Solution
Hamiltonian is given by
H=E0(σz+ασx)H=E_0(\sigma_z+\alpha\sigma_x)H=E0(σz+ασx)
Using pauli matrices
H=E0(100−1)++E0α(0101)H=E_0\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}++ E_0 \alpha\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}H=E0(100−1)++E0α(0011)
H=E0(1αα−1)H=E_0\begin{pmatrix} 1 & \alpha \\ \alpha & -1 \end{pmatrix}H=E0(1αα−1)
if eigenvalue is λ\lambdaλ then
H−λI=0H-\lambda I=0H−λI=0
E0((1−λ)αα−(1+λ))=0E_0\begin{pmatrix} (1-\lambda) & \alpha\\ \alpha & -(1+\lambda) \end{pmatrix}=0E0((1−λ)αα−(1+λ))=0
λ=+E01+α2\lambda=+E_0\sqrt{1+\alpha^2}λ=+E01+α2
or λ=−E01+α2\lambda=-E_0\sqrt{1+\alpha^2}λ=−E01+α2
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment