Suceptibility
Given a free energy functional, \(\mathcal{F} = \mathcal{F}_0 + \mathcal{F}_f\) compare to Thermodynamics: \(\tilde{\mathcal{F}} = \mathcal{F} - fx\) with a generalized force and displacement \(f,x\). So, \(\mathcal{F}_f = \int d\vec{x}f(\vec{x},t)s(\vec{x},t)\).
Examples:
| \(f\) | \(S\) |
| \(H\) | \(M\) |
| \(E\) | \(P\) |
For \(t'
Fourier transforming this \(t\to\omega,x\to k\).
\(\tilde{s}(\vec{k},\omega) = \tilde{\chi}(\vec{k},\omega)\tilde{f}(\vec{k},\omega)\).
\(\chi(\vec{k},\omega) = \int d\vec{x}\int dt\exp(i\omega t)\exp(-i\vec{k}\cdot\vec{x})\chi(\vec{x},t)\) note the signs only are due to us wanting the plane wave to look like \(\exp i(\omega t-kx)\).
Then we get, \(\tilde{\chi}(\vec{k},\omega) = \chi'(\vec{k},\omega) + i\chi''(\vec{k},\omega)\).
So, \(\tilde{\chi}^* = \tilde{\chi}(-\vec{k},-\omega)\).
A system with inversion symmetry, \(\vec{x}=-\vec{x}\), \(\tilde{\chi}^*(\vec{k},\omega) = \tilde{\chi}(\vec{k},-\omega)\).
\(\chi'\) is even in \(\omega\) and \(\chi''\) is odd in \(\omega\).
Then, \(\tilde{\chi}(\vec{k},\omega) = \int d\vec{x}\int dt\exp(i\omega t)\exp(-i\vec{k}\cdot\vec{x})\chi(\vec{x},t)\).
C.f. the AC suceptibility \(\tilde{P}(\vec{k},\omega) = \tilde{\chi}(\vec{k},\omega)\tilde{E}(\vec{k},\omega)\) and \(\tilde{M}(\vec{k},\omega) = \tilde{\chi}(\vec{k},\omega)\tilde{H}(\vec{k},\omega)\).
Obtaining the work, \(W = \int \vec{F}\cdot d\vec{S} = \int \vec{F}\frac{\partial\vec{s}}{\partial t}dt = \int\vec{F}\cdot\vec{v}dt\).
Power: \(P = \lim_{T\to\infty}\frac{1}{T}\int_0^Tf(t)\frac{\partial s}{\partial t}dt = \lim_{t\to\infty}\frac{1}{T}\left\{\int_0^T\left(-s(t)\frac{\partial f}{\partial t}dt\right) - s(t)f(t)|_0^T\right\}\), with our generalized force and displacement.
The boundary terms go to zero, \(f(t) = \Re\{f\omega\exp(i\omega t)\} = \frac{1}{2}\left(f_\omega\exp(-i\omega t) + f^*_\omega\exp(i\omega t)\right) = \Re f_\omega\cos\omega t + \Im f_\omega\sin\omega t\).
\(f(t) = \cos\omega_0 t, f(\omega) = \pi(\delta(\omega-\omega_0) + \delta(\omega+\omega_0))\Rightarrow f(t) = \frac{1}{2\pi}\int f(\omega)\exp(-i\omega t)d\omega\).
Then, \(s(\omega) = \tilde{\chi}(\omega)f(\omega)\).
\(s(t) = \frac{1}{2\pi}\int s(\omega)\exp(-i\omega t)d\omega = \frac{1}{2}\left(\exp(i-\omega_0 t)\chi(\omega_0) + \exp(i\omega_0 t)\chi(-\omega_0)\right) = \frac{1}{2}\left(\exp(i\omega_0 t)(\chi'(\omega_0) + i\chi''(\omega_0)) + \exp(i\omega_0t)(\chi'(\omega_0)-i\chi''(\omega_0))\right) = \chi'(\omega_0)\cos(\omega_0 t) + \chi''(\omega_0)\sin(\omega_0 t)\).
Then the power becomes, \(P = \lim_{T\to\infty}\frac{1}{T}\int_0^Tdt s(t) \frac{1}{2}(i\omega) (f_\omega\exp(-i\omega t) - f_\omega^*\exp(i\omega t))dt\).
