Two State Quantum System with a given Hamiltonian

[bgrape]

Homework Statement

A two state system has the following hamiltonian
<br /> <br /> H=E \left( \begin{array}{cc} 0 &amp; 1 \\<br /> 1 &amp; 0 \end{array} \right)<br /> <br /> <br />


The state at t = 0 is given to be

<br /> <br /> \psi(0)=\left( \begin{array}{cc} 0 \\ 1 \end{array} \right)<br /> <br /> <br />

• Find Ψ(t).

• What is the probability to observe the ground state energy at t = T ?

• What is the probability that the state is

<br /> <br /> \left( \begin{array}{cc} 1 \\ 0 <br /> \end{array} \right)<br /> <br /> <br />

at t = T ?

Homework Equations

<br /> i\hbar\frac{\partial\psi}{\partial t} = \frac{\hbar^2}{2m}\nabla^2\psi + V(\mathbf{r})\psi<br />

The Attempt at a Solution

My main problem is with the notation. I understand that this system has two states with the same energies. I think this system can be an electron in an energy level, it has up or down spin possibilities. I think since both the states have the same energy both are considered ground state therefore we have 100% possibility to observe the ground state energy at t=T and the probability that the state is <br /> <br /> \left( \begin{array}{cc} 1 \\ 0 <br /> \end{array} \right)<br /> <br /> <br /> at t=T seems to be 1/2 intuitively because there is no energy difference between the states. However I don't know the correct notation for Ψ(t). I am thinking of

<br /> \psi(t)=\left( \begin{array}{cc} e^{(-iEt/\hbar)} \\<br /> e^{(-iEt/\hbar)} \end{array} \right)<br />

but this does not define that the system is in the particular state at t=0. Any help is greatly appriciated.

 

[csopi]

Note that \psi(0) is not an eigenstate. So you need to solve the time dependent Schrödinger equation. The wave function at time t is given by:

\psi (t)=\exp({-i/\hbar Ht})\psi (0)

You need to calculate the matrix exponential function!

 

[bgrape]

Thanks for your reply csopi. I have corrected my Ψ(t) as below, what I though of first was a fatal mistake.

<br /> \psi(t)=e^{(-iEt/\hbar)}\left( \begin{array}{cc} 0 \\<br /> 1 \end{array} \right)<br />

Is my assumption that since both energies are the same both can be considered as ground state and probability to observe the state at t=T will always be 1 correct?

for the next part

What is the probability that the state is

\left( \begin{array}{cc} 1 \\ 0 <br /> \end{array} \right)

at t = T ?


I think i will have to evaluate
| \langle \phi | \psi \rangle |^2


and i think \phi = \left( \begin{array}{cc} 1 \\ 0 <br /> \end{array} \right)

and

| \langle \phi | \psi \rangle |^2 = (\left( \begin{array}{cc} 1 &amp; 0 <br /> \end{array} \right)\left( \begin{array}{cc} 0 \\<br /> 1 \end{array} \right)e^{(-iET/\hbar)})^2 =0

Am I on the right track? I am not sure on how to calculate the probability of a state and an energy at a specific time, any help or tip is more than welcome.