# Consistency of time-dependent and time-independent perturbation theory

I am confused as to how time-independent and time-dependent perturbation theories in QM give consistent results at the instant the perturbation is switched on. Suppose I have a two-level system which obeys the time-independent Schrödinger equation,

$$\hat{h}_0 \psi_{0,1}=\epsilon_{0,1}\psi_{0,1}$$

which is initially in the ground state $$\psi_0$$. Then suppose I switch on a time-dependent perturbation $$\delta v(t)$$ at some time $$t_0$$. The system will then be described by a mixed state $$\psi(t)$$, with

$$\psi(t)=c_0(t)\psi_0 + c_1(t)\psi_1$$.

Up to 1st order in $$\delta v(t)$$, the coefficients $$c_{1,2}(t)$$ are given by

$$c_0^{(1)}(t)=1$$

$$c_1^{(1)}(t)=-i\int_{t_0}^t h'_{1,0}(t')e^{i\omega_0 t'} dt'$$, with

$$h'_{1,0}(t)=\langle \psi_1 | h_0 + \delta v(t) | \psi_0 \rangle$$ and $$\omega_0=\epsilon_1-\epsilon_0$$.

(These expressions are taken from David Griffith's book on quantum mechanics and I've ignored $$\hbar$$'s). This implies that the system remains (up to first order) completely in its ground state at time $$t_0$$ when the perturbation is turned on, since the integral expression for $$c^{(1)}_1(t)$$ vanishes at $$t_0$$.

However, if we consider the static Schrödinger equation at $$t_0$$, then could we not use time-independent perturbation theory (with the perturbation given by $$\delta v(t_0)$$) to determine the state $$\psi(t_0)$$? In which case, $$\psi(t_0)$$ would be given by

$$\psi(t_0) = \psi_0 + \frac{h'_{1,0}(t_0)}{\epsilon_0-\epsilon_1}\psi_1$$.

So it appears the state is already a mixed state of both $$\psi_0$$ and $$\psi_1$$. In my mind, this seems to be in contradiction with the result from time-dependent perturbation theory, where $$\psi(t_0)=\psi_0$$.

I'm guessing there is some flaw in using time-independent perturbation theory at the initial time, but can someone explain to me why this is the case or what the problem is if it's not that?

• To put it simple: Time-independent perturbation theory tells you how the new stationary states look like. Time-dependent perturbation theory tells you how the state evolves (approximately) in terms of stationary states (i.e. basis) of the initial unperturbed system. There is no contradiction, it's just 2 separate problems. Jan 28, 2019 at 18:45

Time-dependent perturbation theory pertains to the time-dependent Schrodinger equation and tells you how the time-dependent state $$|\psi(t) \rangle$$ evolves.