- #1

Figaro

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## Homework Statement

The spin components of a beam of atoms prepared in the state ##|\psi_{in}\rangle## are measured and the following experimental probabilities are obtained,

$$P_{+}=\frac{1}{2}, \quad P_{+x}=\frac{3}{4}, \quad P_{+y}=0.067$$

$$P_{-}=\frac{1}{2}, \quad P_{-x}=\frac{1}{4}, \quad P_{-y}=0.933$$

## Homework Equations

##P_+=|\langle+|\psi_{in}\rangle|^2 \quad## Probability of the input state to be in the up state (z-direction)

## The Attempt at a Solution

I can write ##|\psi_{in}\rangle## as a linear combination of the up and down state,

$$|\psi_{in}\rangle = a|+\rangle + b|-\rangle$$

by projecting the input state to the up state, we get ##a=\frac{1}{2}\,## and similarly for the down state, ##\, b=\frac{1}{2}##.

Note that the coefficients are complex numbers so there is a phase term, but the overall phase is not physically interesting and we only consider the phase difference of the coefficients. So,

$$|\psi_{in}\rangle = \sqrt{\frac{1}{2}}(|+\rangle + e^{i\theta}|-\rangle)$$

to determine ##\theta##, we project the previous equation to the ##|+\rangle_x = \sqrt\frac{1}{2}(|+\rangle + |-\rangle)## and ##|-\rangle_x = \sqrt\frac{1}{2}(|+\rangle - |-\rangle)##

$$|_x\langle+|\psi_{in}\rangle|^2 = \frac{3}{4} = \frac{1}{4}(1+e^{i\theta})(1+e^{-i\theta}) \rightarrow 1 = e^{i\theta} + e^{-i\theta} = 2cos(\theta)$$

This shows that ##\theta = \frac{\pi}{3}##, by similar projection for ##|-\rangle_x##, we get ##\theta = \frac{2\pi}{3}##

In this case, I have to know which ##\theta## I should choose so I think I should project it to ##|+\rangle_y## and ##|-\rangle_y##. But should I just do trial and error and see where the results make sense?

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