# Spin 1 Particle in a time dependant magnetic field

1. Dec 7, 2013

### richyw

1. The problem statement, all variables and given/known data

https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-prn2/1483071_10151813160569290_1331827985_n.jpg

2. Relevant equations

$$\left|\psi ,t\right\rangle=\sum_i c_i(t)e^{-\frac{i E_n t}{\hbar}}\left|n\right\rangle$$

3. The attempt at a solution

I'm quite lost on how to even start this. I can work out that the hamiltonian is$$H(t)=\gamma S_z B(t)$$Then I tried to put that into the time-dependant shrodinger equation. $$-i\hbar \frac{d}{dt}\left|\psi(t)\right\rangle=H(t)\left|\psi(t)\right\rangle$$but I am not really sure if this is correct or even how to begin solving this.

2. Dec 7, 2013

### TSny

I think you're on the right track. After substituting the given expression for |ψ(t)> into the Schrodinger equation, you should be able to get a differential equation for each coefficient ci(t).

3. Dec 8, 2013

### richyw

I'm still stuck on this one actually. I don't think I am substituting $\left|\psi ,t \right\rangle$ into the equation properly. How to I get $\psi (t)$ from $\left|\psi ,t \right\rangle$?

4. Dec 8, 2013

### richyw

wait, can I say$$\left| \psi ,t \right\rangle=c_1(t)\left| 1,1 \right\rangle+c_2(t)\left| 1,0 \right\rangle+c_3(t)\left| 1,-1 \right\rangle$$$$\left| \psi ,t \right\rangle=c_1(t)\left(\begin{matrix}1 \\ 0 \\ 0\end{matrix}\right)+c_2(t)\left(\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right)+c_3(t)\left(\begin{matrix}0 \\ 0 \\ 1\end{matrix}\right)$$$$\left| \psi ,t \right\rangle=\left(\begin{matrix} c_1(t) \\ c_2(t) \\ c_3(t)\end{matrix}\right)$$

5. Dec 8, 2013

### richyw

and then $$i\hbar\frac{d}{dt}\left| \psi ,t \right\rangle=\gamma B(t) S_z\left| \psi ,t \right\rangle$$

6. Dec 8, 2013

### richyw

$$i\hbar \left(\begin{matrix} c'_1(t) \\ c'_2(t) \\ c'_3(t)\end{matrix}\right)=\hbar\gamma B(t)\left(\begin{matrix}1 & 0&0\\0&0&0\\0&0&-1\\\end{matrix}\right)\left(\begin{matrix} c_1(t) \\ c_2(t) \\ c_3(t)\end{matrix}\right)$$

7. Dec 8, 2013

### richyw

and then get $$c'_1(t)=-i\gamma B(t) c_1(t)$$$$c'_2(t)=0$$$$c'_3(t)=i\gamma B(t) c_3(t)$$

8. Dec 8, 2013

### TSny

Yes, that looks good. Just have to solve each of these equations.