Given the entangled state

\begin{equation} |\Phi^+\rangle = \frac{1}{\sqrt 2} |00\rangle + \frac{1}{\sqrt 2} |11\rangle \end{equation}

I am trying to calculate the probability that the two qubits end up being the same when measured in different bases and getting different answers with 2 different methods.

Method 1 is given by Section 2.2 p. 21 of this lecture note and reproduced below for convenience:

enter image description here

The answer according to this is $\cos^2 \theta$

Method 2: Consider the expected value of the product of the qubit spins (measuring qubit is technically measuring its spin).

\begin{equation} \begin{split} E[s_a \times s_b] & = (s_a \times s_b) \textrm{Prob}(s_a = s_b) + (s_a \times s_b) \textrm{Prob}(s_a \neq s_b) \\ & = +1 \cdot \textrm{Prob}(s_a = s_b) - 1 \cdot \textrm{Prob}(s_a \neq s_b) \\ & = p - (1 - p) \\ & = 2 p - 1 \end{split} \end{equation}

which gives:

\begin{equation} \textrm{Prob}(s_a = s_b) = \frac{1 + E[s_a \times s_b]}{2} \end{equation}

and $E[s_a \times s_b]$ turns out to be equal to $a_x b_x - a_y b_y + a_z b_z$ where $\vec{a}$ and $\vec{b}$ are the axes chosen for measurement of the spins. See equation 11 in this for reference. I checked by hand and the formula is correct.

Method 1 $\neq$ Method 2. What gives?

  • 2
    $\begingroup$ why do you think that the expectation value of the product of the spins should equal the probability of them being equal? $\endgroup$
    – glS
    Jun 30 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.