Exercise 5

Solution using trigonometry

My first solution is straightforward, but involves some amount of trigonometry.

We know the eigenvector for the eigenvalue \(+1\) from exercise 3.4. It is \[ \ket{\lambda_1} = \begin{pmatrix} \mathrm{e}^{-\mathrm{i} \Phi} \cos \frac{\Theta}{2} \\ \sin \frac{\Theta}{2} \end{pmatrix} \]

If the spin is prepared in the \(\hat{m}\) direction, it is in the state of the corresponding eigenvector: \[ \ket{\lambda_1^m} = \begin{pmatrix} \mathrm{e}^{-\mathrm{i} \Phi_m} \cos \frac{\Theta_m}{2} \\ \sin \frac{\Theta_m}{2} \end{pmatrix} \]

Likewise, if the spin is measured to be \(+1\) in the \(\hat{n}\) direction, the state is: \[ \ket{\lambda_1^n} = \begin{pmatrix} \mathrm{e}^{-\mathrm{i} \Phi_n} \cos \frac{\Theta_n}{2} \\ \sin \frac{\Theta_n}{2} \end{pmatrix} \]

The probability to measure \(\lambda_1 = +1\) for the spin in \(\hat{n}\) direction then is \[ P(\lambda_1) = \braket{\lambda_1^n|\lambda_1^m} \braket{\lambda_1^m|\lambda_1^n} \tag{1}\]

\[ \begin{flalign*} \braket{\lambda_1^m | \lambda_1^n} &= \begin{pmatrix} \mathrm{e}^{\mathrm{i} \Phi_m} \cos \frac{\Theta_m}{2} & \sin \frac{\Theta_m}{2} \end{pmatrix} \begin{pmatrix} \mathrm{e}^{-\mathrm{i} \Phi_n} \cos \frac{\Theta_n}{2} \\ \sin \frac{\Theta_n}{2} \end{pmatrix} \\ &= \mathrm{e}^{\mathrm{i} (\Phi_m - \Phi_n)} \cos \frac{\Theta_n}{2} \cos \frac{\Theta_m}{2} + \sin \frac{\Theta_n}{2} \sin \frac{\Theta_m}{2} \end{flalign*} \]

\[ \begin{flalign*} \braket{\lambda_1^n | \lambda_1^m} &= \braket{\lambda_1^m | \lambda_1^n}^* \\ &= \mathrm{e}^{-\mathrm{i} (\Phi_m - \Phi_n)} \cos \frac{\Theta_n}{2} \cos \frac{\Theta_m}{2} + \sin \frac{\Theta_n}{2} \sin \frac{\Theta_m}{2} \end{flalign*} \]

Multiplication of these two numbers and using the equation \[ \mathrm{e}^{\mathrm{i} \Phi} + \mathrm{e}^{-\mathrm{i} \Phi} = 2 \cos \Phi \] gives us a really long expression:

\[ P(\lambda_1) = \cos^2 \frac{\Theta_m}{2} \cos^2 \frac{\Theta_n}{2} + \sin^2 \frac{\Theta_m}{2} \sin^2 \frac{\Theta_n}{2} + \cos \frac{\Theta_m}{2} \cos \frac{\Theta_n}{2} \sin \frac{\Theta_m}{2} \sin \frac{\Theta_n}{2} 2 \cos (\Phi_m - \Phi_n) \]

Using the following identities: \[ \begin{flalign*} \cos^2 \phi &= \frac{1}{2}(1+\cos 2 \phi) \\ \sin^2 \phi &= \frac{1}{2}(1-\cos 2 \phi) \\ \sin \phi \cos \phi &= \frac{1}{2} \sin 2\phi \end{flalign*} \]

we get: \[ \begin{flalign} P(\lambda_1) &= \frac{1}{2}(1+\cos \Theta_m \cos \Theta_n) + \frac{1}{2} \sin \Theta_m \sin \Theta_n (\cos \Phi_m \cos \Phi_n + \sin \Phi_m \sin \Phi_n) \\ &= \frac{1}{2}(1+\cos \Theta_m \cos \Theta_n + \sin \Theta_m \sin \Theta_n \cos \Phi_m \cos \Phi_n + \sin \Theta_m \sin \Theta_n \sin \Phi_m \sin \Phi_n) \end{flalign} \tag{2}\]

To get an expression for the angle \(\varphi\) between \(\hat{m}\) and \(\hat{n}\) we use the scalar product of these two vectors: \[ \begin{flalign*} \hat{m} \cdot \hat{n} &= \begin{pmatrix} m_x \\ m_y \\ m_z \end{pmatrix} \cdot \begin{pmatrix} n_x \\ n_y \\ n_z \end{pmatrix} = \cos \varphi \end{flalign*} \] since both vectors are normalized.

