# Confusion About Time Dependent Perturbation Theory

Say for simplicity we are in a two state system with a Hamiltonian:

$$H=H_0+V(t)$$

Where $$H_0$$ is: $$\begin{pmatrix} \epsilon_0&0\\ 0&\epsilon_1 \end{pmatrix}$$ While $$V(t)$$ is: $$\begin{pmatrix} 0&V_{01}(t)\\ V_{10}(t)&0 \end{pmatrix}$$ We can then write out state kets as: $$|\psi\rangle=c_0(t)|0\rangle+c_1(t)|1\rangle$$ And then Sakurai tells us that these coefficients can be found to first order to be: $$c_n(t)=\langle n|U_I(t,t_0)|i\rangle$$ $$c_n^{(1)}(t)=\delta_{ni}-\frac{i}{\hbar}\int_{t_0}^te^{i\omega_{ni}t}V_{ni}(t)d{t}$$ Where: $$\omega_{ni}=\frac{E_n-E_i}{\hbar}$$ And then the probability of transitions form $$|0\rangle$$ to $$|1\rangle$$ is given by: $$P(i\rightarrow n)\approx|c_n^{(1)}|^2$$ However, what if I want to calculate the probability that it stays in the state $$|0\rangle$$? That would reduce to: $$c_0(t)=\langle 0|U_I|0\rangle$$ $$c_0(t)=1-\frac{i}{\hbar}\int_{t_0}^te^{i\omega_{10}t}V_{00}(t)$$ $$= 1$$ $$\Rightarrow P(0\rightarrow 0)= 1$$ Which seems like a contradiction as I feel like it should be $$1-|c_1|^2$$, so what is happening here? Why do we get what probability transistion and then an entirely different probability for no transition?

As you indicated, it is only an approximation that $$P(i\rightarrow n) \approx |c_n^{(1)}|^2$$. It only holds, if $$|\sum_j \lambda^j c_n^{(j)}|^2 = |c_n|^2 \ll 1$$. Since you have $$|c_n^{(1)}|^2 = 1$$, you have to include higher order terms for the approximation to be valid (also might be not valid at all; depends on the perturbation).
A quick note: I went through my lecture notes because I never saw the $$\delta_{ni}$$-term in the expression for $$c_n^{(1)}$$, and I also didn't have that there. I'm also a bit confused what the $$U_I(t)$$ is here. Are you confusing Interaction picture with the Schrödinger picture?
The correct expression for for the $$c_n$$'s can be derived from their differential equations: $$\frac{d}{dt}c_n^{(j)} = \frac{1}{i\hbar}\sum_m V_{nm}(t)\ c_m^{(j-1)} e^{i\omega_{nm}t}$$ So for a given initial state $$|i\rangle$$: $$c_n^{(0)} = \delta_{ni}$$, and therefore $$c_f^{(1)}(t) = \frac{1}{i\hbar} \int_{t_0}^t V_{fi} e^{i\omega_{fi}t'}dt'$$