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Hi, i have been trying to figure how to solve this problem, i just don't know how to proceed

A particle with spin s=1 has due to rotational invariance all three states [itex]|s_z\rangle=1[/itex], [itex]|s_z\rangle=0[/itex] and [itex]|s_z\rangle=-1[/itex] degenated with energy [itex]E_0>0[/itex]. Then its introduced an interaction non perturbative V

[itex]V=-\Delta\left(\begin{array}{ccc}

1 & 1 & 1\\

1 & 1 & 1\\

1 & 1 & 1

\end{array}\right)

[/itex]

a) Obtain stationary states and energies of the system

b) If in [itex]t=0[/itex], the particle has its third component with spin [itex]+\hbar[/itex] wich is the probability that after some time its [itex]-\hbar[/itex]

a)

So what i tried is this, since all three states have the same energy and those are eigenstates of the hamiltonian, the hamiltonian must have the form

[itex]H=E_0\left(\begin{array}{ccc}

1 & 0 & 0\\

0 & 1 & 0\\

0 & 0 & 1

\end{array}\right)

[/itex]

So if we add H'=H+V and diagonalize we get

[itex]\left|\begin{array}{ccc}

E_0-\Delta-\lambda & -\Delta & -\Delta\\

-\Delta & E_0-\Delta-\lambda & -\Delta\\

-\Delta & -\Delta & E_0-\Delta-\lambda

\end{array}\right|

[/itex]

So the characteristic equation is

[itex](E_0-\lambda)(E_0-3\Delta-\lambda)=0[/itex]

So we get the energies and the states

[itex](E_0,{-1,0,1})[/itex]

[itex](E_0,{-1,1,0})[/itex]

[itex](E_0-3\Delta,{1,1,1})[/itex]

I just want to know if this is the correct move or not

b)

i don't know how to aproach this, but i guess all i have to do is add the evolution term on each eigenstate [itex]e^{-i E_0 t/\hbar}[/itex] and do [itex]v^\dagger v[/itex]

Thanks for your help :)

## Homework Statement

A particle with spin s=1 has due to rotational invariance all three states [itex]|s_z\rangle=1[/itex], [itex]|s_z\rangle=0[/itex] and [itex]|s_z\rangle=-1[/itex] degenated with energy [itex]E_0>0[/itex]. Then its introduced an interaction non perturbative V

[itex]V=-\Delta\left(\begin{array}{ccc}

1 & 1 & 1\\

1 & 1 & 1\\

1 & 1 & 1

\end{array}\right)

[/itex]

a) Obtain stationary states and energies of the system

b) If in [itex]t=0[/itex], the particle has its third component with spin [itex]+\hbar[/itex] wich is the probability that after some time its [itex]-\hbar[/itex]

## Homework Equations

## The Attempt at a Solution

a)

So what i tried is this, since all three states have the same energy and those are eigenstates of the hamiltonian, the hamiltonian must have the form

[itex]H=E_0\left(\begin{array}{ccc}

1 & 0 & 0\\

0 & 1 & 0\\

0 & 0 & 1

\end{array}\right)

[/itex]

So if we add H'=H+V and diagonalize we get

[itex]\left|\begin{array}{ccc}

E_0-\Delta-\lambda & -\Delta & -\Delta\\

-\Delta & E_0-\Delta-\lambda & -\Delta\\

-\Delta & -\Delta & E_0-\Delta-\lambda

\end{array}\right|

[/itex]

So the characteristic equation is

[itex](E_0-\lambda)(E_0-3\Delta-\lambda)=0[/itex]

So we get the energies and the states

[itex](E_0,{-1,0,1})[/itex]

[itex](E_0,{-1,1,0})[/itex]

[itex](E_0-3\Delta,{1,1,1})[/itex]

I just want to know if this is the correct move or not

b)

i don't know how to aproach this, but i guess all i have to do is add the evolution term on each eigenstate [itex]e^{-i E_0 t/\hbar}[/itex] and do [itex]v^\dagger v[/itex]

Thanks for your help :)

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