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Determine the state $|psi rangle$

Physics Asked by rand on November 9, 2020

The question is:

The angular momentum components of an atom prepared in the state $|psirangle$ 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:
$$
P(+hat{z}) = 1/2 = P(−hat{z})
$$

$$
|langle {uparrow}_z|psirangle|^2=1/2 =|langle {downarrow}_z|psirangle|^2
$$

$$
|psi rangle =alpha |{uparrow}_zrangle+ e^{iδb} beta|{downarrow}_zrangle
$$

$$
|langle {uparrow}_z|psirangle|= alpha = 1/sqrt(2).
$$

Similarly,
$$
|langle {downarrow}_z|psirangle|= beta = 1/sqrt(2).
$$

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

One Answer

REFERENCE : My answer here Understanding the Bloch sphere

$=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=$

Equation (24) in my answer above is

begin{equation} vertpsirangle boldsymbol{=}cosleft(dfrac{theta_3}{2}right)vert u_3rangle boldsymbol{+} e^{iphi_3}sinleft(dfrac{theta_3}{2}right)vert d_3rangle tag{24}label{24} end{equation} where $vert u_3rangle ,vert d_3rangle $ are yours $|{uparrow}_zrangle,|{downarrow}_zrangle$ 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

Correct answer by Frobenius on November 9, 2020

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