Exercise 1

If two state vectors are orthogonal, then their inner product is zero. \[\ket{r} = \frac{1}{\sqrt{2}}\ket{u} + \frac{1}{\sqrt{2}}\ket{d}\] \[\ket{l} = \frac{1}{\sqrt{2}}\ket{u} - \frac{1}{\sqrt{2}}\ket{d}\] \[ \begin{flalign} \braket{r|l} &=\\ &= \braket{\frac{1}{\sqrt{2}}\bra{u} + \frac{1}{\sqrt{2}}\bra{d} | \frac{1}{\sqrt{2}}\ket{u} - \frac{1}{\sqrt{2}}\ket{d} }\\ &= \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \braket{u|u} - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \braket{u|d} + \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \braket{d|u} - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \braket{d|d}\\ &= \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \cdot 1 - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \cdot 0 + \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \cdot 0 - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \cdot 1\\ &= \frac{1}{2}-\frac{1}{2}\\ &=0 \end{flalign} \]

Comment: How to find \(\ket{l}\)

Vector \(\ket{l}\) is given in Eq. 2.6 in the textbook. As we have seen in this exercise, the orthogonality of \(\ket{r}\) and \(\ket{l}\) is easy to prove. But how do we find \(\ket{l}\)?

Let’s take \(\ket{r}\) as given in Eq. 2.5 in the textbook. Finding \(\ket{l}\) means finding the probability amplitudes \(\alpha_u\) and \(\alpha_d\) in the general equation \[\ket{l} = \alpha_u \ket{u} + \alpha_d \ket{d}\]

\(\alpha_u\) and \(\alpha_d\) are complex numbers and they can be written as \(\alpha_u = a_u + \mathrm{i}b_u\) and \(\alpha_d = a_d + \mathrm{i}b_d\). Hence \(\ket{l}\) is specified by four real numbers \(a_u\), \(b_u\), \(a_d\) and \(b_d\).

As stated in the textbook, when the spin has been prepared in state \(\ket{l}\), the probability to measure \(\sigma_z = 1\) or \(\sigma_z = -1\) is \(\frac{1}{2}\), respectively. Hence \[ \braket{l|u}\braket{u|l} = \alpha_u^*\alpha_u = a_u^2+b_u^2 =\frac{1}{2} \tag{1}\]

\[ \braket{l|d}\braket{d|l} = \alpha_d^*\alpha_d = a_d^2+b_d^2 = \frac{1}{2} \tag{2}\]

\(\ket{r}\) and \(\ket{l}\) are orthogonal, hence \(\braket{r|l} = \braket{l|r} = 0\). \[ \begin{flalign*} \braket{r|l} &=\\ &= \braket{\frac{1}{\sqrt{2}}\bra{u} + \frac{1}{\sqrt{2}}\bra{d} | \alpha_u \ket{u} + \alpha_d \ket{d} } \\ &= \frac{1}{\sqrt{2}}\alpha_u + \frac{1}{\sqrt{2}} \alpha_d = 0 \Rightarrow \alpha_u = - \alpha_d\\ \end{flalign*} \]

In the same way we get \[ \braket{l|r} = \frac{1}{\sqrt{2}} \alpha_u^* + \frac{1}{\sqrt{2}} \alpha_d^* = 0 \Rightarrow \alpha_u^* = - \alpha_d^* \]

Hence \[ \begin{flalign*} a_u + \mathrm{i}b_u = - (a_d + \mathrm{i}b_d)\\ a_u - \mathrm{i}b_u = - (a_d - \mathrm{i}b_d) \end{flalign*} \]

Adding and subtracting these equations yields \[a_u = -a_d \tag{3}\] \[b_u = -b_d \tag{4}\]

If you plug in Equation 3 and Equation 4 in Equation 1, you get Equation 2. This means we have only three equations to determine four unknowns. There is still one degree of freedom which is called the phase ambiguity. To keep things simple we can choose \(b_u = 0\). Then it follows the form of \(\ket{l}\) given by Eq. 2.6 in the textbook.

A more general form for \(\ket{l}\) including the phase factor would be \[ \ket{l} = \frac{\mathrm{e}^{\mathrm{i}\Theta}}{\sqrt{2}} \ket{u} - \frac{\mathrm{e}^{\mathrm{i}\Theta}}{\sqrt{2}} \ket{d} \]