Linear response

From the generalized susceptibility

The static linear response can be obtained from the frequency dependent generalized susceptibility \(\chi^{PH}_{\bar{a}b\bar{c}d}(\omega_0, \nu_n, \nu_m)\) or the one-time dependent bosonic propagator \(\langle (\bar{a}b)(\tau) (\bar{c}d) \rangle\) by summing/integrating over frequency/time

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

From an applied external field

The linear response of a system is related to its 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

\[R_{\bar{a}b\bar{c}d} \equiv \partial_{F_{\bar{a}b}} \langle \bar{c} d \rangle \vert_{F_{\bar{a}b} \rightarrow 0}.\]

However, the central difference between the system response \(R_{\bar{a}b\bar{c}d}\) and 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

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

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

Can we reconstruct \(\chi\) from \(R\)? We use the (anti-)periodicity and anticommutation relations yielding

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

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

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

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