In its time-dependent form, the Schrödinger’s equation reads:
H ^ Ψ ( r ⃗ ) = E Ψ ( r ⃗ ) , (1) \hat H \Psi(\vec r) = E \Psi(\vec r)
\tag{1}, H ^ Ψ ( r ) = E Ψ ( r ) , ( 1 ) where the Hamiltonian operator $\hat H$ is $-\frac{\hbar^2}{2m} \nabla^2 + V(\vec r)$.
Let’s first consider the free particle situation where $V=0$, we write the solution as $\Psi_0(\vec r)$:
[ − ℏ 2 2 m ∇ 2 ] Ψ 0 ( r ⃗ ) = E Ψ 0 ( r ⃗ ) . (3) \left[-\frac{\hbar^2}{2m} \nabla^2 \right] \Psi_0(\vec r) = E \Psi_0(\vec r). \tag{3} [ − 2 m ℏ 2 ∇ 2 ] Ψ 0 ( r ) = E Ψ 0 ( r ) . ( 3 ) After some rearranging, we get:
[ E + ℏ 2 2 m ∇ 2 ] Ψ 0 ( r ⃗ ) = [ E − H ^ 0 ( r ⃗ ) ] Ψ 0 ( r ⃗ ) = 0 (4) \begin{aligned}
\left[ E + \frac{\hbar^2}{2m} \nabla^2\right] \Psi_0 (\vec r) =\left[ E - \hat H_0 (\vec r) \right] \Psi_0 (\vec r) = 0 \tag{4}
\end{aligned} [ E + 2 m ℏ 2 ∇ 2 ] Ψ 0 ( r ) = [ E − H ^ 0 ( r ) ] Ψ 0 ( r ) = 0 ( 4 ) and here, the differential equation is homogeneous. The Green’s function for the operator $\left[ E - \hat H_0 (\vec r) \right]$ can be written as:
[ E − H ^ 0 ( r ⃗ ) ] g ( r ⃗ , r ⃗ ′ ) = δ ( r − r ′ ) (5) \left[ E - \hat H_0 (\vec r) \right] g(\vec r, \vec r' ) = \delta(r-r') \tag{5} [ E − H ^ 0 ( r ) ] g ( r , r ′ ) = δ ( r − r ′ ) ( 5 ) Now, let’s add back the potential term (time-independent) as perturbation, we write the solution to this Schrödinger equation as $\Psi(\vec r)$:
[ H ^ 0 + V ( r ⃗ ) ] Ψ ( r ⃗ ) = E Ψ ( r ⃗ ) [ E − H ^ 0 ( r ⃗ ) ] Ψ ( r ⃗ ) = V ( r ⃗ ) Ψ ( r ⃗ ) . (6) \begin{aligned}
\left[\hat H_0+V(\vec r)\right]\Psi(\vec r) &= E \Psi(\vec r)\\
\left[ E - \hat H_0 (\vec r) \right] \Psi (\vec r) &= V(\vec r) \Psi(\vec r).
\end{aligned} \tag{6} [ H ^ 0 + V ( r ) ] Ψ ( r ) [ E − H ^ 0 ( r ) ] Ψ ( r ) = E Ψ ( r ) = V ( r ) Ψ ( r ) . ( 6 ) Here, we are treating $V(\vec r) \Psi(\vec r)$ as the source of the “inhomogeneous” and because $V$ is treated as perturbation, we have kept the energy constant as $E$.
Since we alreay know the Green’s function of the unperturbed operator $E - \hat H_0 (\vec r)$, we can express the solution to this equation as:
Ψ ( r ⃗ ) = Ψ 0 ( r ⃗ ) + ∫ d r ⃗ ′ g ( r ⃗ , r ⃗ ′ ) V ( r ⃗ ′ ) Ψ ( r ⃗ ′ ) . (7) \Psi (\vec r) = \Psi_0 (\vec r) + \int d \vec r' g(\vec r, \vec r' ) V(\vec r') \Psi(\vec r'). \tag{7} Ψ ( r ) = Ψ 0 ( r ) + ∫ d r ′ g ( r , r ′ ) V ( r ′ ) Ψ ( r ′ ) . ( 7 ) At this point, we have two ways to get the Dyson equation.
Using operator
Alternatively, we can still express $\Psi (\vec r, t)$ more directly from the Green’s function of the perturbed system with $G(\vec r, \vec r’)$:
( E − H ^ ) G ( r ⃗ , r ⃗ ′ ) = δ ( r ⃗ − r ⃗ ′ ) (8) (E-\hat H) G(\vec r, \vec r' ) = \delta(\vec r - \vec r') \tag{8} ( E − H ^ ) G ( r , r ′ ) = δ ( r − r ′ ) ( 8 ) with $\hat H = \hat H_0 + V$.
