# Constructing the input state given the probabilities

• Figaro
In summary: Yes, I think that's the correct answer. It should give you the right probability for ##|-\rangle_x## also.
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?

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Figaro said:
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##

TSny said:
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.

TSny said:
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.

Figaro said:
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##.

TSny said:
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?

Figaro said:
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##.

TSny said:
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.

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

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##.

Figaro said:
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##.

TSny said:
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))##
Thus ##\theta = ±\frac{\pi}{3}##

Figaro said:
##|-\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))##
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.

TSny said:
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!

## 1. What is the purpose of constructing the input state given probabilities?

The purpose of constructing the input state given probabilities is to create a representation of a system or experiment that takes into account the likelihood of different outcomes. This allows for more accurate predictions and analysis of the system.

## 2. How is the input state constructed from probabilities?

The input state is constructed by assigning each possible outcome of the system a probability value, which can then be used to calculate the overall probability of the system. This information is then used to create a vector or matrix representation of the input state.

## 3. What factors influence the construction of the input state?

The construction of the input state is influenced by the number of possible outcomes, the probabilities assigned to each outcome, and any constraints or relationships between the outcomes. Other factors such as external variables or experimental conditions may also play a role.

## 4. Can the input state be constructed for any type of system or experiment?

Yes, the input state can be constructed for any type of system or experiment as long as the probabilities of the outcomes can be determined. This includes both simple and complex systems, and can also be applied to quantum mechanical systems.

## 5. How is the input state used in scientific research?

The input state is used in scientific research to model and analyze various systems and experiments. It allows for the calculation of probabilities and prediction of outcomes, which can then be compared to experimental results to validate or refine the model. The input state is also a key component in quantum computing and other applications of quantum mechanics.

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