Two State Quantum System with a given Hamiltonian

[bgrape]

Homework Statement

A two state system has the following hamiltonian
[tex]

H=E \left( \begin{array}{cc} 0 & 1 \\
1 & 0 \end{array} \right)


[/tex]


The state at t = 0 is given to be

[tex]

\psi(0)=\left( \begin{array}{cc} 0 \\ 1 \end{array} \right)


[/tex]

• Find Ψ(t).

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

• What is the probability that the state is

[tex]

\left( \begin{array}{cc} 1 \\ 0
\end{array} \right)


[/tex]

at t = T ?

Homework Equations

[tex]
i\hbar\frac{\partial\psi}{\partial t} = \frac{\hbar^2}{2m}\nabla^2\psi + V(\mathbf{r})\psi
[/tex]

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 [tex]

\left( \begin{array}{cc} 1 \\ 0
\end{array} \right)


[/tex] 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

[tex]
\psi(t)=\left( \begin{array}{cc} e^{(-iEt/\hbar)} \\
e^{(-iEt/\hbar)} \end{array} \right)
[/tex]

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

 

[csopi]

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

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

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.

[tex]
\psi(t)=e^{(-iEt/\hbar)}\left( \begin{array}{cc} 0 \\
1 \end{array} \right)
[/tex]

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

[tex] \left( \begin{array}{cc} 1 \\ 0
\end{array} \right) [/tex]

at t = T ?


I think i will have to evaluate
[tex]| \langle \phi | \psi \rangle |^2 [/tex]


and i think [tex]\phi = \left( \begin{array}{cc} 1 \\ 0
\end{array} \right)[/tex]

and

[tex]| \langle \phi | \psi \rangle |^2 = (\left( \begin{array}{cc} 1 & 0
\end{array} \right)\left( \begin{array}{cc} 0 \\
1 \end{array} \right)e^{(-iET/\hbar)})^2 =0 [/tex]

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.