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