In the paper

"C. H. Bennett, H. J. Bernstein, S. Popescu, and B. Schumacher. Concentrating partial entanglement by local operations. Physical Review A, 53:2046–2052, 1996, quant-ph/9511030"

It is mentioned that if $n$ partly-entangled pairs of 2-state particles are shared between Alice and Bob, then the initial state is

$$\psi(A, B) = \prod\limits_{i=1}^{n} (\cos\theta |\alpha_{1}(i)\beta_{1}(i) \rangle + \sin\theta |\alpha_{2}(i) \beta_{2}(i)\rangle).$$

Let one of the parties (say Alice) perform an incomplete von Neumann measurement projecting the initial state into one of $n + 1$ orthogonal subspaces corresponding to the power $k = 0...n$ to which $\sinθ$ appears in the coefficient. The probability of outcomes is binomially distributed, with outcome $k$ having probability

$$p_{k} = \binom{n}{k}(\cos^{2}\theta)^{(n-k)}(\sin^{2}\theta)^{k}$$

I am not clear why $\cos^{2}\theta$ and $\sin^{2}\theta$ are used in finding the probability instead of $\cos\theta$ and $\sin\theta$

Is it because the probability is the square of amplitude and the amplitude is defined in terms of $\cos\theta$ and $\sin\theta$?

  • $\begingroup$ yes, that's the reason. I'm not sure there's much more to say about it. If $|\psi\rangle=\sum_k \alpha_k |k\rangle$ then the outcome probabilities are $|\alpha_k|^2$ $\endgroup$
    – glS
    Nov 25, 2022 at 9:08


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.