Question

Problem 6.6 Let the two "good" unperturbed states be

${\mathbf{\psi }}_{\mathbf{\pm }}^{\mathbf{0}}\mathbf{=}\mathbf{\alpha }\mathbf{\pm }{\mathbf{\psi }}_{\mathbf{a}}^{\mathbf{0}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{\pm }}{\mathbf{\psi }}_{\mathbf{b}}^{\mathbf{0}}$

where${\mathbf{\alpha }}_{\mathbf{\pm }}$and${\mathbf{\beta }}_{\mathbf{\pm }}$are determined (up to normalization) by Equation 6.22(orEquation6.24). Show explicitly that

(a)are orthogonal;role="math" localid="1655966589608" $\mathbf{(}\mathbf{⟨}{\mathbf{\psi }}_{\mathbf{+}}^{\mathbf{0}}\mathbf{\mid }{\mathbf{\psi }}_{\mathbf{-}}^{\mathbf{0}}\mathbf{⟩}\mathbf{=}\mathbf{0}\mathbf{)}\mathbf{;}$

(b) $\mathbf{⟨}{\mathbf{\psi }}_{\mathbf{+}}^{\mathbf{0}}\mathbf{\left|}{\mathbf{H}}^{\mathbf{\text{'}}}\mathbf{\right|}{\mathbf{\psi }}_{\mathbf{-}}^{\mathbf{0}}\mathbf{⟩}\mathbf{=}\mathbf{0}\mathbf{;}$

(c)$\mathbf{⟨}{\mathbf{\psi }}_{\mathbf{\pm }}^{\mathbf{0}}\mathbf{\left|}{\mathbf{H}}^{\mathbf{\text{'}}}\mathbf{\right|}{\mathbf{\psi }}_{\mathbf{\pm }}^{\mathbf{0}}\mathbf{⟩}\mathbf{=}{\mathbf{E}}_{\mathbf{\pm }}^{\mathbf{1}}\mathbf{,}$with${\mathbf{E}}_{\mathbf{\pm }}^{\mathbf{1}}$given by Equation 6.27.

Short answer

a) It is shown explicitly that$⟨{\mathrm{\psi }}_{+}^0\mid {\mathrm{\psi }}_{-}^0⟩=0$

b) It is shown explicitly that$⟨{\mathrm{\psi }}_{+}^0\left|{\mathrm{H}}^{\text{'}}\right|{\mathrm{\psi }}_{-}^0⟩=0$

c) It is shown explicitly that_{role="math" localid="1655967130676" $⟨{\mathrm{\psi }}_{\pm }^0\left|{\mathrm{H}}^{\text{'}}\right|{\mathrm{\psi }}_{\pm }^0⟩={\mathrm{E}}_{\pm }^1$}

Step-by-step solution

Show ψ±0 are orthogonal (⟨ψ+0∣ψ-0⟩=0)

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

${\mathrm{\psi }}_{\pm }^0={\mathrm{\alpha }}_{\pm }{\mathrm{\psi }}_{\mathrm{a}}^0+{\mathrm{\beta }}_{\pm }{\mathrm{\psi }}_{\mathrm{b}}^0$

where

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

Its given that

$⟨{\mathrm{\psi }}_{+}^0\mid {\mathrm{\psi }}_{-}^0⟩=0$

This gives

$$\begin{gathered} ⟨{\mathrm{\psi }}_{+}^0\mid {\mathrm{\psi }}_{-}^0⟩ \\ =\left({\mathrm{\alpha }}_{+}^{*}{\mathrm{\psi }}_{\mathrm{a}}^0\right|+{\mathrm{\beta }}_{-}^{*}⟨{\mathrm{\psi }}_{\mathrm{b}}^0\left|\right)\left({\mathrm{\alpha }}_{-}\right|{\mathrm{\psi }}_{\mathrm{a}}^0⟩+{\mathrm{\beta }}_{-}|{\mathrm{\psi }}_{\mathrm{b}}^0⟩)⟩ \\ ={\mathrm{\alpha }}_{-}^{*}{\mathrm{\alpha }}_{+}+{\mathrm{\beta }}_{-}^{*}{\mathrm{\beta }}_{+} \end{gathered}$$

where it has been used$⟨{\mathrm{\psi }}_{\mathrm{a}}^0\mid {\mathrm{\psi }}_{\mathrm{b}}^0⟩=0=⟨{\mathrm{\psi }}_{\mathrm{b}}^0\mid {\mathrm{\psi }}_{\mathrm{a}}^0⟩$ and$⟨{\mathrm{\psi }}_{\mathrm{a}}^0\mid {\mathrm{\psi }}_{\mathrm{a}}^0⟩=1=⟨{\mathrm{\psi }}_{\mathrm{b}}^0\mid {\mathrm{\psi }}_{\mathrm{b}}^0⟩$ .

