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 $$G_{\nu \nu'}^R(t, t')$$, we simply take the $$t$$-derivative (we could do the same with the advanced function $$G_{\nu \nu'}^A$$):

\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 $$\Theta$$ is a Dirac delta function $$\delta$$. Also, from the second quantization, $$\expval**{\comm*{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}$$ for $$t = t'$$ is zero when $$\nu \neq \nu'$$.

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

\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 $$\hat{H}_0$$ and the interaction part $$\hat{H}_\mathrm{int}$$ are assumed to be time-independent in the Schrödinger picture. We thus get:

\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 $$\hat{H}_0$$, for any basis, is as follows, where $$u_{\nu' \nu''}$$ are constants:

\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}

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

\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:

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

\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 $$D_{\nu \nu'}^R$$ might still not be doable due to $$\hat{H}_\mathrm{int}$$. The key idea of equation-of-motion theory is to either approximate $$D_{\nu \nu'}^R$$ now, or to differentiate it again $$i \hbar \dv*{D_{\nu \nu'}^R}{t}$$, 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 $$\hat{H}_\mathrm{int} = 0$$, so clearly $$D_{\nu \nu'}^R$$ trivially vanishes then. Let us assume that $$\hat{H}_0$$ is also time-independent, such that $$G_{\nu'' \nu'}^R$$ only depends on the difference $$t - t'$$:

\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 $$(t \!-\! t') \to (\omega + i \eta)$$, where $$\eta \to 0^+$$ ensures convergence:

\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 $$u_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}$$, this reduces to the following:

\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:

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

References

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

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.