Consider the observable q_u = u_1q_1+u_2q_2+u_3q_3 ,
where the q_i are the Pauli matrices and
where u = (u_1,u_2,u_3) in R^3
and the Hamiltonian is H = q3.
Compute the observable (q_u)_H (t) derived from q_u in the Heisenberg picture.
The Heisenberg picture is obtained as follows:
A_H = Exp[i H t/h-bar] A_S Exp[-i H t/h-bar]
Here A_S is the operator in the Schrodinger picture and H is the Hamiltonian. The Hamiltonian is given by sigma_z. We thus need to exponentiate sigma_z. This is not difficult, because sigma_z is already diagonal in the standard basis. I.m.o., this problem is too easy because of this. So, after you do this problem you should generalize the Hamiltonian. Instead of sigma_z you take:
H = h_x sigma_x + h_y sigma_y + h_z sigma_z = h dot sigma
Your problem corresponds to the special case: h_x = h_y = 0 and h_z = 1.
You now need to exponentiate this matrix. If we define |h| as the norm of the vector (h_x,h_y,h_z), then we have:
Exp[iaH] = cos(a|h|) + i h/|h| dot sigma sin(a |h|) = ...
We calculate the Heisenberg representation of a linear combination of the Pauli matrices in the case where the Hamiltonian is proportional to sigma_z