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