Categories: Perturbation, Physics, Quantum mechanics.

Kubo formula

Consider the following quantum Hamiltonian, split into a main time-independent term H^0,S\hat{H}_{0,S} and a small time-dependent perturbation H^1,S\hat{H}_{1,S}, which is turned on at t=t0t = t_0:

H^S(t)=H^0,S+H^1,S(t)\begin{aligned} \hat{H}_S(t) = \hat{H}_{0,S} + \hat{H}_{1,S}(t) \end{aligned}

And let ψS(t)\Ket{\psi_S(t)} be the corresponding solutions to the Schrödinger equation. Then, given a time-independent observable A^\hat{A}, its expectation value A^\expval{\hat{A}} evolves like so, where the subscripts SS and II respectively refer to the Schrödinger and interaction pictures:

A^(t)=ψS(t)A^SψS(t)=ψI(t)A^I(t)ψI(t)=ψI(t0)K^I(t,t0)A^I(t)K^I(t,t0)ψI(t0)\begin{aligned} \expval{\hat{A}}(t) = \matrixel{\psi_S(t)}{\hat{A}_S}{\psi_S(t)} &= \matrixel{\psi_I(t)}{\hat{A}_I(t)}{\psi_I(t)} \\ &= \matrixel{\psi_I(t_0)\,}{\,\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)\,}{\,\psi_I(t_0)} \end{aligned}

Where the time evolution operator K^I(t,t0)\hat{K}_I(t, t_0) is as follows, which we Taylor-expand:

K^I(t,t0)=T{exp ⁣(1it0tH^1,I(t)dt)}1it0tH^1,I(t)dt\begin{aligned} \hat{K}_I(t, t_0) = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\} \approx 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \end{aligned}

With this, the following product of operators (as encountered earlier) can be written as:

K^IA^IK^I(1+it0tH^1,I(t)dt)A^I(t)(1it0tH^1,I(t)dt)A^I(t)it0tA^I(t)H^1,I(t)dt+it0tH^1,I(t)A^I(t)dt\begin{aligned} \hat{K}_I^\dagger \hat{A}_I \hat{K}_I &\approx \bigg( 1 + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \hat{A}_I(t) \bigg( 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \\ &\approx \hat{A}_I(t) - \frac{i}{\hbar} \int_{t_0}^t \hat{A}_I(t) \hat{H}_{1,I}(t') \dd{t'} + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{A}_I(t) \dd{t'} \end{aligned}

Where we have dropped the last term, because H^1\hat{H}_{1} is assumed to be so small that it only matters to first order. Here, we notice a commutator, so we can rewrite:

K^IA^IK^I=A^I(t)it0t[A^I(t),H^1,I(t)]dt\begin{aligned} \hat{K}_I^\dagger \hat{A}_I \hat{K}_I &= \hat{A}_I(t) - \frac{i}{\hbar} \int_{t_0}^t \Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')} \dd{t'} \end{aligned}

Returning to A^\expval{\hat{A}}, we have the following formula, where \Expval{} is the expectation value for ψ(t)\Ket{\psi(t)}, and 0\Expval{}_0 is the expectation value for ψI(t0)\Ket{\psi_I(t_0)}:

A^(t)=K^IA^IK^I0=A^I(t)0it0t[A^I(t),H^1,I(t)]0dt\begin{aligned} \expval{\hat{A}}(t) = \expval{\hat{K}_I^\dagger \hat{A}_I \hat{K}_I}_0 = \expval{\hat{A}_I(t)}_0 - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'} \end{aligned}

Now we define δ ⁣A^ ⁣(t)\delta\!\expval{\hat{A}}\!(t) as the change of A^\expval{\hat{A}} due to the perturbation H^1\hat{H}_1, and insert A^(t)\expval{\hat{A}}(t):

δ ⁣A^ ⁣(t)A^(t)A^I0=it0t[A^I(t),H^1,I(t)]0dt\begin{aligned} \delta\!\expval{\hat{A}}\!(t) \equiv \expval{\hat{A}}(t) - \expval{\hat{A}_I}_0 = - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'} \end{aligned}

Finally, we introduce a Heaviside step function Θ\Theta and change the integration limit accordingly, leading to the Kubo formula describing the response of A^\expval{\hat{A}} to first order in H^1\hat{H}_1:

δ ⁣A^ ⁣(t)=t0CAH1R(t,t)dt\begin{aligned} \boxed{ \delta\!\expval{\hat{A}}\!(t) = \int_{t_0}^\infty C^R_{A H_1}(t, t') \dd{t'} } \end{aligned}

Where we have defined the retarded correlation function CAH1R(t,t)C^R_{A H_1}(t, t') as follows:

CAH1R(t,t)iΘ(t ⁣ ⁣t)[A^I(t),H^1,I(t)]0\begin{aligned} \boxed{ C^R_{A H_1}(t, t') \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 } \end{aligned}

Note that observables are bosonic, because in the second quantization they consist of products of even numbers of particle creation/annihiliation operators. Therefore, this correlation function is a two-particle Green’s function.

A common situation is that H^1\hat{H}_1 consists of a time-independent operator B^\hat{B} and a time-dependent function f(t)f(t), allowing us to split CAH1RC^R_{A H_1} as follows:

H^1,S(t)=B^Sf(t)    CAH1R(t,t)=CABR(t,t)f(t)\begin{aligned} \hat{H}_{1,S}(t) = \hat{B}_S \: f(t) \quad \implies \quad C^R_{A H_1}(t, t') = C^R_{A B}(t, t') f(t') \end{aligned}

Since CABRC_{AB}^R is a Green’s function, we know that it only depends on the difference ttt - t', as long as the system was initially in thermodynamic equilibrium, and H^0,S\hat{H}_{0,S} is time-independent:

CABR(t,t)=CABR(tt)\begin{aligned} C^R_{A B}(t, t') = C^R_{A B}(t - t') \end{aligned}

With this, the Kubo formula can be written as follows, where we have set t0=t_0 = - \infty:

δ ⁣A ⁣(t)=CABR(tt)f(t)dt=(CABRf)(t)\begin{aligned} \delta\!\expval{A}\!(t) = \int_{-\infty}^\infty C^R_{A B}(t - t') f(t') \dd{t'} = (C^R_{A B} * f)(t) \end{aligned}

This is a convolution, so the convolution theorem states that the Fourier transform of δ ⁣A^ ⁣(t)\delta\!\expval{\hat{A}}\!(t) is simply the product of the transforms of CABRC^R_{AB} and ff:

δ ⁣A^ ⁣(ω)=C~ABR(ω)f~(ω)\begin{aligned} \boxed{ \delta\!\expval{\hat{A}}\!(\omega) = \tilde{C}{}^R_{A B}(\omega) \: \tilde{f}(\omega) } \end{aligned}


  1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.
  2. K.S. Thygesen, Advanced solid state physics: linear response theory, 2013, unpublished.