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 \begin{align} p(\omega) &= \int_{-\infty}^\infty d\tau \frac{i\omega\tau}{4}\lim_{T\to\infty}\frac{1}{T}\int_0^Tdt\left(f_\omega^2\exp(-i\omega(2t-\tau)) - f_\omega^{*2}\exp(i\omega(2t-\tau)) + |f_\omega^*|^2\left(\exp(-i\omega\tau) - \exp(i\omega\tau)\right)\right) \\ &= \int_{-\infty}^\infty d\tau \frac{i\omega\tau}{4}\lim_{T\to\infty}\frac{1}{T}(-1)\int_0^Tdt\left(|f_\omega^*|^2\left(\exp(i\omega\tau) - \exp(-i\omega\tau)\right)\right) \\ &= \frac{\omega}{2}|f_\omega|^2\int d\tau \chi(\tau)\sin(\omega\tau) \\ &= \frac{\omega}{2}|f_\omega|^2\Im\{\mathcal{F}(\chi(\tau))\} \\ &= \frac{\omega}{2}|f_\omega|^2\Im\{\tilde{\chi}(\tau)\} \\ &= \frac{\omega}{2}|f_\omega|^2\tilde''{\chi}(\tau) > 0. \end{align} 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: $$H\varphi = -\frac{i}{\hbar}\frac{\partial}{\partial t}\varphi\Rightarrow \varphi(\vec{x},t) = \int dt'\int d\vec{x}' U(\vec{x}-\vec{x}',t-t')\varphi(\vec{x}-\vec{x}')$$.

Static Susceptibility

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, $$\chi = \langle(s-\langle s\rangle)^2\rangle$$.

Fluctuation-Dissipation Theorem

$$p(\omega) = \frac{\omega|f_\omega|^2}{2}\chi''(\omega)$$. 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 $$t>0$$,

\begin{align} \chi(\vec{x},t) = -\beta\frac{\partial C^{id}(\vec{x},t)}{\partial t} \end{align}

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,

\begin{align} \tilde{\chi}''(\omega) = \frac{\omega\beta}{2}\tilde{C}(\omega) \end{align}

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,

\begin{align} \tilde{\chi}(\vec{k},\omega) = \frac{1}{2\hbar}(1-\exp(-\beta\hbar\omega))\tilde{C}(\vec{k},\omega). \end{align}

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 $$\langle\hat{A}\rangle = \frac{\text{Tr}[\hat{\rho}\hat{A}]}{\text{Tr}[\hat{\rho}]}$$.

Kramers-Kronig Relations

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, $$\tilde{\chi}''(\omega) = \frac{-2\omega}{\pi}\int_0^\infty \frac{\chi'(\omega)}{\omega'^2-\omega^2}d\omega'$$.

Created: 2023-06-25 Sun 02:31

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