- #1

Muh. Fauzi M.

- 17

- 1

$$ \textbf{B}=\hat{i}B_1 cos(\omega t) + \hat{j} B_2 sin(\omega t) + \hat{k} B_z ,$$

Using the relation ## H = - \frac{e\hbar}{2mc} \mathbf \sigma \cdot \mathbf B ##, then I got the form

$$ H = H_0 + H' , $$

where

$$ H'= - \frac{e \hbar}{2 m c} (B_1 cos(\omega t) \sigma_1 + B_2 sin(\omega t) \sigma_2 ) .$$

To find the spin state in the interaction representation, I substitute the ## H' ## in

$$ i\hbar\frac{\partial}{\partial t} | \psi \rangle = H' | \psi \rangle , $$

with ## | \psi \rangle = \begin{pmatrix} a \\ b \end{pmatrix} .##

Next, I tried to solving it, e.g. for ## a ##

$$\frac{\partial^2 a}{{\partial t}^2} + \lambda^2 a = 0 ,$$

with

$$ \lambda^2 = \big(\frac{e}{2mc}\big)^2[B_1^2 cos^2(\omega t)+B_2^2(sin^2(\omega t))] .$$

Then I'm heading for the boundary condition.

But then I realized that the ## \lambda^2 ## is depend on ## t ##.

So, my question, is it "ok" to continue to the boundary condition, or there is something wrong with my attempt?