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


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

  • 1
    $\begingroup$ 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. $\endgroup$
    – valdo
    Jan 28, 2019 at 18:45

1 Answer 1


You're mixing up the time-dependent and time-independent Schrodinger equations.

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

All states can be written as a linear combination of energy eigenstates, which are solutions of the time-independent Schrodinger equation. Time-independent perturbation theory tells you how the energy eigenstates are modified when the Hamiltonian is.

Suppose a system is originally in an energy eigenstate. When a perturbation instantly turns on, time-dependent perturbation theory tells us the energy eigenstates have changed. This doesn't mean the state of the system has instantly changed, it just means that the state isn't an energy eigenstate anymore. To actually compute the evolution of the state, you use time-dependent perturbation theory.


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