Spin Time Development: Solving for Initial State in B-Field | Homework Help

[Lawrencel2]

Homework Statement

I have a B field initially in the x-direction, and its constant:
$\widehat{H}= - (\dfrac{e}{mc})\overrightarrow{B} \dot{} \overrightarrow{S}$

At t=0 it was prepared so that S_z has an eigenvalue of + hbar/2
I want to show the time evolution of the initial state.

Homework Equations

$\widehat{H}= - (\dfrac{e}{mc})\overrightarrow{B} \dot{} \overrightarrow{S}$
$$\widehat{U} = e^{\dfrac{-i H}{\hbar}t}$$ Time evolution

The Attempt at a Solution

It would evolve as so:
$$|\alpha, t> = c_{1} e^{\dfrac{-i H}{\hbar}t} |+> +c_{2} e^{\dfrac{-i H}{\hbar}t} |->$$

But we know at t = 0 it should give us a spin up of hbar/2..
So evaluate at t=0:

$$|\alpha, 0> = c_{1} e^{\dfrac{-i H}{\hbar}0} |+> +c_{2} e^{\dfrac{-i H}{\hbar}0} |->$$
$$|\alpha, 0> = c_{1} |+>+c_{2} |-> = 1|+>$$
Then that means that C1 = 1 and C2 = 0..?

But intuitively I know that it precesses in the ZY plane since B is in the X-direction.. If I just got rid of the C2 coefficient, then i will never be able to show how it precesses into the spin down? Ahh i feel like I'm missing something here.. Any tips on what I am doing wrong?

 

[DrClaude]

It would evolve as so:
$$|\alpha, t> = c_{1} e^{\dfrac{-i H}{\hbar}t} |+> +c_{2} e^{\dfrac{-i H}{\hbar}t} |->$$

If $| + \rangle$ and $| - \rangle$ are eigenstates of $S_z$, then they are not eigenstates of the Hamiltonian. This equation is therefore not correct.

 

[Lawrencel2]

If $| + \rangle$ and $| - \rangle$ are eigenstates of $S_z$, then they are not eigenstates of the Hamiltonian. This equation is therefore not correct.

Then how can I express the state? I need to find the development of the state up to some arbitrary time T when the b field switches into the z direction...

 

[DrClaude]

You have to write it in terms of the eigenstates of $S_x$.

 

[Lawrencel2]

You have to write it in terms of the eigenstates of $S_x$.

AH. well I'm confused now.
$$\vert\alpha, t \rangle = \dfrac{a}{\sqrt{2}} e^{i t\dfrac{H}{\hbar}} \vert +\rangle\pm\dfrac{b}{\sqrt{2}} e^{i t\dfrac{H}{\hbar}} \vert -\rangle$$
the summed state represents spin up in z and difference terms represent spin down?

 

[DrClaude]

Let me introduce a notation to make things clearer:
$$
\hat{S}_z | \pm \rangle_z = \pm \frac{\hbar}{2} | \pm \rangle_z \\
\hat{S}_x | \pm \rangle_x = \pm \frac{\hbar}{2} | \pm \rangle_x
$$
What you are writing is
$$
|\alpha, t \rangle = c_1 e^{-i \hat{H} t / \hbar} | + \rangle_z + c_2 e^{-i \hat{H} t / \hbar} | - \rangle_z
$$
The problem is that $| + \rangle_z$ and $| - \rangle_z$ are not eigenstates of $\hat{H}$ (you should show that). That means that $e^{-i \hat{H} t / \hbar}$ mixes the two spin states, and you can't treat it as a simple complex number. You also cannot easily calculate the exponential of a matrix that is not diagonal.

You therefore need to go to a basis where $\hat{H}$ is diagonal, i.e., a basis of eigenstates of $\hat{H}$, namely $| + \rangle_x$ and $| - \rangle_x$ (you can show that also). Then you need to take the correct initial state in this basis, look at its time evolution, and if you calculate for instance $\langle \hat{S}_z \rangle$, you will find that it is time-dependent, hence you will indeed see a precession in the yz plane.

 

[Lawrencel2]

So I started over and found the EigenVectors of the hamiltonian at t<T (Interval where B points in X-direction)..
$$\mid\alpha\rangle = \dfrac{1}{\sqrt{2}} \mid +\rangle_{x} e^{\dfrac{\imath e B_{x} t}{2mc}}+ \dfrac{1}{\sqrt{2}} \mid -\rangle_{x} e^{\dfrac{-\imath e B_{x} t}{2mc}}\\ Where: \\ \mid +\rangle_{x} =\binom{1}{1} \mid -\rangle_{x} =\binom{1}{-1}$$But I don't know what X is initially, I only know where it is in Z. When I evaluate it for t=0 I get:
$$\mid\alpha, t=0 \rangle = \dfrac{1}{\sqrt{2}} \mid +\rangle_{x} + \dfrac{1}{\sqrt{2}} \mid -\rangle_{x}$$
Which gives us: //
$$\mid\alpha, t=0 \rangle = \binom{1}{0}$$
Which is the spin up Z ket correct?
So when t -> T
My hamiltonian then changes to B in the Z direction