Response & Correlation

• Calculating eigenstates and eigenenergies is one thing, but what do experimentalists actually measure?

• Want to understand how many body systems respond dynamically to an external probe

• Time evolution of spontaneous fluctuations (thermal or quantum) obeys similar dynamics $\longrightarrow$ fluctuation–dissipation relation

Quantum fluctuations: one oscillator

• For a single (undamped) oscillator we have ground state fluctuations $$H = \frac{p^2}{2m} + \frac{1}{2}m\omega_0^2 y^2$$ $$\bra{0} y^2 \ket{0} = \frac{1}{2m \omega_0}$$
• At finite temperature $$\langle\langle y^2\rangle\rangle=\tr\left[\rho\\, y^2\right] = \frac{\coth(\beta\omega_0/2)}{2m\omega_0}$$ $\rho = e^{-\beta H}/Z$ is equilibrium density matrix, and $Z=\tr[e^{-\beta H}]$ is partition function

• The double angular brackets $\langle\langle (\cdots)\rangle\rangle$ denote that we are taking quantum and thermal expectations.

• What about time dependent fluctuations? A natural candidate: $$\langle\langle y(t)y(0)\rangle\rangle,$$ $y(t) = e^{iHt} y e^{-iHt}$ (Heisenberg picture)

• This gives the quantum noise spectrum

$$S(\omega) = \int_{-\infty}^\infty \langle\langle y(t)y(0)\rangle\rangle e^{i\omega t}\,dt.$$

• BUT: since $y(0)$ and $y(t)$ do not commute with each other $\langle\langle y(t)y(0)\rangle\rangle\neq \langle\langle y(0)y(t)\rangle\rangle=\langle\langle y(-t)y(0)\rangle\rangle$, so

$$S(\omega)\neq S(-\omega)$$

• Many of the properties of $S(\omega)$ are most easily understood from spectral representation

• Insert complete set of energy eigenstates between $y(0)$ and $y(t)$

$$S(\omega) = 2\pi\sum_{m,n} \frac{e^{-\beta E_n}}{Z} |\bra{n}y\ket{m}|^2 \delta(\omega-E_m+E_n)$$

• Reason for asymmetry in $S(\omega)$ is that term with $\delta(\omega-E_m+E_n)$ is weighted by $e^{-\beta E_n}$, while term with $\delta(\omega-E_n+E_m)$ has $e^{-\beta E_m}$

$$S(\omega) = S(-\omega) e^{\beta\omega}.$$

• At zero temperature $\beta\to\infty$ and we can see that $S(\omega<0)=0$.

• Let’s evaluate $S(\omega)$ for oscillator. The matrix elements are $$\bra{n}y\ket{m} = \frac{1}{\sqrt{2m\omega_0}}\begin{cases} \sqrt{m+1} & \text{if } n=m+1 \\ \sqrt{m} & \text{if } n=m-1. \end{cases}$$ \begin{align*} S(\omega)&=\frac{\pi}{m\omega_0} \sum_n \frac{e^{-\beta E_n}}{Z} \left[n\delta(\omega+\omega_0)+(n+1)\delta(\omega-\omega_0)\right]\nonumber\\ & = \frac{\pi}{m\omega_0}\left[n_\text{B}(\omega_0)\delta(\omega+\omega_0)+(n_\text{B}(\omega_0)+1)\delta(\omega-\omega_0)\right] \end{align*} $$n_\text{B}(\omega)\equiv \frac{1}{\exp\left(\beta\omega\right)-1}$$ is Bose distribution function

\begin{align*} S(\omega)&=\frac{\pi}{m\omega_0} \sum_n \frac{e^{-\beta E_n}}{Z} \left[n\delta(\omega+\omega_0)+(n+1)\delta(\omega-\omega_0)\right]\nonumber\\ & = \frac{\pi}{m\omega_0}\left[n_\text{B}(\omega_0)\delta(\omega+\omega_0)+(n_\text{B}(\omega_0)+1)\delta(\omega-\omega_0)\right] \end{align*}

• Shows the predicted asymmetry between positive and negative frequencies. We can check that $$\int S(\omega)\frac{d\omega}{2\pi} = \langle\langle y^2\rangle\rangle= \frac{\coth(\beta\omega_0/2)}{2m\omega_0}$$ as we found before.

