3) + 4 p(b=4) =$ $3 \left(p(a=1 \land b=3) + p(a=2 \land b=3)\right) + 4 \left(p(a=1 \land b=4) + p(a=2 \land b=4)\right)$.) – celtschk Apr 26 '14 at 10:49
• What i get from what are you saying is that the measure of A and B is done on the same state for both observables, therebefore the only way of measuring 2 and 4 is if we are measuring on $|v_2\rangle$ , since there is a $\frac{4}{7}$ weight on this state, the probability is just the probability of measuring $|v_2\rangle$ is $p({a=2}$ ^ ${b=4}) = \frac{4}{7}$ or we could just look at $\rho$ and see the probability of measuring that as you said. – spectator Apr 26 '14 at 14:16
• The state you're measuring is given by $\rho$. There may, or may not, be a $|v_1\rangle$ or $|v_2\rangle$ "underlying" that mixed state $\rho$. Quantum mechanics simply doesn't care how you produced it. But yes, if $\rho$ is created by randomly choosing between $|v_1\rangle$ and $|v_2\rangle$, then two measurements done at the same time necessarily are done also on the same of the two states. But again, it doesn't really matter. You can get the very same mixed state in infinitely many ways, and all of them are indistinguishable by quantum measurements. What you measure is just $\rho$. – celtschk Apr 26 '14 at 15:40