Quantum Fundamentals: Winter-2025
Eigenvectors Practice (SOLUTION): Due Day 12 Tu 2/25 Math Bits

1. Eigenvectors of Pauli Matrices S1 5196S
   1. Find the eigenvalues and normalized eigenvectors of the Pauli matrices \(\sigma_x\), \(\sigma_y\), and \(\sigma_z\) (see the Spins Reference Sheet posted on the course website).

      \[\sigma_x= \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda&-1\\-1&\lambda\\ \end{matrix} \right\vert=0=\lambda^2-1 \Rightarrow\lambda=1,-1\] For \(\lambda=1\), \[ \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=x \Rightarrow \vert v\rangle_1= \begin{pmatrix} 1\\ 1\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_1 =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ 1\\ \end{pmatrix} \] For \(\lambda=-1\), \[ \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=-x \;\Rightarrow \; \vert v\rangle_{-1} = \begin{pmatrix} -1\\ 1\\ \end{pmatrix} \;\Rightarrow\; \vert\hat v\rangle_{-1} =\frac{1}{ \sqrt 2} \begin{pmatrix} -1\\ 1\\ \end{pmatrix} \]

      \[\sigma_y= \begin{pmatrix} 0&-i\\ i&0\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda&i\\-i&\lambda\\ \end{matrix} \right\vert=0=\lambda^2-1 \Rightarrow\lambda=1,-1\] For \(\lambda=1\), \[ \begin{pmatrix} 0&-i\\ i&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow -iy=x \Rightarrow \vert v\rangle_1= \begin{pmatrix} 1\\ i\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_1 =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ i\\ \end{pmatrix} \] For \(\lambda=-1\), \[ \begin{pmatrix} 0&-i\\ i&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow iy=x \;\Rightarrow \; \vert v\rangle_{-1} = \begin{pmatrix} 1\\ -i\\ \end{pmatrix} \;\Rightarrow\; \vert\hat v\rangle_{-1} =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ -i\\ \end{pmatrix} \]

      \[\sigma_z= \begin{pmatrix} 1&0\\ 0&-1\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda-1&0\\0&\lambda+1\\ \end{matrix} \right\vert=0=\lambda^2-1 \Rightarrow\lambda=1,-1\] For \(\lambda=1\), \[ \begin{pmatrix} 1&0\\ 0&-1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=x, y=-y \Rightarrow \vert\hat v\rangle_1 = \begin{pmatrix} 1\\ 0\\ \end{pmatrix} \] For \(\lambda=-1\), \[ \begin{pmatrix} 1&0\\ 0&-1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=-x, y=y \;\Rightarrow \; \vert\hat v\rangle_{-1} = \begin{pmatrix} 0\\ 1\\ \end{pmatrix} \]

2. Eigen Practice S1 5196S Find the eigenvectors and eigenvalues of the matrices from the Linear Transformations small group activity from Tuesday's class. Keep working until you are fluent. Make up some \(2\times 2\) and \(3\times 3\) matrices of your own if you need more practice.

   \[A_1= \begin{pmatrix} 0&1\\ -1&0\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda&-1\\1&\lambda\\ \end{matrix} \right\vert=0=\lambda^2+1 \Rightarrow\lambda=i,-i\] For \(\lambda=i\), \[ \begin{pmatrix} 0&1\\ -1&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =i \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=ix \Rightarrow \vert v\rangle_i= \begin{pmatrix} 1\\ i\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_i =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ i\\ \end{pmatrix} \] For \(\lambda=-i\), \[ \begin{pmatrix} 0&1\\ -1&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-i \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=-ix \;\Rightarrow \; \vert v\rangle_{-i} = \begin{pmatrix} 1\\ -i\\ \end{pmatrix} \;\Rightarrow\; \vert\hat v\rangle_{-i} =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ -i\\ \end{pmatrix} \]

   \[A_2= \begin{pmatrix} 0&-1\\ 1&0\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda&1\\-1&\lambda\\ \end{matrix} \right\vert=0=\lambda^2+1 \Rightarrow\lambda=i,-i\] For \(\lambda=i\), \[ \begin{pmatrix} 0&-1\\ 1&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =i \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=-ix \Rightarrow \vert v\rangle_i= \begin{pmatrix} 1\\ -i\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_i =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ -i\\ \end{pmatrix} \] For \(\lambda=-i\), \[ \begin{pmatrix} 0&-1\\ 1&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-i \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=ix \Rightarrow \vert v\rangle_{-i} = \begin{pmatrix} 1\\ i\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_{-i} =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ i\\ \end{pmatrix} \]

