QM: Groundstates and changing Hamiltonians

[Niles]

Homework Statement

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

<br /> H = \left( {\begin{array}{*{20}c}<br /> {\varepsilon _1 } &amp; { - V} \\<br /> { - V} &amp; {\varepsilon _2 } \\<br /> \end{array}} \right)<br />

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:


<br /> H = \left( {\begin{array}{*{20}c}<br /> {\varepsilon _1 } &amp; { - V} \\<br /> { - V} &amp; {\varepsilon _2 } \\<br /> \end{array}} \right)<br />

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

 

[Irid]

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

 

[Niles]

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?

 

[Irid]

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".

 

[Niles]

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.

 

[Niles]

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:

<br /> P = |\langle m_0 | v_-\rangle|^2.<br />

Thanks.