Categories: Physics, Quantum mechanics.

Equation-of-motion theory

In many-body quantum theory, equation-of-motion theory is a method to calculate the time evolution of a system’s properties using Green’s functions.

Starting from the definition of the retarded single-particle Green’s function GννR(t,t)G_{\nu \nu'}^R(t, t'), we simply take the tt-derivative (we could do the same with the advanced function GννAG_{\nu \nu'}^A):

iGννR(t,t)t=Θ(t ⁣ ⁣t)t[c^ν(t),c^ν(t)]+Θ(t ⁣ ⁣t)t[c^ν(t),c^ν(t)]=δ(t ⁣ ⁣t)[c^ν(t),c^ν(t)]+Θ(t ⁣ ⁣t)[dc^ν(t)dt,c^ν(t)]\begin{aligned} i \hbar \pdv{G^R_{\nu \nu'}(t, t')}{t} &= \pdv{\Theta(t \!-\! t')}{t} \Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} + \Theta(t \!-\! t') \pdv{}{t}\Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} \\ &= \delta(t \!-\! t') \Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} + \Theta(t \!-\! t') \Expval{\Comm{\dv{\hat{c}_\nu(t)}{t}}{\hat{c}_{\nu'}^\dagger(t)}_{\mp}} \end{aligned}

Where we have used that the derivative of a Heaviside step function Θ\Theta is a Dirac delta function δ\delta. Also, from the second quantization, [c^ν(t),c^ν(t)]\expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} for t=tt = t' is zero when νν\nu \neq \nu'.

Since we are in the Heisenberg picture, we know the equation of motion of c^ν(t)\hat{c}_\nu(t):

dc^ν(t)dt=i[H^0(t),c^ν(t)]+i[H^int(t),c^ν(t)]\begin{aligned} \dv{\hat{c}_\nu(t)}{t} = \frac{i}{\hbar} \comm{\hat{H}_0(t)}{\hat{c}_\nu(t)} + \frac{i}{\hbar} \comm{\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)} \end{aligned}

Where the single-particle part of the Hamiltonian H^0\hat{H}_0 and the interaction part H^int\hat{H}_\mathrm{int} are assumed to be time-independent in the Schrödinger picture. We thus get:

iGννRt=δννδ(t ⁣ ⁣t)+iΘ(t ⁣ ⁣t)[[H^0,c^ν]+[H^int,c^ν],c^ν]\begin{aligned} i \hbar \pdv{G^R_{\nu \nu'}}{t} &= \delta_{\nu \nu'} \delta(t \!-\! t')+ \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\comm{\hat{H}_0}{\hat{c}_\nu} + \comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}} \end{aligned}

The most general form of H^0\hat{H}_0, for any basis, is as follows, where uννu_{\nu' \nu''} are constants:

H^0=ννuννc^νc^ν    [H^0,c^ν]=νuννc^ν\begin{aligned} \hat{H}_0 = \sum_{\nu' \nu''} u_{\nu' \nu''} \hat{c}_{\nu'}^\dagger \hat{c}_{\nu''} \quad \implies \quad \comm{\hat{H}_0}{\hat{c}_\nu} = - \sum_{\nu''} u_{\nu \nu''} \hat{c}_{\nu''} \end{aligned}

Using the commutator identity for [AB,C]\comm{A B}{C}, we decompose it like so:

[H^0,c^ν]=ννuνν[c^νc^ν,c^ν]=ννuνν(c^ν[c^ν,c^ν]+[c^ν,c^ν]c^ν)\begin{aligned} \comm{\hat{H}_0}{\hat{c}_\nu} &= \sum_{\nu' \nu''} u_{\nu \nu''} \comm{\hat{c}_{\nu'}^\dagger \hat{c}_{\nu''}}{\hat{c}_\nu} = \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{c}_{\nu'}^\dagger \comm{\hat{c}_{\nu''}}{\hat{c}_\nu} + \comm{\hat{c}_{\nu'}^\dagger}{\hat{c}_\nu} \hat{c}_{\nu''} \Big) \end{aligned}

Bosons have well-known commutation relations, so the result follows directly:

[H^0,b^ν]=ννuνν(b^ν[b^ν,b^ν]+[b^ν,b^ν]b^ν)=νuννb^ν\begin{aligned} \comm{\hat{H}_0}{\hat{b}_\nu} &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{b}_{\nu'}^\dagger \comm{\hat{b}_{\nu''}}{\hat{b}_\nu} + \comm{\hat{b}_{\nu'}^\dagger}{\hat{b}_\nu} \hat{b}_{\nu''} \Big) = - \sum_{\nu''} u_{\nu \nu''} \hat{b}_{\nu''} \end{aligned}

Fermions only have anticommutation relations, so a bit more work is necessary:

