Two State System Described by a Time-Dependent Hamiltonian

[sharrington3]

Homework Statement

A two state system is described the time dependent Hamiltonian
$$\hat{H}=|\psi\rangle E\langle\psi|+|\phi\rangle E \langle\phi|+|\psi\rangle V(t)\langle\phi|+|\phi\rangle V(t)\langle\psi|$$
Where
$$\langle \psi|\psi \rangle = 1=\langle \phi|\phi\rangle, \langle \phi|\psi \rangle=0=\langle \psi|\phi \rangle \\V(t)→0, t→±∞$$
Given that as t→-∞ the system was in the state $$|\psi\rangle$$ find the probability that it will end up in state $$|\phi\rangle$$ as t→+∞.

Homework Equations

$$\hat{H}_\psi|\psi\rangle = E|\psi\rangle$$
$$\hat{H}_\phi|\phi\rangle = E|\phi\rangle$$
$$\hat{H}_0= \hat{H}_\psi+\hat{H}_\phi$$
$$\hat{H}(t)=\hat{H}_0+\hat{V}(t)$$

The Attempt at a Solution

To be completely honest, I don't even really know were to begin. I can somewhat grasp the fact that this time dependent Hamiltonian can be written as the sum of the time independent Hamiltonian and some perturbation V(t), but beyond that, I'm at a loss.

 

[mark726]

I understand the notation and how to solve for an energy eigenvalue, but I'm not sure how to go about finding the probability of the system ending up in state |\phi\rangle.

your first step would be to understand the problem and the given information. Let's break down the problem and see what we know:

1. We have a two-state system described by a time-dependent Hamiltonian.
2. The Hamiltonian can be written as the sum of a time-independent part (corresponding to the two states) and a perturbation V(t).
3. The system starts in the state |\psi\rangle at t=-∞.
4. As t→+∞, the perturbation V(t) approaches 0.

Now, let's think about what we want to find: the probability that the system will end up in state |\phi\rangle as t→+∞. To find this, we need to use the time evolution of the system, which is governed by the Schrodinger equation:

i\hbar\frac{d}{dt}|\psi(t)\rangle = \hat{H}(t)|\psi(t)\rangle

From the given information, we know that the time-independent Hamiltonian \hat{H}_0 has two energy eigenvalues E_\psi and E_\phi, corresponding to the states |\psi\rangle and |\phi\rangle respectively. So, we can write the solution to the Schrodinger equation as:

|\psi(t)\rangle = c_\psi(t)e^{-iE_\psi t/\hbar}|\psi\rangle + c_\phi(t)e^{-iE_\phi t/\hbar}|\phi\rangle

where c_\psi(t) and c_\phi(t) are time-dependent coefficients that we need to solve for.

Now, we can use the initial condition given in the problem to solve for these coefficients. At t=-∞, we know that the system is in the state |\psi\rangle, so c_\psi(-∞)=1 and c_\phi(-∞)=0. Plugging this into our solution, we get:

|\psi(t)\rangle = e^{-iE_\psi t/\hbar}|\psi\rangle

At t=+∞, we want to find the probability amplitude for the system to be in the state |\phi\rangle, which we can write as:

\langle\phi|\