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

[Dreak]

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?

 

[wotanub]

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.