[H^0,f^ ⁣ν]=ννuνν(f^ ⁣ν[f^ ⁣ν,f^ ⁣ν]+[f^ ⁣ν,f^ ⁣ν]f^ ⁣ν)=ννuνν(f^ ⁣ν{f^ ⁣ν,f^ ⁣ν}2f^ ⁣νf^ ⁣νf^ ⁣ν+{f^ ⁣ν,f^ ⁣ν}f^ ⁣ν2f^ ⁣νf^ ⁣νf^ ⁣ν)=ννuνν(δννf^ ⁣ν2{f^ ⁣ν,f^ ⁣ν}f^ ⁣ν)=νuννf^ ⁣ν\begin{aligned} \comm{\hat{H}_0}{\hat{f}_{\!\nu}} &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \comm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}} + \comm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big) \\ &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \acomm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}} - 2 \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu} \hat{f}_{\!\nu''} + \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} - 2 \hat{f}_{\!\nu} \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu''} \Big) \\ &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \delta_{\nu \nu'} \hat{f}_{\!\nu''} - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big) = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''} \end{aligned}

Substituting this into GννRG_{\nu \nu'}^R’s equation of motion, we recognize another Green’s function GννRG_{\nu'' \nu'}^R:

iGννRt=δννδ(t ⁣ ⁣t)+iΘ(t ⁣ ⁣t)([[H^int,c^ν],c^ν]νuνν[c^ν,c^ν])=δννδ(t ⁣ ⁣t)+iΘ(t ⁣ ⁣t)[[H^int,c^ν],c^ν]+νuννGννR(t,t)\begin{aligned} i \hbar \pdv{G^R_{\nu \nu'}}{t} &= \delta_{\nu \nu'} \delta(t \!-\! t') + \frac{i}{\hbar} \Theta(t \!-\! t') \bigg( \Expval{\comm{\comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}} - \sum_{\nu''} u_{\nu \nu''} \Expval{\comm{\hat{c}_{\nu''}}{\hat{c}_{\nu'}^\dagger}_{\mp}} \bigg) \\ &= \delta_{\nu \nu'} \delta(t \!-\! t') + \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\comm{\comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}} + \sum_{\nu''} u_{\nu \nu''} G_{\nu''\nu'}^R(t, t') \end{aligned}

Rearranging this as follows yields the main result of equation-of-motion theory:

ν(iδννtuνν)GννR(t,t)=δννδ(t ⁣ ⁣t)+DννR(t,t)\begin{aligned} \boxed{ \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t, t') = \delta_{\nu \nu'} \delta(t \!-\! t') + D_{\nu \nu'}^R(t, t') } \end{aligned}

Where DννRD_{\nu \nu'}^R represents a correction due to interactions H^int\hat{H}_\mathrm{int}, and also has the form of a retarded Green’s function, but with c^ν\hat{c}_{\nu} replaced by [H^int,c^ν]\comm{-\hat{H}_\mathrm{int}}{\hat{c}_\nu}:

DννR(t,t)iΘ(t ⁣ ⁣t)[[H^int(t),c^ν(t)],c^ν(t)]\begin{aligned} \boxed{ D^R_{\nu'' \nu'}(t, t') \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\comm{\comm{-\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)}}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} } \end{aligned}

Unfortunately, calculating DννRD_{\nu \nu'}^R might still not be doable due to H^int\hat{H}_\mathrm{int}. The key idea of equation-of-motion theory is to either approximate DννRD_{\nu \nu'}^R now, or to differentiate it again idDννR/dti \hbar \idv{D_{\nu \nu'}^R}{t}, and try again for the resulting corrections, until a solvable equation is found. There is no guarantee that that will ever happen; if not, one of the corrections needs to be approximated.

For non-interacting particles H^int=0\hat{H}_\mathrm{int} = 0, so clearly DννRD_{\nu \nu'}^R trivially vanishes then. Let us assume that H^0\hat{H}_0 is also time-independent, such that GννRG_{\nu'' \nu'}^R only depends on the difference ttt - t':

ν(iδννtuνν)GννR(tt)=δννδ(tt)\begin{aligned} \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t - t') = \delta_{\nu \nu'} \delta(t - t') \end{aligned}

We take the Fourier transform (t ⁣ ⁣t)(ω+iη)(t \!-\! t') \to (\omega + i \eta), where η0+\eta \to 0^+ ensures convergence:

ν(δνν(ω+iη)uνν)GννR(ω)=δνν\begin{aligned} \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega) = \delta_{\nu \nu'} \end{aligned}

If we assume a diagonal basis uνν=ενδννu_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}, this reduces to the following:

δνν=ν(δνν(ω+iη)ενδνν)GννR(ω)=((ω+iη)εν)GννR(ω)\begin{aligned} \delta_{\nu \nu'} &= \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - \varepsilon_\nu \delta_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega) \\ &= \Big( \hbar (\omega + i \eta) - \varepsilon_\nu \Big) G^R_{\nu \nu'}(\omega) \end{aligned}

For a non-interacting, time-independent Hamiltonian, we therefore arrive at:

GννR(ω)=δνν(ω+iη)εν\begin{aligned} \boxed{ G^R_{\nu \nu'}(\omega) = \frac{\delta_{\nu \nu'}}{\hbar (\omega + i \eta) - \varepsilon_\nu} } \end{aligned}


  1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.