   \[A_3= \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda&-1\\-1&\lambda\\ \end{matrix} \right\vert=0=\lambda^2-1 \Rightarrow\lambda=1, -1\] For \(\lambda=1\), \[ \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=x \Rightarrow \vert v\rangle_1= \begin{pmatrix} 1\\ 1\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_1 =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ 1\\ \end{pmatrix} \] For \(\lambda=-1\), \[ \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=-x \Rightarrow \vert v\rangle_{-1} = \begin{pmatrix} 1\\ -1\\ \end{pmatrix} \;\Rightarrow\; \vert\hat v\rangle_{-1} =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ -1\\ \end{pmatrix} \]

   \[A_4= \begin{pmatrix} 1&0\\0&-1\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda-1&0\\ 0&\lambda+1\\ \end{matrix} \right\vert=0 =(\lambda+1)(\lambda-1) \Rightarrow\lambda=1, -1\] For \(\lambda=1\), \[ \begin{pmatrix} 1&0\\ 0&-1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=x,~y=-y \Rightarrow \vert v\rangle_1=\vert\hat v\rangle_1 = \begin{pmatrix} 1\\ 0\\ \end{pmatrix} \] For \(\lambda=-1\), \[ \begin{pmatrix} 1&0\\ 0&-1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=-x,~y=y \Rightarrow \vert v\rangle_{-1} =\vert\hat v\rangle_{-1} = \begin{pmatrix} 0\\ 1\\ \end{pmatrix} \]

   \[A_5= \begin{pmatrix} -1&0\\ 0&-1\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda+1&0\\0&\lambda+1\\ \end{matrix} \right\vert=0=(\lambda+1)^2 \Rightarrow\lambda=-1, {\rm degenerate}\] For \(\lambda=-1\), \[ \begin{pmatrix} -1&0\\ 0&-1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=x,~y=y \Rightarrow \vert v\rangle_{-1} =\vert\hat v\rangle_{-1} = \begin{pmatrix} 1\\ 0\\ \end{pmatrix} ,\; \begin{pmatrix} 0\\ 1\\ \end{pmatrix} \] Any linear combination of these two eigenvectors is also an eigenvector. Because of the degeneracy, the eigenvectors form a two-dimensional space. \[A_6= \begin{pmatrix} 1&2\\ 1&2\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda-1&-2\\ -1&\lambda-2\\ \end{matrix} \right\vert=0 =(\lambda-1)(\lambda-2)-2 \Rightarrow\lambda=0, 3\] For \(\lambda=0\), \[ \begin{pmatrix} 1&2\\ 1&2\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =0 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=-2y \Rightarrow \vert v\rangle_0= \begin{pmatrix} -2\\ 1\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_0 =\frac{1}{ \sqrt 5} \begin{pmatrix} -2\\ 1\\ \end{pmatrix} \] For \(\lambda=3\), \[ \begin{pmatrix} 1&2\\ 1&2\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =3 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=y \Rightarrow \vert v\rangle_3= \begin{pmatrix} 1\\ 1\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_3 =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ 1\\ \end{pmatrix} \]

   \[A_7= \begin{pmatrix} 1&2\\ 9&4\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda-1&-2\\ -9&\lambda-4\\ \end{matrix} \right\vert=0 =(\lambda-1)(\lambda-4)-18 \Rightarrow\lambda=7, -2\] For \(\lambda=7\), \[ \begin{pmatrix} 1&2\\ 9&4\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =7 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow y=3x \Rightarrow \vert v\rangle_7= \begin{pmatrix} 1\\ 3\\ \end{pmatrix} \Rightarrow \vert\hat v\rangle_7 =\frac{1}{ \sqrt{10}} \begin{pmatrix} 1\\ 3\\ \end{pmatrix} \] For \(\lambda=-2\), \[ \begin{pmatrix} 1&2\\ 9&4\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =-2 \begin{pmatrix} x\\ y\\ \end{pmatrix} \;\Rightarrow \; 2y=-3x \Rightarrow \vert v\rangle_{-2} = \begin{pmatrix} 2\\ -3\\ \end{pmatrix} \;\Rightarrow\; \vert\hat v\rangle_{-2} =\frac{1}{ \sqrt{13}} \begin{pmatrix} 2\\ -3\\ \end{pmatrix} \]

