Finding ##S_x## eigenstate using experiments

In summary, the author is discussing the Stern Gerlach experiment and how the results can be used to determine the eigenstates of a particle in terms of the eigenstates of its spin partner. There is some sign ambiguity in the use of the experimental results, and two other conventions are introduced to resolve the issue.
  • #1
Kashmir
468
74
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?
 
Physics news on Phys.org
  • #2
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.
 
  • Like
Likes Kashmir, Doc Al and PeroK
  • #3
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 ?
 
  • #4
DrClaude said:
but a choice is made to recover a right-handed system of coordinates
Can you also please explain this?
 
  • #5
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##.
 
  • Like
Likes Kashmir

Similar threads

  • Quantum Physics
Replies
9
Views
977
Replies
1
Views
570
Replies
9
Views
1K
Replies
24
Views
2K
  • Quantum Physics
Replies
4
Views
695
  • Quantum Physics
Replies
3
Views
977
  • Quantum Physics
Replies
8
Views
880
  • Quantum Physics
Replies
2
Views
838
Replies
2
Views
603
  • Quantum Physics
Replies
31
Views
1K
Back
Top