Constructing the input state given the probabilities

[Figaro]

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?

 

[TSny]

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 $\theta = \frac{\pi}{3}$

OK

by similar projection for $|-\rangle_x$, we get $\theta = \frac{5\pi}{6}$

I get something else here.

I think I should project it to $|+\rangle_y$ and $|-\rangle_y$ but I'm not sure how to do it and interpret the result.

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

 

[Figaro]

OK

I get something else here.

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

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

 

[Figaro]

OK

I get something else here.

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

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.

 

[TSny]

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

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

 

[Figaro]

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

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

 

[TSny]

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

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 $|-\rangle_x$.

 

[Figaro]

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 $|-\rangle_x$.

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.

 

[TSny]

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

 

[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 $|-\rangle_y$.

 

[TSny]

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'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 $|-\rangle_x$.

 

[Figaro]

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 $|-\rangle_x$.

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

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

 

[TSny]

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

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

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.

 

[Figaro]

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.

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