Classical limit $\beta\omega\to 0$

\begin{align*} S(\omega)&= \frac{\pi}{m\omega_0}\left[n_\text{B}(\omega_0)\delta(\omega+\omega_0)+(n_\text{B}(\omega_0)+1)\delta(\omega-\omega_0)\right]\\\\ &\to \frac{k_\text{B}T}{2m\omega_0^2} \times 2\pi\left[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)\right] \end{align*}

• Consistent with equipartition: $\frac{1}{2}m\omega_0^2 \langle y^2\rangle=\frac{1}{2}k_\text{B}T$

Many oscillators

• We can express $y(t)$ in terms of normal modes as $$y(t) = \sum_k \left[c^{}_k \adop_k(t) + c_k^* \aop_k(t)\right]$$
• Time evolution of mode operators is

$$\adop_k(t) = e^{i\omega_k t}\adop_k,\quad \aop_k(t) = e^{-i\omega_k t}\aop_k$$

• Repeating calculation of $S(\omega)$ gives

\begin{align*} S(\omega)= 2\pi\sum_k |c_k|^2\left[n_\text{B}(\omega_k)\delta(\omega+\omega_k)+(n_\text{B}(\omega_k)+1)\delta(\omega-\omega_k)\right] \end{align*}

• In continuum limit sum of $\delta(\omega\pm\omega_k)\longrightarrow$ smooth $S(\omega)$

Response of oscillator system

• Solve Heisenberg equations of motion $$H = \sum_k \omega_k \adop_k\aop_k - f(t)y,$$ where $y$ is written in terms of normal modes $$\frac{d\aop_k}{dt} = -i\omega_k \aop_k +i c_k f(t),$$

• Solution is $\aop_k(t) = e^{-i\omega_k t}\aop_k(0)+a_{k,f}(t)$ with $$\aop_{k,f}(\omega) \equiv \frac{c_k}{\omega_k-\omega-i0} f(\omega).$$ ($-i0$ introduced for response analytic in UHP i.e. causal)

• Response of $\aop_k(t)\longrightarrow y(t)$ gives

$$y(\omega) = \sum_k |c_k|^2\left[\frac{1}{\omega_k-\omega-i0}+\frac{1}{\omega_k+\omega+i0}\right]f(\omega).$$

• This defines the response function $\chi(\omega)$ $$\chi(\omega)\equiv \frac{y(\omega)}{f(\omega)} = \sum_k |c_k|^2\left[\frac{1}{\omega_k-\omega-i0}+\frac{1}{\omega_k+\omega+i0}\right]$$

• Then use…

$$\text{Im} \frac{1}{x\mp i0} = \pm\pi\delta(x),$$

$$\operatorname{Im}\chi(\omega) =\sgn(\omega)\pi\sum_k |c_k|^2\delta(\omega_k-\omega)$$

• $S(\omega)$ and $\operatorname{Im}\chi(\omega)$ are then related by

$$S(\omega) = 2\operatorname{Im}\chi(\omega)\left[n_\text{B}(\omega)+1\right]$$

• This is a quantum fluctuation dissipation relation

Check that in the classical limit ($k_\text{B}T\gg \hbar \omega$) $$S(\omega) = 2\operatorname{Im}\chi(\omega)\left[n_\text{B}(\omega)+1\right]$$ reduces to $$S(\omega)=\frac{2k_\text{B}T}{\omega} \operatorname{Im}\chi(\omega)$$ Consistent with classical equipartition

Golden Rule and Dissipation

• We saw that classically $\text{Im}\,\chi(\omega)$ related to dissipation

• FDT relates $\operatorname{Im}\chi(\omega)$ and $S(\omega)$. How is $S(\omega)$ related to dissipation?

• Regard driving force as perturbation that causes transitions between energy eigenstates $$H_\text{pert} = - f(t)y$$ with $f(t)=f_0\cos\omega t$

• In lowest order perturbation theory, system can make a transitions $\pm\omega$ in energy. Rates found from Fermi’s golden rule

$$\Gamma_{n\to m} = 2\pi \left(\frac{f_0}{2}\right)^2|\bra{n}y\ket{m}|^2 \delta(\pm\omega+E_m-E_n).$$

• Total rate is sum over all initial ($\ket{n}$) and final states ($\ket{m}$), including probability $e^{-\beta E_n}/Z$ of finding the system initially in $\ket{n}$

• c.f. spectral representation of $S(\omega)$

$$S(\omega) = 2\pi\sum_{m,n} \frac{e^{-\beta E_n}}{Z} |\bra{n}y\ket{m}|^2 \delta(\omega-E_m+E_n)$$

$$\Gamma(\omega) = S(\omega)\left(\frac{f_0}{2}\right)^2$$

• $S(\omega)$ measures rate of transitions absorbing energy $\omega$; $S(-\omega)$ the rate for emitting energy $\omega$.

