# Homework Help: Particle with spin 1 in a non perturtuvative interaction

1. May 31, 2013

### WarDieS

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

1. The problem statement, all variables and given/known data
A particle with spin s=1 has due to rotational invariance all three states $|s_z\rangle=1$, $|s_z\rangle=0$ and $|s_z\rangle=-1$ degenated with energy $E_0>0$. Then its introduced an interaction non perturbative V

$V=-\Delta\left(\begin{array}{ccc} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{array}\right)$

a) Obtain stationary states and energies of the system
b) If in $t=0$, the particle has its third component with spin $+\hbar$ wich is the probability that after some time its $-\hbar$

2. Relevant equations

3. 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

$H=E_0\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)$

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

$\left|\begin{array}{ccc} E_0-\Delta-\lambda & -\Delta & -\Delta\\ -\Delta & E_0-\Delta-\lambda & -\Delta\\ -\Delta & -\Delta & E_0-\Delta-\lambda \end{array}\right|$
So the characteristic equation is
$(E_0-\lambda)(E_0-3\Delta-\lambda)=0$
So we get the energies and the states
$(E_0,{-1,0,1})$
$(E_0,{-1,1,0})$
$(E_0-3\Delta,{1,1,1})$

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 $e^{-i E_0 t/\hbar}$ and do $v^\dagger v$

Last edited: May 31, 2013
2. May 31, 2013

### TSny

Yes, that looks good!

I think you might have the right idea. Can you "expand" the $|s_z=+\hbar \rangle$ state in terms of your eigenstates of the full Hamiltonian?

Last edited: May 31, 2013
3. May 31, 2013

### WarDieS

I am lost on this, what i think may work is this
I have to solve the equation

$i\hbar\frac{\partial}{\partial x} \chi = H \chi$

where
$\chi=\left(\begin{array}{ccc} \alpha(t) \\ \beta(t) \\ \gamma(t) \end{array}\right)$

The problem with this is that this gives a really complicated result with the system
$i \hbar \dot{\alpha}(t)=(E_0-\Delta)\alpha(t)-\Delta(\beta(t)+\gamma(t))$
$i \hbar \dot{\beta}(t)=(E_0-\Delta)\beta(t)-\Delta(\alpha(t)+\gamma(t))$
$i \hbar \dot{\gamma}(t)=(E_0-\Delta)\gamma(t)-\Delta(\beta(t)+\alpha(t))$

And this seems wrong to me

Thanks for help :)

4. May 31, 2013

### TSny

At time t = 0, you know the state is $|\psi(0)\rangle = |s_z = +\hbar\rangle$. How would you represent this state as a column vector?

Can you expand this column vector in terms of the eigenvetors
$|u1\rangle = \left(\begin{array}{ccc} -1 \\ 0 \\ 1 \end{array}\right)$ $,|u2\rangle = \left(\begin{array}{ccc} -1 \\ 1 \\ 0 \end{array}\right)$ $,|u3\rangle = \left(\begin{array}{ccc} 1 \\ 1 \\ 1 \end{array}\right)$ ?

5. May 31, 2013

### WarDieS

So if if we normalize the vector we have

$|u_1\rangle = \frac{1}{\sqrt{2}} \left(\begin{array}{ccc} -1 \\ 0 \\ 1\end{array}\right)$

$|u_2\rangle = \frac{1}{\sqrt{2}} \left(\begin{array}{ccc} -1 \\ 1 \\ 0\end{array}\right)$

$|u_3\rangle = \frac{1}{\sqrt{3}} \left(\begin{array}{ccc} 1 \\ 1 \\ 1\end{array}\right)$
so $a|u_1\rangle+b|u_2\rangle+c|u_3\rangle=\left(\begin{array}{ccc} 1 \\ 0 \\ 0\end{array}\right)$

gives us a system, when we solve this we have the vector

$|s_z{_h}\rangle=\left(\begin{array}{ccc} \frac{-\sqrt{2}}{3} \\ \frac{-\sqrt{2}}{3} \\ \frac{1}{3}\end{array}\right)$

Now its turn for the time evolution
$|s_z=\hbar\rangle=\left(\begin{array}{ccc} \frac{-\sqrt{2}}{3}\exp{\frac{-i E_1 t}{\hbar}} \\ \frac{-\sqrt{2}}{3}\exp{\frac{-i E_2 t}{\hbar}} \\ \frac{1}{3}\exp{\frac{-i E_3 t}{\hbar}}\end{array}\right)$

Since i am being asked for $-\hbar$, i think i need to do this with that vector

$|s_z=-\hbar\rangle=\left(\begin{array}{ccc} \frac{2\sqrt{2}}{3}\exp{\frac{-i E_1 t}{\hbar}} \\ \frac{-\sqrt{2}}{3}\exp{\frac{-i E_2 t}{\hbar}} \\ \frac{1}{3}\exp{\frac{-i E_3 t}{\hbar}}\end{array}\right)$

Now ot obtain the evolution with time i have to get the expectation value i assume

$\langle s_z\rangle=\langle s_z=-\hbar|S_z|s_z=-\hbar\rangle$

Is this good?

