Particle's spin when subject to a constant magnetic field

[Jalo]

Homework Statement

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?

Homework Equations

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

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

 

[TSny]

Hello, Daniel.

First, there appears to be a problem with a factor of 2 in your expression

\widehat{H}|\psi>=-\frac{qB\hbar}{4m}(\beta (t)|+>_{z}+\alpha (t) |->_{z}).

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

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

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

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

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.

 

[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!