Exercise 3

The eigenvalue equation is: \[ \begin{flalign*} \begin{pmatrix} \cos \Theta & \sin \Theta \\ \sin \Theta & -\cos \Theta \end{pmatrix} \begin{pmatrix} \cos \alpha \\ \sin \alpha \end{pmatrix} = \lambda \begin{pmatrix} \cos \alpha \\ \sin \alpha \end{pmatrix} \end{flalign*} \] \[ \begin{flalign*} \cos \Theta \cos \alpha + \sin \Theta \sin \alpha &= \lambda \cos \alpha \\ \sin \Theta \cos \alpha - \cos \Theta \sin \alpha &= \lambda \sin \alpha \end{flalign*} \]

Compare these equations with the trigonometric addition formulas, this yields:

\[ \cos (\Theta - \alpha) = \lambda \cos \alpha \tag{1}\] \[ \sin (\Theta - \alpha) = \lambda \sin \alpha \tag{2}\]

Square these equations and add them: \[ \begin{flalign*} \cos^2 (\Theta - \alpha) &= \lambda^2 \cos^2 \alpha \\ \sin^2 (\Theta - \alpha) &= \lambda^2 \sin^2 \alpha \\ \cos^2 (\Theta - \alpha) + \sin^2 (\Theta - \alpha) &= \lambda^2 ( \cos^2 \alpha +\sin^2 \alpha) \\ \Rightarrow \lambda^2 = 1 \\ \Rightarrow \lambda_1 = 1, \lambda_2 = -1 \end{flalign*} \]

Now solve Equation 1 and Equation 2 for the two eigenvalues. The solution for \(\lambda_1=1\) is very simple: \[ \begin{flalign*} \cos (\Theta - \alpha) &= \cos \alpha \\ \sin (\Theta - \alpha) &= \sin \alpha \\ \Rightarrow \Theta - \alpha &= \alpha \\ \Rightarrow \alpha &= \frac{\Theta}{2} \end{flalign*} \]

\(\ket{\lambda_1}\) takes the form \[ \ket{\lambda_1} = \begin{pmatrix} \cos \frac{\Theta}{2} \\ \sin \frac{\Theta}{2} \end{pmatrix} \]

To solve Equation 1 and Equation 2 for \(\lambda_2=-1\) we need the identities \[ \begin{flalign*} \sin (\uppi + \alpha) &= - \sin \alpha \\ \cos (\uppi + \alpha) &= - \cos \alpha \end{flalign*} \]

Thus leading to \[ \begin{flalign*} \cos (\Theta - \alpha) &= - \cos \alpha =\cos (\uppi + \alpha) \\ \sin (\Theta - \alpha) &= - \sin \alpha =\sin (\uppi + \alpha) \\ \Rightarrow \Theta - \alpha &= \uppi + \alpha \\ \Rightarrow \alpha &= \frac{\Theta}{2} - \frac{\uppi}{2} \end{flalign*} \] \[ \begin{flalign*} \cos \left( \frac{\Theta}{2} - \frac{\uppi}{2} \right) &= \sin \frac{\Theta}{2}\\ \sin \left( \frac{\Theta}{2} - \frac{\uppi}{2} \right) &= - \cos \frac{\Theta}{2} \end{flalign*} \]

So \(\ket{\lambda_2}\) takes the form \[ \ket{\lambda_2} = \begin{pmatrix} \sin \frac{\Theta}{2}\\ - \cos \frac{\Theta}{2} \end{pmatrix} \]

In the book \(\ket{\lambda_2}\) is given as \[ \ket{\lambda_2} = \begin{pmatrix} - \sin \frac{\Theta}{2}\\ \cos \frac{\Theta}{2} \end{pmatrix} \]

This is just the negative eigenvector of our solution. Both solutions are right, because of the linearity of the operator \(\mathbf L\). If \(\ket{\lambda}\) is an eigenvector of operator \(\mathbf L\) with eigenvalue \(\lambda\), then \(-\ket{\lambda}\) is also an eigenvector with eigenvalue \(\lambda\): \[ \begin{flalign*} \mathbf L\ket{\lambda} &= \lambda \ket{\lambda} \\ \mathbf L \ket{-\lambda} &= -\mathbf L \ket{\lambda} = - \lambda \ket{\lambda} = \lambda \ket{-\lambda} \end{flalign*} \]