Using Eq. 5, we can write Eq. 8 as:
G ( r ⃗ , r ⃗ ′ ) = [ E − H ^ ] − 1 δ ( r ⃗ − r ⃗ ′ ) = [ E − H ^ ] − 1 [ E − H ^ 0 ( r ⃗ ) ] g ( r ⃗ , r ⃗ ′ ) = [ E − H ^ ] − 1 [ E − H ^ + V ( r ⃗ ) ] g ( r ⃗ , r ⃗ ′ ) = g ( r ⃗ , r ⃗ ′ ) + [ E − H ^ ] − 1 V ( r ⃗ ) g ( r ⃗ , r ⃗ ′ ) (9) \begin{aligned}
G\left(\vec r, \vec r^{\prime}\right) &=[E-\hat H]^{-1} \delta\left(\vec r-\vec r^{\prime}\right) \\
&=[E-\hat H]^{-1}[ E-\hat H_{0}(\vec r) ] g\left(\vec r, \vec r^{\prime} \right) \\
&=[E-\hat H]^{-1} [E-\hat H+V(\vec r)] g\left(\vec r, \vec r^{\prime} \right) \\
&=g\left(\vec r, \vec r^{\prime} \right)+ [ E-\hat H ]^{-1} V(\vec r) g\left(\vec r, \vec r^{\prime} \right) \tag{9}
\end{aligned} G ( r , r ′ ) = [ E − H ^ ] − 1 δ ( r − r ′ ) = [ E − H ^ ] − 1 [ E − H ^ 0 ( r )] g ( r , r ′ ) = [ E − H ^ ] − 1 [ E − H ^ + V ( r )] g ( r , r ′ ) = g ( r , r ′ ) + [ E − H ^ ] − 1 V ( r ) g ( r , r ′ ) ( 9 ) and according to the inverse operator notation :
… ( λ − L ^ ) − 1 = ∫ … G ( x , x ′ ) d x ′ (10) \ldots(\lambda-\hat{L})^{-1}=\int \ldots G\left(x, x^{\prime}\right) d x^{\prime} \tag{10} … ( λ − L ^ ) − 1 = ∫ … G ( x , x ′ ) d x ′ ( 10 ) we can express Eq. 9 as:
G ( r ⃗ , r ⃗ ′ ) = g ( r ⃗ , r ⃗ ′ ) + ∫ G ( r ⃗ , r ⃗ 1 ) V ( r ⃗ 1 ) g ( r ⃗ 1 , r ⃗ ′ ) d r ⃗ 1 . (11) G\left(\vec r, \vec r^{\prime} \right)=g\left(\vec r, \vec r^{\prime} \right)+\int G\left(\vec r, \vec r_{1}\right) V\left(\vec r_{1}\right) g\left(\vec r_{1}, \vec r^{\prime}\right) d \vec r_{1}. \tag{11} G ( r , r ′ ) = g ( r , r ′ ) + ∫ G ( r , r 1 ) V ( r 1 ) g ( r 1 , r ′ ) d r 1 . ( 11 ) If we drop the integration symbols:
G = g + g V G G = g + gVG G = g + g V G This result is called the Dyson equation, and it allows us to express the Green’s function of the perturbed system in term of the unperturbed one.
Using wavefunctions
Similarly, we can find $\Psi(\vec r)$ from $G\left(\vec r, \vec r^{\prime} \right)$, to do this, we re-express the un-perturbed equation as:
E Ψ 0 ( r ⃗ ) − H ^ 0 Ψ 0 ( r ⃗ ) = 0 E Ψ 0 ( r ⃗ ) − ( H ^ 0 + V ) Ψ 0 ( r ⃗ ) = − V Ψ 0 ( r ⃗ ) ( E − H ^ ) Ψ 0 ( r ⃗ ) = − V Ψ 0 ( r ⃗ ) (12) \begin{aligned}
E\Psi_0(\vec r) - \hat H_0 \Psi_0(\vec r) &= 0\\
E\Psi_0(\vec r) - (\hat H_0 + V) \Psi_0(\vec r) &= -V \Psi_0(\vec r)\\
(E - \hat H )\Psi_0(\vec r) &= -V \Psi_0(\vec r)\\ \tag{12}
\end{aligned} E Ψ 0 ( r ) − H ^ 0 Ψ 0 ( r ) E Ψ 0 ( r ) − ( H ^ 0 + V ) Ψ 0 ( r ) ( E − H ^ ) Ψ 0 ( r ) = 0 = − V Ψ 0 ( r ) = − V Ψ 0 ( r ) ( 12 ) Using Eq. 8 and noting that V is treated as perturbation so that we can keep $E$, we get:
Ψ 0 ( r ⃗ ) = Ψ ( r ⃗ ) − ∫ G ( r ⃗ , r ⃗ ′ ) V ( r ⃗ ′ ) Ψ 0 ( r ⃗ ′ ) d r ⃗ ′ (13) \Psi_0(\vec r) =\Psi(\vec r) - \int G(\vec r, \vec r') V(\vec r') \Psi_0(\vec r') d \vec r' \tag{13} Ψ 0 ( r ) = Ψ ( r ) − ∫ G ( r , r ′ ) V ( r ′ ) Ψ 0 ( r ′ ) d r ′ ( 13 ) Note that the homogeneous solution to operator $(E - \hat H )$ in Eq. 12 is just $\Psi(\vec r)$.
