Exercise 4

Eq. 3.23 for spherical coordinates is \[ \sigma_n = \begin{pmatrix} \cos \Theta & \sin \Theta \, \mathrm{e}^{-\mathrm{i} \Phi} \\ \sin \Theta \, \mathrm{e}^{\mathrm{i} \Phi} & - \cos \Theta \end{pmatrix} \]

To compute the eigenvalues we set the determinant to zero, this yields the characteristic polynomial: \[ \det \begin{vmatrix} (\cos \Theta - \lambda) & \sin \Theta \, \mathrm{e}^{-\mathrm{i} \Phi}\\ \sin \Theta \, \mathrm{e}^{\mathrm{i} \Phi} & -(\cos \Theta + \lambda) \end{vmatrix} = 0 \] \[ \begin{flalign*} - (\cos \Theta - \lambda) (\cos \Theta + \lambda) - \sin ^2 \Theta &= 0 \\ -(\cos ^2 \Theta + \sin^2 \Theta) + \lambda^2 &= 0 \\ \Rightarrow \lambda^2 &= 1 \\ \Rightarrow \lambda &= \pm 1 \end{flalign*} \]

which is no surprise.

We choose \(\lambda = +1\) and write the equations for the eigenvector \(\ket{\lambda} = \begin{pmatrix} a \\ b \end{pmatrix}, a, b \in \mathbb{C}\).

\[ a \cos \Theta + b \sin \Theta \, \mathrm{e}^{-\mathrm{i} \Phi} = a \tag{1}\] \[ a \sin \Theta \, \mathrm{e}^{\mathrm{i} \Phi} - b \cos \Theta = b \tag{2}\] \[ a^2 + b^2 = 1 \tag{3}\]

The brute force attack, as used in the comment of exercise 3.3, turns out to be a little bit too hard.

In exercise 3.3 we used \(\ket{\lambda} = \begin{pmatrix} \cos \alpha \\ \sin \alpha \end{pmatrix}\). Let’s see whether we succeed with this and plug this in Equation 1:

\[ \cos \Theta \cos \alpha + \sin \Theta \, \mathrm{e}^{-\mathrm{i} \Phi} \sin \alpha = \cos \alpha \]

Look carefully at the left side of this equation: it looks like the addition theorem for \(\cos (\Theta - \alpha)\), apart from the factor \(\mathrm{e}^{-\mathrm{i} \Phi}\). Let’s use \(\mathrm{e}^{-\mathrm{i} \Phi} \cos \alpha\) instead of \(\cos \alpha\):

\[ \begin{flalign*} \cos \Theta \, \mathrm{e}^{-\mathrm{i} \Phi} \cos \alpha + \sin \Theta \, \mathrm{e}^{-\mathrm{i} \Phi} \sin \alpha &= \mathrm{e}^{-\mathrm{i} \Phi}\cos \alpha \\ \mathrm{e}^{-\mathrm{i} \Phi} (\cos \Theta \cos \alpha + \sin \Theta \sin \alpha) &= \mathrm{e}^{-\mathrm{i} \Phi}\cos \alpha \\ \cos (\Theta - \alpha) &= \cos \alpha \\ \Rightarrow \alpha &= \frac{\Theta}{2} \end{flalign*} \]

Eigenvector \(\ket{\lambda_1}\) has the form \[ \ket{\lambda_1} = \begin{pmatrix} \mathrm{e}^{-\mathrm{i} \Phi} \cos \frac{\Theta}{2} \\ \sin \frac{\Theta}{2} \end{pmatrix} \]

The eigenvector \(\ket{\lambda_{2}}\) for the eigenvalue \(\lambda=-1\) can be found in a very similar way.

A more easy approach is to use the orthogonality of the eigenvectors: \[ \braket{\lambda_1 | \lambda_2} = 0 \]

Assume \[ \ket{\lambda_{2}} = \begin{pmatrix} a \\ b \end{pmatrix}, a, b \in \mathbb{C} \]

Then \[ \begin{flalign*} a \mathrm{e}^{\mathrm{i} \Phi} \cos \frac{\Theta}{2} + b \sin \frac{\Theta}{2} &= 0 \\ \Rightarrow a &= \sin \frac{\Theta}{2} \\ \Rightarrow b &= - \mathrm{e}^{\mathrm{i} \Phi} \cos \frac{\Theta}{2} \end{flalign*} \]

The Eigenvector \(\ket{\lambda_{2}}\) has the form \[ \ket{\lambda_{2}} = \begin{pmatrix} \sin \frac{\Theta}{2} \\ -\mathrm{e}^{\mathrm{i} \Phi} \cos \frac{\Theta}{2} \end{pmatrix} \]