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 [itex]\vec{B}=B\vec{e_{x}}[/itex].
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
[itex]\widehat{H}=-\vec{u_{B}}.\widehat{S}[/itex]
[itex]u_{B}=\frac{q}{2m_{e}}[/itex]
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:
[itex]|\psi>(t=0)=|+>_{z}[/itex]
Then I assumed that, while being under the influence of the constant magnetic field along the x direction, the state is:
[itex]|\psi>=\alpha (t)|+>_{z} + \beta (t) |->_{z}[/itex]
Next I applied the hamiltonian to my state [itex]|\psi>[/itex]. Since [itex]\vec{B} = B\vec{e_{x}}[/itex]:
[itex]\widehat{H}|\psi>=-\frac{qB}{2m}\widehat{S_{x}}|\psi>[/itex]
[itex]\widehat{H}|\psi>=-\frac{qB}{2m}\widehat{S_{x}}(\alpha (t)|+>_{z}+\beta (t) |->_{z})[/itex]
Since [itex]\widehat{S_{x}}=\hbar \sigma _{x}[/itex], applying it to [itex]|\psi>[/itex] returns:
[itex]\widehat{H}|\psi>=-\frac{qB\hbar}{4m}(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]
Defining [itex]\omega=\frac{qB}{4m}[/itex]:
[itex]\widehat{H}|\psi>=-w\hbar(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]
Now, using Schrodinger's equation we get:
[itex]\widehat{H}|\psi>=i\hbar \frac{d}{dt}|\psi>[/itex]
[itex]\frac{d}{dt}(\alpha (t)|+>_{z} + \beta (t) |->_{z}) = iw(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]
Separating this we get:
[itex]\frac{d}{dt}\alpha (t) = iw\beta (t)[/itex]
[itex]\frac{d}{dt}\beta (t) = iw\alpha (t)[/itex]
Applying another derivative to the first differential equation we get:
[itex]\frac{d^{2}}{dt^{2}}\alpha (t) = iw\frac{d}{dt}\beta (t)[/itex]
[itex]\frac{d^{2}}{dt^{2}}\alpha (t) = -w^{2}\alpha[/itex]
Doing the same to the second achieves a similar result:
[itex]\frac{d^{2}}{dt^{2}}\beta (t) = -w^{2}\beta[/itex]
Solving both I got:
[itex]\alpha (t) = Ae^{iwt} + Be^{-iwt}[/itex]
[itex]\alpha (t) = Ce^{iwt} + De^{-iwt}[/itex]
Now, since I know that [itex]|\psi>(t=0) = |+>_{z}[/itex], I know that:
[itex]A+B=1[/itex]
[itex]C = -D[/itex]
From this I can conclude that:
[itex]\beta (t) = Fsin(wt)[/itex]
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
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 [itex]\vec{B}=B\vec{e_{x}}[/itex].
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
[itex]\widehat{H}=-\vec{u_{B}}.\widehat{S}[/itex]
[itex]u_{B}=\frac{q}{2m_{e}}[/itex]
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:
[itex]|\psi>(t=0)=|+>_{z}[/itex]
Then I assumed that, while being under the influence of the constant magnetic field along the x direction, the state is:
[itex]|\psi>=\alpha (t)|+>_{z} + \beta (t) |->_{z}[/itex]
Next I applied the hamiltonian to my state [itex]|\psi>[/itex]. Since [itex]\vec{B} = B\vec{e_{x}}[/itex]:
[itex]\widehat{H}|\psi>=-\frac{qB}{2m}\widehat{S_{x}}|\psi>[/itex]
[itex]\widehat{H}|\psi>=-\frac{qB}{2m}\widehat{S_{x}}(\alpha (t)|+>_{z}+\beta (t) |->_{z})[/itex]
Since [itex]\widehat{S_{x}}=\hbar \sigma _{x}[/itex], applying it to [itex]|\psi>[/itex] returns:
[itex]\widehat{H}|\psi>=-\frac{qB\hbar}{4m}(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]
Defining [itex]\omega=\frac{qB}{4m}[/itex]:
[itex]\widehat{H}|\psi>=-w\hbar(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]
Now, using Schrodinger's equation we get:
[itex]\widehat{H}|\psi>=i\hbar \frac{d}{dt}|\psi>[/itex]
[itex]\frac{d}{dt}(\alpha (t)|+>_{z} + \beta (t) |->_{z}) = iw(\beta (t)|+>_{z}+\alpha (t) |->_{z})[/itex]
Separating this we get:
[itex]\frac{d}{dt}\alpha (t) = iw\beta (t)[/itex]
[itex]\frac{d}{dt}\beta (t) = iw\alpha (t)[/itex]
Applying another derivative to the first differential equation we get:
[itex]\frac{d^{2}}{dt^{2}}\alpha (t) = iw\frac{d}{dt}\beta (t)[/itex]
[itex]\frac{d^{2}}{dt^{2}}\alpha (t) = -w^{2}\alpha[/itex]
Doing the same to the second achieves a similar result:
[itex]\frac{d^{2}}{dt^{2}}\beta (t) = -w^{2}\beta[/itex]
Solving both I got:
[itex]\alpha (t) = Ae^{iwt} + Be^{-iwt}[/itex]
[itex]\alpha (t) = Ce^{iwt} + De^{-iwt}[/itex]
Now, since I know that [itex]|\psi>(t=0) = |+>_{z}[/itex], I know that:
[itex]A+B=1[/itex]
[itex]C = -D[/itex]
From this I can conclude that:
[itex]\beta (t) = Fsin(wt)[/itex]
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
0 commentaires:
Enregistrer un commentaire