   \[A_8= \begin{pmatrix} 1&1\\ -1&1\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda-1&-1\\ 1&\lambda-1\\ \end{matrix} \right\vert=0 =(\lambda-1)^2+1 \Rightarrow\lambda=1\pm i\] For \(\lambda=1+i\), \[ \begin{pmatrix} 1&1\\ -1&1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =(1+i) \begin{pmatrix} x\\ y\\ \end{pmatrix} \;\Rightarrow \; y=ix \Rightarrow \vert v\rangle_{(1+i)} = \begin{pmatrix} 1\\ i\\ \end{pmatrix} \;\Rightarrow\; \vert\hat v\rangle_{(1+i)} =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ i\\ \end{pmatrix} \] For \(\lambda=1-i\), \[ \begin{pmatrix} 1&1\\ -1&1\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =(1-i) \begin{pmatrix} x\\ y\\ \end{pmatrix} \,\Rightarrow \, y=-ix \,\Rightarrow \, \vert v\rangle_{(1-i)} = \begin{pmatrix} 1\\ -i\\ \end{pmatrix} \,\Rightarrow\, \vert\hat v\rangle_{(1-i)} =\frac{1}{ \sqrt 2} \begin{pmatrix} 1\\ -i\\ \end{pmatrix} \]

   \[A_9= \begin{pmatrix} 2&0\\ 0&2\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda-2&0\\ 0&\lambda-2\\ \end{matrix} \right\vert=0 =(\lambda-2)^2 \Rightarrow\lambda=2, {\rm degenerate}\] For \(\lambda=2\), \[ \begin{pmatrix} 2&0\\ 0&2\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =2 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=x,~y=y \Rightarrow \vert v\rangle_2=\vert\hat v\rangle_2 = \begin{pmatrix} 1\\ 0\\ \end{pmatrix} \; \begin{pmatrix} 0\\1\\ \end{pmatrix} \] Any linear combination of these two eigenvectors is also an eigenvector. Because of the degeneracy, the eigenvectors form a two-dimensional space. \[A_{10} = \begin{pmatrix} 1&0\\ 0&0\\ \end{pmatrix} \Rightarrow \left\vert \begin{matrix} \lambda-1&0\\ 0&\lambda\\ \end{matrix} \right\vert=0 =(\lambda-1)(\lambda) \Rightarrow\lambda=0, 1\] For \(\lambda=0\), \[ \begin{pmatrix} 1&0\\ 0&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =0 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=0 \Rightarrow \vert v\rangle_0=\vert\hat v\rangle_0= \begin{pmatrix} 0\\ 1\\ \end{pmatrix} \] For \(\lambda=1\), \[ \begin{pmatrix} 1&0\\ 0&0\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ \end{pmatrix} =1 \begin{pmatrix} x\\ y\\ \end{pmatrix} \Rightarrow x=x,~y=0 \Rightarrow \vert v\rangle_1=\vert\hat v\rangle_1= \begin{pmatrix} 1\\ 0\\ \end{pmatrix} \]

3. Diagonalization S1 5196S
   1. Let \[|\alpha\rangle \doteq \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1 \end{pmatrix} \qquad \rm{and} \qquad |\beta\rangle \doteq \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ -1 \end{pmatrix}\] Show that \(\left|{\alpha}\right\rangle \) and \(\left|{\beta}\right\rangle \) are orthonormal. (If a pair of vectors is orthonormal, that suggests that they might make a good basis.)

      Orthonormal means both orthogonal and normalized.

