Linear Response Theory

A small perturbation act on a system in equilibrium may produce an effect related to this external field, which can be treat as a linear response up to first order approximation.

We assume the system is in equilibrium with Hamiltonian H before time t_0. The perturbation is turned on after t_0, and the Hamiltonian becomes

(1)   \begin{equation*} H(t)=H+H^{ex}(t)=H-B(t)h(t) \end{equation*}

where h(t) is the external field which coupled with the quantum operator B(t). The Shrodinger is thus

(2)   \begin{equation*} i\hbar\frac{\partial}{\partial t}|j\rangle_S =\left[H+H^{ex}(t)\right]|j\rangle_S \end{equation*}

and we seek a solution in the form

(3)   \begin{equation*} |j(t)\rangle_S =e^{-iHt/\hbar}U(t)|j(0)\rangle_S \end{equation*}

where U(t), by taking the above form into the Schordinger equation, satisfy

(4)   \begin{equation*} i\hbar\frac{\partial}{\partial t}U(t) =H^{ex}_H(t)U(t) \end{equation*}

with the help of Dyson series, we can get

(5)   \begin{equation*} U(t) =1-\frac{i}{\hbar}\int_{t_0}^{t}dt'H^{ex}_H(t)+\cdots \end{equation*}

The expectation value of A(t) at stare |j\rangle is thus

(6)   \begin{align*} \langle j(t)|_S A_S(t) |j(t)\rangle_S \nonumber &=\langle j(0)|_S U(t)^{-1}e^{iHt/\hbar}A_S(t)e^{-iHt/\hbar}U(t)|j(0)\rangle_S\\ &=\langle j(0)|_H A_H(t) |j(0)\rangle_H +\frac{i}{\hbar}\langle j(0)|_H \int_{t_0}^{t}dt'\left[H^{ex}_H(t'), A_H(t)\right] |j(0)\rangle_H+\cdots \end{align*}

The fist term is just the equilibrium value, while the LHS is nonequilibrium value, thus

(7)   \begin{align*} \delta \langle A(t) \rangle \nonumber &=\langle A(t) \rangle_{ne}-\langle A(t) \rangle_{eq}\\ \nonumber &\approx \frac{i}{\hbar}\langle j(0)|_H \int_{t_0}^{t}dt'\left[H^{ex}_H(t'), A_H(t)\right] |j(0)\rangle_H \\  &=\frac{i}{\hbar}\langle j(0)|_H \int_{t_0}^{t}dt'\left[A_H(t), B_H(t')\right] |j(0)\rangle_H h(t') \end{align*}

At finite temperature, the average of A(t) is

(8)   \begin{align*} \langle A(t) \rangle_{ne} =Tr\rho_{ne} A(t) =\sum_{j}\frac{1}{Z}e^{-i\beta E_j(t)} \langle j(t)|_S A_S(t) |j(t)\rangle_S \end{align*}

In the first order approximation, we can still use the unperturbed density operator, \rho_{ne}=\rho_{eq}=\rho. Using 7 in the above equation, we can get

(9)   \begin{equation*} \delta \langle A(t) \rangle =\frac{i}{\hbar} \int_{t_0}^{t}dt' Tr\{\rho \left[A_H(t), B_H(t')\right]\} h(t') \end{equation*}

Define the dynamic susceptibility as

(10)   \begin{equation*} \chi_{AB}(t,t')=\frac{i}{\hbar}\theta(t-t') Tr\{\rho \left[A_H(t), B_H(t')\right]\} \end{equation*}

Considering \theta(t-t')=0 if t<t' and h(t')=0 for t'<t_0, the response is thus

(11)   \begin{equation*} \delta \langle A(t) \rangle =\frac{i}{\hbar} \int_{-\infty}^{\infty}dt' \chi_{AB}(t,t') h(t') \end{equation*}

In fact, the dynamic susceptibility has time translation symmetry, that is

(12)   \begin{equation*} \chi_{AB}(t,t') =\chi_{AB}(t-t') \end{equation*}

which can be easily proved with the trace property Tr(XY)=Tr(YX), take the term Tr\left[\rho A_H(t) B_H(t')\right] for example,

(13)   \begin{align*} Tr\left[\rho A_H(t) B_H(t')\right] \nonumber &=Tr\left[\rho U(t, 0)^{-1}A_S(0)U(t,0)U^{-1}(t',0) B_S(t')U(t',0)\right]\\ \nonumber &=Tr\left[\rho U(t',0)U(t, 0)^{-1}A_S(0)U(t,0)U^{-1}(t',0) B_S(0)\right]\\ \nonumber &=Tr\left[\rho U(t, t')^{-1}A_S(0)U(t,t') B_S(0)\right]\\ &=Tr\left[\rho A_S(t-t') B_S(0)\right] \end{align*}

The response

(14)   \begin{equation*} \delta \langle A(t) \rangle =\frac{i}{\hbar} \int_{-\infty}^{\infty}dt' \chi_{AB}(t-t') h(t') \end{equation*}

is obviously the convolution of \chi_{AB}(t) and h(t). After Fourier transformation, it will give

(15)   \begin{equation*} \delta \langle A(\omega) \rangle =\frac{i}{\hbar} \chi_{AB}(\omega) h(\omega) \end{equation*}

That’s why the response is called linear.



[1] Fetter, Alexander L., and John Dirk Walecka. Quantum theory of many-particle systems. Courier Corporation, 2012.
[2] Mazenko, Gene F. Nonequilibrium statistical mechanics. John Wiley & Sons, 2008.

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