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On the orthogonality of the spin operator in a $z$-basis

Physics Asked on January 18, 2021

After, in the $S_z$-basis $|S_{z,pm}rangle$ denoted by $|pmrangle$ and in units $frac{hbar}{2}$, finding the spin operator in a general direction $vec{n} = (sinthetacosphi,sinthetasinphi,costheta)$:

$$hat{vec{S}}cdotvec{n} = sinthetacosphibegin{bmatrix}
0 & 1
1 & 0
end{bmatrix} + sinthetasinphibegin{bmatrix}
0 & -i
i & 0
end{bmatrix} + costhetabegin{bmatrix}
1 & 0
0 & -1
end{bmatrix} = begin{bmatrix}
costheta & e^{-iphi}sintheta
e^{iphi} sintheta & -costheta
end{bmatrix}$$

I find the eigenvectors
$$|S_{vec{n},+}rangle = begin{bmatrix}
costheta/2
e^{iphi}sintheta/2
end{bmatrix}text{ and }|S_{vec{n},-}rangle = begin{bmatrix}
sintheta/2
-e^{iphi}costheta/2
end{bmatrix}$$

Now it is normally quite easy to see that these should be orthogonal but for some reason I just don’t get $langle S_{vec{n},+}|S_{vec{n},-}rangle$ to equal zero:
$$langle S_{vec{n},+}|S_{vec{n},-}rangle = cosfractheta2 sinfractheta2left(1-e^{2iphi}right)$$
what is going wrong? thanks in advance.

One Answer

Take the complex conjugate when you go from bra to ket. So $e^{i phi} to e^{-i phi}$ and the bracket is $(1-1)=0$

Correct answer by RogerJBarlow on January 18, 2021

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