Linear Response Theory of Entanglement Entropy

Consider a quantum system which is weakly coupled to an external perturbation $F(t)\hat{H}_{1}$ at $t=t_{0}$ with time dependence $F(t)$. The Hamiltonian for such system is given by

$\displaystyle\hat{H}(t)=\hat{H}_{0}+F(t)\hat{H}_{1}$

(1)

where $\hat{H}_{0}$ is the Hamiltonian of unperturbed system. In interaction picture (we denote an operator or density matrix in interaction picture through adding ”I” to its subscript), the average of any operator $\hat{O}_{I}(t)$ over state $\rho_{I}(t)$ reads

$\displaystyle\langle\hat{O}_{I}(t)\rangle$

$\displaystyle\equiv\text{tr}[\rho_{I}(t)\hat{O}_{I}(t)]$

(2)

where $\rho_{I}(t)=\hat{U}_{I}(t,t_{0})\rho_{I}(t_{0})\hat{U}_{I}^{\dagger}(t,t_{0})$ with the time evolution operator $\hat{U}_{I}(t,t_{0})=\mathcal{T}\exp[-i\int_{t_{0}}^{t}dt^{\prime}F(t^{\prime})\hat{H}_{1,I}(t^{\prime})]$. If $F(t)$ is so small such that $F(t)\hat{H}_{1}$ is much smaller than $\hat{H}_{0}$, we can describe $\langle\hat{O}_{I}(t)\rangle\equiv\text{tr}[\rho_{I}(t)\hat{O}_{I}(t)]=\text{tr}[\rho(t)\hat{O}(t)]$ by performing an expansion in powers of $F(t)$, working to the first order solution to $\hat{U}_{I}(t)$ (we assume $\hbar=1$ hereafter):

$\displaystyle\delta\langle\hat{O}_{I}(t)\rangle=\int_{t_{0}}^{\infty}dt^{\prime}R(t,t^{\prime})F(t^{\prime}))+\mathcal{O}(F^{2})$

(3)

where $\delta\langle\hat{O}_{I}(t)\rangle\equiv\langle\hat{O}_{I}(t)\rangle_{F}-\langle\hat{O}_{I}(t)\rangle_{F=0}$ and $R(t,t^{\prime})=-i\theta(t-t^{\prime}))\langle[\hat{O}_{I}(t),\hat{H}_{1,I}(t^{\prime})]\rangle_{0}$, with $\langle\cdot\rangle_{0}\equiv\text{tr}\left[\cdot\rho(t_{0})\right]$ an average with respect to the initial state $\rho(t_{0})$. The real function $R(t,t^{\prime})$ is the well-known Kubo formula of observable $\hat{O}$ for system which is subject to a perturbation $F(t)\hat{H}_{1}$, which is also referred to as the linear response function. It is readily to see that the Kubo formula can be written in terms of two-point correlation function $C(t,t^{\prime})\equiv\langle\hat{O}_{I}(t)\hat{H}_{1,I}(t^{\prime})\rangle_{0}$, if $\rho(t_{0})$ is an equilibrium state, the correlation function provides a measure of equilibrium fluctuations in the system, therefore, it is revealed by Eq.(3) that although the response of an equilibrium system to external perturbation should move the system away from equilibrium, if the perturbation is weak enough, the response is dictated by the equilibrium fluctuations. We may conclude that the knowledge of the equilibrium state is useful in predicting the behaviour of nonequilibrium process.

Stationary. If the initial state is equilibrium state, its density matrix then commutes with $\hat{H}_{0}$, inserting $\hat{O}_{I}(t)=e^{i\hat{H}_{0}t}\hat{O}e^{-i\hat{H}_{0}t}$ and $\hat{H}_{1,I}(t)=e^{i\hat{H}_{0}t^{\prime}}\hat{H}_{1}e^{-i\hat{H}_{0}t^{\prime}}$ into the expression of correlation function and using the cyclic property of trace, we have

$\displaystyle C(t,t^{\prime})$

$\displaystyle=\langle\hat{O}_{I}(t-t^{\prime})\hat{H}_{1}\rangle$

(4)

that is, the equilibrium state correlation function is invariant under time translations: $C(t,t^{\prime})=C(t-t^{\prime})$, which results in translation invariant Kubo formula: $R(t,t^{\prime})=R(t-t^{\prime})$.

