Exercise 3

Exercise 3 a)

\[ \begin{flalign*} \braket{o|u} &= \braket{ \gamma^* \bra{u} + \delta^* \bra{d}| \ket{u} } = \gamma^*\\ \braket{u|o} &= \braket{ \bra{u} | \gamma \ket{u} + \delta \ket{d} } = \gamma \end{flalign*} \] \[ \Rightarrow \braket{o|u} \braket{u|o} = \gamma^* \gamma = \frac{1}{2} \]

\[ \begin{flalign*} \braket{o|d} &= \braket{ \gamma^* \bra{u} + \delta^* \bra{d}| \ket{d} } = \delta^*\\ \braket{d|o} &= \braket{ \bra{d} | \gamma \ket{u} + \delta \ket{d} } = \delta \end{flalign*} \]

\[ \Rightarrow \braket{o|d} \braket{d|o} = \delta^* \delta = \frac{1}{2} \]

\[ \begin{flalign*} \braket{i|u} &= \braket{ \alpha^* \bra{u} + \beta^* \bra{d}| \ket{u} } = \alpha^*\\ \braket{u|i} &= \braket{ \bra{u} | \alpha \ket{u} + \beta \ket{d} } = \alpha \end{flalign*} \]

\[ \Rightarrow \braket{i|u} \braket{u|i} = \alpha^* \alpha = \frac{1}{2} \]

\[ \begin{flalign*} \braket{i|d} &= \braket{ \alpha^* \bra{u} + \beta^* \bra{d}| \ket{d} } = \beta^*\\ \braket{d|i} &= \braket{ \bra{d} | \alpha \ket{u} + \beta \ket{d} } = \beta \end{flalign*} \] \[ \Rightarrow \braket{i|d} \braket{d|i} = \beta^* \beta = \frac{1}{2} \]

Exercise 3 b)

From Eqs. 2.9 we know: \[ \begin{flalign*} \braket{o|r}\braket{r|o} &= \frac{1}{2}\\ &= \braket{\gamma^*\bra{u} + \delta^* \bra{d} | \frac{1}{\sqrt{2}} \ket{u} + \frac{1}{\sqrt{2}} \ket{d} } \braket{ \frac{1}{\sqrt{2}} \bra{u} + \frac{1}{\sqrt{2}} \bra{d} | \gamma \ket{u} + \delta \ket{d} }\\ &= \left( \frac{1}{\sqrt{2}}\gamma^* + \frac{1}{\sqrt{2}}\delta^*\right) \left( \frac{1}{\sqrt{2}}\gamma + \frac{1}{\sqrt{2}}\delta\right) \\ &= \frac{1}{2} \gamma^*\gamma + \frac{1}{2} \gamma^*\delta + \frac{1}{2} \delta^*\gamma + \frac{1}{2} \delta^*\delta\\ &= \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \gamma^*\delta + \frac{1}{2} \delta^*\gamma + \frac{1}{2} \cdot \frac{1}{2}\\ \frac{1}{2}&= \frac{1}{2} + \frac{1}{2} \gamma^*\delta + \frac{1}{2} \delta^*\gamma \\ 0 &= \frac{1}{2} ( \gamma^*\delta + \delta^*\gamma) \end{flalign*} \] \[ \Rightarrow \gamma^*\delta + \delta^*\gamma = 0 \]

\[ \begin{flalign*} \braket{i|r}\braket{r|i} &= \frac{1}{2}\\ &= \braket{\alpha^*\bra{u} + \beta^* \bra{d} | \frac{1}{\sqrt{2}} \ket{u} + \frac{1}{\sqrt{2}} \ket{d} } \braket{ \frac{1}{\sqrt{2}} \bra{u} + \frac{1}{\sqrt{2}} \bra{d} | \alpha \ket{u} + \beta \ket{d} }\\ &= \left( \frac{1}{\sqrt{2}}\alpha^* + \frac{1}{\sqrt{2}}\beta^*\right) \left( \frac{1}{\sqrt{2}}\alpha + \frac{1}{\sqrt{2}}\beta\right) \\ &= \frac{1}{2} \alpha^*\alpha + \frac{1}{2} \alpha^*\beta + \frac{1}{2} \beta^*\alpha + \frac{1}{2} \beta^*\beta\\ &= \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \alpha^*\beta + \frac{1}{2} \beta^*\alpha + \frac{1}{2} \cdot \frac{1}{2}\\ \frac{1}{2}&= \frac{1}{2} + \frac{1}{2} \alpha^*\beta + \frac{1}{2} \beta^*\alpha \\ 0 &= \frac{1}{2} ( \alpha^*\beta + \beta^*\alpha) \end{flalign*} \]

\[ \Rightarrow \alpha^*\beta + \beta^*\alpha = 0 \]

Exercise 3 c)

Let’s write \(\alpha\) and \(\beta\) as complex numbers with a real and imaginary part: \[ \begin{flalign*} \alpha &= a + \mathrm{i} b \\ \beta &= c + \mathrm{i} d \end{flalign*} \]

Now \[ \begin{flalign*} \alpha^* \beta + \alpha \beta^* &= (a - \mathrm{i} b) (c + \mathrm{i} d) + (a + \mathrm{i} b) (c - \mathrm{i} d)\\ &= 2ac + 2bd = 0 \\ \Rightarrow ac+bd &= 0 \end{flalign*} \]

\[ \begin{flalign*} \alpha^* \beta &= (a - \mathrm{i} b) (c + \mathrm{i} d)\\ &= \underbrace{ac +bd}_{=0} + \mathrm{i} (ad-bc) \end{flalign*} \]

\(\alpha^*\beta\) is a complex number whose real part is zero. These numbers are pure imaginary. The same holds true for \(\gamma^*\delta\).