Exam 2018 Ex1

Exam 2018 Ex1

par Stefan Eric,
Number of replies: 1

Hello,

I am a bit confused in the first question when we calculate the probabilities. Assume we are looking for P(yi=xi=0|ei=di=0), since ei=di=0 the qubit sent by Alice will be measure by Bob in the basis {|alpha>, |alpha_p>}. We assume that xi=0 so Bob receives the qubit |0> from Alice and wants to perform a measurement. So I assumed that it has to used the projector P = |alpha><alpha| + |alpha_p><alpha_p| and will get the state P|0>. And then we we calculate the probability to have yi=0 and this happens when the measured state is |alpha>, and the following operation would give us that solution : |<alpha|P|0>|^2 and this results in |<alpha|0>|^2 = |cos(alpha)<0|0> + sin(alpha)<1|0>|^2 = cos(alpha)^2. Is this the correct reasoning or am I missing something?

Thank you.

In reply to Stefan Eric

Re: Exam 2018 Ex1

par Nicolas Macris,
Hi, not sure your reasoning is correct. First when you write
P = |alpha><alpha| + |alpha_p><alpha_p| well this is not a measurement projector here and its just the identity matrix: the sum of two orthogonal projectors in a 2-dim space. There are two projectors in this situation P_1  = |alpha><alpha| and P_2 = |alpha_p> <alpha_2 |.

Here is a simpler way of looking at things. When Bob receives |0> from Alice the prob to get y_i=0 is the prob that the state is projected to |alpha> after Bob's measurement. There are two ways to compute this depending how you apply Born's rule:

prob = |<final state | initial state >|^2 = |<alpha | 0>|^2 = (cos alpha)^2

simply !

Your expression |<alpha|P|0>|^2 is meaningless... you are confusing with the fact that

prob = |<alpha | 0>|^2 = <0|alpha> <alpha | 0> = <0| P_1|0>

which is another way of writing the Born rule.

Hope all this clarifies.
N.M