Causal. The system should not respond before the external perturbation is applied, therefore $R(t)=0$ for $t<0$, this is enforced by the step function in the expression of $R(t)$. The causality will result in analyticity of $\chi(\omega)$ in the upper half complex plane, we will show it in the following. The Fourier expansion of $R(t)$ reads

$\displaystyle R(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\chi(\omega)e^{-i\omega t}d\omega$

(5)

if $t<0$ we can perform the integral by completing the contour in the upper half plane, the complementary integral has to be zero since the real part of the exponent $-i(i|\omega_{i}|)\cdot(-|t|)\rightarrow-\infty$ along the contour in the upper half plane, with $\omega_{i}$ the imaginary part of $\omega$. On the other hand, according to Cauchy’s residue theorem, the integral along the contour is given by the sum of the residues inside the contour. So there is no poles of $\chi(\omega)$ for $\text{Im}\chi(\omega)>0$, which means that $\chi(\omega)$ is analytic in the upper half plane.  
Kramers-Krönig relation. Because $\chi(\omega)$ is analytic in the upper half complex plane, it is found that the imaginary and real parts of $\chi(\omega)$ is not independent, they are connected by the Kramers-Krönig relationToll (1956); de L. Kronig (1926); Kramers (1927):

	$\displaystyle\text{Re}[\chi(\omega)]$	$\displaystyle=\frac{1}{\pi}\mathcal{P}\int_{-\infty}^{\infty}\frac{\text{Im}[\chi(\omega)]}{\omega^{\prime}-\omega}d\omega^{\prime}$		(6)
	$\displaystyle\text{Im}[\chi(\omega)]$	$\displaystyle=-\frac{1}{\pi}\mathcal{P}\int_{-\infty}^{\infty}\frac{\text{Re}[\chi(\omega)]}{\omega^{\prime}-\omega}d\omega^{\prime}.$		(7)

where $\mathcal{P}$ denotes the Cauchy principal value. Using Kramers-Krönig relation, the response function can be reconstructed given just real or imaginary part.

Susceptibility. According to the stationary property, for equilibrium initial state, the Kubo formula is time-translation invariant: $R(t,t^{\prime})=R(t-t^{\prime})$, Eq.(3) becomes a convolution, its Fourier transform reads

$\displaystyle\langle\delta\hat{O}_{I}(\omega)\rangle=\chi(\omega)F(\omega)+\mathcal{O}(F^{2})$

(8)

where $F(\omega)$ is the Fourier transforms of $F(t)$, $\chi(\omega)$ is the Fourier transform of Kubo formula and is oftern referred to as generalized susceptibility. We learn from Eq.(8) that the response of observable $\hat{O}$ to time-dependent perturbation $F(t)\hat{H}_{1}$ is ’local’ in frequency space, this is a feature of equilibrium linear response. Thanks to this locality, we can directly infer the generalized susceptibility $\chi(\omega)$ of $\hat{O}$ for a known perturbation $F(t)\hat{H}_{1}$ through measuring the time evolution of $\langle\hat{O}_{I}(t)\rangle$. The susceptibility of some observable to certain perturbation is essential in investigating the properties of equilibrium systems (e.g. classifying different materials using magnetic susceptibility). Therefore the conventional LRT is an important tool in investigating the properties of unknown systems.

Fluctuation-dissipation relation. Denoting the imaginary part of generalized susceptibility $\chi(\omega)$ as $\chi^{\prime\prime}(\omega)$, it can be written as

$\displaystyle\chi^{\prime\prime}(\omega)=\frac{1}{2i}\int_{-\infty}^{\infty}dte^{i\omega t}[R(t)-R(-t)]$

(9)

which is not invariant under time-reversal $t\rightarrow-t$. Hence, $\chi^{\prime\prime}(\omega)$ is called the dissipative or absorptive part of the generalized susceptibility. The fluctuation–dissipation theorem says that the dissipation effects are determined by natural fluctuation in thermal equilibrium state. To be more specific, for canonical ensemble with the inverse temperature $\beta$, $\rho=e^{-\beta\hat{H}_{0}}/Z$, the imaginary part of susceptibility $\chi^{\prime\prime}(\omega)$ which describe the dissipation and (the Fourier transform of) the correlation function $\tilde{C}(\omega)$ are connected by the Fluctuation-dissipation relation:

$\displaystyle\tilde{C}(\omega)=-2[n_{B}(\omega)+1]\chi^{\prime\prime}(\omega)$

(10)

where $n_{B}(\omega)=1/(e^{\beta\omega}-1)$ is the Bose-Einstein distribution function.

