# Calculating probability that two entangled qubits are the same when measured in different bases

Given the entangled state

$$$$|\Phi^+\rangle = \frac{1}{\sqrt 2} |00\rangle + \frac{1}{\sqrt 2} |11\rangle$$$$

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:

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).

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

which gives:

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

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?

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