Exercise 2

\[ \begin{flalign*} \ket{i} &= \frac{1}{\sqrt{2}}\ket{u} + \frac{\mathrm{i}}{\sqrt{2}}\ket{d}\\ \ket{o} &= \frac{1}{\sqrt{2}}\ket{u} - \frac{\mathrm{i}}{\sqrt{2}}\ket{d} \end{flalign*} \]

It is somewhat tedious to show that all the conditions are satisfied. I picked only one sample of the conditions 2.7, 2.8 and 2.9, the proofs for the remaining conditions work the same way.

Condition 2.7

\[ \begin{flalign*} \braket{i|o} &= \braket{ \frac{1}{\sqrt{2}}\bra{u} + \frac{\mathrm{-i}}{\sqrt{2}}\bra{d} | \frac{1}{\sqrt{2}}\ket{u} - \frac{\mathrm{i}}{\sqrt{2}}\ket{d} }\\ &= \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\braket{u|u} - \frac{1}{\sqrt{2}}\frac{\mathrm{i}}{\sqrt{2}}\braket{u|d} - \frac{\mathrm{i}}{\sqrt{2}}\frac{1}{\sqrt{2}}\braket{d|u} + \frac{\mathrm{i}}{\sqrt{2}}\frac{\mathrm{i}}{\sqrt{2}}\braket{d|d}\\ &= \frac{1}{2}+ \frac{\mathrm{i}^2}{(\sqrt{2})^2}\\ &=\frac{1}{2}-\frac{1}{2}\\ &=0 \end{flalign*} \]

Condition 2.8

\[ \begin{flalign*} \braket{i|u} \braket{u|i}&= \braket{ \frac{1}{\sqrt{2}}\bra{u} + \frac{\mathrm{-i}}{\sqrt{2}}\bra{d} | \ket{u}} \braket{ \bra{u} | \frac{1}{\sqrt{2}}\ket{u} + \frac{\mathrm{i}}{\sqrt{2}}\ket{d}} \\ &= \left( \frac{1}{\sqrt{2}} \braket{u|u} + \frac{\mathrm{-i}}{\sqrt{2}} \braket{d|u} \right) \left( \frac{1}{\sqrt{2}} \braket {u|u} + \frac{\mathrm{i}}{\sqrt{2}} \braket {u|d} \right) \\ &= \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\\ &= \frac{1}{2} \end{flalign*} \]

Condition 2.9

\[ \begin{flalign*} \braket{i|r} &= \braket{ \frac{1}{\sqrt{2}}\bra{u} + \frac{\mathrm{-i}}{\sqrt{2}}\bra{d} | \frac{1}{\sqrt{2}}\ket{u} + \frac{1}{\sqrt{2}}\ket{d} } \\ &= \frac{1}{2} - \frac{\mathrm{i}}{2} \end{flalign*} \]

\[ \begin{flalign*} \braket{r|i} = \braket{i|r}^* = \frac{1}{2} + \frac{\mathrm{i}}{2} \end{flalign*} \]

\[ \begin{flalign*} \braket{i|r}\braket{r|i} = \left(\frac{1}{2} - \frac{\mathrm{i}}{2}\right) \left( \frac{1}{2} + \frac{\mathrm{i}}{2} \right) = \left( \frac{1}{2} \right)^2 - \left( \frac{\mathrm{i}}{2} \right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \end{flalign*} \]

The form of \(\ket{i}\) and \(\ket{o}\) is not unique since a multiplication with a phase factor \(\mathrm{e}^{\mathrm{i}\Theta}\) leads to a form which also satisfies all the conditions. See also the comment for exercise 2, lecture 1.