Rearranging Eq. 13, we get:
Ψ ( r ⃗ ) = Ψ 0 ( r ⃗ ) + ∫ G ( r ⃗ , r ⃗ ′ ) V ( r ⃗ ′ ) Ψ 0 ( r ⃗ ′ ) d r ⃗ ′ (13) \Psi(\vec r) =\Psi_0(\vec r) + \int G(\vec r, \vec r') V(\vec r') \Psi_0(\vec r') d \vec r' \tag{13} Ψ ( r ) = Ψ 0 ( r ) + ∫ G ( r , r ′ ) V ( r ′ ) Ψ 0 ( r ′ ) d r ′ ( 13 ) Remembering Eq. 7, dropping the integration, with Eq. 13, we can re-write it as:
Ψ = Ψ 0 + g V Ψ = Ψ 0 + g V ( Ψ 0 + g V Ψ ) = Ψ 0 + g V Ψ 0 + g V g V ( Ψ 0 + g V Ψ ) = Ψ 0 + ( g + g V g + g V g V g + ⋯ ) V Ψ 0 (14) \begin{aligned}
\Psi &= \Psi_0 + gV\Psi\\
&= \Psi_0 + gV (\Psi_0 + gV \Psi)\\
&= \Psi_0 + gV\Psi_0 + gVgV(\Psi_0 + gV \Psi)\\
&= \Psi_0 + (g + gVg + gVgVg+ \cdots)V\Psi_0\\
\end{aligned} \tag{14} Ψ = Ψ 0 + g V Ψ = Ψ 0 + g V ( Ψ 0 + g V Ψ ) = Ψ 0 + g V Ψ 0 + g V g V ( Ψ 0 + g V Ψ ) = Ψ 0 + ( g + g V g + g V g V g + ⋯ ) V Ψ 0 ( 14 ) Comparing Eq. 13 with Eq. 13, we see that:
G = g + g V G G = g + gVG G = g + g V G or symbolically:
G − 1 = g − 1 − V . G^{-1} = g^{-1} -V. G − 1 = g − 1 − V . Which, again, gives us the so-called Dyson equation.
Inverse operator notation
Because:
( λ − L ^ ) G ( x , x ′ ) = δ ( x , x ′ ) ∫ ( λ − L ^ ) G ( x , x ′ ) f ( x ′ ) d x ′ = ∫ δ ( x , x ′ ) f ( x ′ ) d x ′ ( λ − L ^ ) ∫ G ( x , x ′ ) f ( x ′ ) d x ′ = f ( x ) ∫ G ( x , x ′ ) f ( x ′ ) d x ′ = ( λ − L ^ ) − 1 f ( x ) (A.1) \begin{aligned}
(\lambda - \hat L) G(x,x') &= \delta (x,x')\\
\int (\lambda - \hat L) G(x,x') f(x') dx' &= \int \delta (x,x') f(x') dx'\\
(\lambda - \hat L) \int G(x,x') f(x') dx' &= f(x)\\
\int G(x,x') f(x') dx' &= (\lambda - \hat L)^{-1} f(x)\\
\end{aligned} \tag{A.1} ( λ − L ^ ) G ( x , x ′ ) ∫ ( λ − L ^ ) G ( x , x ′ ) f ( x ′ ) d x ′ ( λ − L ^ ) ∫ G ( x , x ′ ) f ( x ′ ) d x ′ ∫ G ( x , x ′ ) f ( x ′ ) d x ′ = δ ( x , x ′ ) = ∫ δ ( x , x ′ ) f ( x ′ ) d x ′ = f ( x ) = ( λ − L ^ ) − 1 f ( x ) ( A.1 ) we can infer:
… ( λ − L ^ ) − 1 = ∫ … G ( x , x ′ ) d x ′ (A.2) \ldots(\lambda-\hat{L})^{-1}=\int \ldots G\left(x, x^{\prime}\right) d x^{\prime} \tag{A.2} … ( λ − L ^ ) − 1 = ∫ … G ( x , x ′ ) d x ′ ( A.2 ) Where $\ldots$ can be any operator or functions.
Reference: Green’s functions in quantum mechanics courses
Reference: An introduction to Green’s function in many-body condensed-matter quantum systems