Time-Dependent Linear Response Theory

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    <p>–&gt; <!-- Note that here, unlike in most text books, I have opted to not use the interacting picture.  --> &lt;!–</p>

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References

Ref: https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/89ef6d5958ee59bae9a91345c3d8c8e4_MIT8_06S18ch4.pdf
Ref: https://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_dep.pdf

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 time-evolution of stationary eigenstate $\ket{n}$ follows energy conservation:

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

We start by 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}

we can do this when the perturbation $\lambda V(t)$ is small and $\ket{m}$ (approximately) span the entire Hilbert space.

$\lambda$ dependence

Note that since $\ket{\psi(t)}$ is the full time dependent wavefunction which depends on $\lambda$, the expansion coefficients $c_m(t)$ (implicitly) depends on $\lambda$ as well.

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 \left [ \dot{c}_m(t) e^{-iE_mt/\hbar} \ket{m} - iE_m/\hbar c_m(t) e^{-iE_mt/\hbar} \ket{m} \right] &= [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}

which is the equation of motion of the states under the stationary eigenstate basis. It can be re-write as

\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 $V_{nm}(t)=\braket{n| V(t)|m}$.

Maclaurin series

The Maclaurin is Taylor series expanded at zero $f(x)|_{x\to 0} = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x^n)$

Remembering that $c_m(t)$ also depends on $\lambda$ via $\ket{\psi(t)}$, we can expand the time-dependent coefficients $c_m(t)$ in powers of the parameter $\lambda$ (Maclaurin series, ignoring the $1/n!$ as it does not matter in this case)

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

If we also expand $\dot{c}_n(t)$ as a Maclaurin series of $\lambda$,

\begin{aligned} \dot{c}_n(t) &= \dot{c}_n(t)^{(0)} + \lambda \dot{c}_n(t)^{(1)} + \lambda^2 \dot{c}_n(t)^{(2)} + \cdots \tag{5} \end{aligned}

substituting Eq. 4 into Eq.3, we see that

\begin{aligned} \dot{c}_n(t) &=\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 \tag{5.1} \end{aligned}

Comparing Eq. 5.1 to Eq. 5, we can get the expression of $\dot{c}_n(t)$ at each order of $\lambda$ using $c_n(t)$.

Focusing on the zero-th order term of $\lambda$

\dot{c}_n(t)^{(0)} = 0

hence we know that

c_n(t)^{(0)} = \mathrm{constant}

Initial state

</p> with this: -->

Assuming we are starting from time $t_0$ and systems is prepared to be in state $i$ (for multi-Fermion, non-interacting system, a series of occupied initial states), i.e.,

\ket{\psi(t_0)} = \sum_m c_m(t_0) e^{-iE_mt_0/\hbar} \ket{m} = c_i(t_0) e^{-iE_it_0/\hbar} \ket{i}=e^{-iE_it_0/\hbar} \ket{i},

the $\lambda$ expansion of $c_m(t)$ at $t_0$ should be stationary and we have

c_m(t_0) = \delta_{mi} = c_m^{(0)}(t_0) \tag{4.1}

and (again because at $t_0$ no matter how we change $\lambda$, we have system prepared in $\ket{i}$),

$c_m^{(1)}(t_0) = c_m^{(2)}(t_0) = \cdots =c_m^{(n)}(t_0) =0 \tag{4.2}$

Combining with Eq. 4.1, we see

c_n(t)^{(0)} = \delta_{ni}

First order

Focusing on the first order term 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)}

Remembering Eq. 4.2, ${c}_n(t_0)^{(1)}=0$, we can write

{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 up to first order

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

This coefficient represents the amplitude (to the first order) of a system initialised to be in state $\ket{i}$ that ended 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 $\ket{n}$ ($n\neq i$) at time $t$ 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{8}

and the corresponding transition probability is

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

Second order

Focusing on the second order term

\dot{c}_n^{(2)}(t) = \frac{1}{i\hbar}\sum_m c^{(1)}_m(t) e^{i\omega_{nm}t} V_{nm}(t)

integrating from $t_0$ to $t$

\begin{aligned} c_n^{(2)}(t) &= \frac{1}{i\hbar}\sum_m \int_{t_0}^t c^{(1)}_m(t) e^{i\omega_{nm}t} V_{nm}(t)\\ &= \frac{1}{i\hbar}\sum_m \int_{t_0}^{t} \delta t'' \left[ \frac{1}{i\hbar} \int_{t_0}^{t''} \delta t' e^{i\omega_{mi}t'} V_{mi}(t') \right] e^{i\omega_{nm}t''} V_{nm}(t'')\\ &= \frac{-1}{\hbar} \sum_m \int_{t_0}^{t} \delta t'' \int_{t_0}^{t''} \delta t' V_{nm}(t'')V_{mi}(t') e^{i\omega_{mi}t'} e^{i\omega_{nm}t''} \end{aligned}

Normalisation of $\phi(t)$

Next, let’s investigate the norm of the time-dependent wavefunction $\ket{\psi(t)}$ when it is expanded to different orders of $\lambda$.