• Asymmetry of $S(\omega)$ is asymmetry between emission and absorption of radiation

• The rate of energy absorption is

$$\omega\Gamma(\omega) = \omega S(\omega)\left(\frac{f_0}{2}\right)^2 = \frac{1}{2}\omega\operatorname{Im}\chi(\omega)\left[n_\text{B}(\omega)+1\right]f_0^2$$

$$\omega\Gamma(\omega) = \omega S(\omega)\left(\frac{f_0}{2}\right)^2 = \frac{1}{2}\omega\operatorname{Im}\chi(\omega)\left[n_\text{B}(\omega)+1\right]f_0^2$$

Compare with \begin{align*} W_\text{diss} = \langle f(t) \dot y(t)\rangle = \frac{1}{2}\omega\operatorname{Im}\chi(\omega)f_0^2 \end{align*} They agree when $n_\text{B}(\omega)\to 0$: energy of transition being much larger than thermal energy $\hbar\omega\gg k_B T$.

Linear Response: Formal Theory

• So far, considered only response of linear systems!

• How do we talk about linear response in general?

Kubo Formula

• How does observable $A$ depend on $\lambda_t$, which appears (for small variations) linearly in the Hamiltonian

$$H_t = H_0 - \lambda_t B?$$

• $B=-\frac{\partial H}{\partial \lambda}$ is generalized force; $\lambda$ generalized displacement
• In the interaction picture $\ket{\Psi_I(t)} \equiv e^{iH_0 t}\ket{\Psi(t)}$ $$i\frac{\partial \ket{\Psi_I(t)}}{\partial t} = -\lambda_t B_I(t) \ket{\Psi_I(t)},$$ where $B_I(t) = e^{iH_0 t}B e^{-iH_0 t}$

• Result of first order time dependent perturbation theory is $\ket{\Psi_I(t)}=\ket{\Psi(0)}+\ket{\Psi^{(1)}_I(t)}+\cdots$, with

$$\ket{\Psi^{(1)}_I(t)} = i\int_0^t dt' \lambda_{t'} B_I(t') \ket{\Psi(0)}$$ \begin{align*} \braket{\Psi(t)|A|\Psi(t)} &= \braket{\Psi_I(t)|A_I(t) |\Psi_I(t)} \nonumber\\ &=\braket{\Psi(0)|A_I(t)|\Psi(0)} +i \int_0^t dt' \lambda_{t'}\braket{\Psi(0)| \left[A_I(t),B_I(t')\right] |\Psi(0)}. \end{align*}

• Mixed states can be treated by averaging over a distribution of quantum states $\langle\cdots \rangle \longrightarrow \langle\langle\cdots\rangle\rangle$

• $\chi_{AB}(t)$ of $A$ due to perturbation that couples to $B$ $$\chi_{AB}(t) = i\langle\langle\left[A_I(t),B_I(0)\right]\rangle\rangle,\quad t>0.$$ (Normally write $A(t)$ rather than $A_I(t)$: Heisenberg picture for the unperturbed problem)

• This is the Kubo formula. It expresses the response of a system in terms of the dynamics of the unperturbed system.

Fluctuation Dissipation Theorem (general)

• Start from the correlation function

$$S_{AB}(t) \equiv \langle\langle A_I(t)B_I(0)\rangle\rangle$$

• Recalling that $$\langle\langle\cdots \rangle\rangle = \frac{1}{Z}\tr\left[e^{-\beta H}\cdots\right]$$ you should be able to show that $$S_{AB}(t) = S_{BA}(-t-i\beta).$$

Hint: use the cyclic property of the trace.

$$S_{AB}(t) = S_{BA}(-t-i\beta).$$

• Fourier transforming, we arrive at $$S_{AB}(\omega) = e^{\beta\omega} S_{BA}(-\omega).$$ (we’ve seen this before!)

