# 11.2: Quantum Linear Response Functions


To develop a quantum description of the linear response function, we start by recognizing that the response of a system to an applied external agent is a problem we can solve in the interaction picture. Our time-dependent Hamiltonian is

\begin{align} H (t) &= H _ {0} - f (t) \hat {A} \\[4pt] &= H _ {0} + V (t) \label{10.48} \end{align}

$$H_o$$ is the material Hamiltonian for the equilibrium system. The external agent acts on the equilibrium system through $$\hat{A}$$, an operator in the system states, with a time-dependence $$f(t)$$. We take $$V(t)$$ to be a small change, and treat this problem with perturbation theory in the interaction picture.

We want to describe the nonequilibrium response $$\overline {A(t)}$$, which we will get by ensemble averaging the expectation value of $$\hat{A}$$, i.e. $$\overline {\langle A (t) \rangle}$$. Remember the expectation value for a pure state in the interaction picture is

\begin{align} \langle A (t) \rangle & = \left\langle \psi _ {I} (t) \left| A _ {I} (t) \right| \psi _ {I} (t) \right\rangle \\[4pt] & = \left\langle \psi _ {0} \left| U _ {I}^{\dagger} A _ {I} U _ {I} \right| \psi _ {0} \right\rangle \label{10.49} \end{align}

The interaction picture Hamiltonian for Equation \ref{10.48} is

\left.\begin{aligned} V _ {I} (t) & = U _ {0}^{\dagger} (t) V (t) U _ {0} (t) \\[4pt] & = - f (t) A _ {I} (t) \end{aligned} \right. \label{10.50}

To calculate an ensemble average of the state of the system after applying the external potential, we recognize that the nonequilibrium state of the system characterized by described by $$| \psi _ {I} (t) \rangle$$ is in fact related to the initial equilibrium state of the system $$| \psi _ o\rangle$$ through a time-propagator, as seen in Equation \ref{10.49}. So the nonequilibrium expectation value $$\overline {A (t)}$$ is in fact obtained by an equilibrium average over the expectation value of $$U _ {I}^{\dagger} A _ {I} U _ {I}$$:

$\overline {A (t)} = \sum _ {n} p _ {n} \left\langle n \left| U _ {I}^{\dagger} A _ {I} U _ {I} \right| n \right\rangle \label{10.51}$

Again $$| n \rangle$$ are eigenstates of $$H_o$$. Working with the first order solution to $$U_I(t)$$

$U _ {I} \left( t - t _ {0} \right) = 1 + \dfrac {i} {\hbar} \int _ {t _ {0}}^{t} d t^{\prime} f \left( t^{\prime} \right) A _ {I} \left( t^{\prime} \right)\label{10.52}$

we can now calculate the value of the operator $$\hat{A}$$ at time $$t$$, integrating over the history of the applied interaction $$f(t')$$:

\left.\begin{aligned} A (t) & = U _ {I}^{\dagger} A _ {I} U _ {I} \\ & = \left\{1 - \frac {i} {\hbar} \int _ {t _ {0}}^{t} d t^{\prime} f \left( t^{\prime} \right) A _ {I} \left( t^{\prime} \right) \right\} A _ {I} (t) \left\{1 + \frac {i} {\hbar} \int _ {t _ {0}}^{t} d t^{\prime} f \left( t^{\prime} \right) A _ {I} \left( t^{\prime} \right) \right\} \end{aligned} \right. \label {10.53}

Here note that $$f$$ is the time-dependence of the external agent. It does not involve operators in $$H_o$$ and commutes with $$A$$. Working toward the linear response function, we just retain the terms linear in

\left.\begin{aligned} A (t) & \cong A _ {I} (t) + \dfrac {i} {\hbar} \int _ {t _ {0}}^{t} d t^{\prime}\, f \left( t^{\prime} \right) \left\{A _ {I} (t) A _ {I} \left( t^{\prime} \right) - A _ {I} \left( t^{\prime} \right) A _ {I} (t) \right\} \\[4pt] & = A _ {I} (t) + \dfrac {i} {\hbar} \int _ {t _ {0}}^{t} d t^{\prime} \,f \left( t^{\prime} \right) \left[ A _ {I} (t) , A _ {I} \left( t^{\prime} \right) \right] \end{aligned} \right. \label{10.54}

