1
$\begingroup$

I have been reading up on Bernoulli factories(if you do not know what that is don't worry the question is not related to that). The algorithm by Dale et. al proposes using conditional probability for the measurement of bell states. I am unable to understand how this is happening. The original state is $$|p> = \sqrt p|0> + \sqrt {1-p}|1>$$

Here is a snippet from the paper enter image description here

If one evaluates the measurement operators for bell basis states we have 4 operators, $M_0 = |\psi^+><\psi^+|$, $M_1 = |\psi^-><\psi^-|$, $M_2 = |\phi^+><\phi^+|$ and $M_3 = |\phi^-><\phi^-|$

it can be seen that $$<pp|M_0|pp> = 2p(1-p)$$ $$<pp|M_3|pp> = \frac{(2p-1)^2}{2}$$

which is not equalent to half. Where am I going wrong?

$\endgroup$

1 Answer 1

2
$\begingroup$

Since you don't give your calculations, it's a bit hard to say where you're going wrong! If I do the calculations, I get \begin{align*} \langle pp|M_0|pp\rangle&=2p(1-p) \\ \langle pp|M_3|pp\rangle&=(2p-1)^2/2 \end{align*} which seem to match with the results that you state. But it's also consistent with what they say (for a certain interpretation. I'll concede I understood it the way you understood it on first reading):

the total probability of getting either answer $M_0$ or $M_3$ is $2p(1-p)+(2p-1)^2/2=\frac12$

So, now assume that you've done the measurement, and you've had the 50% of cases where you got either $M_0$ or $M_3$, but assume you don't know which. What are the chances that you got $M_0$? $$ \frac{p_0}{p_0+p_3}=2p_0=4p(1-p). $$ Sorted!

$\endgroup$
2
  • $\begingroup$ Thanks @DaftWullie. I see how I got lost in the words(The calculations that you and I did are similar, considering we are getting same results). Follow up question - let's say I were to implement this(maybe in qiskit) and I would get the counts for all the basis(let's call them m0,m1,m2 and m3) then we can simply do m0/(m0+m3). Just to be clear there is no way to restrict the actual measurment operation on m0 and m3, right? Like we do changing the basis or something? $\endgroup$ Commented May 28 at 7:41
  • $\begingroup$ Technically, you could do something like amplitude amplification, but it's too much effort. You're much better off just performing the measurement of all 4 operators (although there are only 3 possible outcomes) and post-selecting on the cases where your desired measurement results come out. $\endgroup$
    – DaftWullie
    Commented May 28 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.