Categories: Physics, Quantum mechanics.

Dyson equation

Consider the time-dependent Schrödinger equation, describing a wavefunction Ψ0(r,t)\Psi_0(\vb{r}, t):

itΨ0(r,t)=H^0(r)Ψ0(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 G0(r,t;r,t)G_0(\vb{r}, t; \vb{r}', t') satisfies the following:

(itH^0(r))G0(r,t;r,t)=δ(rr)δ(tt)\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 G^01(r,t)\hat{G}{}_0^{-1}(\vb{r}, t) as follows, so that G^01G0=δ(r ⁣ ⁣r)δ(t ⁣ ⁣t)\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t'):

G^01(r,t)itH^0(r)\begin{aligned} \hat{G}{}_0^{-1}(\vb{r}, t) &\equiv i \hbar \pdv{}{t}- \hat{H}_0(\vb{r}) \end{aligned}

Note that G^01\hat{G}{}_0^{-1} is an operator, while G0G_0 is a function. For the sake of consistency, we thus define the operator G^0(r,t)\hat{G}_0(\vb{r}, t) as a multiplication by G0G_0 and integration over r\vb{r}' and tt':

G^0(r,t)fG0(r,t;r,t)f(r,t)drdt\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(r,t)f(\vb{r}, t), so that G^01G^0=G^0G^01=1\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 G^01\hat{G}{}_0^{-1}:

G^01(r,t)Ψ0(r,t)=0\begin{aligned} \hat{G}{}_0^{-1}(\vb{r}, t) \: \Psi_0(\vb{r}, t) = 0 \end{aligned}

Let us assume that H^0\hat{H}_0 is simple, such that G0G_0 and G^01\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 H^1(r,t)\hat{H}_1(\vb{r}, t), in which case the corresponding fundamental solution G(r,r,t,t)G(\vb{r}, \vb{r}', t, t') satisfies:

δ(rr)δ(tt)=(itH^0(r)H^1(r,t))G(r,t;r,t)=(G^01(r,t)H^1(r,t))G(r,t;r,t)\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 GG. The perturbed wavefunction Ψ(r,t)\Psi(\vb{r}, t) satisfies the Schrödinger equation:

(G^01(r,t)H^1(r,t))Ψ(r,t)=0\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 G^01Ψ0=0\hat{G}{}_0^{-1} \Psi_0 = 0, which we put on the right, and then we apply G^0\hat{G}_0 in front:

G^01ΨH^1Ψ=G^01Ψ0    ΨG^0H^1Ψ=Ψ0\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:

Ψ=Ψ0+G^0H^1Ψ=Ψ0+G^0H^1Ψ0+G^0H^1G^0H^1Ψ=Ψ0+G^0H^1Ψ0+G^0H^1G^0H^1Ψ0+G^0H^1G^0H^1G^0H^1Ψ0+...=Ψ0+(G^0+G^0H^1G^0+G^0H^1G^0H^1G^0+...)H^1Ψ0\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 G^\hat{G} and write the so-called Dyson equation:

G^=G^0+G^0H^1G^\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 G^(r,t)\hat{G}(\vb{r}, t). Once a satisfactory accuracy is obtained, the perturbed wavefunction Ψ\Psi can be calculated from:

Ψ=Ψ0+G^H^1Ψ0\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 G^(r,t)\hat{G}(\vb{r}, t), but what about the fundamental solution function G(r,t;r,t)G(\vb{r}, t; \vb{r}', t')? Let us take its definition, multiply it by an arbitrary f(r,t)f(\vb{r}, t), and integrate over GG’s second argument pair:

(G^01 ⁣ ⁣H^1)G(r,t)f(r,t)drdt=δ(r ⁣ ⁣r)δ(t ⁣ ⁣t)f(r,t)drdt=f\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 (r,t)(\vb{r}, t) for brevity. We now apply G^0(r,t)\hat{G}_0(\vb{r}, t) to this equation (which contains an integral over tt'' independent of tt'):

G^0f=(G^0G^01G^0H^1)G(r,t)f(r,t)drdt=(1G^0H^1)G(r,t)f(r,t)drdt\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 G^\hat{G} is defined as multiplication by the function GG followed by integration:

G^(r,t)f(r,t)G(r,t;r,t)f(r,t)drdt\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.