Comment: How to find \(\ket{\lambda}\) without assuming a specific form

Let’s drop the assumption that \(\ket{\lambda}\) has the form \(\begin{pmatrix}\cos \alpha\\ \sin \alpha \end{pmatrix}\). Let’s assume the form \(\begin{pmatrix}a\\ b \end{pmatrix}\) with \(a, b \in \mathbb{C}\). The eigenvalue equation is: \[ \begin{flalign*} \begin{pmatrix} \cos \Theta & \sin \Theta \\ \sin \Theta & -\cos \Theta \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \lambda \begin{pmatrix} a \\ b \end{pmatrix} \end{flalign*} \]

or \[ \begin{flalign*} a \cos \Theta + b \sin \Theta &= \lambda a \\ a \sin \Theta - b \cos \Theta &= \lambda b \end{flalign*} \]

This is a system of two linear equations with three unknown variables: \(a, b, \lambda\). Since the eigenvector should be normalized, we add another condition: \[ a^2 +b^2 = 1 \]

To find the eigenvalues, the usual way is to build the characteristic polynomial by evaluating the determinant of the linear equation system. \[ \begin{flalign*} a (\cos \Theta - \lambda) + b \sin \Theta &= 0 \\ a \sin \Theta - b (\cos \Theta + \lambda) &= 0 \end{flalign*} \] \[ \det \begin{vmatrix} (\cos \Theta - \lambda) & \sin \Theta \\ \sin \Theta & -(\cos \Theta + \lambda) \end{vmatrix} = 0 \] \[ \begin{flalign*} - (\cos \Theta - \lambda) (\cos \Theta + \lambda) - \sin^2 \Theta &= 0 \\ (\cos \Theta - \lambda) (\cos \Theta + \lambda) + \sin^2 \Theta &= 0 \\ \cos^2 \Theta - \lambda^2 + \sin^2 \Theta &= 0 \\ -\lambda^2 + 1 &= 0\\ \Rightarrow \lambda^2 &= 1 \\ \Rightarrow \lambda &= \pm 1 \end{flalign*} \]

First we choose \(\lambda=1\) and try to find the eigenvector for this eigenvalue: \[ \begin{flalign*} a \cos \Theta + b \sin \Theta &= a \\ a \sin \Theta - b \cos \Theta &= b \end{flalign*} \]

We solve the first equation for \(a\) in terms of \(b\): \[ a = \frac{-b \sin \Theta}{\cos \Theta -1} \tag{3}\]

and plug this in the condition \(a^2+b^2=1\): \[ \frac{b^2 \sin^2 \Theta}{(\cos \Theta-1)^2} +b^2 = 1 \] Solving for \(b^2\) yields (after a little bit of algebra) \[ b^2 = \frac{1}{2}( 1 - \cos \Theta ) \] The following equations are very helpful: \[ \cos \Theta = \cos \left( 2 \frac{\Theta}{2} \right) = \cos^2 \frac{\Theta}{2} - \sin^2 \frac{\Theta}{2} \]

\[ \sin \Theta = \sin \left( 2 \frac{\Theta}{2} \right) = 2 \sin \frac{\Theta}{2} \cos \frac{\Theta}{2} \] \[ \Rightarrow b^2 = \frac{1}{2}( 1 - \cos^2 \frac{\Theta}{2} + \sin^2 \frac{\Theta}{2} ) \] \[ b^2 = \frac{1}{2}( \sin^2 \frac{\Theta}{2} + \sin^2 \frac{\Theta}{2} ) \] \[ b^2 = \sin^2 \frac{\Theta}{2} \] \[ \Rightarrow b = \pm \sin \frac{\Theta}{2} \] We choose the plus sign and plug this into Equation 3: \[ \begin{flalign*} a &= \frac{- \sin \frac{\Theta}{2} \sin \Theta}{\cos \Theta -1 } \\ &= \frac{\sin \frac{\Theta}{2} \sin \Theta}{1 - \cos \Theta } \\ &= \frac{2 \sin^2 \frac{\Theta}{2} \cos \frac{\Theta}{2}}{1 - \cos^2 \frac{\Theta}{2} + \sin^2 \frac{\Theta}{2} } \\ &= \frac{2 \sin^2 \frac{\Theta}{2} \cos \frac{\Theta}{2}} {2 \sin^2 \frac{\Theta}{2} } \\ &= \cos \frac{\Theta}{2} \end{flalign*} \] For \(\lambda_1=1\) we found \(\ket{\lambda_1} = \begin{pmatrix} \cos \frac{\Theta}{2} \\ \sin \frac{\Theta}{2} \end{pmatrix}\).

The procedure for \(\lambda_2 = -1\) is very similar.