# Determine the state $|\psi \rangle$

The question is:

The angular momentum components of an atom prepared in the state $$|\psi\rangle$$ are measured and the following experimental probabilities are obtained: $$$$P(+\hat{z}) = 1/2, P(−\hat{z}) = 1/2,$$$$ $$$$P(\hat{x}) = 3/4, P(−\hat{x}) = 1/4,$$$$ $$$$P(+\hat{y}) = 0.067, P(−\hat{y}) = 0.933.$$$$ From this experimental data, determine the state $$|\psi \rangle$$. Note that in performing the measurements, the state $$|\psi \rangle$$ is prepared again and again.

My attempt: $$P(+\hat{z}) = 1/2 = P(−\hat{z})$$ $$|\langle {\uparrow}_z|\psi\rangle|^2=1/2 =|\langle {\downarrow}_z|\psi\rangle|^2$$ $$|\psi \rangle =\alpha |{\uparrow}_z\rangle+ e^{iδb} \beta|{\downarrow}_z\rangle$$ $$|\langle {\uparrow}_z|\psi\rangle|= \alpha = 1/\sqrt(2).$$ Similarly, $$|\langle {\downarrow}_z|\psi\rangle|= \beta = 1/\sqrt(2).$$

However, I don't know how to find $$e^{iδb}$$ term. Could someone please give a hint?

• Why haven't you used the rest of the information given to you? Nov 6, 2020 at 20:12
• @rand do the same thing, but with $\langle \uparrow_x|\psi\rangle$ etc... , you will just have to calculate overlaps such as $\langle \uparrow_x|\uparrow_z\rangle$ etcetera and you'll be done Nov 6, 2020 at 20:16
• @Qmechanic : IMO, this question must not be tagged as homework-and-exercises since it's a chance for users to learn about Bloch sphere, a tool beyond the narrow frame of an exercise. If you don't agree I'll change my answer to a hint removing the results or I'll delete it at all. Nov 7, 2020 at 21:55
• @Frobenius Very relatedly, it's an excellent way to introduce the concept of quantum tomography and to illustrate how the phases store the information about probabilities of non-commuting operators (non-commuting w.r.t. the operator in whose eigenbasis we are expanding).
– user87745
Dec 29, 2020 at 1:27
• @Frobenius To clarify, I think the homework-and-exercises tag should stay because it is clearly a type of question that would fall under that category. Just that it should not be deleted because it is a good question.
– user87745
Dec 30, 2020 at 1:50

REFERENCE : My answer here Understanding the Bloch sphere

$$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$$

Equation (24) in my answer above is

$$$$\vert\psi\rangle \boldsymbol{=}\cos\left(\dfrac{\theta_3}{2}\right)\vert u_3\rangle \boldsymbol{+} e^{i\phi_3}\sin\left(\dfrac{\theta_3}{2}\right)\vert d_3\rangle \tag{24}\label{24}$$$$ where $$\vert u_3\rangle ,\vert d_3\rangle$$ are yours $$|{\uparrow}_z\rangle,|{\downarrow}_z\rangle$$ respectively.

From the given probabilities $$P(+\hat{z}) = 1/2, P(−\hat{z}) = 1/2$$ the state lies on the "equator" of the Bloch sphere. So from Figure-01 in my REFERENCED answer $$\theta_3=\pi/2$$. The angle $$\phi_3=\boldsymbol{-}\pi/3$$ could be found from one only of the probabilities $$P(\hat{x}) = 3/4, P(−\hat{x}) = 1/4,P(+\hat{y}) = 0.067$$, $$P(−\hat{y}) = 0.933$$ and Figure-02 in my REFERENCED answer.

Note : I suggest you to "study" the excellent @CR Drost's answer about the Bloch sphere in above link.

See a 3d view of Figure-03 here