Be similar to the conventional case, supposing at time $t=t_{0}$, a time-dependent perturbation $F(t)\hat{H}_{1}$ is applied to a composite system which consists of subsystems A and B, its Hamiltonian has the same form as (1). Solving the dynamics of this composite system using perturbative method (see appendix A for detail), we have

$\displaystyle\delta\rho(t)=-i\int_{t_{0}}^{t}d\tau[\hat{H}_{1,I}(\tau-t),\rho_{0}]F(\tau)+\mathcal{O}(F^{2}),$

(11)

where $\rho_{0}\equiv\rho(t_{0})$, $\delta\rho(t)\equiv\rho(t)-\rho_{0}$, and $\hat{H}_{1,I}(t)=\hat{U}_{0}^{\dagger}(t,t_{0})\hat{H}_{\text{1}}\hat{U}_{0}(t,t_{0})$, with $\hat{U}_{0}(t,t_{0})=\exp[-i\hat{H}_{0}(t-t_{0})]$.

The entanglement entropy of subsystem A is given by $S_{A}(t)=-\text{tr}[\rho_{A}(t)\ln\rho_{A}(t)]$, with $\rho_{A}(t)=\text{tr}_{B}[\rho(t)]$. For infinitesimal change $\delta\rho_{A}(t)\equiv\rho_{A}(t)-\rho_{A}(t_{0})$, the time-dependent change in entanglement entropy readsMa et al. (2019): $\delta S_{A}(t)\equiv S_{A}(t)-S_{A}(t_{0})\approx-\text{tr}_{A}\left[\delta\rho_{A}(t)\ln\rho_{A}(t)\right]$, furthermore, we can expand the operator $\ln\rho_{A}(t)$ in terms of $\delta\rho_{A}(t)$ as $\ln\rho_{A}(t)=\ln\rho_{A}(t_{0})|+\mathcal{O}(\delta\rho_{A})$, then to the first order in $\delta\rho_{A}(t)$, we have

$\displaystyle\delta S_{A}(t)=-\text{tr}_{A}\left[\delta\rho_{A}(t)\ln\rho_{A}(t_{0})\right]+\mathcal{O}(\delta\rho_{A}).$

(12)

What’s interesting about this equation is that the right-hand side of (12) has the form of the change in the expectation value of an Hermitian operator $\ln\rho_{A}(t_{0})$. Similar result can be seen in Ref. Strasberg et al. (2017).

Inserting Eq. (11) into Eq. (12), we obtain the time evolution of EE under weak driving (see appendix B for detail):

$\displaystyle\delta S_{A}(t)=\int_{t_{0}}^{\infty}R_{E}(t,\tau)F(\tau)d\tau+\mathcal{O}(F^{2})$

(13)

where $R_{E}(t,\tau)$ is given by the following expression

$\displaystyle R_{E}(t,\tau)=-i\theta(t-\tau)\langle[\hat{s}_{A},\hat{H}_{1,I}(\tau-t)]\rangle_{0}.$

(14)

where $\hat{s}_{A}\equiv-\ln\rho_{A}(t_{0})$, Comparing Eq. (13) and Eq. (14) with Eq. (3), we found that to first order in $F(t)$, the change in EE is equivalent to the change in the expectation of observable $\langle\hat{s}_{A}(t)\rangle$:

$\displaystyle\delta S_{A}(t)\approx\delta\langle\hat{s}_{A}(t)\rangle$

(15)

Eq. (13), Eq. (14) and Eq. (15) are key results of this work, they revealed that: for equilibrium open system A, the response of EE is determined by the linear response of observable $\hat{s}_{A}=-\ln\rho_{A}(t_{0})$. Eq. (14) has the same form as conventional linear response function, it describes how EE and the observable $\hat{s}_{A}$ response to weak external force $F(t)\hat{H}_{1}$, we refer to $R_{E}(t,\tau)$ as the Kubo formula of EE.

The Kubo formula of EE is also the Kubo formula of an observable, this fact implies that $R_{E}(t,\tau)$ has the same properties of conventional linear response function.

