I Finding ##S_x## eigenstate using experiments

Kashmir
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Quantum mechanics, McIntyre, pg 62
IMG_20220426_181555.JPG

For above spin ##1## Stern Gerlach experiment a set of results is
"##
\begin{array}{c}
\mathcal{P}_{1 x}=\left.\left.\right|_{x}\langle 1 \mid 1\rangle\right|^{2}=\frac{1}{4} \\
\mathcal{P}_{0 x}=\left.\left.\right|_{x}\langle 0 \mid 1\rangle\right|^{2}=\frac{1}{2} \\
\mathcal{P}_{-1 x}=\left.\right|_{x}\left(-\left.1|1\rangle\right|^{2}=\frac{1}{4},\right.
\end{array}
##
as illustrated in Fig. 2.12. These experimental results can be used to determine the ##S_{x}## eigenstates in terms of the ##S_{z}## basis
##
\begin{array}{l}
|1\rangle_{x}=\frac{1}{2}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{2}|-1\rangle \\
|0\rangle_{x}=\frac{1}{\sqrt{2}}|1\rangle-\frac{1}{\sqrt{2}}|-1\rangle \\
|-1\rangle_{x}=\frac{1}{2}|1\rangle-\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{2}|-1\rangle
\end{array}
##"To find the ##S_{x}## eigenstates in terms of the ##S_{z}## basis I need two more similar experiments in which the input to Sx analyzer are ##0,-1## spin particles respectively.

However I am getting a sign ambiguity while using the experimental results.

Below are two expressions which both are in line with the experiments:

* ##|1\rangle_{x}=\frac{1}{2}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{2}|-1\rangle##
##|0\rangle_{x}=\frac{1}{\sqrt{2}}|1\rangle-\frac{1}{\sqrt{2}}|-1\rangle##
##|-1\rangle_{x}=\frac{1}{2}|1\rangle-\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{2}|-1\rangle ##

* ##|1\rangle_{x}=\frac{1}{2}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{2}|-1\rangle##
##|0\rangle_{x}=\frac{1}{\sqrt{2}}|1\rangle+\frac{1}{\sqrt{2}}|-1\rangle##
##|-1\rangle_{x}=\frac{1}{2}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{2}|-1\rangle ##How do we resolve this?
 
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Kashmir said:
* ##|1\rangle_{x}=\frac{1}{2}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{2}|-1\rangle##
##|0\rangle_{x}=\frac{1}{\sqrt{2}}|1\rangle+\frac{1}{\sqrt{2}}|-1\rangle##
##|-1\rangle_{x}=\frac{1}{2}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{2}|-1\rangle ##
These states are not orthogonal.

Note that there is also some convention coming in. It is possible to find other linear combinations that satisfy the experiment, but a choice is made to recover a right-handed system of coordinates. It is also conventional to take spin along x to have real coefficients in the Sz basis.
 
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DrClaude said:
These states are not orthogonal.

Note that there is also some convention coming in. It is possible to find other linear combinations that satisfy the experiment, but a choice is made to recover a right-handed system of coordinates. It is also conventional to take spin along x to have real coefficients in the Sz basis.
Here are another orthogonal vectors ##|1\rangle_{x}=\frac{1}{2}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{2}|-1\rangle##

##|0\rangle_{x}=\frac{1}{\sqrt{2}}|1\rangle+\frac{1}{\sqrt{2}}|-1\rangle##

##|-1\rangle_{x}=\frac{1}{2}|1\rangle-\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{2}|-1\rangle ##

So the other conventions as you mentioned are used to pick out a convenient set of orthogonal expressions ?
 
DrClaude said:
but a choice is made to recover a right-handed system of coordinates
Can you also please explain this?
 
Kashmir said:
Here are another orthogonal vectors ##|1\rangle_{x}=\frac{1}{2}|1\rangle+\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{2}|-1\rangle##

##|0\rangle_{x}=\frac{1}{\sqrt{2}}|1\rangle+\frac{1}{\sqrt{2}}|-1\rangle##

##|-1\rangle_{x}=\frac{1}{2}|1\rangle-\frac{1}{\sqrt{2}}|0\rangle-\frac{1}{2}|-1\rangle ##
Yes, this would be a possibility.

Kashmir said:
Can you also please explain this?
You want, for instance, that a rotation by ##\pi/2## around the y-axis will transform ##\ket{1}_x## into ##\ket{1}_z##.
 
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