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^S∣ψS(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⁡ ⁣(1iℏ∫t0tH^1,I(t′)dt′)}≈1−iℏ∫t0tH^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^I†A^IK^I≈(1+iℏ∫t0tH^1,I(t′)dt′)A^I(t)(1−iℏ∫t0tH^1,I(t′)dt′)≈A^I(t)−iℏ∫t0tA^I(t)H^1,I(t′)dt′+iℏ∫t0tH^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^I†A^IK^I=A^I(t)−iℏ∫t0t[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^I†A^IK^I⟩0=⟨A^I(t)⟩0−iℏ∫t0t⟨[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^I⟩0=−iℏ∫t0t⟨[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)=∫t0∞CAH1R(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′)≡−iℏΘ(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^S f(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 t−t′t - 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(t−t′)\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(t−t′)f(t′)dt′=(CABR∗f)(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}

References

  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.