\(s(t) = \int_{-\infty}^t dt'\chi(t-t') f(t')\).
Since there is no response from the future, \(\chi(t-t')= 0\) for \(t
Then, \(\omega\chi''(\omega)> 0\) which means each one must be greater than zero.
Know that \(\chi''(\omega)\) is odd.
Note: \(\chi(k,\omega) = \chi(\omega)\) shorthand was used.
Example: Calculate the current through a wire.
Suppose we have \(V(t) = V_\omega\cos(\omega t)\).
\(\langle p\rangle = \frac{\langle V^2\rangle}{R} = \frac{V_\omega^2\langle\cos^2\omega t\rangle}{R} = \frac{V_\omega^2}{2R}\).
Suppose we have a wire of length \(L\) and cross-section \(A\), conductivity \(\sigma = \frac{1}{\rho}\).
\(R = \frac{\rho L}{A} = \frac{L}{\sigma A}\)
The force (electric field) due to the voltage is then \(E(t) = E_\omega\cos\omega t = \frac{V_\omega}{L}\cos\omega t\).
The power density, power per unit volume, \(p = \frac{\langle p\rangle}{V} = \frac{V_\omega^2}{2R(L\cdot A)} = \frac{V_\omega^2}{2\rho L^2} = \frac{V_\omega^2\sigma}{2L^2} = \frac{1}{2}\sigma E_\omega^2 = \frac{1}{2}\omega |f_\omega|^2\chi''(\omega)\).
Thus, the suceptibity is \(chi''(\omega) = \frac{\sigma}{\omega}\) and \(|f_\omega|^2 = E_\omega^2\).
\(\alpha''(\omega)\), the polarizability is, \(\alpha'' = \frac{\sigma}{\omega}\).
\(p = \lim_{\omega\to0} p(\omega) \Rightarrow \sigma = \lim_{\omega\to 0}\omega\alpha''(\omega)\) gives the DC conductivity on the left and the AC polarizability on the right.
Suppose we have an order parameter \(s_{macro}(\vec{x},t) = \overline{s} - \int d\vec{x}'G(\vec{x}-\vec{x}',t) (s_{macro}(\vec{x},t) - \overline{s})\).
Transforming this, \(\tilde{s}_{macro}(\vec{k}',\omega)
= \overline{\tilde{s}} - \int d\vec{k}'\tilde{G}(\vec{k}-\vec{k},\omega)(\tilde{s}_{macro}(\vec{k}',\omega)-\overline{\tilde{s}})\)
Compare if we have a Schrodinger equation and a propogator: \(H\varphi = -\frac{i}{\hbar}\frac{\partial}{\partial t}\varphi\Rightarrow \varphi(\vec{x},t) = \int dt'\int d\vec{x}' U(\vec{x}-\vec{x}',t-t')\varphi(\vec{x}-\vec{x}')\).
Removing the time dependence.
\(\chi_0(\vec{r})\).
The response to this continuous force (a lean rather than a kick) is then \(s(\vec{x}) = \int d\vec{x}'\chi_0(\vec{x}-\vec{x}')f(\vec{x}')\).
We will get: \(\tilde{\chi}(\vec{k}) = \frac{\partial\langle s_k\rangle_{eq}}{\partial f_k}|_{f=0}\).
\(F_f(t) = -\int d\vec{x}f(\vec{x},t)s(\vec{x},t)\).
For a static system, \(F_f = -\vec{x}f(\vec{x})s(\vec{x})\).
Then, \(\langle \tilde{s}_k\rangle_{eq} = \frac{\text{Tr}(\tilde{s}_k\exp(-\beta(F_0 - F_f)))}{\text{Tr}(\exp(-\beta(F_0 - F_f)))} = \frac{1}{\beta V}\frac{\partial \ln Z}{\partial f_k}\).
\(\tilde{\chi}_0(\vec{k}) = \frac{\partial\langle s_k\rangle_{eq}}{\partial f_k}|_{f=0} = \beta\hat{C}(\vec{k},0)\).
The hat here means just a fourier transform for space.
For the uniform case, thermodynamically, \(\chi = \langle(s-\langle s\rangle)^2\rangle\).
\(p(\omega) = \frac{\omega|f_\omega|^2}{2}\chi''(\omega)\).
Want \(\chi\leftrightarrow C\), connect fluctuations.
