How can I find the S_{x} operator using spin base transformation?

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SUMMARY

The discussion centers on the transformation of spin operators, specifically the S_{x} operator, using the Z spin base. The Z spin base is defined with the matrix representation of 1 0; 0 1, while the X spin states are given as |s_x, +> and |s_x, ->. The confusion arises when attempting to derive the X matrix, which is expected to be 0 1; 1 0, but does not align with the Z basis. The solution involves using an S matrix for transformation and determining the operator from its eigenvalues and eigenvectors before rotating it to the Z basis.

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Dreak
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There is something I'm struggling with and I can't seem to find the problem.


We have the Z spinbase with:
z = (1/sqrt(2))² <BRA|*(|s_z,+> + |s_z,->)
which gives following z matrix:

1 0
0 1


and we have for X:

|s_x, +> = 1/sqrt(2) |s_z,+> + |s_z,->)
|s_x, -> = 1/sqrt(2) |s_z,+> - |s_z,->)

Now I have a problem with making the x matrix.
this one is equal to

0 1
1 0

but this doesn't fit with the base above?
for example the first component:
<s_x,+|s_x,+> = 1/2 {<s_z,+|s_z,+> + <s_z,+|s_z,-> + <s_z,-|s_z,+> + <s_z,-|s_z,-> }

<s_z,+|s_z,+> = <s_z,-|s_z,-> = 1
<s_z,+|s_z,-> = <s_z,-|s_z,+> = 0 because of orthogonality,


so we get that <s_x,+|s_x,+> = 1 instead of 0?

What do I do wrong?
 
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It's not that simple...

To transform the operator to a different basis, you need to use an "S matrix." I'll represent it by [itex]\mathbb{S}[/itex] so it's less confusing because the spin operator normally use S.

[itex]S_{z} \stackrel{→}{_{x}} \mathbb{S}^{\dagger}S_z\mathbb{S}[/itex]

where [itex]\mathbb{S} → \bigl(\begin{smallmatrix} \langle +z\:|+x \rangle&\langle +z\:|-x \rangle\\ \langle -z\:|+x \rangle&\langle -z\:|-x \rangle \end{smallmatrix} \bigr)[/itex]

But this won't get you the [itex]S_{x}[/itex] operator... To do this you need to determine the operator from its eigenvalues and eigenvectors, then rotate it to the z basis.
 
Last edited:

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