Question

Quantum Mechanics : Given a Matrix (Hamiltonian) of the form ſa b a) Find the Eigenvalues b) Find the Eigenvectors c) Use the above Eigenvectors to find the spin polarization vector given by st= |x112 – [X212

Answer 1

To find the eigenvalues you need to compute,
det (H - XI2 = 0.12
is the indentity matrix of dimension 2. So on solving this we get an algebric equation (quadratic equation) in $\lambda$ , $a d-a \lambda -b^2-d \lambda +\lambda ^2 = 0.$ Now solving for $\lambda$ we get, $\lambda_{1, 2} = \frac{1}{2} \left(-\sqrt{(a-d)^2+4 b^2}+a+d\right),\frac{1}{2} \left(\sqrt{(a-d)^2+4 b^2}+a+d\right).$

Now to compute the eigenvectors one needs to plug in
1.29
into $H - \lambda_{1, 2} I_2 = (x_1, x_2)^{T}.$ Solving for will lead to pair of linear equations which on solving will lead to two different pairs of values of which leads to the eigenvectors. They are, $x_1, x_2 = \left\{-\frac{\sqrt{a^2-2 a d+4 b^2+d^2}-a+d}{2 b},1\right\}^{T},\left\{-\frac{-\sqrt{a^2-2 a d+4 b^2+d^2}-a+d}{2 b},1\right\}^{T}.$

Now, for the spin polarisation part, $S_{\pm} = \left\{\frac{\left(\sqrt{a^2-2 a d+4 b^2+d^2}-a+d\right)^2}{4 b^2}-\frac{\left(-\sqrt{a^2-2 a d+4 b^2+d^2}-a+d\right)^2}{4 b^2},0\right\}^{T}.$