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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: \begin{equation} P(+\hat{z}) = 1/2, P(−\hat{z}) = 1/2, \end{equation} \begin{equation} P(\hat{x}) = 3/4, P(−\hat{x}) = 1/4, \end{equation} \begin{equation} P(+\hat{y}) = 0.067, P(−\hat{y}) = 0.933. \end{equation} 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?

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  • $\begingroup$ Why haven't you used the rest of the information given to you? $\endgroup$
    – J. Murray
    Commented Nov 6, 2020 at 20:12
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    $\begingroup$ @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 $\endgroup$
    – user2723984
    Commented Nov 6, 2020 at 20:16
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    $\begingroup$ @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. $\endgroup$
    – Voulkos
    Commented Nov 7, 2020 at 21:55
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    $\begingroup$ @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). $\endgroup$
    – user87745
    Commented Dec 29, 2020 at 1:27
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    $\begingroup$ @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. $\endgroup$
    – user87745
    Commented Dec 30, 2020 at 1:50

1 Answer 1

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REFERENCE : My answer here Understanding the Bloch sphere

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

Equation (24) in my answer above is

\begin{equation} \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} \end{equation} 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.

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See a 3d view of Figure-03 here

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