• $\chi_{AB}(t)$ (a commutator) can be written in terms of $S_{AB}(\omega)$

\begin{align*} \chi_{AB}(t) &= \begin{cases} i\left[S_{AB}(t)-S_{BA}(-t)\right] & t>0\\ 0 & t<0. \end{cases}\\ &= i\theta(t)\left[S_{AB}(t)-S_{BA}(-t)\right] \end{align*}

$$\chi_{AB}(t)=i\theta(t)\left[S_{AB}(t)-S_{BA}(-t)\right]$$

• $\chi_{AB}(\omega)$ can then be expressed as the convolution. The Fourier transform of the step function is

$$\tilde\theta(\omega) = \frac{i}{\omega+i0},$$ \begin{align*} \chi_{AB}(\omega)&= -\int \frac{d\omega'}{2\pi}\frac{S_{AB}(\omega')-S_{BA}(-\omega')}{\omega-\omega'+i0}\\ &=-\int \frac{d\omega'}{2\pi}\frac{S_{AB}(\omega')\left[1-e^{-\beta\omega'}\right]}{\omega-\omega'+i0} \end{align*}

• To make sense of this formula, use the Kramers–Kronig relation \begin{align*} \chi_{AB}(\omega) &=\operatorname{Re}\chi_{AB}(\omega) + i\operatorname{Im}\chi_{AB}(\omega)\\ &= \cp \int_{-\infty}^\infty \frac{d\omega'}{\pi}\frac{\operatorname{Im}\chi_{AB}(\omega')}{\omega'-\omega} + i\operatorname{Im}\chi_{AB}(\omega)\\ &= \int_{-\infty}^\infty \frac{d\omega'}{\pi}\frac{\operatorname{Im}\chi_{AB}(\omega')}{\omega'-\omega-i0}, \end{align*} where we used $$\frac{1}{x+i0} = \mathcal{P}\frac{1}{x} -i\pi\delta(x).$$
• Finally! $$S_{AB}(\omega) = 2\operatorname{Im}\chi_{AB}(\omega)\left[n_\text{B}(\omega)+1\right]$$

Spectral Representation

The quantites $\chi_{AB}(\omega)$ and $S_{AB}(\omega)$ have spectral representations in terms of energy eigenstates and eigenvalues

$$S_{AB}(\omega) = 2\pi\sum_{m,n} \frac{e^{-\beta E_m}}{Z} \bra{m}A\ket{n}\bra{n}B\ket{m} \delta(\omega-E_n+E_m).$$

Use the spectral representation to prove the fluctuation dissipation relation

• $S_{AA}(\omega)$ can be interpreted in terms of Fermi golden rule, as we saw for oscillator. Notice that $S_{AA}(\omega)>0$.

Response of Matter

• Back to many body physics!

Density Response

• Suppose system is subject to time dependent potential corresponding to a term in Hamiltonian

$$H_\text{pert} = \sum_{j=1}^N V(\br_i,t) = \int V(\br,t)\rho(\br), d\br = \frac{1}{L^3}\sum_\bq V_\bq(t) \rho_{-\bq}$$

• Perturbation couples to the density: how is density affected? In a translationally invariant system $$\langle\langle \rho_\bq(t)\rangle\rangle = -\frac{1}{L^3} \int_{-\infty}^t \chi^{\rho}_\bq(t-t') V_\bq(t)\,dt',$$ where the density response function $\chi^\rho_\bq(t)$ is $$\chi_\rho(\bq,t) = i\langle\langle\left[\rho_\bq(t),\rho_{-\bq}(0)\right]\rangle\rangle$$
• General theory applies with $A=\rho_\bq$ and $B=\rho_{-\bq}$. At $T=0$ $$S_\rho(\bq,\omega) = 2\pi\sum_{n} |\bra{0}\rho_\bq\ket{n}|^2 \delta(\omega-E_n+E_0)$$

• This is the dynamical structure factor, on account of its importance in scattering experiments.

