# Derivations about Time-Dependent Perturbation Theory (Griffiths)

I've been reading the Time-Dependent Perturbation Theory in Griffiths' book. There are many places that puzzle me a lot.

The system is two-level($$\psi_a,\psi_b,E_a). Suppose the particle starts out in the lower state: $$c_a(0)=1, \quad c_b(0)=0$$. There is a time-dependent perturbation $$\hat{H}^{\prime}$$.

I can understand the zeroth $$c_a^{(0)}(t)=1, \quad c_b^{(0)}(t)=0$$, and the first order is $$\frac{d c_a^{(1)}}{d t}=0 \Rightarrow c_a^{(1)}(t)=1$$ $$\frac{d c_b^{(1)}}{d t}=-\frac{i}{\hbar} H_{b a}^{\prime} e^{i \omega_0 t} \Rightarrow c_b^{(1)}=-\frac{i}{\hbar} \int_0^t H_{b a}^{\prime}\left(t^{\prime}\right) e^{i \omega_0 t^{\prime}} d t^{\prime} \, .$$ I also understand that the error in the first-order approximation is evident in the fact that $$\left|c_a^{(1)}(t)\right|^2+\left|c_b^{(1)}(t)\right|^2 \neq 1 \, .$$

I don't understand when Griffiths mentions:

"the error in the first-order approximation is evident in the fact that $$\left|c_a^{(1)}(t)\right|^2+\left|c_b^{(1)}(t)\right|^2 \neq 1$$, However, $$\left|c_a^{(1)}(t)\right|^2+\left|c_b^{(1)}(t)\right|^2$$ is equal to 1 to first order in $$\hat{H}^{\prime}$$, which is all we can expect from a first-order approximation. And the same goes for the higher orders."

Why is $$\left|c_a^{(1)}(t)\right|^2+\left|c_b^{(1)}(t)\right|^2$$ equal to 1 again? What does the "first order in $$\hat{H}^{\prime}$$" specifically indicate?

He just means that $$\left|c^{(1)}_a(t)\right|^2+\left|c^{(1)}_b(t)\right|^2=1+O(t^2)$$ A rather trivial observation.
• I dont think that this is what the author means here. Why would the author write "is equal to one to first order in $H^\prime$" if he meant this?
• First order always means $O(t^1)$, and you can see that there is no first order change to the normalisation here. Jun 24 at 5:54
First order means that your solutions contains powers of $$H'$$ only up to one. If you work it out properly you can check that the normalization of your solution is indeed equal to one plus terms which contains TWO powers of $$H'$$ which are beyond the "accuracy" of your solution. You can push your solution to any order in powers of $$H'$$ (with increasing computational complexity) and you always find that the normalization is equal to one plus terms which contain powers of $$H'$$ higher than the order of expansion you have achieved (if your solution contain terms up to $$(H')^3$$ the normalization will be one plus terms with at least FOUR powers of $$H$$'). Formally this works for ANY $$H'$$, it is useful only if the contribution of $$H'$$ terms becomes smaller for larger power of $$H'$$ so that the error you are making is small. You can check Cohen Tannoudji chapter XIII section B where everything is treated in detail. Other books like Sakurai approach time dependent perturbation theory using Dyson series but for non realtivistic quantum mechanics it is an unnecessary complication.