Linear response from generalized susceptibility

The static linear response can be obtained from the frequency dependent generalized susceptibility \(\chi^{(PH)}_{\bar{a}b\bar{c}d}\) or the one-time dependent bosonic propagator \(\langle (\bar{a}b)(\tau) (\bar{c}d) \rangle\) by summing/integrating over freqiency/time

(1)\[\chi^{(PH)}_{\bar{a}b\bar{c}d} = \frac{1}{\beta} \int_0^\beta d\tau \, \langle (\bar{a}b)(\tau) (\bar{c}d) \rangle - \langle \bar{a} b \rangle \langle \bar{c} d \rangle = \frac{1}{\beta^2} \sum_{nm} \chi^{(PH)}_{\bar{a}b\bar{c}d}(\omega_0, \nu_n, \nu_m)\]

Linear response from applied external field

The linear response of a system is related to the systems response to the application of an external field. Hence, a brute force approach to compute the static linear response is to apply a general quadratic field \(F_{\bar{a}b} ( \bar{a} b + \bar{b} a)\) to the system and compute the derivative of the single particle density matrix \(\rho_{\bar{a}b} \equiv \langle \bar{a}b \rangle\) with respect to the field \(F_{\bar{a}b}\) in the limit \(F_{\bar{a}b} \rightarrow 0\). In other words we can compute the system response \(R_{\bar{a}b\bar{c}d}\) as

(2)\[R_{\bar{a}b\bar{c}d} \equiv \partial_{F_{\bar{a}b}} \langle \bar{c} d \rangle |_{F_{\bar{a}b} = 0}\]

However, the central difference between the system response \(R_{\bar{a}b\bar{c}d}\) and the static the generalized susceptibility \(\chi^{(PH)}_{\bar{a}b\bar{c}d}\) is that the applied field has to be Hermitian. Whence, they are related by

(3)\[R_{\bar{a}b\bar{c}d} = \chi^{(PH)}_{\bar{a}b\bar{c}d} + \chi^{(PH)}_{\bar{b}a\bar{c}d}\]

and the response \(R_{\bar{a}b\bar{c}d}\) is symmetric in \(\bar{a}b\) transposes \(R_{\bar{a}b\bar{c}d} = R_{\bar{b}a\bar{c}d}\).

Reconstructing \(\chi\) from \(R\)

from the symmetry relation

(4)\[\langle (\bar{a}b)(\tau) (\bar{c}d) \rangle = [ \xi \langle (\bar{d}c)(\beta - \tau) (\bar{b}a) \rangle ]^*\]

we can compute four relations involving pair-wise permutations of \(\chi^{(PH)}_{\bar{a}b\bar{c}d}\) from \(R_{\bar{a}b\bar{c}d}\)

(5)\[\begin{split}R_{\bar{a}b\bar{c}d} = \chi^{(PH)}_{\bar{a}b\bar{c}d} + \chi^{(PH)}_{\bar{b}a\bar{c}d} \\ R_{\bar{a}b\bar{d}c} = \chi^{(PH)}_{\bar{a}b\bar{d}c} + \chi^{(PH)}_{\bar{b}a\bar{d}c} \\ [R_{\bar{c}d\bar{a}b}]^* = \chi^{(PH)}_{\bar{b}a\bar{d}c} + \chi^{(PH)}_{\bar{b}a\bar{c}d} \\ [R_{\bar{c}d\bar{b}a}]^* = \chi^{(PH)}_{\bar{a}b\bar{d}c} + \chi^{(PH)}_{\bar{a}b\bar{c}d} \\\end{split}\]

However, these relations are not linearly independent and can not be solved for the \(\chi^{(PH)}\) components.