Cases: \(f(\vec{x},t) \to \chi(\vec{x},t)\) and \(f(\vec{x})\to\chi(\vec{x})\).
Starting from forced equilibrium at \(t=0\) due to \(f(\vec{x})\).
Now we turn off the force.
So, \(f(\vec{x},t>0) = 0\) so what is our \(\chi\)?
From our response, \(s(\vec{x},t-t_0) = \int d\vec{x} G(\vec{x}-\vec{x}',t-t_0)s(\vec{x}',t_0)\).
For \(t_0 = 0\), \(s(\vec{x},t) = \int d\vec{x} G(\vec{x}-\vec{x}',t)s(\vec{x}',0)\).
Then, \(s(x,t) = \int d\vec{x}\int_{-\infty}^t dt'\chi(\vec{x}-\vec{x}',t-t')f(\vec{x},t')\).
We choose, \(f(\vec{x},t) = f(\vec{x})\) for \(t<0\) and 0 otherwise.
Then, \(s(x,t\geq 0) = \int d\vec{x}\int_{-\infty}^0 dt'\chi(\vec{x}-\vec{x}',t-t')f(\vec{x})\).
Let \(\tau=t-t'\), \(dt = -d\tau\), \(t'\in\{-\infty,0\}\) so \(\tau\in\{+\infty, t\}\).
Then, \(s(\vec{x},t) = \int dx f(\vec{x})\int_t^\infty d\tau\chi(\vec{x}-\vec{x}',\tau)\) [2b].
For the ideal gas, \(\chi_0(\vec{r}) = \frac{\delta(\vec{r})}{\alpha}\).
Then, inserting this, \(s(x,0) = \rho(\vec{x},0) = \frac{f(\vec{x})}{\alpha}\).
Which gives, \(\rho(\vec{x},t) = \int d\vec{x}'\frac{f(\vec{x}')}{\alpha}G(\vec{x}-\vec{x}',t)\).
\(C^{id} (\vec{x},t) = \frac{G(\vec{x},t)}{\beta\alpha}\).
Then, \(\rho(\vec{x},t) = \int d\vec{x}' \frac{f(\vec{x}')}{\alpha}\beta\alpha C^{id}(\vec{x}-\vec{x}',t) = \int d\vec{x}' f(\vec{x}')\beta C^{id}(\vec{x}-\vec{x}',t)\).
Comparing this to the last statement in the previous paragraph [2b], \(\beta C^{id}(\vec{x}-\vec{x}',t)\int_t^\infty d\tau\chi(\vec{x}-\vec{x}',\tau)\).
Then, \(\beta\frac{\partial C^{id}}{\partial t} = \chi(\vec{x},\infty) - \chi(\vec{x},t) = -\chi(\vec{x},t)\).
Which gives us the FD theorem in the time domain, for \(t>0\),
And is zero for \(t<0\).
Note that \(C(\vec{x},t) = C(\vec{x},-t)\).
Now we can calculate \(\chi''(\omega)\).
\(\overline{\chi}(\vec{r},\omega) = \tilde{\chi}(\omega) = \int_{\pm\infty}dt\chi(t)\exp(i\omega t)\) which is zero for \(t<0\).
So, \(\tilde{\chi}(\omega) = \int_0^\infty dt\chi(t)\exp(i\omega t) = -\beta\int_0^\infty \frac{\partial C(t)}{\partial t}\exp(i\omega t)\).
Integrating by parts,
\(= -\beta\left(\left.C(t)\exp(i\omega t)\right|_0^\infty - \int_0^\infty dt C(t)(i\omega)\exp(i\omega t)\right) = \tilde{\chi}' + i\tilde{\chi}''\).
The first term is real, thus, \(\tilde{\chi}'' = \Im \beta\int_0^\infty dtC(t)(i\omega)\exp(i\omega t) = \beta\omega\int_0^\infty dtC(t)\cos(\omega t)\).
Since \(C(t) = C(-t)\) and \(\cos(t) = \cos(-t)\), thus the integrand is even.
