# Homework Help: Particle's spin when subject to a constant magnetic field

1. Jan 31, 2014

### Jalo

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

An alkali atom, on it's fundamental state, passes through a Stern-Gerlach apparatum, which will only transmit atoms with their spins aligned along the +z direction. After that the atoms travel, during a finite time τ, through a region of constant magnetic field $\vec{B}=B\vec{e_{x}}$.

After that time τ the atoms pass through a new Stern-Gerlach apparatum, which only allows atoms with spin along -z to pass. What's the probability that they will pass?

2. Relevant equations

Pauli matrices
$\widehat{H}=-\vec{u_{B}}.\widehat{S}$
$u_{B}=\frac{q}{2m_{e}}$

3. The attempt at a solution

From the problem it's easy to see that the state of the system at the instant t=0 is:

$|\psi>(t=0)=|+>_{z}$

Then I assumed that, while being under the influence of the constant magnetic field along the x direction, the state is:

$|\psi>=\alpha (t)|+>_{z} + \beta (t) |->_{z}$

Next I applied the hamiltonian to my state $|\psi>$. Since $\vec{B} = B\vec{e_{x}}$:

$\widehat{H}|\psi>=-\frac{qB}{2m}\widehat{S_{x}}|\psi>$
$\widehat{H}|\psi>=-\frac{qB}{2m}\widehat{S_{x}}(\alpha (t)|+>_{z}+\beta (t) |->_{z})$

Since $\widehat{S_{x}}=\hbar \sigma _{x}$, applying it to $|\psi>$ returns:
$\widehat{H}|\psi>=-\frac{qB\hbar}{4m}(\beta (t)|+>_{z}+\alpha (t) |->_{z})$

Defining $\omega=\frac{qB}{4m}$:

$\widehat{H}|\psi>=-w\hbar(\beta (t)|+>_{z}+\alpha (t) |->_{z})$

Now, using Schrodinger's equation we get:

$\widehat{H}|\psi>=i\hbar \frac{d}{dt}|\psi>$
$\frac{d}{dt}(\alpha (t)|+>_{z} + \beta (t) |->_{z}) = iw(\beta (t)|+>_{z}+\alpha (t) |->_{z})$

Separating this we get:

$\frac{d}{dt}\alpha (t) = iw\beta (t)$
$\frac{d}{dt}\beta (t) = iw\alpha (t)$

Applying another derivative to the first differential equation we get:
$\frac{d^{2}}{dt^{2}}\alpha (t) = iw\frac{d}{dt}\beta (t)$
$\frac{d^{2}}{dt^{2}}\alpha (t) = -w^{2}\alpha$

Doing the same to the second achieves a similar result:

$\frac{d^{2}}{dt^{2}}\beta (t) = -w^{2}\beta$

Solving both I got:

$\alpha (t) = Ae^{iwt} + Be^{-iwt}$
$\alpha (t) = Ce^{iwt} + De^{-iwt}$

Now, since I know that $|\psi>(t=0) = |+>_{z}$, I know that:
$A+B=1$
$C = -D$

From this I can conclude that:
$\beta (t) = Fsin(wt)$

This is as far as I can get.. I don't know what to do from here. Am I approaching the problem wrongly? Any help would be appreciated.

Thanks.
Daniel

2. Feb 1, 2014

### TSny

Hello, Daniel.

First, there appears to be a problem with a factor of 2 in your expression
I don't believe the 4 in the denominator is correct. Did you include the "g-factor" of the electron? Anyway, you might see if you can track down the error. (Hope I'm not the one in error, I always get a headache with the factors of 2 in this type of problem!)

Later, you have

In the second equation I think you meant to type $\beta (t)$ instead of $\alpha (t)$.

This looks good (but your $\omega$ will need to be corrected for the factor of 2 problem mentioned above).

But note, C and D are not independent of A and B. Remember you have

So, you can determine $\beta (t)$ from $\alpha (t)$.

You should be able to find A and B from the initial conditions and from the fact that you want your state to be normalized.

3. Feb 2, 2014

### Jalo

I forgot that I was only interested in the $|->_{z}$ component.
Regarding the factor of 2, I'll correct it. However my major issue has been resolved, so thanks!