The norm of $\psi (t)$ reads

\begin{aligned} |\psi (t)|^2 &=\left | \sum_m c_m(t) e^{-iE_mt/\hbar} \ket{m} \right |^2\\ &=\sum_{mn} \bra{n} c^*_n(t) e^{iE_n t/\hbar} c_m(t) e^{-iE_mt/\hbar} \ket{m}\\ &=\sum_{mn} c^*_n(t)c_m(t) e^{i(E_n - E_m) t/\hbar} \delta_{nm}\\ &=\sum_{n} c^*_n(t)c_n(t) e^{i(E_n - E_n) t/\hbar} \\ &=\sum_{n} c^*_n(t)c_n(t) \\ &= \sum_{n} \left( \delta_{ni} + {c}_n(t)^{(1)*} \right )\left (\delta_{ni} + {c}_n(t)^{(1)}\right) \\ &= \sum_{n} \left( |\delta_{ni}|^2 + \delta_{ni}{c}_n(t)^{(1)*} + \delta_{ni}^* {c}_n(t)^{(1)} + |{c}_n(t)^{(1)}|^2 \right ) \\ &= 1 +{c}_i(t)^{(1)*} + {c}_i(t)^{(1)} + \sum_n |{c}_n(t)^{(1)}|^2\\ \end{aligned}

we notice that

{c}_i(t)^{(1)} = \frac{1}{i\hbar} \int_{t_0}^t \delta t' e^{i\omega_{ii}t'} V_{ii}(t') = \frac{1}{i\hbar} \int_{t_0}^t \delta t'V_{ii}(t')

is either zero ($V_{ii}=0$) or pure imaginary (because $V_{ii}$ should be real as $V$ is hermitian) , hence

{c}_i(t)^{(1)*} + {c}_i(t)^{(1)} =0

so that

|\psi (t)|^2 = 1 + \sum_n |{c}_n(t)^{(1)}|^2

where we do see an increase in the norm, and the difference being the possibility of finding the system in state $n$, i.e., $\sum_n |{c}_n(t)^{(1)}|^2$ !

Now, if we expand $c$ to second order with respect to $\lambda$, the norm would go like

1 + \left[ {c}_i(t)^{(2)} + {c}_i(t)^{(2)*} \right] + \sum_n |{c}_n(t)^{(1)}|^2 + O(\lambda^3) + O(\lambda^4)

where

\begin{aligned} {c}_i(t)^{(2)} + {c}_i(t)^{(2)*} &= -\frac{1}{\hbar^2} \sum_m \int_{t_0}^{t} dt' \int_{t_0}^{t'} dt'' V_{im}(t') V_{mi}(t'') e^{i\omega_{im}(t'-t'')} \\& + -\frac{1}{\hbar^2} \sum_m \int_{t_0}^{t} dt' \int_{t_0}^{t'} dt'' V_{mi}(t') V_{im}(t'') e^{-i\omega_{im}(t'-t'')}\\ &= -\frac{1}{\hbar^2} \sum_m \int_{t_0}^{t} dt' \int_{t_0}^{t'} dt'' V_{im}(t') V_{mi}(t'') e^{i\omega_{im}(t'-t'')} \\& + -\frac{1}{\hbar^2} \sum_m \int_{t_0}^{t} dt'' \int_{t_0}^{t''} dt' V_{mi}(t'') V_{im}(t') e^{i\omega_{im}(t''-t')}\\ % &= - \sum_n |{c}_n(t)^{(1)}|^2 \end{aligned}

here in the second line, I have swapped dummy variables $t’$ and $t’’$ and used the fact that, noticing that we are integrating over two triangles (variables $t, t’’$) of the same equation, they can be combined into a single square integration

\begin{aligned} {c}_i(t)^{(2)} + {c}_i(t)^{(2)*} &= -\frac{1}{\hbar^2} \sum_m \int_{t_0}^{t} dt' \int_{t_0}^{t} dt'' V_{im}(t') V_{mi}(t'') e^{i\omega_{im}(t'-t'')} \\ \end{aligned}

At the same time, we have

\sum_m |c_m^{(1)}|^2 = \frac{1}{\hbar^2} \int_{t_0}^{t}\!\! dt' \int_{t_0}^{t}\!\! dt'' \sum_m V_{mi}(t') V_{im}(t'') e^{i\omega_{mi}(t'-t'')} = -\left[ {c}_i(t)^{(2)} + {c}_i(t)^{(2)*} \right]

This means that the second order expansion of amplitude generates a decrease in the possibility of finding the system in the original state $i$, and this decrease in possibility exactly compensates for the added norm from the first order expansion of the amplitude. However, doing so also introduces additional higher-order deviation of the norm which will get canceled by even higher-order amplitude expansion. And in an infinite order expansion, the norm is recovered.

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    <p>This is information intended to draw attention.</p>

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