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
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
$$ S(\omega) = \int_{-\infty}^\infty \langle\langle y(t)y(0)\rangle\rangle e^{i\omega t}\,dt. $$
$$ S(\omega)\neq S(-\omega) $$
Many of the properties of $S(\omega)$ are most easily understood from spectral representation
Insert complete set of energy eigenstates between $y(0)$ and $y(t)$
$$ S(\omega) = 2\pi\sum_{m,n} \frac{e^{-\beta E_n}}{Z} |\bra{n}y\ket{m}|^2 \delta(\omega-E_m+E_n) $$
$$ S(\omega) = S(-\omega) e^{\beta\omega}. $$
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
$$ \begin{align*} S(\omega)&=\frac{\pi}{m\omega_0} \sum_n \frac{e^{-\beta E_n}}{Z} \left[n\delta(\omega+\omega_0)+(n+1)\delta(\omega-\omega_0)\right]\nonumber\\ & = \frac{\pi}{m\omega_0}\left[n_\text{B}(\omega_0)\delta(\omega+\omega_0)+(n_\text{B}(\omega_0)+1)\delta(\omega-\omega_0)\right] \end{align*} $$
$$ \begin{align*} S(\omega)&= \frac{\pi}{m\omega_0}\left[n_\text{B}(\omega_0)\delta(\omega+\omega_0)+(n_\text{B}(\omega_0)+1)\delta(\omega-\omega_0)\right]\\\\ &\to \frac{k_\text{B}T}{2m\omega_0^2} \times 2\pi\left[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)\right] \end{align*} $$
$$ \adop_k(t) = e^{i\omega_k t}\adop_k,\quad \aop_k(t) = e^{-i\omega_k t}\aop_k $$
$$ \begin{align*} S(\omega)= 2\pi\sum_k |c_k|^2\left[n_\text{B}(\omega_k)\delta(\omega+\omega_k)+(n_\text{B}(\omega_k)+1)\delta(\omega-\omega_k)\right] \end{align*} $$
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)
$$ y(\omega) = \sum_k |c_k|^2\left[\frac{1}{\omega_k-\omega-i0}+\frac{1}{\omega_k+\omega+i0}\right]f(\omega). $$
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…
$$ \text{Im} \frac{1}{x\mp i0} = \pm\pi\delta(x), $$
$$ \operatorname{Im}\chi(\omega) =\sgn(\omega)\pi\sum_k |c_k|^2\delta(\omega_k-\omega) $$
$$ S(\omega) = 2\operatorname{Im}\chi(\omega)\left[n_\text{B}(\omega)+1\right] $$
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
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 $f(t)=f_0\cos\omega t$
$$ \Gamma_{n\to m} = 2\pi \left(\frac{f_0}{2}\right)^2|\bra{n}y\ket{m}|^2 \delta(\pm\omega+E_m-E_n). $$
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 $\hbar\omega\gg k_B T$.
So far, considered only response of linear systems!
How do we talk about linear response in general?
$$ H_t = H_0 - \lambda_t B? $$
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.
$$ S_{AB}(t) \equiv \langle\langle A_I(t)B_I(0)\rangle\rangle $$
$$ \langle\langle\cdots \rangle\rangle = \frac{1}{Z}\tr\left[e^{-\beta H}\cdots\right] $$
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 $S_{AB}(\omega)$
$$ \begin{align*} \chi_{AB}(t) &= \begin{cases} i\left[S_{AB}(t)-S_{BA}(-t)\right] & t>0\\ 0 & t<0. \end{cases}\\ &= i\theta(t)\left[S_{AB}(t)-S_{BA}(-t)\right] \end{align*} $$
$$ \chi_{AB}(t)=i\theta(t)\left[S_{AB}(t)-S_{BA}(-t)\right] $$
$$
\tilde\theta(\omega) = \frac{i}{\omega+i0},
$$
$$ \begin{align*} \chi_{AB}(\omega)&= -\int \frac{d\omega'}{2\pi}\frac{S_{AB}(\omega')-S_{BA}(-\omega')}{\omega-\omega'+i0}\\ &=-\int \frac{d\omega'}{2\pi}\frac{S_{AB}(\omega')\left[1-e^{-\beta\omega'}\right]}{\omega-\omega'+i0} \end{align*} $$
$$ \begin{align*} \chi_{AB}(\omega) &=\operatorname{Re}\chi_{AB}(\omega) + i\operatorname{Im}\chi_{AB}(\omega)\\ &= \cp \int_{-\infty}^\infty \frac{d\omega'}{\pi}\frac{\operatorname{Im}\chi_{AB}(\omega')}{\omega'-\omega} + i\operatorname{Im}\chi_{AB}(\omega)\\ &= \int_{-\infty}^\infty \frac{d\omega'}{\pi}\frac{\operatorname{Im}\chi_{AB}(\omega')}{\omega'-\omega-i0}, \end{align*} $$
where we used
$$
\frac{1}{x+i0} = \mathcal{P}\frac{1}{x} -i\pi\delta(x).
