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11.2: Quantum Linear Response Functions

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    107287
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    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.

    Readings

    1. Mukamel, S., Principles of Nonlinear Optical Spectroscopy. Oxford University Press: New York, 1995; Ch. 5.

    Contributors and Attributions


    11.2: Quantum Linear Response Functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Andrei Tokmakoff via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.