• The static structure factor is $$S_\rho(\bq) = \int S_\rho(\bq,\omega) \frac{d\omega}{2\pi} = \langle\langle\rho_\bq\rho_{-\bq}\rangle\rangle$$

Sum rules

• $S_\rho(\bq,\omega)$ obeys certain general relations irrespective of the particular model under consideration

• If interaction depends only on density, it commutes with $\rho_\bq$, so the commutator is determined by kinetic energy

$$T = -\frac{1}{2m}\sum_{j=1}^N \nabla_i^2$$

• Taking $\rho_\bq =\sum_{j=1}^N e^{-i\bq\cdot\br_j}$, we find

$$[[H,\rho_\bq],\rho_{-\bq}] = -\frac{N\bq^2}{m}$$

• Evaluate by introducing resolution of the identity $\sum_n \ket{n}\bra{n}=1$

\begin{align*} \braket{0}{[[H,\rho_\bq],\rho_{-\bq}]}{0}&=\braket{0|H\rho_\bq \rho_{-\bq}- \rho_\bq H\rho_{-\bq}-\rho_{-\bq} H\rho_\bq+\rho_{-\bq}\rho_\bq H|0}\\ &=2\sum_n|\bra{0}\rho_\bq\ket{n}|^2\left(E_0-E_n\right) \end{align*}

• We can relate this to $S(\bq,\omega)$ using spectral representation $$S_\rho(\bq,\omega) = 2\pi\sum_{n} |\bra{0}\rho_\bq\ket{n}|^2 \delta(\omega-E_n+E_0),$$ to give the f-sum rule

$$\int_{-\infty}^\infty \omega S(\bq,\omega) \frac{d\omega}{2\pi}= \frac{N\bq^2}{2m}$$

Compressibility sum rule

• Compressibility $$\beta=-\frac{1}{V}\frac{\partial V}{\partial p}.$$ (inverse of Bulk modulus)

• $p = -\frac{\partial E_0}{\partial V}$ at $T=0$. If $E_0 = V \epsilon(\rho)$, then

$$\beta^{-1} = \rho^2 \epsilon''(\rho).$$

$$\beta^{-1} = \rho^2 \epsilon''(\rho).$$

• In the presence of a potential $V(\br)$, energy density is

$$\epsilon(\rho_0+\delta\rho) = \frac{1}{2\beta\rho_0^2} \left[\delta\rho\right]^2 + V(\br)\delta\rho$$

• Minimizing with respect to $\delta\rho$ gives

$$\epsilon(V(\br)) = - \frac{\beta\rho_0^2}{2} \left[V(\br)\right]^2$$

• Compare with perturbation theory
• Change in energy due to perturbation

$$\sum_j V_0 \cos(\bq\cdot \br_j) = \frac{V_0}{2}\left[\rho_\bq+\rho_{-\bq}\right]$$

• Second order perturbation theory for ground state $$E^{(2)} =\frac{V_0^2}{4} \sum_{n\neq 0} \frac{|\braket{0|\rho_\bq|n}|^2}{E_0-E_n} =-\frac{V_0^2}{4}\int_0^\infty \frac{S(\bq,\omega)}{\omega}\frac{d\omega}{2\pi}$$ (first order vanishes)
• Compare with

$$\epsilon(V(\br)) = - \frac{\beta\rho_0^2}{2} \left[V(\br)\right]^2$$

• This gives the compressibility sum rule at zero temperature

$$\lim_{\bq\to 0}\int_0^\infty \frac{S(\bq,\omega)}{\omega}\frac{d\omega}{2\pi} = \frac{N\rho\beta}{2}.$$

• Compressibility sum rule often written in terms of speed of sound $c = (\beta m \rho)^{-1/2}$

$$\lim_{\bq\to 0}\int_0^\infty \frac{S(\bq,\omega)}{\omega}\frac{d\omega}{2\pi} = \frac{N}{2mc^2}.$$

Single Mode Approximation

• Some systems (e.g Bose gases), are well described by $$S_\rho(\bq,\omega) \sim 2\pi S_\rho(\bq) \delta(\omega - \omega(\bq)),$$ for low $\bq$, where $\omega(\bq)$ is dispersion relation of collective excitations (e.g. Bogoliubov modes)

• In this approximation, f-sum rule tells us that

$$S_\rho(\bq) = \frac{N\bq^2}{2m\omega(\bq)}.$$

Example 1: BEC (no interactions)

• Excitations out of the condensate are free particles $\omega(\bq) = \frac{\bq^2}{2m}$ $$S_\rho(\bq) = N.$$

• Completely uncorrelated particle positions (Poisson statistics).

Example 2: interacting BEC / elastic chain

• $\omega(\bq) = c|\bq|$ i.e. linear dispersion with finite speed of sound

$$S_\rho(\bq) = \frac{N|\bq|}{2mc}.$$

• Density fluctuations vanish as wavevector goes to zero, indicating long-range correlations between positions in the ground state.

Check compressibility sum rule