QM: Groundstates and changing Hamiltonians

Homework Statement

At time t<0 I have a Hamiltonian given by:

$$H = \left( {\begin{array}{*{20}c} {\varepsilon _1 } & { - V} \\ { - V} & {\varepsilon _2 } \\ \end{array}} \right)$$

with the ground state $$\left| {v_ - } \right\rangle = A\left( {\sqrt 2 - 1} \right)\left| 1 \right\rangle + \left| 2 \right\rangle$$, where A is some constant.

Now at t=0 the Hamiltonain changes, since $$\varepsilon_1=\varepsilon_2=0$$ for t>0.

I have to find the probability of finding the particle in the groundstate of the new Hamiltonian at the time t=0.

The Attempt at a Solution

Ok, first I have found the groundstate $$\left| {m_0 } \right\rangle$$ of the new Hamiltonian. Since I have to find the probability of the particle being in the state $$\left| {m_0 } \right\rangle$$ at t=0, I still need the Hamiltonian for t<0, i.e. I need:

$$H = \left( {\begin{array}{*{20}c} {\varepsilon _1 } & { - V} \\ { - V} & {\varepsilon _2 } \\ \end{array}} \right)$$

I am a little uncertain of what should be done now. Can you push me in the right direction?

I'm also uncertain, but my guess is that you should first normalize your ground states and then take the square modulus of their inner product to get the probability:

$$P = |\langle v_- | m_0\rangle|^2$$

I think we're on the right track, Irid.

I'm quite sure I have to use $$P = |\langle v_- | \Psi\rangle|^2$$, but how can I extract \Psi from the Hamiltonian at t=0?

You said that you have Psi, only named it m_0. Anyway, I think one needs to set up an eigenvalue equation and you'll obtain eigenvectors. The one which corresponds to the lowest energy is termed "the ground state".

I have already found (and normalized) the groundstate $$\left| {m_0 } \right\rangle$$. But the groundstate is not the same as the wavefunction itself, i.e. the solution of the time-dependant Schrödinger-equation?

Please correct me if I am wrong.

Ok, I worked it out. The particle is in the groundstate, so $$\Psi(x,0) = \left| {v__- } \right\rangle$$ and the new stationary state is $$\psi(x) = \left| {m_0 } \right\rangle$$. That is, we can find the probability as:

$$P = |\langle m_0 | v_-\rangle|^2.$$

Thanks.