Hydrogen in Magnetic Field, Interaction Representation

[Muh. Fauzi M.]

The hydrogen is placed in the external magnetic field:
$$ \textbf{B}=\hat{i}B_1 cos(\omega t) + \hat{j} B_2 sin(\omega t) + \hat{k} B_z ,$$

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?

 

[kuruman]

How did you get the differential equation for a? It seems to me you should get two coupled differential equations involving a and b. Is what you show the decoupled result? Also, should you not include the term $B_z \sigma_3$ in your expression for $H'$?

 

[Muh. Fauzi M.]

How did you get the differential equation for a? It seems to me you should get two coupled differential equations involving a and b. Is what you show the decoupled result? Also, should you not include the term $B_z \sigma_3$ in your expression for $H'$?

I don't include the $B_z \sigma_3$ because it time independent.

I realized that I don't differentiate with time the ${B_1 cos (\omega t) \mp i B_2 sin (\omega t)}$ term in the $\partial a/\partial t$ and $\partial b/\partial t$.

Here's my result after differentiating it (for $\partial a/\partial t$),

$$ \frac{\partial^2 a}{{\partial t}^2} = - \frac{e\hbar}{2mc}\Big[ \frac{ie}{2mc} [B_1^2 cos^2{\omega t}+B_2^2 sin^2{\omega t}] a + \frac{2mc}{e\hbar} \omega i \frac{B_1 cos \omega t + i B_2 sin \omega t}{B_1 cos \omega t - i B_2 sin \omega t} \frac{\partial a}{\partial t} \Big] .$$

 

[kuruman]

I don't include the Bzσ3 B_z \sigma_3 because it time independent.

So what? Is it not a perturbation to $H_0$? As long as $B_z \neq 0$, you cannot ignore it.

 

[Muh. Fauzi M.]

So what? Is it not a perturbation to $H_0$? As long as $B_z \neq 0$, you cannot ignore it.

Thanks for your clue.

I am using Schwinger-Tomonaga equation, where the Hamiltonian that evolve with time is only the time dependent. CMIIW

 

[Muh. Fauzi M.]

update, I have finish this work.