Now use the relation

${\mathrm{\beta }}_{\pm }={\mathrm{\alpha }}_{\pm }({\mathrm{E}}_{\pm }^1-{\mathrm{W}}_{\mathrm{aa}})/{\mathrm{W}}_{\mathrm{ab}}$

which gives

$⟨{\mathrm{\psi }}_{+}^0\mid {\mathrm{\psi }}_{-}^0⟩={\mathrm{\alpha }}_{-}^{*}{\mathrm{\alpha }}_{+}\left(1+\frac{({\mathrm{E}}_{+}^1-{\mathrm{W}}_{\mathrm{aa}})({\mathrm{E}}_{-}^1-{\mathrm{W}}_{\mathrm{aa}})}{|{\mathrm{W}}_{\mathrm{ab}}{|}^2}\right)$

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

$⟨{\mathrm{\psi }}_{+}^0\mid {\mathrm{\psi }}_{-}^0⟩=\frac{\left({\mathrm{\alpha }}_{-}^{*}{\mathrm{\alpha }}_{+}\right)}{|{\mathrm{W}}_{\mathrm{ab}}{|}^2}\left(\right|{\mathrm{W}}_{\mathrm{ab}}{|}^2-\left|{\mathrm{W}}_{\mathrm{ab}}{|}^2\right)=0$

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

Show that ⟨ψ+0|H'|ψ-0⟩=0

Now show that

$⟨{\mathrm{\psi }}_{+}^0\left|{\mathrm{H}}^{\text{'}}\right|{\mathrm{\psi }}_{-}^0⟩=0$

It is given that

$$\begin{gathered} ⟨{\psi }_{+}^0\left|H^{\text{'}}\right|{\psi }_{-}^0⟩ \\ ={\alpha }_{+}^{*}{\alpha }_{-}⟨{\psi }_a^0\left|H^{\text{'}}\right|{\psi }_a^0⟩+{\alpha }_{+}^{*}{\beta }_{-}⟨{\psi }_a^0\left|H^{\text{'}}\right|{\psi }_b^0⟩+{\beta }_{+}^{*}{\alpha }_{-}⟨{\psi }_b^0\left|H^{\text{'}}\right|{\psi }_a^0⟩+{\beta }_{+}^{*}{\beta }_{-}⟨{\psi }_b^0\left|H^{\text{'}}\right|{\psi }_b^0⟩ \\ ={\alpha }_{+}^{*}{\alpha }_{-}{W}_{aa}+{\alpha }_{+}^{*}{\beta }_{-}{W}_{ab}+{\beta }_{+}^{*}{\alpha }_{-}{W}_{ba}+{\beta }_{+}^{*}{\beta }_{-}{W}_{bb} \\ ={\alpha }_{+}^{*}{\alpha }_{-}\left[W_{aa}+W_{ab}\frac{\left(E_{-}^1-W_{aa}\right)}{W_{ab}}+W_{ba}\frac{\left(E_{+}^1-W_{aa}\right)}{W_{ba}}+W_{bb}\frac{(E_{+}^1-W_{aa})}{W_{b}a}\frac{(E_{-}^1-W_{a}a)}{W_{ab}}\right] \\ ={\alpha }_{+}^{*}{\alpha }_{-}\left[W_{aa}+E_{-}^1-W_{aa}+E_{+}^1-W_{aa}+W_{bb}\frac{(E_{+}^1-W_{aa})(E_{-}^1-W_{aa}}{|W_{ab}{|}^2}\right] \\ ={\alpha }_{+}^{*}{\alpha }_{-}\left[W_{aa}+W_{bb}-W_{aa}+W_{bb}\frac{-|W_{ab}{|}^2}{|W_{ab}{|}^2}\right]=0 \end{gathered}$$

Show that ⟨ψ±0|H'|ψ±0⟩=E±1

Now calculate$⟨{\mathrm{\psi }}_{\pm }^0\left|{\mathrm{H}}^{\text{'}}\right|{\mathrm{\psi }}_{\pm }^0⟩$

This gives

$$\begin{gathered} ⟨{\psi }_{\pm }^0\left|H^{\text{'}}\right|{\psi }_{\pm }^0⟩ \\ ={\alpha }_{\pm }^{*}{\alpha }_{\pm }⟨{\psi }_a^0\left|H^{\text{'}}\right|{\psi }_a^0⟩+{\alpha }_{\pm }^{*}{\beta }_{\pm }⟨{\psi }_a^0\left|H^{\text{'}}\right|{\psi }_b^0⟩+{\beta }_{\pm }^{*}{\alpha }_{\pm }⟨{\psi }_b^0\left|H^{\text{'}}\right|{\psi }_a^0⟩+{\beta }_{\pm }^{*}{\beta }_{\pm }⟨{\psi }_b^0\left|H^{\text{'}}\right|{\psi }_b^0⟩ \\ =|{\alpha }_{\pm }{|}^2\left[W_{aa}+W_{ab}\frac{(E_{\pm }^1-W_{aa})}{W_{ab}}\right]+|{\beta }_{\pm }{|}^2\left[W_{ba}\frac{(E_{\pm }^1-W_{bb})}{W_{ba}+W_{bb}}\right] \\ =|{\alpha }_{\pm }{|}^2{E}_{\pm }^1+|{\beta }_{\pm }{|}^2{E}_{\pm }^1 \\ =\left(\right|{\alpha }_{\pm }{|}^2+\left|{\beta }_{\pm }{|}^2\right)E_{\pm }^1=E_{\pm }^1 \end{gathered}$$

where this relation has been used.

role="math" localid="1655972198230" ${\alpha }_{\pm }={\beta }_{\pm }(E_{\pm }-W_{bb})/W_{ba}$

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

$|{\alpha }_{\pm }{|}^2+|{\beta }_{\pm }{|}^2=1.$