What is "to the first order in H"?

[Haorong Wu]

I'm learning Griffiths' QM (3rd edn).

In Chapter 11 (Quantum Dynamics), there is an expression I'm not familiar with:

$\left| C_b \right|^2 = \left[ - \frac i \hbar \int_0^t H'_{ba} \left( t' \right) e^{i \omega_0 t'} \, dt' \right] \left[ \frac i \hbar \int_0^t H'_{ba} \left( t' \right) e^{-i \omega_0 t'} \, dt' \right] =0$ (to first order in $H'$),
where $H'$ is a time-dependent perturbation of a two-level system.

There are some other places where the expression of "to the first order in H" appears. I can't remember anywhere I have learnd the expressions in calculus or other mathematics courses.

 

[DrClaude]

It means that in deriving the expression, only terms of first order in $H'$ where kept, and that higher order terms ($H'^2$, $H'^3$, etc.) have been dropped.

 

[Haorong Wu]

It means that in deriving the expression, only terms of first order in $H'$ where kept, and that higher order terms ($H'^2$, $H'^3$, etc.) have been dropped.

Hi, DrClaude. I'm still confused. Shouldn't $\left[ \int H'_{ba} \left( t' \right) \cdot \int H'_{ba} \left( t' \right) \right]$ mean second order in $H'$?

 

[DrClaude]

I only have the 2nd ed. of Griffiths, so I can't check, but this looks like perturbation theory, so it would be $c_b$ that is expressed to 1st order in $H'$.

 

[Haorong Wu]

I only have the 2nd ed. of Griffiths, so I can't check, but this looks like perturbation theory, so it would be $c_b$ that is expressed to 1st order in $H'$.

Yes, it is the perturbation theory. Well, the expression is from the solutions of the 2nd edn. The problem is 9.4.

Here is the solution.

Would you mind look at it in your freetime?

Thank you so much!

 

[DrClaude]

I get it now. It is $|c_b|^2 = 0$ to first order in $H'$.

 

[Haorong Wu]

I get it now. It is $|c_b|^2 = 0$ to first order in $H'$.

I'm sorry I still can not see how can it be first order in $H'$. How does $\left[ \int H'_{ba} \left( t' \right) \cdot \int H'_{ba} \left( t' \right) \right]$ relate to the first order in $H'$?

Wait, is it because $\left| H' \right| =H_{aa} H_{bb} -H_{ab} H_{ba}$ and here $H_{aa}=H_{bb}=0$?

 

[DrClaude]

I'm sorry I still can not see how can it be first order in $H'$. How does $\left[ \int H'_{ba} \left( t' \right) \cdot \int H'_{ba} \left( t' \right) \right]$ relate to the first order in $H'$?

Wait, is it because $\left| H' \right| =H_{aa} H_{bb} -H_{ab} H_{ba}$ and here $H_{aa}=H_{bb}=0$?

No. As you noticed yourself

Shouldn't $\left[ \int H'_{ba} \left( t' \right) \cdot \int H'_{ba} \left( t' \right) \right]$ mean second order in $H'$?

This expression is 2nd order in $H'$, therefore, to first order, $|c_b|^2 = 0$.

 

[Haorong Wu]

No. As you noticed yourself

This expression is 2nd order in $H'$, therefore, to first order, $|c_b|^2 = 0$.

Ah...I understand it now. Thanks!