Exercise 5

Spherical Coordinates

With \[ \sigma_n = \sigma \cdot \vec{n} \]

the time-independent Schrödinger Equation is \[ \frac{\hbar \omega}{2} \sigma_n \ket{E_j} = E_j \ket{E_j} \]

We already solved a very similar equation in Exercise 3.4: \[ \sigma_n \ket{\lambda_j} = \lambda_j \ket{\lambda_j} \]

using spherical coordinates and found the eigenvalues to be \(\lambda_1 = 1\) and \(\lambda_2 = -1\). If we multiply this equation with the factor \(\frac{\hbar \omega}{2}\), we get \[ \frac{\hbar \omega}{2} \sigma_n \ket{\lambda_j} = \frac{\hbar \omega}{2} \lambda_j \ket{\lambda_j} \]

Hence \[ \begin{flalign*} E_j &= \frac{\hbar \omega}{2} \lambda_j \Rightarrow E_1 = \frac{\hbar \omega}{2}, E_2 = -\frac{\hbar \omega}{2} \\ \ket{E_j} &= \ket{\lambda_j} \\ \Rightarrow \ket{E_1} &= \begin{pmatrix} \mathrm{e}^{-\mathrm{i} \Phi} \cos \frac{\Theta}{2} \\ \sin \frac{\Theta}{2} \end{pmatrix}, \ket{E_2} = \begin{pmatrix} \sin \frac{\Theta}{2} \\ -\mathrm{e}^{\mathrm{i} \Phi} \cos \frac{\Theta}{2} \end{pmatrix} \end{flalign*} \]

The eigenvalues are just scalars and are independent of the coordinate system.

The eigenvectors have a different form in cartesian coordinates. This solution will be shown here for completeness.

Cartesian Coordinates

Let’s assume the eigenvector \(\ket{E_1}\) with eigenvalue \(E_1 = \frac{\hbar \omega}{2}\) has the form \[ \ket{E_1} = \begin{pmatrix} z_1 \\ z_2 \end{pmatrix} = \begin{pmatrix} a_1 + \mathrm{i} b_1 \\ a_2 + \mathrm{i} b_2 \end{pmatrix} \]

The time-independent Schrödinger Equation is \[ \frac{\hbar \omega}{2} \begin{pmatrix} n_z & (n_x - \mathrm{i} n_y) \\ (n_x + \mathrm{i} n_y) & -n_z \end{pmatrix} \begin{pmatrix} a_1 + \mathrm{i} b_1 \\ a_2 + \mathrm{i} b_2 \end{pmatrix} = \frac{\hbar \omega}{2} \begin{pmatrix} a_1 + \mathrm{i} b_1 \\ a_2 + \mathrm{i} b_2 \end{pmatrix} \]

The factor \(\frac{\hbar \omega}{2}\) cancels on both sides. We get the following equations: \[ \begin{flalign*} (a_1 + \mathrm{i} b_1)(n_z - 1) + (n_x - \mathrm{i} n_y) (a_2 + \mathrm{i} b_2) &= 0 \\ (n_x + \mathrm{i} n_y)(a_1 + \mathrm{i} b_1) - (a_2+\mathrm{i} b_2)(n_z +1) &= 0 \end{flalign*} \]

The second equation doesn’t give us more information than the first one. Proof: multiply the second equation with a factor \(\frac{n_z-1}{n_x + \mathrm{i} n_y}\) and you get the normalization condition for \(\vec{n}\): \(n_x^2 + n_y^2 +n_z^2 =1\). Hence we’ll consider only the first equation.

We can separate this equation into two equations for the real and imaginary part: \[ \begin{flalign*} a_1 (n_z -1) + a_2 n_x + b_2 n_y &= 0 \\ b_1 (n_z -1) + b_2 n_x - a_2 n_y &= 0 \end{flalign*} \]

Since the eigenvector should be normalized, we add another condition: \[ a_1^2 + b_1^2 + a_2^2 +b_2^2 = 1 \]

Now we have 3 equations for 4 unknown real numbers, one variable can be chosen freely. We will choose \[ b_2 = 0 \]

and the equations are reduced to \[ \begin{flalign*} a_1 (n_z -1) + a_2 n_x &= 0 \\ b_1 (n_z -1) - a_2 n_y &= 0 \\ a_1^2 + b_1^2 + a_2^2 &= 1 \end{flalign*} \]

We solve the 2nd equation for \(a_2\): \[ a_2 = \frac{n_z-1}{n_y}\,b \]

plug this in the 1st equation and solve for \(a_1\): \[ a_1 = - \frac{n_x}{n_y}\, b \]

From the third equation we get \[ \begin{flalign*} \frac{n_x^2}{n_y^2}\,b_1^2 + b_1^2 + \frac{(n_z-1)^2}{n_y^2}\, b_1^2 &= 1 \\ b_1^2 \left( \frac{n_x^2}{n_y^2} + \frac{n_y^2}{n_y^2} + \frac{(n_z-1)^2}{n_y^2} \right) &= 1 \end{flalign*} \]

and since \[ |\vec{n}| =1 \Rightarrow n_x^2 + n_y^2 +n_z^2 = 1 \]

we eventually get \[ \begin{flalign*} b_1^2 = \frac{n_y^2}{2(1-n_z)} & \Rightarrow b_1 = n_y \frac{1}{\sqrt{2(1-n_z)}} \\ & \Rightarrow a_1 = -n_x \frac{1}{\sqrt{2(1-n_z)}} \\ & \Rightarrow a_2 = (n_z-1) \frac{1}{\sqrt{2(1-n_z)}} \end{flalign*} \]

The complete form of the eigenvector is \[ \ket{E_1} = \frac{1}{\sqrt{2(1-n_z)}} \begin{pmatrix} -n_x + \mathrm{i} n_y \\ n_z -1 \end{pmatrix} \]

We can find \(\ket{E_2}\) in the same way by solving the equation with the eigenvalue \(E_2 = -\frac{\hbar \omega}{2}\). The result is \[ \ket{E_2} = \frac{1}{\sqrt{2(1 + n_z)}} \begin{pmatrix} -n_x + \mathrm{i} n_y \\ n_z + 1 \end{pmatrix} \]

It is easy to prove that \[ \braket{E_2 | E_1 } = 0 \]

Keep in mind that we chose \(b_2=0\), which is somewhat arbitrary. The solution is not unique, there are other pairs of eigenvectors with a different form.