      To show that the vectors are normalized, calculate \begin{align} \left\langle {\alpha}\middle|{\alpha}\right\rangle &=\frac{1}{\sqrt{2}} \begin{pmatrix}1&1\end{pmatrix}\;\frac{1}{\sqrt{2}} \begin{pmatrix}1\\1\end{pmatrix}\\ &= \frac{1}{{2}} \left( 1*1 + 1*1 \right) \\ &=1\\ \left\langle {\beta}\middle|{\beta}\right\rangle &=\frac{1}{\sqrt{2}} \begin{pmatrix}1&-1\end{pmatrix}\; \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\\ &= \frac{1}{{2}} \left( 1*1 + (-1)*(-1) \right) \\ &=1 \end{align}

      To show that the vectors are orthogonal, calculate \begin{align} \langle\beta|\alpha\rangle &= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}\\ &= \frac{1}{{2}} \left( 1*1 + (-1)*1 \right) \\ &= 0 \end{align}

   2. Consider the matrix \[C\doteq \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \] Show that the vectors \(|\alpha\rangle\) and \(|\beta\rangle\) are eigenvectors of C and find the eigenvalues. (Note that showing something is an eigenvector of an operator is far easier than finding the eigenvectors if you don't know them!)

      \(C|\alpha\rangle \doteq \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 4|\alpha\rangle\)
      \(C|\beta\rangle \doteq \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 2 \\ -2 \end{pmatrix} = 2|\beta\rangle\)

      So \(\left|{\alpha}\right\rangle \) is an eigenvector with eigenvalue 4 and \(\left|{\beta}\right\rangle \) is an eigenvector with eigenvalue 2.

   3. A operator is always represented by a diagonal matrix if it is written in terms of the basis of its own eigenvectors. What does this mean? Find the matrix elements for a new matrix \(E\) that corresponds to \(C\) expanded in the basis of its eigenvectors, i.e. calculate \(\langle\alpha|C|\alpha\rangle\), \(\langle\alpha|C|\beta\rangle\), \(\langle\beta|C|\alpha\rangle\) and \(\langle\beta|C|\beta\rangle\) and arrange them into a sensible matrix \(E\). Explain why you arranged the matrix elements in the order that you did.

      \(\langle\alpha|C|\alpha\rangle = \frac{1}{{2}} \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{{2}} \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 4\)
      \(\langle\alpha|C|\beta\rangle = \frac{1}{{2}} \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{{2}} \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ -2 \end{pmatrix} = 0\)
      \(\langle\beta|C|\alpha\rangle = \frac{1}{{2}} \begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{{2}} \begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 0\)
      \(\langle\beta|C|\beta\rangle = \frac{1}{{2}} \begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{{2}} \begin{pmatrix} 1 & -1 \end{pmatrix} \begin{pmatrix} 2 \\ -2 \end{pmatrix} = 2\)

      \[E=\begin{pmatrix} 4&0\\0&2 \end{pmatrix}\]

   4. Find the determinants of \(C\) and \(E\). How do these determinants compare to the eigenvalues of these matrices?

      \(\left| \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \right| = 9 - 1 = 8 \\ \left| \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \right| = 8\\\) The determinants for the two matrices are the same, and each is equal to the product of the eigenvalues.

4. Eigen Spin Challenge S1 5196S Consider the arbitrary Pauli matrix \(\sigma_n=\hat n\cdot\vec \sigma\) where \(\hat n\) is the unit vector pointing in an arbitrary direction.
   1. Find the eigenvalues and normalized eigenvectors for \(\sigma_n\). The answer is: \[ \begin{pmatrix} \cos\frac{\theta}{2}e^{-i\phi/2}\\{} \sin\frac{\theta}{2}e^{i\phi/2}\\ \end{pmatrix} \begin{pmatrix} -\sin\frac{\theta}{2}e^{-i\phi/2}\\{} \cos\frac{\theta}{2}e^{i\phi/2}\\ \end{pmatrix} \] It is not sufficient to show that this answer is correct by plugging into the eigenvalue equation. Rather, you should do all the steps of finding the eigenvalues and eigenvectors as if you don't know the answer. Hint: \(\sin\theta=\sqrt{1-\cos^2\theta}\).