\[ \begin{flalign*} \cos \varphi &= \begin{pmatrix} \sin \Theta_m \cos \Phi_m \\ \sin \Theta_m \sin \Phi_m \\ \cos \Theta_m \end{pmatrix} \cdot \begin{pmatrix} \sin \Theta_n \cos \Phi_n \\ \sin \Theta_n \sin \Phi_n \\ \cos \Theta_n \end{pmatrix} \\ &= \sin \Theta_m \cos \Phi_m \sin \Theta_n \cos \Phi_n + \sin \Theta_m \sin \Phi_m \sin \Theta_n \sin \Phi_n + \cos \Theta_m \cos \Theta_n \end{flalign*} \]

Plug this in Equation 2: \[ P(\lambda_1) = \frac{1}{2}(1 + \cos \varphi) = \cos^2 \frac{\varphi}{2} \]

Solution using a coordinate transformation

A more smart approach would be to rotate the coordinate system so that \(\hat{m}\) points in \(z\) direction.

The rotation of the coordinate system from the old coordinates \(r\), \(\Theta\), \(\Phi\) to the new coordinates \(r'\), \(\Theta'\) and \(\Phi'\) is specified by the coordinate transformation equations as follows: \[ \begin{flalign*} \Theta' &= \Theta - \Theta_m \\ \Phi' &= \Phi - \Phi_m \\ r' &= r \end{flalign*} \tag{3}\]

Since all vectors are normalized (\(r=1\)), we can safely ignore this coordinate.

The coordinates of \(\hat{m}\) in the old coordinate system are \[ \begin{flalign*} \hat{m}_x &= \sin \Theta_m \cos \Phi_m \\ \hat{m}_y &= \sin \Theta_m \sin \Phi_m \\ \hat{m}_z &= \cos \Theta_m \end{flalign*} \]

The coordinates of \(\hat{m}\) in the new coordinate system are \[ \begin{flalign*} \hat{m}'_x &= \sin \Theta'_m \cos \Phi'_m \\ \hat{m}'_y &= \sin \Theta'_m \sin \Phi'_m \\ \hat{m}'_z &= \cos \Theta'_m \end{flalign*} \]

Using the transformation equations Equation 3 for \(\Theta = \Theta_m\) and \(\Phi=\Phi_m\) we get: \[ \begin{flalign*} \hat{m}'_x &= \sin (\Theta_m - \Theta_m) \cos (\Phi_m - \Phi_m ) = 0 \\ \hat{m}'_y &= \sin (\Theta_m - \Theta_m) \sin (\Phi_m - \Phi_m) = 0 \\ \hat{m}'_z &= \cos (\Theta_m - \Theta_m) = 1 \end{flalign*} \]

Note that we didn’t change any vector. We only changed the coordinate system, which can be chosen freely.

\(\hat{m}\) points in \(z'\) direction, as desired. After preparing the spin its state is the eigenvector of \(\sigma_z'\), namely \(\ket{\lambda_{1}^{m'}}=\ket{u}\).

We write down \(\hat{n}\) using the new coordinates: \[ \begin{flalign*} \hat{n}'_x &= \sin \Theta'_n \cos \Phi'_n = \sin (\Theta_n - \Theta_m) \cos (\Phi_n - \Phi_m )\\ \hat{n}'_y &= \sin \Theta'_n \sin \Phi'_n = \sin (\Theta_n - \Theta_m) \sin (\Phi_n - \Phi_m)\\ \hat{n}'_z &= \cos \Theta'_n = \cos (\Theta_n - \Theta_m) \end{flalign*} \]

In exercise 4 we already calculated the eigenvectors and eigenvalues for the spin operator in direction \(\hat{n}\), for \(\lambda_1^{n'} = +1\) the eigenvector becomes

\[ \ket{\lambda_1^{n'}} = \begin{pmatrix} \mathrm{e}^{-\mathrm{i} \Phi_n'} \cos \frac{\Theta_n'}{2} \\ \sin \frac{\Theta_n'}{2} \end{pmatrix} \]

The probability to measure \(\lambda_1 = +1\) for the spin in \(\hat{n}'\) direction then is (refer to Equation 1) \[ P(\lambda_1) = \braket{\lambda_1^{n'}|\lambda_1^{m'}} \braket{\lambda_1^{m'}|\lambda_1^{n'}} \]

which turns out to be

\[P(\lambda_1) = \cos^2 \frac{\Theta'_n}{2}\]

\(\Theta'_n\) is defined as the angle between \(\hat{n}'\) and the \(z'\) direction, which is the direction of \(\hat{m}'\). Since \(\Phi\) has no meaning for a vector in \(z\) direction, \(\Theta'_n\) is the angle between \(\hat{m}'\) and \(\hat{n}'\), so we get the same result as in the first solution using trigonometry.