How does one determine the possible outcomes of a set of successive measurements in the bases $Y_a$, $X_b$, $Y_c$ for a set of entangled states, for example of the form:

$$\psi = \frac{1}{\sqrt{2}}(|0\rangle_A|0\rangle_B|0\rangle_C-|1\rangle_A|1\rangle_B|1\rangle_C)?$$

I know that the y-matrix has form $\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$ with eigenstates $\begin{align} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ \pm i \end{bmatrix} \end{align}$ while the x-matrix has form $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ with eigenvectors $\begin{align} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ \pm 1 \end{bmatrix} \end{align}$, but I am not sure how measurement works with a state of entangled qubits.

Is the ideal approach to first express the overall state in terms of the eigenstates of the upcoming measurement basis, then to determine all possible outcomes for measuring a given bit A, B, or C? Or is there a more efficient way? Intuitively, it seems like since the initial state $\psi$ is a superposition of the Y eigenstates that the measurement could yield either result after measuring $Y_a$ and $Y_c$, but I know the result must contain an odd number of ones.

  • $\begingroup$ Just to make clear a point you just mention in passing: $\newcommand{\ket}[1]{\mid #1 \rangle} \newcommand{\bra}[1]{\langle #1 \mid} \newcommand{\bra}[1]{\langle #1 \mid} \newcommand{\braket}[2]{\langle #1 \mid #2 \rangle}\ket{0}_X$ is an eigenstate of $Y_X$? $\endgroup$ – user154997 Oct 31 '17 at 12:47
  • $\begingroup$ Sorry for the confusion, $\psi$ is actually written in the eigenstates of $Z$! $\endgroup$ – Alekxos Oct 31 '17 at 15:36

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