Since our system is initially at equilibrium, we set $$t _ {0} = - \infty$$ and switch variables to the time interval $$\tau = t - t^{\prime}$$ and using

$A _ {I} (t) = U _ {0}^{\dagger} (t) A U _ {0} (t)$

to obtain

$A (t) = A _ {I} (t) + \dfrac {i} {\hbar} \int _ {0}^{\infty} d \tau \,f ( t - \tau ) \left[ A _ {I} ( \tau ) , A _ {I} ( 0 ) \right] \label{10.55}$

We can now calculate the expectation value of $$A$$ by performing the ensemble-average described in Equation \ref{10.51}. Noting that the force is applied equally to each member of ensemble, we have

$\overline {A (t)} = \langle A \rangle + \dfrac {i} {\hbar} \int _ {0}^{\infty} d \tau f ( t - \tau ) \left\langle \left[ A _ {I} ( \tau ) , A _ {I} ( 0 ) \right] \right\rangle \label{10.56}$

The first term is independent of $$f$$, and so it comes from an equilibrium ensemble average for the value of $$A$$.

$\langle A (t) \rangle = \sum _ {n} p _ {n} \left\langle n \left| A _ {I} \right| n \right\rangle = \langle A \rangle \label{10.57}$

The second term is just an equilibrium ensemble average over the commutator in $$A_I(t)$$:

$\left\langle \left[ A _ {I} ( \tau ) , A _ {I} ( 0 ) \right] \right\rangle = \sum _ {n} p _ {n} \left\langle n \left| \left[ A _ {I} ( \tau ) , A _ {I} ( 0 ) \right] \right| n \right\rangle \label{10.58}$

Comparing Equation \ref{10.56} with the expression for the linear response function, we find that the quantum linear response function is

$\left. \begin{array} {r l} {R ( \tau )} & {= - \dfrac {i} {\hbar} \left\langle \left[ A _ {I} ( \tau ) , A _ {I} ( 0 ) \right] \right\rangle} & {\tau \geq 0} \\[4pt] {} & {= 0} & {\tau < 0} \end{array} \right. \label{10.59}$

or as it is sometimes written with the unit step function in order to enforce causality:

$R ( \tau ) = - \dfrac {i} {\hbar} \Theta ( \tau ) \left\langle \left[ A _ {I} ( \tau ) , A _ {I} ( 0 ) \right] \right\rangle \label{10.60}$

The important thing to note is that the time-development of the system with the applied external potential is governed by the dynamics of the equilibrium system. All of the time-dependence in the response function is under $$H_o$$.

The linear response function is therefore the sum of two correlation functions with the order of the operators interchanged, which is the imaginary part of the correlation function $$C''(\tau)$$

\left.\begin{aligned} R ( \tau ) & = - \dfrac {i} {\hbar} \Theta ( \tau ) \left\{\left\langle A _ {I} ( \tau ) A _ {I} ( 0 ) \right\rangle - \left\langle A _ {I} ( 0 ) A _ {I} ( \tau ) \right\rangle \right\} \\[4pt] & = - \dfrac {i} {\hbar} \Theta ( \tau ) \left( C _ {A A} ( \tau ) - C _ {A A}^{*} ( \tau ) \right) \\[4pt] & = \dfrac {2} {\hbar} \Theta ( \tau ) C^{\prime \prime} ( \tau ) \end{aligned} \right.\label{10.61}

As we expect for an observable, the response function is real. If we express the correlation function in the eigenstate description:

$C (t) = \sum _ {n , m} p _ {n} \left| A _ {m n} \right|^{2} e^{- i \omega _ {m n} t} \label{10.62}$

then

$R (t) = \dfrac {2} {\hbar} \Theta (t) \sum _ {n , m} p _ {n} \left| A _ {m n} \right|^{2} \sin \omega _ {m n} t \label{10.63}$

$$R(t)$$ can always be expanded in sines—an odd function of time. This reflects that fact that the impulse response must have a value of 0 (the deviation from equilibrium) at $$t = t_o$$, and move away from 0 at the point where the external potential is applied.