# What are the outcome probability in entanglement distillation?

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 $$\sin\theta$$?

• 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 $|\alpha_k|^2$
– glS
Nov 25, 2022 at 9:08