Time-Dependent Linear Response Theory

We are considering a Hamiltonian split as:

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

where $H_0$ is the unperturbed time-independent Hamiltonian (which we know how to solve), and $\lambda V(t)$ is the time-dependent perturbation where $\lambda$ is a parameter that controls how small the perturbation is.

The full Schrödinger equation can be written as:

$$i\hbar \frac{\partial}{\partial t} \ket{\psi(t)} = H(t)\ket{\psi(t)} = [H_0 + \lambda V(t)]\ket{\psi(t)} \tag{1}$$

The eigenstates $\ket{n}$ and eigenvalues $E_n$ of the stationary $H_0$ satisfy

$$H_0\ket{n} = E_n \ket{n}$$

Without perturbation, the stationary time-evolution of eigenstate $\ket{n}$ is

$$\ket{n(t)} = e^{-iE_nt/\hbar} \ket{n}$$

We expand the full solution $\ket{\psi(t)}$ in terms of the unperturbed eigenstates $\ket{m(t)}$:

$$\ket{\psi(t)} = \sum_m c_m(t) e^{-iE_mt/\hbar} \ket{m} \tag{2}$$

assuming the perturbation $\lambda V(t)$ is small, the expansion coefficients $c_m(t)$ should vary very slow.

Now plug Eq. 2 into Eq. 1, we get

$$\begin{aligned} i\hbar \frac{\partial}{\partial t} \ket{\psi(t)} &= [H_0 + \lambda V(t)]\ket{\psi(t)} \\ i\hbar \frac{\partial}{\partial t} \sum_m c_m(t) e^{-iE_mt/\hbar} \ket{m} &= [H_0 + \lambda V(t)] \sum_m c_m(t) e^{-iE_mt/\hbar} \ket{m}\\ i\hbar \sum_m (\dot{c}_m(t) e^{-iE_mt/\hbar} \ket{m} - iE_m/\hbar c_m(t) e^{-iE_mt/\hbar} \ket{m}) &= [H_0 + \lambda V(t)] \sum_m c_m(t) e^{-iE_mt/\hbar} \ket{m}\\ i\hbar \sum_m \dot{c}_m(t) e^{-iE_mt/\hbar} \ket{m} +\sum_m E_m c_m(t) e^{-iE_mt/\hbar} \ket{m} &= [H_0 + \lambda V(t)] \sum_m c_m(t) e^{-iE_mt/\hbar} \ket{m}\\ i\hbar \sum_m \dot{c}_m(t) e^{-iE_mt/\hbar} \ket{m} &= \lambda V(t) \sum_m c_m(t) e^{-iE_mt/\hbar} \ket{m}\\ \end{aligned}$$

left multiply by $\bra{n}$ and use the orthonormal condition of the stationary eigenstates $\braket{n|m}=\delta_{mn}$

$$\begin{aligned} i\hbar \dot{c}_n(t) e^{-iE_nt/\hbar} &= \sum_m c_m(t) e^{-iE_mt/\hbar} \braket{n| \lambda V(t)|m}\\ \dot{c}_n(t) &= \frac{1}{i\hbar}\sum_m c_m(t) e^{-i(E_m-E_n)t/\hbar} \braket{n| \lambda V(t)|m}\\ \end{aligned}$$

and we can rewrite it to

$$\dot{c}_n(t) = \frac{1}{i\hbar}\sum_m c_m(t) e^{i\omega_{nm}t} \lambda V_{nm}(t) \tag{3}$$

where $\omega_{nm}=\frac{E_n-E_m}{\hbar}$ and $V_{nm}(t)=\braket{n|V(t)|m}$.

We can expand the time-dependent coefficients $c_m(t)$ in powers of the parameter $\lambda$

$$c_m(t) = c_m^{(0)}(t) + \lambda c_m^{(1)}(t) + \lambda^2 c_m^{(2)}(t) + \cdots \tag{4}$$

Assuming we are starting from time $t_0$ and systems is prepared to be in state $i$, the zeroth order expansion should be stationary coefficients at $t_0$ (as $\lambda=0$, we turn off the interaction and system becomes time independent)

$$c_m(t_0) = c_m^{(0)}(t) = \delta_{mi}$$

which means the zeroth coefficients are constants and $\dot{c}_m^{(0)}(t) = 0$, combining with Eq. 4, the expansion of Eq.3 can be written as

$$\begin{aligned} \dot{c}_n(t)^{(0)} &+ \lambda \dot{c}_n(t)^{(1)} + \lambda^2 \dot{c}_n(t)^{(2)} + \cdots\\ &=\frac{1}{i\hbar}\sum_m \lambda c^{(0)}_m(t) e^{i\omega_{nm}t} V_{nm}(t) + \frac{1}{i\hbar}\sum_m \lambda^2 c^{(1)}_m(t) e^{i\omega_{nm}t} V_{nm}(t) \cdots \end{aligned}$$

To first order of $\lambda$, we have

$$\begin{aligned} \dot{c}_n(t)^{(1)} &= \frac{1}{i\hbar}\sum_m c^{(0)}_m(t) e^{i\omega_{nm}t} V_{nm}(t) \\ &= \frac{1}{i\hbar}\sum_m \delta_{mi} e^{i\omega_{nm}t} V_{nm}(t)\\ &= \frac{1}{i\hbar} e^{i\omega_{ni}t} V_{ni}(t)\tag{6} \end{aligned}$$

Using fundamental theorem of calculus

$${c}_n(t)^{(1)} - {c}_n(t_0)^{(1)} = \int_{t_0}^t \delta t' \dot{c}(t')^{(1)}$$

since the system is initialized at $c_m(t_0) =\delta_{mi}$, we have ${c}_n(t_0)^{(1)}=0$, hence

$${c}_n(t)^{(1)} = \frac{1}{i\hbar} \int_{t_0}^t \delta t' e^{i\omega_{ni}t'} V_{ni}(t') \tag{7}$$

Combining with the zeroth order $c_n(t)^{(0)}=\delta_{ni}$ , we see that to first order, $c_n(t) = c_n(t)^{(1)}+\delta_{ni}$. This is the amplitude of possibility (to the first order) of system (initialized to be in state $\ket{i}$) end up in state $\ket{n}$ ($n\neq i$) after time $t$, in other words, the first-order transition amplitude for going from state $\ket{i}$ to $\ket{n}$ ($n\neq i$) is

$$a_{i\to n}(t) = {c}_n(t) = \frac{1}{i\hbar} \int_{t_0}^t \delta t' e^{i\omega_{ni}t'} V_{ni}(t') \tag{5}$$

and the corresponding transition probability is

$$P_{i\to n}(t) = |a_{i\to n}(t) |^2$$