Quantum Measuring Sequence

If $A$ is measured and $a_1$ is obtained, the state of the system after the measurement is $\left |\psi_1\right\rangle$ . Then, $B$ is now measured with outcomes $b_1$ of probability $|\left\langle \phi_1 \middle| \psi_1\right\rangle|^2 = \frac{9}{25}$ and $b_2$ of probability $|\left\langle \phi_2 \middle| \psi_1\right\rangle|^2 = \frac{16}{25}$ .

Now $A$ is measured again, we can express $\left |\phi_1\right\rangle$ and $\left |\phi_2\right\rangle$ as:

$\left |\phi_1\right\rangle = \frac{3}{5} \left |\psi_1\right\rangle + \frac{4}{5} \left |\psi_2\right\rangle$

$\left |\phi_2\right\rangle = \frac{4}{5} \left |\psi_1\right\rangle - \frac{3}{5} \left |\psi_2\right\rangle$

The probability of obtaining $a_1$ if we knew the outcome of the measurement of $B$ as $b_1$ and the system is in state $\left |\phi_1\right\rangle$ is $|\left\langle \psi_1 \middle| \phi_1\right\rangle|^2 = \frac{9}{25}$ . The probability of obtaining $a_1$ if we knew the outcome of the measurement of $B$ as $b_2$ and the system is in state $\left |\phi_2\right\rangle$ is $|\left\langle \psi_1 \middle| \phi_2\right\rangle|^2 = \frac{16}{25}$ . The total probability of obtaining $a_1$ in the series of measurement $A \rightarrow B \rightarrow A$ is:

$P =$ Probability of obtaining $a_1$ after obtaining $b_1$ + Probability of obtaining $a_1$ after obtaining $b_2$

$P = |\left\langle \phi_1 \middle| \psi_1\right\rangle|^2 |\left\langle \psi_1 \middle| \phi_1\right\rangle|^2 + |\left\langle \phi_2 \middle| \psi_1\right\rangle|^2 |\left\langle \psi_1 \middle| \phi_2\right\rangle|^2$

$P = \frac{9}{25}^2 + \frac{16}{25}^2$ = 0.539

Hey. i guess you should've stated the initial state of the system. Or else we won't have a definite answer.

Assuming that the first measurement of $A$ yields $a_{1}$ .

Thus the system is in $\psi_{1}$ .

So, the probability of getting $a_{1}$ is...

$(\frac{3}{5})^{2}(\frac{3}{5})^{2} + (\frac{4}{5})^{2}(\frac{4}{5})^{2} = \frac{337}{625} \approx \boxed{0.539}$