Recent content by rustyrusty

thank you guys. Im normally not this dense, I swear! Just been at it all week and I think my brain is fried..

Well, with the Time evolution operator. When it acts on an eigenstate, it will simply replace the Hamiltonian in the exp to its corresponding eigenvalue. We use eigenstates so we can express the initial state as a linear combination of the two eigenstates. So, If I were to express the initial...

Not too sure? I know it gives us states that naturally describe the system. But i know that d & |2> ISNT an eigenvalue/state of the system. So how do these eigenstates allow me to say that it is initially in state two?

These states give us the energy eigenvalues possible from the full hamiltonian.

The old basis 1&2 where eigenstates of the first hamiltonian. Since we added in the new hamiltonian, It changed the eigenstates.

The point is to get the correct eigenkets of the system given the hamiltonian. Then the eigenkets form a complete set and i can express any state as a linear combination of these eigenkets I found earlier

Well, the problem says to find the state at time t>0. I know that it is initially in a state |2> at t=0.

H=\left( \begin{array}{cc} \text{a} & c e^{-i w t} \\ c e^{i w t} & \text{d} \end{array} \right); Then eigenvalues of: \text{$\lambda $1}=\frac{1}{2} \left(d+a-\sqrt{d^2-2a d+a^2+4 c^2}\right)\\ \text{$\lambda $2}=\frac{1}{2} \left(d+a+\sqrt{d^2-2a d+a^2+4 c^2}\right) Plugged those back in...

Homework Statement I have a hamiltonian: \begin{pmatrix} a &0 \\ 0&d \end{pmatrix} + \begin{pmatrix} 0 &ce^{i w t} \\ ce^{-iwt}&0 \end{pmatrix}=\begin{bmatrix} a & c e^{i w t} \\ c e^{-i w t}&d \\ \end{bmatrix} Where the first hamiltonian can be labeled with states |1> and |2>...