Like conventional LRT, if $[\rho_{0},\hat{H}_{0}]=0$, the EE Kubo formula will be stationary: $R_{E}(t,\tau)=R_{E}(t-\tau)$, then the Fourier transform(FT) of Eq.(13) reads

$\displaystyle\delta S_{A}(\omega)=\chi_{E}(\omega)F(\omega)+\mathcal{O}(F^{2}),$

(16)

where $\chi_{E}(\omega)$ and $F(\omega)$ are FT of $R_{E}(t)$ and $F(t)$ respectively. Eq. (16) has the same form as (8) except for that $S_{A}(t)$ is not an expectation value of an observable. $R_{E}(t)$ and $\chi_{E}(\omega)$ also satisfy properties such as causality, analyticity of $\chi_{E}(\omega)$ in the upper-half plane and there is a Kramers-Kröning relation between real and imaginary part of $\chi_{E}(\omega)$.

Entanglement susceptibility. We refer to $\chi_{E}(\omega)$ as entanglement susceptibility. According to Eq. (13), the time evolution of EE can be determined by measuring an observable $\hat{s}_{A}(t)$. $F(t)$ is normally known to the one who applies the perturbation, we thus can infer the Kubo formula $R_{E}(t)$ from (16) and hence the entanglement susceptibility $\chi_{E}(\omega)$. The Kubo formula is independent of $F(t)$, for a given $\hat{H}_{1}(t)$, once we determine its Kubo formula, we may predict the response for any $F(t)$. Although $\hat{s}_{A}$ is hard to measure, for a system with few degrees of freedom, (e. g., a spin-1/2) we can evaluate it using state tomography.

Canonical ensemble EE linear response If the initial state of subsystem A is described by canonical ensemble $\rho_{A}(t_{0})=e^{-\beta\hat{H}_{A}}/Z_{A}$ with inverse temperature $\beta=1/k_{B}T$, $\hat{H}_{A}$ is the Hamiltonian of subsystem A and $Z_{A}$ is the corresponding partition function, then $\ln\rho_{A}(t_{0})=-\beta\hat{H}_{A}-\ln Z_{A}$, inserting this equation into (13), we have

$\displaystyle-k_{B}T\delta S_{A}(t)=\int_{t_{0}}^{\infty}(1/\beta)R_{E}(t,\tau)F(\tau)d\tau+\mathcal{O}(F^{2})$

(17)

it is readily to see that $(1/\beta)R_{E}(t,\tau)=-i\theta(t-\tau)\langle[\hat{H}_{A},\hat{H}_{1,I}(\tau-t)]\rangle_{0}$, which is the Kubo formula of observable $\hat{H}_{A}$ and perturbation $\hat{H}_{1}(t)$, $\delta Q_{A}(t)\equiv-k_{B}T\delta S_{A}$ has a physical meaning of the heat change of subsystem A during a process with fixed temperature $T$. We can conclude from Eq. (17) that $\delta Q_{A}(t)=\langle\hat{H}_{A}(t)\rangle$ to the first order in $F(t)$, this implies that for an open quantum system which is described by the canonical ensemble, if we know how its internal energy responds to a known perturbation, we then know how its EE responds to the same perturbation.

Given an LRT of EE, a natural question arises: the sign of response is determined by the sign of driving, or the sign of $\alpha$, consider a separable initial state, then $S_{A}(t_{0})=0$, we can always choose an appropriate sign of $\alpha$ such that the linear correction of EE is negative, that is $\delta S_{A}(t)<0$, then $S_{A}(t)=S_{A}(t_{0})+\delta S_{A}(t)<0$, this is, of course, not true. We may have a similar question for maximally entangled states. Viewed from another perspective, the EE of product states and maximally entangled states must be extreme values, therefore $\delta S_{A}=0$ for these states. To address these problems, we propose Theorem 1.

Proof. We here give the proof of the statement about separable states in Theorem 1, the proof about maximally states is similar and we show it in appendix C. The basic idea of this proof is to prove that $\ln\rho_{A}(t_{0})$ is diagonal while any diagonal element of $\delta\rho_{A}(t)$ is zero, then $\text{tr}_{A}[\delta\rho_{A}(t)\ln\rho_{A}(t_{0})]=0$.

