An electron in an oscillating magnetic field

[Haorong Wu]

Homework Statement

An electron is at rest in an oscillating magnetic field
$$ \mathbf B = B_0 cos\left( ωt \right) \hat k $$
where $B_0$ and $ω$ are constants.

What is the minimum field ($B_0$) required to force a complete flip in $S_x$?

Homework Equations

$$H=- γ \mathbf B \cdot \mathbf S $$
$$ c^{(x)}_{-} = χ^{(x)†}_{-} χ$$

The Attempt at a Solution

I have solved that
$$ χ(t) = \begin{pmatrix} \frac 1 {\sqrt 2} exp \left( -i \frac {γ B_0} {2ω} sin ( ωt ) \right) \\ \frac 1 {\sqrt 2} exp \left( i \frac {γ B_0} {2ω} sin ( ωt ) \right) \end{pmatrix} $$
Also, the probability of getting $- \frac \hbar 2$ by measuring $S_x$ is
$$ P(- \frac \hbar 2 ) = {| c^{(x)}_{-} |}^2 = sin^2 ( \frac {γ B_0 sin(ωt)} {2ω}) $$

The statement of "to force a complete flip in $S_x$" troubles me. Since English is not my first language, I failed to find out the meaning of the statement at google. From the solution, it seems that the probability $P(- \frac \hbar 2 ) =1$ relates to something that forces a complete flip.
So, what is a complete flip?

 

[DrClaude]

This is definitely not clear, but I would consider a "complete flip" to mean going from spin-up to spin-down (or vice versa). In this case, it would be spin-up or down along the x axis.

 

[Haorong Wu]

This is definitely not clear, but I would consider a "complete flip" to mean going from spin-up to spin-down (or vice versa). In this case, it would be spin-up or down along the x axis.

But how does it relate to the probability $P(- \frac \hbar 2)=1$ ?

Thanks!

 

[DrClaude]

But how does it relate to the probability $P(- \frac \hbar 2)=1$ ?

Reading that as $P(S_x = -\hbar/2)=1$, that corresponds to a spin-down state along x. So starting from spin-up, you have $P(S_x = \hbar/2)=1$, and when $P(S_x = -\hbar/2)=1$ is attained, that means it is a "complete flip."

 

[Haorong Wu]

Reading that as $P(S_x = -\hbar/2)=1$, that corresponds to a spin-down state along x. So starting from spin-up, you have $P(S_x = \hbar/2)=1$, and when $P(S_x = -\hbar/2)=1$ is attained, that means it is a "complete flip."

Thank you very much! It makes sense to me now.