Homework Help: Constructing the input state given the probabilities

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1. Feb 11, 2017

Figaro

1. The problem statement, all variables and given/known data
The spin components of a beam of atoms prepared in the state $|\psi_{in}\rangle$ are measured and the following experimental probabilities are obtained,
$$P_{+}=\frac{1}{2}, \quad P_{+x}=\frac{3}{4}, \quad P_{+y}=0.067$$
$$P_{-}=\frac{1}{2}, \quad P_{-x}=\frac{1}{4}, \quad P_{-y}=0.933$$

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

3. 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?

Last edited: Feb 12, 2017
2. Feb 12, 2017

TSny

OK
I get something else here.

Try the same way you did for the projection onto $|+\rangle_x$ and $|-\rangle_x$

3. Feb 12, 2017

Figaro

I'm sorry, it should be $\theta = \frac{2\pi}{3}$, I edited my post.

4. Feb 12, 2017

Figaro

What I'm thinking is to project the input state to $|+\rangle_y$ and $|-\rangle_y$ then use the angles I've got and see which one would correspond to the probability given in the y measurement. Do you think this is a valid way, so far $\theta = \frac{2\pi}{3}$ matches the results.

5. Feb 12, 2017

TSny

The value of $\theta$ for the projection onto $|-\rangle_x$ must be consistent with $\theta$ for projection onto $|+\rangle_x$.

6. Feb 12, 2017

Figaro

So you mean $\theta = \frac{2\pi}{3}$ is not the angle I want?

7. Feb 12, 2017

TSny

I don't think so. For the projection onto $|+\rangle_x$ there is another possibility. That is, $\cos \theta = 1/2$ has more than one solution. Similarly, for $|-\rangle_x$ there will be more than one solution for $\theta$. Hopefully, one of the solutions for $|+\rangle_x$ matches one of the solutions for $|-\rangle_x$.

8. Feb 12, 2017

Figaro

I think to identify the correct angle is to test the angles on the projection of $|\psi_{in}\rangle$ onto $|+\rangle_y$ and $|-\rangle_y$
I've found that if I use $\theta = \frac{\pi}{3}$. The probability turns out to be what the experimental results say.

Last edited: Feb 12, 2017
9. Feb 12, 2017

TSny

Yes, I think that's the correct answer. It should give you the right probability for $|-\rangle_x$ also.

10. Feb 13, 2017

Figaro

Yes, I just need to correct myself in my initial post, the angles that I should get for both $|+\rangle_x$ and $|-\rangle_x$ are $\theta = ±\frac{\pi}{3}$, I made a lousy mistake when projecting $|\psi_{in}\rangle$ onto $|-\rangle_x$. It should be the same for both states, and I should verify which angle is the right one by projecting $|\psi_{in}\rangle$ onto $|+\rangle_y$ and $|-\rangle_y$.

11. Feb 13, 2017

TSny

I'm not quite getting that. I'm geting that it's $\theta = ±\frac{\pi}{3}$ for $|+\rangle_x$, but it's $\theta = \frac{\pi}{3}$ and $\theta =$ something else for $|-\rangle_x$.

12. Feb 13, 2017

Figaro

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

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

13. Feb 13, 2017

TSny

Yes, you are right. I didn't bother to look back at my notes and I was confusing the case of $|-\rangle_x$ with $|+\rangle_y$ and $|-\rangle_y$ (where I think you get $\theta = \pi/3$ and $\theta = 2\pi/3$). Sorry, for causing you to have to take the time to correct me.

Good work.

14. Feb 13, 2017

Figaro

At least we got the chance to discuss this. Thanks for the discussion!