- #1
RJLiberator
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Homework Statement
Suppose that two measurements are made on a qubit in rapid succession. The first is δz and the second is δn. Suppose the first results is always +1. Calculate the probabilities of obtaining the results +/- 1 for the second measurement in terms of the angle θ between z and n. Show that <σn> = cos(θ) for the second measurement.
Homework Equations
δz = (1, 0), (0,1) In this problem we only use (1,0)
δn = 1/sqrt(2) ( sqrt(1+nz)), sqrt(1-nz) e^(iθz) ), 1/sqrt(2) ( sqrt(1-nz), -sqrt(1+nz)e^(iθz) )
<σn> = average
The Attempt at a Solution
The diagram of the situation and calculations are pretty straight forward here.
The problem is, how do I calculate the probabilities in terms of the angle between the vectors? I don't know how to interpret this.
prob(+1) = (1+nz)/2
prob(-1) = (1-nz)/2
prob(+1)+prob(-1) = 1 which checks out.
The average value <σn> is equal to nz as we take prob(+1)-prob(-1).
So, I have all the answers, but how do I do this in terms of the angle between them? I thought it was a definition that the average aka <σn> is equal to cos(θ)...In my notes, I have that prob(+1) = cos^2(θ/2)
prob(-1) = sin^2(θ/2)
but I'm not sure how this would help.
Any hints on how to do this?