I confused about how to calculate the PROBABILITIES and getting a certain result of measuring Bell's states with Pauli matrices as the operator. When you measure something, the state involved would be projected onto an eigenstate of the observable.

Given $|ψ⟩ = \frac{1}{\sqrt2} (|01⟩ + |10⟩)$ as the state and $σy = \left(\begin{matrix}0&-i\\i&0\\\end{matrix}\right)$ as the observable.

How to calculate probability on the first qubit actually? What is the state after measurement?

  • $\begingroup$ please use meaningful titles for your posts $\endgroup$
    – glS
    Nov 17, 2020 at 16:26
  • 1
    $\begingroup$ this is essentially a duplicate of Probabilities of entangled state $\endgroup$
    – glS
    Nov 17, 2020 at 16:27
  • $\begingroup$ okay, noted. thank you. $\endgroup$ Nov 18, 2020 at 6:28

1 Answer 1


The matrix $\sigma_Y$ has two eigenvalues $\pm 1$, so the one-qubit projectors for the measurement are $$ \frac12(I\pm\sigma_Y). $$ Since you're measuring a two-qubit state, you need to express the measurement projects on both qubits. Since you're doing nothing to the second qubit, that's just the identity matrix: $$ P_{\pm}=\frac12(I\pm\sigma_Y)\otimes I. $$ You can check that $P_++P_-=I\otimes I$.

Now measurement follows the standard formalism. The probabilities of getting either outcome are $$ p_{\pm}=\langle\psi|P_{\pm}|\psi\rangle $$ and the possible states after the measurement are $$ |\psi_{\pm}\rangle=P_{\pm}|\psi\rangle/\sqrt{p_{\pm}}. $$


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