      \begin{eqnarray*} 0 &=& \left\vert \begin{pmatrix} \cos\theta-\lambda &\sin\theta\,e^{-i\phi}\\ \sin\theta\, e^{i\phi}&-\cos\theta-\lambda \end{pmatrix} \right\vert \\ {} &=& -(\cos\theta-\lambda)(\cos\theta+\lambda)-\sin^2\theta \\ {} &=& (\lambda^2-\cos^2\theta)-\sin^2\theta \\ {} &=& \lambda^2-1 \\ {} &\Rightarrow& \lambda=\pm 1 \end{eqnarray*} For \(\lambda=1\): \[ \begin{pmatrix} \cos\theta&\sin\theta\, e^{-i\phi}\\\sin\theta\, e^{i\phi}&-\cos\theta\\ \end{pmatrix} \, \begin{pmatrix} x\\ y\\ \end{pmatrix} = +1\, \begin{pmatrix} x\\ y\\ \end{pmatrix} \] \(\Rightarrow\) \[y=\frac{1-\cos\theta}{\sin\theta}\, e^{i\phi}\, x =\tan\frac{\theta}{2}\, e^{i\phi}\, x =\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\, e^{i\phi}\, x \] If you choose \(x=1\), then the eigenvector is \[\vert v\rangle_1 \doteq \begin{pmatrix} 1\\ \tan\frac{\theta}{2}\, e^{i\phi} \end{pmatrix} \] and the norm squared is \[_1\langle v\vert v\rangle_1 \doteq \begin{pmatrix} 1&\tan\frac{\theta}{2}\, e^{-i\phi}\\ \end{pmatrix} \begin{pmatrix} 1\\ \tan\frac{\theta}{2}\, e^{i\phi}\\ \end{pmatrix} =1+\tan^2\frac{\theta}{2} =\frac{1}{\cos^2\frac{\theta}{2}}\] And the norm is \[\sqrt{_1\langle v\vert v\rangle_1} =\frac{1}{\cos\frac{\theta}{2}}\] Therefore, for normalized vectors, divide by the norm above, and adjust the phase: \[\vert v\rangle_1 \doteq e^\frac{i\phi}{2} \begin{pmatrix} \cos\frac{\theta}{2}e^\frac{-i\phi}{2}\\ (\cos\frac{\theta}{2})(\tan\frac{\theta}{2})\, e^\frac{i\phi}{2} \end{pmatrix} \] \[\vert \hat v\rangle_1 \doteq \begin{pmatrix} x\\ y\\ \end{pmatrix} = \begin{pmatrix} \cos\frac{\theta}{2}\,e^{-i\phi/2}\\ {} \sin\frac{\theta}{2}\,e^{i\phi/2}\\ \end{pmatrix} \] Note that in the last line I've dropped the overall phase. This choice of phase is just convention, albeit a nicely symmetric one. You can multiply both components of this eigenvector by the same overall phase and still have a normalized eigenvector.

      For \(\lambda=-1\): \[ \begin{pmatrix} \cos\theta&\sin\theta\, e^{-i\phi}\\\sin\theta\, e^{i\phi}&-\cos\theta\\ \end{pmatrix} \, \begin{pmatrix} x\\ y\\ \end{pmatrix} = -1\, \begin{pmatrix} x\\ y\\ \end{pmatrix} \] \(\Rightarrow\) \[x=\frac{1-\cos\theta}{-\sin\theta}\, e^{-i\phi}\, y =-\tan\frac{\theta}{2}\, e^{-i\phi}\, y =-\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\, e^{-i\phi}\, y \] If you choose, \(y=1\), then the eigenvector is: \[\vert v\rangle_{-1} \doteq \begin{pmatrix} -\tan\frac{\theta}{2}\, e^{-i\phi}\\1\\ \end{pmatrix} \] and the norm squared is: \[_{-1}\langle v\vert v\rangle_{-1} \doteq \begin{pmatrix} -\tan\frac{\theta}{2}\, e^{i\phi}&1\\ \end{pmatrix} \begin{pmatrix} -\tan\frac{\theta}{2}\, e^{-i\phi}\\ 1\\ \end{pmatrix} =1+\tan^2\frac{\theta}{2} =\frac{1}{\cos^2\frac{\theta}{2}}\] Therefore, for normalized vectors, divide by the norm above, and adjust the phase: \[\vert \hat v\rangle_{-1} \doteq \begin{pmatrix} x\\ y\\ \end{pmatrix} = \begin{pmatrix} -\sin\frac{\theta}{2}\,e^{-i\phi/2}\\ {} \cos\frac{\theta}{2}\,e^{i\phi/2}\\ \end{pmatrix} \] Note that this choice of phase is just convention, albeit a nicely symmetric one. You can multiply both components of this eigenvector by the same overall phase and still have a normalized eigenvector.

