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Information about a quantum system could be drawn from its response to a small perturbation. This is formulated in what is known as linear response theory. In second quantization, consider a perturbation $V$ to the Hamiltonian $H_0$ of an isolated system of $N$ particles. The full Hamiltonian of the perturbed system is given by:

$$H=H_0+V\,.$$

We are now interested in the effect of the perturbation $V$ on the expectation value of a certain observable (local, or not) $A(x)$ at temperature $T=0$. We use the interaction picture for $H$, which is somewhat similar to the Heisenberg picture for $H_0$. Before the perturbation, we have: $$\left\langle A(x)\right\rangle=\left\langle G\right|A_{\mathrm{I}}(x,t)\left|G\right\rangle\,,$$ where $\left|G\right\rangle$ is the ground state of $H_0$, and $A_{\mathrm{I}}(x,t)$ is the operator $A$ in the interaction picture. $\left\langle A(x)\right\rangle$ does not depend on time anyway as we are in a stationary state, but we just keep the time dependence as a matter of principle. Note that in the interaction picture, the expectation value of an observable is given by that operator in the interaction picture sandwiched by the states also in the interaction picture, i.e., $$\left\langle B(t)\right\rangle =\left\langle\psi_{\mathrm{I}}(t)\right|B_{\mathrm{I}}(t)\left|\psi_{\mathrm{I}}(t)\right\rangle\,.$$

In the standard treatment of linear response theory, the state (in interaction picture) is assumed to be $\left|G\right\rangle$ at far away in the past ($t=-\infty$). As the perturbation is turned on adiabatically, it evolves according to: $$\left|\psi_{\mathrm{I}}(t)\right\rangle=U(t)\left|G\right\rangle\,,$$ where $U(t)$ is the (time-ordered) evolution operator of the state in interaction picture, $$U(t)=\mathcal{T}\exp\left(-\frac{\mathrm{i}}{\hbar}\int\limits_{-\infty}^{t}\mathrm{d}t'\,V_{\mathrm{I}}(t')\right)\,.$$

So after the perturbation is introduced, the expectation value of observable $A(x)$ is:

$$\left\langle A'(x,t)\right\rangle=\left\langle G\right|U^{\dagger}(t)A_{\mathrm{I}}(x,t)U(t)\left|G\right\rangle\,.$$

In linear response theory, we are interested in this expectation value before and after the perturbation to the first order in the perturbation $V$. Usually, $V$ is given as a coupling between an external field $f(x)$ and the internal observable $A(x)$. That is:

$$V=\int\mathrm{d}x\,f(x)A(x)\,.$$ The change in the expectation value is: $$\left\langle\delta A(x,t)\right\rangle=\lim_{\eta\to 0^+}e^{\mathrm{i}\eta t}\int\limits_{-\infty}^t\mathrm{d}t'\int\mathrm{d}x'\,\chi(x,x',t-t')f(x')\,,$$ where $\eta$ is an infinitesimal rate to adiabatically turn on the perturbation and $\chi$ is the response function, $$\chi(x,x',t-t')=-\frac{\mathrm{i}}{\hbar}\theta(t-t')e^{\mathrm{i}\eta(t'-t)}\left\langle G\right|\left[A(x,t),A(x',t')\right]\left|G\right\rangle\,,$$ where the Heaviside step function $\theta$ ensures causality is satisfied.

Overall, I am happy with this treatment, but I ask myself, why the expectation value after perturbation is not $$\left\langle A'(x)\right\rangle=\left\langle G'\right|A(x)\left|G'\right\rangle$$

instead, where $\left| G'\right\rangle$ is the ground state of the full Hamiltonian. As the thermodynamic system is in constant fluctuation, it should transition to the ground state of the full Hamiltonian. I could convince myself that $U(t)\left|G\right\rangle$ is approximately $\left|G'\right\rangle$ to the first order in $V$, but this is still somewhat unsatisfactory for me.

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  • $\begingroup$ Hey, I just stumbled over your question, have you found an answer yet? $\endgroup$ – iqopi Aug 1 at 23:31
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    $\begingroup$ I could convince myself that U(t)|G⟩ is approximately |G′⟩ to the first order in V, but this is still somewhat unsatisfactory for me. How is this unsatisfactory ? Maybe you should put a dummy indicator $\epsilon$ of the order of the perturbation in your development. Then you'll see that the $A'$ that you quoted is not the expectation value for the observable after the perturbation but rather the expectation value for the observable at first order. In the end you are right with your conclusion, but your development mixes up 'true' expectation value and expectation value at first order. $\endgroup$ – Naptzer Aug 6 at 15:47

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