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


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?


1 Answer 1


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 $c_f^{(1)}(t) = \frac{1}{i\hbar} \int_{t_0}^t V_{fi} e^{i\omega_{fi}t'}dt'$

  • $\begingroup$ I am working solely in the interaction at the moment, the equations I have are from page 340 of sakurai in the section on time dependent perturbation theory $\endgroup$
    – Chris
    Commented Oct 11, 2021 at 21:06

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