91Ó°ÊÓ

Two vectors are orthogonal in Euclidean space if and only if their dot product is zero, i.e. they form a 90° (/2 radian) angle, or one of the vectors is zero.

In this problem show that for

${\mathrm{ψ}}_{Â\pm }^0={\mathrm{Î\pm }}_{Â\pm }{\mathrm{ψ}}_{\mathrm{a}}^0+{\mathrm{β}}_{Â\pm }{\mathrm{ψ}}_{\mathrm{b}}^0$

where

$$\begin{gathered} {\mathrm{Î\pm }}_{Â\pm }{W}_{aa}+{\mathrm{β}}_{Â\pm }{W}_{ab} \\ ={\mathrm{Î\pm }}_{Â\pm }{E}_{Â\pm }^1{\mathrm{Î\pm }}_{Â\pm }{W}_{ba}+{\mathrm{β}}_{Â\pm }{W}_{bb} \\ ={\mathrm{β}}_{Â\pm }{E}_{Â\pm }^1{E}_{Â\pm }^1 \\ =\frac{1}{2}\left[W_{aa}+W_{bb}Â\pm \sqrt{(W_{aa}-W_{bb}{)}^2+4|W_{ab}{|}^2}\right] \end{gathered}$$

Its given that

$⟨{\mathrm{ψ}}_{+}^0âˆ\pounds {\mathrm{ψ}}_{-}^0âŸ\copyright =0$

This gives

$$\begin{gathered} ⟨{\mathrm{ψ}}_{+}^0âˆ\pounds {\mathrm{ψ}}_{-}^0âŸ\copyright \\ =\left({\mathrm{Î\pm }}_{+}^{*}{\mathrm{ψ}}_{\mathrm{a}}^0\right|+{\mathrm{β}}_{-}^{*}⟨{\mathrm{ψ}}_{\mathrm{b}}^0\left|\right)\left({\mathrm{Î\pm }}_{-}\right|{\mathrm{ψ}}_{\mathrm{a}}^0âŸ\copyright +{\mathrm{β}}_{-}|{\mathrm{ψ}}_{\mathrm{b}}^0âŸ\copyright )âŸ\copyright \\ ={\mathrm{Î\pm }}_{-}^{*}{\mathrm{Î\pm }}_{+}+{\mathrm{β}}_{-}^{*}{\mathrm{β}}_{+} \end{gathered}$$

where it has been used$⟨{\mathrm{ψ}}_{\mathrm{a}}^0âˆ\pounds {\mathrm{ψ}}_{\mathrm{b}}^0âŸ\copyright =0=⟨{\mathrm{ψ}}_{\mathrm{b}}^0âˆ\pounds {\mathrm{ψ}}_{\mathrm{a}}^0âŸ\copyright$ and$⟨{\mathrm{ψ}}_{\mathrm{a}}^0âˆ\pounds {\mathrm{ψ}}_{\mathrm{a}}^0âŸ\copyright =1=⟨{\mathrm{ψ}}_{\mathrm{b}}^0âˆ\pounds {\mathrm{ψ}}_{\mathrm{b}}^0âŸ\copyright$ .

Now use the relation

${\mathrm{β}}_{Â\pm }={\mathrm{Î\pm }}_{Â\pm }({\mathrm{E}}_{Â\pm }^1-{\mathrm{W}}_{\mathrm{aa}})/{\mathrm{W}}_{\mathrm{ab}}$

which gives

$⟨{\mathrm{ψ}}_{+}^0âˆ\pounds {\mathrm{ψ}}_{-}^0âŸ\copyright ={\mathrm{Î\pm }}_{-}^{*}{\mathrm{Î\pm }}_{+}\left(1+\frac{({\mathrm{E}}_{+}^1-{\mathrm{W}}_{\mathrm{aa}})({\mathrm{E}}_{-}^1-{\mathrm{W}}_{\mathrm{aa}})}{|{\mathrm{W}}_{\mathrm{ab}}{|}^2}\right)$

The numerator in the second term is equal to

$$\begin{gathered} \frac{1}{4}\left[{\mathrm{W}}_{\mathrm{a}}\mathrm{a}-{\mathrm{W}}_{\mathrm{bb}}+\sqrt{({\mathrm{W}}_{\mathrm{aa}}-{\mathrm{W}}_{\mathrm{bb}}{)}^2+4|{\mathrm{W}}_{\mathrm{ab}}{|}^2}\right]\left[{\mathrm{W}}_{\mathrm{aa}}-{\mathrm{W}}_{\mathrm{bb}}-\sqrt{({\mathrm{W}}_{\mathrm{a}}\mathrm{a}-{\mathrm{W}}_{\mathrm{b}}\mathrm{b}{)}^2+4|{\mathrm{W}}_{\mathrm{a}}\mathrm{b}{|}^2}\right] \\ =\frac{1}{4}\left(\right({\mathrm{W}}_{\mathrm{aa}}-{\mathrm{W}}_{\mathrm{bb}}{)}^2-({\mathrm{W}}_{\mathrm{aa}}-{\mathrm{W}}_{\mathrm{bb}}{)}^2-4|{\mathrm{W}}_{\mathrm{ab}}{|}^2) \\ =-|{\mathrm{W}}_{\mathrm{ab}}{|}^2 \end{gathered}$$

which gives

$⟨{\mathrm{ψ}}_{+}^0âˆ\pounds {\mathrm{ψ}}_{-}^0âŸ\copyright =\frac{\left({\mathrm{Î\pm }}_{-}^{*}{\mathrm{Î\pm }}_{+}\right)}{|{\mathrm{W}}_{\mathrm{ab}}{|}^2}\left(\right|{\mathrm{W}}_{\mathrm{ab}}{|}^2-\left|{\mathrm{W}}_{\mathrm{ab}}{|}^2\right)=0$

The states ${\mathrm{ψ}}_{Â\pm }^0$are orthogonal.