Thanks again for your help :D

6. May 31, 2013

### TSny

Your expressions for a, b, and c look good.

You didn't quite evolve in time correctly. It's only the eigenstates of H (i.e., |u1>|,u2>,|u3>) that have the simple time evolution of factors of exp(-iEt/$\hbar$), for appropriate values of E.

So, you found ψ(0) = a |u1> + b |u2> + c |u3>.

To find ψ(t), let each of the |u> states evolve in time with it's own exponential factor. Then combine into one column vector.

7. May 31, 2013

### TSny

Your expressions for a, b, and c look good. But, you didn't quite evolve in time correctly. It's only the eigenstates of H (i.e., |u1>|,u2>,|u3>) that have the simple time evolution of factors of exp(-iEt/$\hbar$), for appropriate values of E.

So, you found ψ(0) = a |u1> + b |u2> + c |u3>.

To find ψ(t), let each of the |u> states evolve in time with it's own exponential factor. Then combine into one column vector.

[EDIT: For example, at time t = 0 we have a|u1> = a $\left(\begin{array}{ccc} -1\\ 0\\ 1 \end{array}\right)$. To evolve this in time just multiply by a factor of $e^{-iE_0 t/\hbar}$.]

EDIT 2: Also, as I understand it, you are not looking for $\langle s_z \rangle$. Rather your a looking for the probability that at time t the system will be in the state $|s_z = -\hbar \rangle = \left(\begin{array}{ccc} 0\\ 0\\ 1 \end{array}\right)$

Last edited: May 31, 2013
8. May 31, 2013

### TSny

I have been assuming that we always write our column vectors with respect to the basis in which you wrote your matrices for $H_o$ and $V$. This is the basis where

$\left(\begin{array}{ccc} 1\\ 0\\ 0 \end{array}\right) = |s_{z=\hbar}\rangle$ = state where z-component of spin is $+\hbar$

$\left(\begin{array}{ccc} 0\\ 1\\ 0 \end{array}\right) = |s_{z=0}\rangle$ = state where z-component of spin is 0

$\left(\begin{array}{ccc} 0\\ 0\\ 1 \end{array}\right) = |s_{z=-\hbar}\rangle$ = state where z-component of spin is $-\hbar$

It appears to me that you might be trying to switch to writing column vectors such that

$\left(\begin{array}{ccc} 1\\ 0\\ 0 \end{array}\right) = |u_1\rangle$ , $\left(\begin{array}{ccc} 0\\ 1\\ 0 \end{array}\right) = |u_2\rangle$, $\left(\begin{array}{ccc} 0\\ 0\\ 1 \end{array}\right) = |u_3\rangle$

You can't do that since $|u1\rangle$ and $|u2\rangle$ are not orthogonal.

9. Jun 1, 2013

### WarDieS

Ok, after some thinking i did this
I get this vector after diagonalizing the full hamiltonian and now i make sure they are orthogonal to each other

$|u_1\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc} 1\\ -1\\ 0 \end{array}\right) = |s_{z=\hbar}\rangle$

$|u_1\rangle=\frac{1}{\sqrt{6}}\left(\begin{array}{ccc} 1\\ 1\\ -2 \end{array}\right) = |s_{z=\hbar}\rangle$

$|u_1\rangle=\frac{1}{\sqrt{3}}\left(\begin{array}{ccc} 1\\ 1\\ 1 \end{array}\right) = |s_{z=\hbar}\rangle$

Expanding the initial function and adding the time evolution terms

$|i\rangle = \sum_i \psi_i \langle \psi_i | i\rangle$

$|i\rangle=\frac{1}{\sqrt{2}}|u_1\rangle e^{\frac{-i E_0 t}{\hbar}}+\frac{1}{\sqrt{6}}|u_2\rangle e^{\frac{-i E_0 t}{\hbar}}+\frac{1}{\sqrt{3}}|u_3\rangle e^{\frac{-i (E_0-3\Delta) t}{\hbar}}$

Now we have the initial state expanded using the full hamiltonian eigenstates, we must find the amplitude with the vector

$|s_z=-\hbar\rangle=\left(\begin{array}{ccc} 0\\ 0\\ 1 \end{array}\right)$

that is

$\langle s_z=-\hbar|i\rangle=\left(\frac{-1}{3}+\frac{e^{i 3 \Delta t/\hbar}}{3}\right)e^{-i E_0 t/\hbar} = c_i$

now its time to get the probability
$p(t)=|c_i|^2=\frac{1}{9}\left(2-2\cos(3\Delta t/\hbar\right)$

Wich i really hope its the right answer :/ , atleast $p(0)=0$ wich is a necessary condition.

10. Jun 1, 2013

### TSny

Yes. Very nicely done!

11. Jun 1, 2013

### WarDieS

i am really grateful, thanks for your patience with me :D