Thus, \(\tilde{\chi}'' = \frac{\beta\omega}{2}\int_{-\infty}^\infty dt C(t)\cos\omega t = \frac{\beta\omega}{2}\int_{-\infty}^\infty dt C(t)\cos\omega t + i\frac{\beta\omega}{2}\int_{-\infty}^\infty dt C(t)\sin\omega t = \frac{\beta\omega}{2}\int_{-\infty}^\infty dt C(t)\exp(i\omega t) = \frac{\omega\beta}{2}\tilde{C}(\omega)\).
Then we get the FD theorem in frequency domain for classical systems,
Then, \(p(\omega) = \frac{\beta\omega^2|f_\omega|^2}{4}\tilde{C}(\omega)\).
So in the continuum limit for matter, we get density fluctuations.
For solids (crystals), this would be lattice vibrations.
The QM version,
Starting point is QM version of linear response, Kudo formula,
\(\langle \hat{A}(t)\rangle = \langle \hat{A}\rangle_0 - i\int_{t_0}^t dt'\langle[\hat{A}(t),\hat{V}(t)]\rangle\), \(\langle\cdot\rangle_0\) with respect to equilibrium and \(\langle\hat{A}\rangle = \frac{\text{Tr}[\hat{\rho}\hat{A}]}{\text{Tr}[\hat{\rho}]}\).
We have, and abbreviate, \(\chi(\vec{r},t) \equiv \chi(t)\), real and \(t>0\).
Then, \(\vec{\chi}(\vec{r},\omega) = \chi'(\omega) + i\chi''(\omega)\).
Then, \(\tilde{\chi}(-\omega) = -\tilde{\chi}''(\omega)\).
2 real valued functions on half line, \(\omega > 0\).
Let \(\chi(\omega) \to \chi(u+iv) = \int_0^\infty dt\chi(t)\exp(iut)\exp(-vt)\).
The zero coming from the causality condition.
From Cauchy’s theorem for a function analytic in a region and a contour in the region, \(\oint_C f(z')dz = 0\).
Cauchy’s integral formula for a simple pole is \(\oint_c\frac{f(z')}{z'-z}dz' = 2\pi i f(z)\).
Then, \(\oint_c \frac{\chi(\omega')}{\omega'-\omega}d\omega' = 0\) by making a loop that goes near \(\omega\) but cuts it out with a small half circle.
Taking the off-axis loop to infinity, we get zero.
The integral on the axis, \(\lim\limits_{\varepsilon\to0} = \left(\int_{-\infty}^{\omega-\varepsilon} + \int_{\omega+\varepsilon}^\infty\right)\frac{\tilde{\chi}(\omega')d\omega'}{\omega' - \omega} = PV\int \frac{\chi'(\omega')}{\omega' - \omega}d\omega\)
The semicircle integral, \(\int\frac{\tilde{\chi}(\omega')}{\omega'-\omega}d\omega' = \int\frac{\tilde{\chi}(\omega)}{\omega'-\omega}d\omega' = \chi(\omega)\ln(\omega-\omega)|_{\omega+\epsilon\exp(i\pi)}^{\omega+\epsilon\exp(i0)} = -i\pi\tilde{\chi}(\omega)\).
Then, the principle value of the integral is \(i\pi\tilde{\chi}(\omega)\).
Hence, \(\tilde{\chi}(\omega) = \frac{1}{\pi i}\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega)}{\omega'-\omega}d\omega'\).
So, \(\tilde{\chi}'(\omega) = \Re\{\tilde{\chi}(\omega)\} = \frac{1}{\pi}\int_{-\infty}^\infty \frac{\chi''(\omega')}{\omega'-\omega}d\omega'\).
Since \(\tilde{\chi}(\omega) = -\tilde{\chi}^*(-\omega)\), \(\tilde{\chi}''(\omega) = -\tilde{\chi}''^*(-\omega)\).
So, \(\tilde{\chi}'(\omega) = \frac{1}{\pi}\int_0^\infty\chi''(\omega) \left(\frac{1}{\omega'-\omega} - \frac{1}{-\omega'-\omega}\right) = \frac{2}{\pi}\int_0^\infty\frac{\chi''(\omega)\omega'}{\omega'^2-\omega^2}d\omega'\).
Alternatively, \(\tilde{\chi}''(\omega) = \frac{-2\omega}{\pi}\int_0^\infty \frac{\chi'(\omega)}{\omega'^2-\omega^2}d\omega'\).
Static Susceptibility
Fluctuation-Dissipation Theorem
Kramers-Kronig Relations