   2. Show that the eigenvectors from part (a) above are orthogonal.

      \[_{-1}\langle \hat v\vert \hat v\rangle_1 \doteq \begin{pmatrix} -\sin\frac{\theta}{2}\, e^{i\phi/2}& \cos\frac{\theta}{2}\, e^{-i\phi/2}\\ \end{pmatrix} \begin{pmatrix} \cos\frac{\theta}{2}\, e^{-i\phi/2}\\{} \sin\frac{\theta}{2}\, e^{i\phi/2}\\ \end{pmatrix} =-\sin\frac{\theta}{2}\, \cos\frac{\theta}{2} +\cos\frac{\theta}{2}\, \sin\frac{\theta}{2} =0\]

   3. Simplify your results from part (a) above by considering the three separate special cases: \(\hat n=\hat\imath\), \(\hat n=\hat\jmath\), \(\hat n=\hat k\). In this way, find the eigenvectors and eigenvalues of \(\sigma_x\), \(\sigma_y\), and \(\sigma_z\).

      Notice that the eigenvalues \(\lambda=\pm 1\) are independent of the value of \(\hat n\).

      If \(\hat n =\hat\imath\), then \(\theta=\frac{\pi}{2}\) and \(\phi=0\), and \(\hat n \cdot \vec \sigma=\sigma_x\). \[\vert\hat v\rangle_1 \rightarrow \begin{pmatrix} \cos\frac{\pi}{4}\\ \sin\frac{\pi}{4}\\ \end{pmatrix} =\frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1\\ \end{pmatrix} \] \[\vert\hat v\rangle_{-1} \rightarrow \begin{pmatrix} -\sin\frac{\pi}{4}\\ \cos\frac{\pi}{4}\\ \end{pmatrix} =\frac{1}{\sqrt2} \begin{pmatrix} -1\\ 1\\ \end{pmatrix} \]

      If \(\hat n =\hat\jmath\), then \(\theta=\frac{\pi}{2}\) and \(\phi=\frac{\pi}{2}\), and \(\hat n \cdot \vec \sigma=\sigma_y\). \[\vert\hat v\rangle_1 \rightarrow \begin{pmatrix} \cos\frac{\pi}{4}\, e^{-i\frac{\pi}{4}}\\ \sin\frac{\pi}{4} \end{pmatrix} \, e^{i\frac{\pi}{4}}\\ =\frac{1}{2} \begin{pmatrix} 1-i\\ 1+i\\ \end{pmatrix} =\frac{1}{\sqrt2} \begin{pmatrix} 1\\ i\\ \end{pmatrix} e^\frac{-i\pi}{4} \] \[\vert\hat v\rangle_{-1} \rightarrow \begin{pmatrix} -\sin\frac{\pi}{4}\, e^{-i\frac{\pi}{4}}\\ \cos\frac{\pi}{4}\, e^{i\frac{\pi}{4}}\\ \end{pmatrix} =\frac{1}{2} \begin{pmatrix} -1+i\\ 1+i\\ \end{pmatrix} =\frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ -i\\ \end{pmatrix} e^\frac{3i\pi}{4} \]

      If \(\hat n =\hat k\), then \(\theta=0\) and you can choose \(\phi=0\), and \(\hat n \cdot \vec \sigma=\sigma_z\). \[\vert\hat v\rangle_1 \rightarrow \begin{pmatrix} \cos 0\\ \sin 0\\ \end{pmatrix} = \begin{pmatrix} 1\\ 0\\ \end{pmatrix} \] \[\vert\hat v\rangle_{-1} \rightarrow \begin{pmatrix} -\sin 0\\ \cos 0\\ \end{pmatrix} = \begin{pmatrix} 0\\ 1\\ \end{pmatrix} \] These answers are the same as those in part (a) above. Note, however, that I have adjusted the phases (i.e. multiplied by an overall constant of magnitude 1) to put these answers in the same form as part (a).