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.
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