Now show that

$⟨{\mathrm{ψ}}_{+}^0\left|{\mathrm{H}}^{\text{'}}\right|{\mathrm{ψ}}_{-}^0âŸ\copyright =0$

It is given that

$$\begin{gathered} ⟨{\mathrm{ψ}}_{+}^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_{-}^0âŸ\copyright \\ ={\mathrm{Î\pm }}_{+}^{*}{\mathrm{Î\pm }}_{-}⟨{\mathrm{ψ}}_a^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_a^0âŸ\copyright +{\mathrm{Î\pm }}_{+}^{*}{\mathrm{β}}_{-}⟨{\mathrm{ψ}}_a^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_b^0âŸ\copyright +{\mathrm{β}}_{+}^{*}{\mathrm{Î\pm }}_{-}⟨{\mathrm{ψ}}_b^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_a^0âŸ\copyright +{\mathrm{β}}_{+}^{*}{\mathrm{β}}_{-}⟨{\mathrm{ψ}}_b^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_b^0âŸ\copyright \\ ={\mathrm{Î\pm }}_{+}^{*}{\mathrm{Î\pm }}_{-}{W}_{aa}+{\mathrm{Î\pm }}_{+}^{*}{\mathrm{β}}_{-}{W}_{ab}+{\mathrm{β}}_{+}^{*}{\mathrm{Î\pm }}_{-}{W}_{ba}+{\mathrm{β}}_{+}^{*}{\mathrm{β}}_{-}{W}_{bb} \\ ={\mathrm{Î\pm }}_{+}^{*}{\mathrm{Î\pm }}_{-}\left[W_{aa}+W_{ab}\frac{\left(E_{-}^1-W_{aa}\right)}{W_{ab}}+W_{ba}\frac{\left(E_{+}^1-W_{aa}\right)}{W_{ba}}+W_{bb}\frac{(E_{+}^1-W_{aa})}{W_{b}a}\frac{(E_{-}^1-W_{a}a)}{W_{ab}}\right] \\ ={\mathrm{Î\pm }}_{+}^{*}{\mathrm{Î\pm }}_{-}\left[W_{aa}+E_{-}^1-W_{aa}+E_{+}^1-W_{aa}+W_{bb}\frac{(E_{+}^1-W_{aa})(E_{-}^1-W_{aa}}{|W_{ab}{|}^2}\right] \\ ={\mathrm{Î\pm }}_{+}^{*}{\mathrm{Î\pm }}_{-}\left[W_{aa}+W_{bb}-W_{aa}+W_{bb}\frac{-|W_{ab}{|}^2}{|W_{ab}{|}^2}\right]=0 \end{gathered}$$

Now calculate$⟨{\mathrm{ψ}}_{Â\pm }^0\left|{\mathrm{H}}^{\text{'}}\right|{\mathrm{ψ}}_{Â\pm }^0âŸ\copyright$

This gives

$$\begin{gathered} ⟨{\mathrm{ψ}}_{Â\pm }^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_{Â\pm }^0âŸ\copyright \\ ={\mathrm{Î\pm }}_{Â\pm }^{*}{\mathrm{Î\pm }}_{Â\pm }⟨{\mathrm{ψ}}_a^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_a^0âŸ\copyright +{\mathrm{Î\pm }}_{Â\pm }^{*}{\mathrm{β}}_{Â\pm }⟨{\mathrm{ψ}}_a^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_b^0âŸ\copyright +{\mathrm{β}}_{Â\pm }^{*}{\mathrm{Î\pm }}_{Â\pm }⟨{\mathrm{ψ}}_b^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_a^0âŸ\copyright +{\mathrm{β}}_{Â\pm }^{*}{\mathrm{β}}_{Â\pm }⟨{\mathrm{ψ}}_b^0\left|H^{\text{'}}\right|{\mathrm{ψ}}_b^0âŸ\copyright \\ =|{\mathrm{Î\pm }}_{Â\pm }{|}^2\left[W_{aa}+W_{ab}\frac{(E_{Â\pm }^1-W_{aa})}{W_{ab}}\right]+|{\mathrm{β}}_{Â\pm }{|}^2\left[W_{ba}\frac{(E_{Â\pm }^1-W_{bb})}{W_{ba}+W_{bb}}\right] \\ =|{\mathrm{Î\pm }}_{Â\pm }{|}^2{E}_{Â\pm }^1+|{\mathrm{β}}_{Â\pm }{|}^2{E}_{Â\pm }^1 \\ =\left(\right|{\mathrm{Î\pm }}_{Â\pm }{|}^2+\left|{\mathrm{β}}_{Â\pm }{|}^2\right)E_{Â\pm }^1=E_{Â\pm }^1 \end{gathered}$$

where this relation has been used.

role="math" localid="1655972198230" ${\mathrm{Î\pm }}_{Â\pm }={\mathrm{β}}_{Â\pm }(E_{Â\pm }-W_{bb})/W_{ba}$

in the second equality, and the fact that the sum of all probabilities is equal to one

$|{\mathrm{Î\pm }}_{Â\pm }{|}^2+|{\mathrm{β}}_{Â\pm }{|}^2=1.$