The Hamiltonian of a system is \hat{H}=\varepsilon \vec{\sigma }.\vec{n}, where ε is a constant having the dimensions of energy, \vec{n} is an arbitrary unit vector, and \sigma _{x} ,\sigma _{y}, and \sigma _{z} are the Pauli matrices.
(a) Find the energy eigenvalues and normalized eigenvectors of \hat{H} .
(b) Find a transformation matrix that diagonalizes \hat{H} .
(a) Using the Pauli matrices \sigma _{x}=\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) ,\sigma _{y}=\left(\begin{matrix} 0 & -i \\ i & 0 \end{matrix} \right),\sigma _{z}=\left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right) and the expression of an arbitrary unit vector in spherical coordinates \vec{n}=(\sin \theta \cos \varphi )\vec{i}+(\sin \theta \sin \varphi )\vec{j}+(\cos \theta )\vec{k}, we can rewrite the Hamiltonian
\hat{H}=\varepsilon \vec{\sigma }.\vec{n}=\varepsilon (\sigma _{x}\sin \theta \cos \varphi +\sigma _{y}\sin \theta \sin \varphi + \sigma _{z}\cos \theta ) (5.249)
in the following matrix form:
\hat{H}=\varepsilon \left(\begin{matrix} \cos \theta & \exp (-i\varphi )\sin \theta \\ \exp (i\varphi )\sin \theta & -\cos \theta \end{matrix} \right). (5.250)
The eigenvalues of \hat{H} are obtained by solving the secular equation \det (H-E)=0, or
(\varepsilon \cos \theta -E)(-\varepsilon \cos \theta -E)-\varepsilon ^{2} \sin ^{2} \theta =0, (5.251)
which yields two eigenenergies E_{1} =\varepsilon and E_{2} =-\varepsilon .
The energy eigenfunctions are obtained from
\varepsilon \left(\begin{matrix} \cos \theta & \exp (-i\varphi )\sin \theta \\ \exp (i\varphi )\sin \theta & -\cos \theta \end{matrix} \right)\left(\begin{matrix} x \\ y \end{matrix} \right) =E\left(\begin {matrix} x \\ y \end{matrix} \right). (5.252)
For the case E=E_{1} =\varepsilon, this equation yields
(\cos \theta -1)x+y\sin \theta \exp (-i\varphi )=0, (5.253)
which in turn leads to
\frac{x}{y} =\frac{\sin \theta \exp (-i\varphi )}{1-\cos \theta } =\frac{\cos \theta /2\exp (-i\varphi /2)}{\sin \theta /2\exp (i\varphi /2)}; (5.254)
hence
\left(\begin{matrix} x_{1} \\ y_{1} \end{matrix} \right) =\left (\begin{matrix} \exp (-i\varphi /2)\cos (\theta /2) \\ \exp (i\varphi /2)\sin (\theta /2) \end{matrix} \right) ; (5.255)
this vector is normalized. Similarly, in the case where E=E_{2} =-\varepsilon, we can show that the second normalized eigenvector is
\left(\begin{matrix} x_{2} \\ y_{2} \end{matrix} \right) =\left (\begin{matrix} -\exp (-i\varphi /2)\sin (\theta /2) \\ \exp (i\varphi /2)\cos (\theta /2) \end{matrix} \right). (5.256)
(b) A transformation \hat{U} that diagonalizes \hat{H} can be obtained from the two eigenvectors obtained in part (a): U_{11} =x_{1},U_{21} =y_{1},U_{12} =x_{2},U_{22} =y_{2}. That is,
U=\left(\begin{matrix} \exp (-i\varphi /2)\cos (\theta /2) & -\exp (-i\varphi /2)\sin (\theta /2) \\ \exp (i\varphi /2)\sin (\theta /2) & \exp (i\varphi /2)\cos (\theta /2) \end{matrix} \right) . (5.257)
Note that this matrix is unitary, since U^{\dagger } =U^{-1} and \det (U)=1. We can ascertain that
\hat{U}\hat{H}\hat{U}^{\dagger } =\left(\begin{matrix} \varepsilon & 0 \\ 0 & -\varepsilon \end{matrix} \right). (5.258)