Categories: Physics, Quantum mechanics.

# Dyson equation

Consider the time-dependent Schrödinger equation, describing a wavefunction $$\Psi_0(\vb{r}, t)$$:

\begin{aligned} i \hbar \pdv{t} \Psi_0(\vb{r}, t) = \hat{H}_0(\vb{r}) \: \Psi_0(\vb{r}, t) \end{aligned}

By definition, this equation’s fundamental solution $$G_0(\vb{r}, t; \vb{r}', t')$$ satisfies the following:

\begin{aligned} \Big( i \hbar \pdv{t} - \hat{H}_0(\vb{r}) \Big) G_0(\vb{r}, t; \vb{r}', t') = \delta(\vb{r} - \vb{r}') \: \delta(t - t') \end{aligned}

From this, we define the inverse $$\hat{G}{}_0^{-1}(\vb{r}, t)$$ as follows, so that $$\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$$:

\begin{aligned} \hat{G}{}_0^{-1}(\vb{r}, t) &\equiv i \hbar \pdv{t} - \hat{H}_0(\vb{r}) \end{aligned}

Note that $$\hat{G}{}_0^{-1}$$ is an operator, while $$G_0$$ is a function. For the sake of consistency, we thus define the operator $$\hat{G}_0(\vb{r}, t)$$ as a multiplication by $$G_0$$ and integration over $$\vb{r}'$$ and $$t'$$:

\begin{aligned} \hat{G}_0(\vb{r}, t) \: f \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'} \end{aligned}

For an arbitrary function $$f(\vb{r}, t)$$, so that $$\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$$. Moving on, the Schrödinger equation can be rewritten like so, using $$\hat{G}{}_0^{-1}$$:

\begin{aligned} \hat{G}{}_0^{-1}(\vb{r}, t) \: \Psi_0(\vb{r}, t) = 0 \end{aligned}

Let us assume that $$\hat{H}_0$$ is simple, such that $$G_0$$ and $$\hat{G}{}_0^{-1}$$ can be found without issues by solving the defining equation above.

Suppose we now add a more complicated and possibly time-dependent term $$\hat{H}_1(\vb{r}, t)$$, in which case the corresponding fundamental solution $$G(\vb{r}, \vb{r}', t, t')$$ satisfies:

\begin{aligned} \delta(\vb{r} - \vb{r}') \: \delta(t - t') &= \Big( i \hbar \pdv{t} - \hat{H}_0(\vb{r}) - \hat{H}_1(\vb{r}, t) \Big) G(\vb{r}, t; \vb{r}', t') \\ &= \Big( \hat{G}{}_0^{-1}(\vb{r}, t) - \hat{H}_1(\vb{r}, t) \Big) G(\vb{r}, t; \vb{r}', t') \end{aligned}

This equation is typically too complicated to solve, so we would like an easier way to calculate this new $$G$$. The perturbed wavefunction $$\Psi(\vb{r}, t)$$ satisfies the Schrödinger equation:

\begin{aligned} \Big( \hat{G}{}_0^{-1}(\vb{r}, t) - \hat{H}_1(\vb{r}, t) \Big) \Psi(\vb{r}, t) = 0 \end{aligned}

We know that $$\hat{G}{}_0^{-1} \Psi_0 = 0$$, which we put on the right, and then we apply $$\hat{G}_0$$ in front:

\begin{aligned} \hat{G}_0^{-1} \Psi - \hat{H}_1 \Psi = \hat{G}_0^{-1} \Psi_0 \quad \implies \quad \Psi - \hat{G}_0 \hat{H}_1 \Psi &= \Psi_0 \end{aligned}

This equation is recursive, so we iteratively insert it into itself. Note that the resulting equations are the same as those from time-dependent perturbation theory:

\begin{aligned} \Psi &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi \\ &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi \\ &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi_0 + \: ... \\ &= \Psi_0 + \big( \hat{G}_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \hat{G}_0 + \: ... \big) \hat{H}_1 \Psi_0 \end{aligned}

The parenthesized expression clearly has the same recursive pattern, so we denote it by $$\hat{G}$$ and write the so-called Dyson equation:

\begin{aligned} \boxed{ \hat{G} = \hat{G}_0 + \hat{G}_0 \hat{H}_1 \hat{G} } \end{aligned}

Such an iterative scheme is excellent for approximating $$\hat{G}(\vb{r}, t)$$. Once a satisfactory accuracy is obtained, the perturbed wavefunction $$\Psi$$ can be calculated from:

\begin{aligned} \boxed{ \Psi = \Psi_0 + \hat{G} \hat{H}_1 \Psi_0 } \end{aligned}

This relation is equivalent to the Schrödinger equation. So now we have the operator $$\hat{G}(\vb{r}, t)$$, but what about the fundamental solution function $$G(\vb{r}, t; \vb{r}', t')$$? Let us take its definition, multiply it by an arbitrary $$f(\vb{r}, t)$$, and integrate over $$G$$’s second argument pair:

\begin{aligned} \iint \big( \hat{G}{}_0^{-1} \!-\! \hat{H}_1 \big) G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'} = \iint \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'} = f \end{aligned}

Where we have hidden the arguments $$(\vb{r}, t)$$ for brevity. We now apply $$\hat{G}_0(\vb{r}, t)$$ to this equation (which contains an integral over $$t''$$ independent of $$t'$$):

\begin{aligned} \hat{G}_0 f &= \big( \hat{G}_0 \hat{G}{}_0^{-1} - \hat{G}_0 \hat{H}_1 \big) \iint_{-\infty}^\infty G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'} \\ &= \big( 1 - \hat{G}_0 \hat{H}_1 \big) \iint_{-\infty}^\infty G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'} \end{aligned}

Here, the shape of Dyson’s equation is clearly recognizable, so we conclude that, as expected, the operator $$\hat{G}$$ is defined as multiplication by the function $$G$$ followed by integration:

\begin{aligned} \hat{G}(\vb{r}, t) \: f(\vb{r}, t) \equiv \iint_{-\infty}^\infty G(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'} \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.