For a composite system consists of two subsystem A and B, if its eigenstates are separable or product states of A and B, then the Hamiltonian of this composite system must be an non-interacting Hamiltonian of A and B,

$\displaystyle\hat{H}_{0}=\hat{H}_{A}+\hat{H}_{B}$

(18)

before proceeding, we assume that $\hat{H}_{0}$ is non-degenerate, then both subsystem A and B are non-degenerate, their eigenstates and eigenvalues are given by

	$\displaystyle\hat{H}_{A}|f_{i_{A}}\rangle$	$\displaystyle=f_{i_{A}}|f_{i_{A}}\rangle$		(19)
	$\displaystyle\hat{H}_{B}|g_{j_{B}}\rangle$	$\displaystyle=g_{j_{B}}|g_{j_{B}}\rangle$		(20)

the eigenvectors of $\hat{H}_{0}$ are product states of the eigenvectors of A and B:

$\displaystyle\hat{H}_{0}\left(|f_{i_{A}}\rangle\otimes|g_{j_{B}}\rangle\right)=(f_{i_{A}}+g_{j_{B}})|f_{i_{A}}\rangle\otimes|g_{j_{B}}\rangle$

(21)

denoting $|S_{i_{A},j_{B}}\rangle=|f_{i_{A}}\rangle\otimes|g_{j_{B}}\rangle$, it is readily to see that if the initial state is diagonal with respect to basis $\{|S_{i_{A},j_{B}}\rangle\}$, that is $\rho_{0}=\sum_{i_{A},j_{B}}P_{i_{A},j_{B}}|S_{i_{A},j_{B}}\rangle\langle S_{i_{A},j_{B}}|$, then the reduced density matrix of subsystem A is given by

$\displaystyle\rho_{A}(t_{0})=\text{tr}_{B}[\rho_{0}]=\sum_{i_{A}}\Big{(}\sum_{j_{B}}P_{i_{A},j_{B}}\Big{)}|f_{i_{A}}\rangle\langle f_{i_{A}}|$

(22)

which is diagonal with respect to eigenbasis of $\hat{H}_{A}$, so is $\ln\rho_{A}(t_{0})$.

Given the intial state $\rho_{0}=\sum_{i_{A},j_{B}}P_{i_{A},j_{B}}|S_{i_{A},j_{B}}\rangle\langle S_{i_{A},j_{B}}|$ which is diagonal, if $[\hat{H}_{0},\rho_{0}]=0$ then an arbitrary entry of the commutator in Eq.(11) is given by $\langle S_{i_{A},j_{B}}|[\hat{H}_{1,I}(t-\tau),\rho_{0}]|S_{k_{A},l_{B}}\rangle=(p_{i_{A},j_{B}}-p_{k_{A},l_{B}})\langle S_{i_{A},j_{B}}|\hat{H}_{1,I}(t-\tau)|S_{k_{A},l_{B}}\rangle$, and

	$\displaystyle\langle\delta\rho(t)\rangle_{m,m}$	$\displaystyle=-i\int_{t_{0}}^{t}d\tau\langle S_{m}|[\hat{H}_{1,I}(t-\tau),\rho_{0}]|S_{m}\rangle F(\tau)$
		$\displaystyle\ \ \ \ +\mathcal{O}(F^{2})$
		$\displaystyle=\mathcal{O}(F^{2})$		(23)

where $\langle\delta\rho(t)\rangle_{m,m}\equiv\langle\rho(t)-\rho_{0}\rangle_{m,m}$. The above equation implies that after applying perturbation, any diagonal element of linear correction to initial state is zero, does this means that any diagonal element of $\delta\rho_{A}(t)$ is zero? Consider an arbitary off-diagonal element $|S_{i_{A},j_{B}}\rangle\langle S_{k_{A},l_{B}}|$ of the initial state,

$\displaystyle\text{tr}_{B}(|S_{i_{A},j_{B}}\rangle\langle S_{k_{A},l_{B}}|)=\delta_{j_{B},l_{B}}|f_{i_{A}}\rangle\langle f_{k_{A}}|$

(24)

which does not contribute to the diagonal element of subsystem A with respect to basis ${|f_{i_{A}}}\rangle$. As any diagonal element of $\rho_{0}$ is zero, we may conclude that the $\delta\rho_{A}(t)$ is hollow, that is the diagonal entries are all zero.

In summary, if the initial eigenstate is a separable state of A and B, moreover, if $\hat{H}_{A}+\hat{H}_{B}$ is non-degenerate, then $\ln\rho_{A}(t_{0})$ is diagonal, while $\delta\rho_{A}(t)$ is hollow. Thus the linear response of EE $\delta S_{A}(t)=\mathcal{O}(F^{2})$.