$$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
$$ H_\text{pert} = \sum_{j=1}^N V(\br_i,t) = \int V(\br,t)\rho(\br), d\br = \frac{1}{L^3}\sum_\bq V_\bq(t) \rho_{-\bq} $$
$$ \langle\langle \rho_\bq(t)\rangle\rangle = -\frac{1}{L^3} \int_{-\infty}^t \chi^{\rho}_\bq(t-t') V_\bq(t)\,dt', $$
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 $$ S_\rho(\bq) = \int S_\rho(\bq,\omega) \frac{d\omega}{2\pi} = \langle\langle\rho_\bq\rho_{-\bq}\rangle\rangle $$
$S_\rho(\bq,\omega)$ 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
$$ T = -\frac{1}{2m}\sum_{j=1}^N \nabla_i^2 $$
$\rho_\bq =\sum_{j=1}^N e^{-i\bq\cdot\br_j}$, we find$$ [[H,\rho_\bq],\rho_{-\bq}] = -\frac{N\bq^2}{m} $$
$$ \begin{align*} \braket{0|[[H,\rho_\bq],\rho_{-\bq}]|0}&=\braket{0|H\rho_\bq \rho_{-\bq}- \rho_\bq H\rho_{-\bq}-\rho_{-\bq} H\rho_\bq+\rho_{-\bq}\rho_\bq H|0}\\ &=2\sum_n|\bra{0}\rho_\bq\ket{n}|^2\left(E_0-E_n\right) \end{align*} $$
$S(\bq,\omega)$ 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 $$ \beta=-\frac{1}{V}\frac{\partial V}{\partial p}. $$ (inverse of Bulk modulus)
$p = -\frac{\partial E_0}{\partial V}$ at $T=0$. If $E_0 = V \epsilon(\rho)$, then
$$ \beta^{-1} = \rho^2 \epsilon''(\rho). $$
$$ \beta^{-1} = \rho^2 \epsilon''(\rho). $$
$$ \epsilon(\rho_0+\delta\rho) = \frac{1}{2\beta\rho_0^2} \left[\delta\rho\right]^2 + V(\br)\delta\rho $$
$$ \epsilon(V(\br)) = - \frac{\beta\rho_0^2}{2} \left[V(\br)\right]^2 $$
$$ \sum_j V_0 \cos(\bq\cdot \br_j) = \frac{V_0}{2}\left[\rho_\bq+\rho_{-\bq}\right] $$
$$ \epsilon(V(\br)) = - \frac{\beta\rho_0^2}{2} \left[V(\br)\right]^2 $$
$$ \lim_{\bq\to 0}\int_0^\infty \frac{S(\bq,\omega)}{\omega}\frac{d\omega}{2\pi} = \frac{N\rho\beta}{2}. $$
$$ \lim_{\bq\to 0}\int_0^\infty \frac{S(\bq,\omega)}{\omega}\frac{d\omega}{2\pi} = \frac{N}{2mc^2}. $$
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
$$ S_\rho(\bq) = \frac{N\bq^2}{2m\omega(\bq)}. $$
Excitations out of the condensate are free particles $\omega(\bq) = \frac{\bq^2}{2m}$ $$ S_\rho(\bq) = N. $$
Completely uncorrelated particle positions (Poisson statistics).
$$ S_\rho(\bq) = \frac{N|\bq|}{2mc}. $$
Check compressibility sum rule