Partial Measure Probability

Let be a $$|\psi\rangle = \dfrac{3}{5\sqrt{2}}|00 \rangle- \dfrac{3i}{5\sqrt{2}}|01 \rangle+ \dfrac{2\sqrt{2}}{5}|10 \rangle - \dfrac{2\sqrt{2} i}{5}|11 \rangle$$

state with two qubits. Make a partial measurement, in computational basis, of this state, using $Z_1 = O_A \otimes I_B$ and $Z_2 = I_A \otimes O_B.$

1) Suppose that, after measuring $Z_1$, we measure $Z_2.$

2) Suppose that, after measuring $Z_2$, we measure $Z_1.$

If the intermediate quantum states ate disregarded, is there a difference between $p_0^{1)}$ and $p_0^{2)}$?. Where $p_0^{1)}$ is a probability associate to $|00 \rangle$ for the 1) and $p_0^{2)}$ is a probability associate to $|00 \rangle$ for the 2).

I trying

1) Measure $Z_1$. A probability of obtain the $0\leq k \leq 2-1 = 1$ than outcomes measure is $$p_k=\sum_{j=0}^{2^1-1 = 1} |a_{kj}|^2$$. Then $p_0= |a_{00}|^2 + |a_{01}|^2 = \dfrac{9}{25}$ and $p_1= |a_{10}|^2 + |a_{11}|^2 = \dfrac{16}{25}.$ If the result is $0$ the state after apply $Z_1$ $$\dfrac{1}{\sqrt{p_k}}|k\rangle \left(\sum_{j=0}^{1} a_{kj}|j\rangle\right) = \dfrac{1}{\sqrt{p_0}}|0\rangle \left(a_{00}|0\rangle + a_{01}|1 \rangle\right) = \dfrac{5}{3}|0\rangle (\dfrac{3}{5\sqrt{2}}|0\rangle - \dfrac{3i}{5\sqrt{2}}|1 \rangle).$$

Measure $Z_2$ In this case we measure in the second qubit. A probability of obtain $0\leq k \leq 1$ value than result is $p_k = \sum_{j=0}^{1} |a_{jk}|^2$. From previous state

$p_0^{1)} = |a_{00}|^2 + |a_{10}|^2 = 1/2 + 0 = 1/2$

2) Measure $Z_2$ A probabilty of obtain the $0\leq k \leq 2-1 = 1$ value than outcomes of measure $Z_2$ is $$p_k=\sum_{j=0}^{2^1-1 = 1} |a_{jk}|^2.$$ Then $p_0= |a_{00}|^2 + |a_{10}|^2 = \dfrac{1}{2}$ and $p_1= |a_{01}|^2 + |a_{11}|^2 = \dfrac{1}{2}.$ If outcome after apply $Z_2$ is $0$, the state is $$\dfrac{1}{\sqrt{p_k}} \left(\sum_{j=0}^{1} a_{jk}|j\rangle\right)|k\rangle = \dfrac{1}{\sqrt{p_0}} \left(a_{00}|0\rangle + a_{10}|1 \rangle\right)|0\rangle = \sqrt{2}(\dfrac{3}{5\sqrt{2}}|0\rangle + \dfrac{2\sqrt{2}}{5}|1 \rangle)|0\rangle.$$

Measure $Z_1$ In this case we measure the first qubit. A probability of obtain the $0\leq k \leq 1$ value than measure is $p_k = \sum_{j=0}^{1} |a_{kj}|^2$. From the previous state $p_0^{2)} = |a_{00}|^2 + |a_{01}|^2 = 9/25.$

Question: For me exist such difference, but my professor say this is wrong. For you?

-