# How does the Green's function related the wavefunctions at different space-time points in Schrödinger's equation?

I have been trying to study Quantum Field Theory and have come across Green's Functions for the first time. While referring to Tom Lancaster's book Quantum Field Theory for the Gifted Amateur, the author states in Eqn 16.13 that $$\phi(x,t_x)=\int dy\ G^{+}(x,t_x,y,t_y)\phi(y,t_y)\tag{16.13},$$ where $$G^+$$ is the time-retarded Green's Function.

I understand that in language of Green's Functions, we are trying to solve $$\mathcal{L}\phi=0$$ where $$\mathcal{L} = \hat{H}-i\frac{\partial}{\partial t}$$ is our Linear Operator. I also understand the fact that the Green's Function by definition satisfies $$\mathcal{L}\ G^+(x,t_x,y,t_y)=\delta(x-y)\delta(t_x-t_y).$$ I know that if any differential equation can be expressed as $$\mathcal{L}\ \psi(x)=f(x)$$, then we can express the solution as

$$\psi(x)=\int du\ G(x,u) \ f(u)\ du,$$ however, in this case the source term seems to be 0!, so I cannot understand how exactly we are writing the first expression. It would be really helpful if the derivation can be outlined as the book does not cover it, it just cites that the 16.13 follows from 16.6, I am failing to see the connection here. Operating by $$\mathcal{L}$$ on both sides of the first equation and using the definition of the Green's function seems to give $$\mathcal{L}\ \psi(x,t_x)=\psi(x,t_y)\delta(t_x-t_y),$$ which is another statement that I cannot make sense of.

The author also makes a comment that there is "no need" to do an integral over time coordinates, which implies somehow the dependence on time of the wavefunction and the Green's Function would somehow cancel out, it would also help me if anyone call tell me how to show that explicitly.

Suppose you know the wavefunction $$\phi(t^\prime,x^\prime)=\langle x^\prime |\phi(t^\prime)\rangle \tag{1} \label{1}$$ (in the Schrödinger picture) at some (fixed) initial time $$t^\prime$$ and you want to find the wavefunction $$\phi(t,x)=\langle x|\phi(t)\rangle=\langle x|e^{-iH(t-t^\prime)}|\phi(t^\prime)\rangle \tag{2} \label{2}$$ at a later time $$t\gt t^\prime$$ when the time evolution is governed by the Hamilton operator $$H$$. Inserting the identity operator $$\mathbb{I}$$ in the form $$\mathbb{I}=\!\int\limits_{-\infty}^\infty \!dx^\prime |x \rangle \langle x^\prime|, \tag{3} \label{3}$$ eq. \eqref{2} can be rewritten in the form $$\phi(t,x)=\!\int\limits_{-\infty}^\infty \! dx^\prime \langle x |e^{-iH(t-t^\prime)} |x^\prime\rangle \langle x^\prime |\phi(t^\prime)\rangle= \int\limits_{-\infty}^\infty \! dx^\prime \langle x |e^{-iH(t-t^\prime)}|x^\prime\rangle\, \phi(t^\prime, x^\prime). \tag{4} \label{4}$$ Note that the integral kernel $$\langle x|e^{-iH(t-t^\prime)} |x^\prime \rangle$$ satisfies the homogeneous Schrödinger equation $$\left(i\, \partial/\partial t- H_x\right) \langle x |e^{-iH(t-t^\prime)} |x^\prime \rangle=0, \tag{5} \label{5}$$ where $$H_x$$ denotes the action of the Hamilton operator in the $$x$$-representation. As we have assumed $$t\gt t^\prime$$, we could equally well determine $$\phi(t,x)$$ from the initial wave function $$\phi(t^\prime,x^\prime)$$ by $$\phi(t,x) =\int\limits_{-\infty}^\infty\! dx^\prime \, G^+(t,x;t^\prime,x^\prime) \phi(t^\prime, x^\prime) \tag{6} \label{6}$$ with $$G^+(t,x;t^\prime,x^\prime)= \langle x |e^{-iH(t-t^\prime)} |x^\prime \rangle \, \theta(t-t^\prime), \tag{7} \label{7}$$ where $$\theta(t-t^\prime)$$ denotes the Heaviside step function. Using \eqref{5} and $$\frac{\partial}{\partial t}\theta(t-t^\prime) =\delta(t-t^\prime)$$, we find \begin{align}\left(i \partial /\partial t -H_x \right) G^+(t,x;t^\prime,x^\prime) &= i\langle x |e^{-iH(t-t^\prime)} |x^\prime \rangle \delta(t-t^\prime) \\[5pt] &= i \langle x | x^\prime \rangle \delta(t-t^\prime)\\[5pt] &= i \delta(x-x^\prime) \delta(t-t^\prime), \end{align} \tag{8} \label{8} showing that $$G^+(t,x;t^\prime,x^\prime)$$ is indeed the retarded Green function of the differential operator $$\partial /\partial t +i H_x$$.

As an example, let us consider the Hamilton operator of a free particle, $$H=\frac{P^2}{2m}, \qquad H_x=-\frac{1}{2m} \frac{\partial^2}{\partial x^2}. \tag{9} \label{9}$$ $$G^+$$ can either be obtained by computing \begin{align} G^+(t,x;t^\prime,x^\prime) &=\langle x |e^{-iP^2(t-t^\prime)/2m} |x^\prime \rangle \theta(t-t^\prime) \\[5pt]&= \int\limits_{-\infty}^\infty \! \! dp\, e^{-ip^2(t-t^\prime)/2m} \langle x|p\rangle\langle p|x^\prime \rangle \theta(t-t^\prime) \\[5pt] &= \int\limits_{-\infty}^\infty \! \! \frac{dp}{2\pi}\, e^{-ip^2(t-t^\prime)/2m} e^{ip(x-x^\prime)} \theta(t-t^\prime), \end{align} \tag{10} \label{10} or, alternatively, by inserting the Fourier representation (using translation invariance of the differential operator) $$G^+(t,x;t^\prime,x^\prime) = \int\limits_{\mathbb{R}^2} \frac{d \omega \,dp}{2\pi} e^{-i\omega (t-t^\prime)} e^{ip(x-x^\prime)}\tilde{G}(\omega,p) \tag{11}$$ into \eqref{8}. The form of $$\tilde{G}(\omega,p)$$ is easily obtained and the $$\omega$$-integration can be performed by complex integration choosing the path in such a way that the boundary condition $$G^+(t,x;t^\prime,x^\prime) =0$$ for $$t \lt t^\prime$$ is satisfied. In both cases one finds the final result $$G^+(t,x;t^\prime,x^\prime)= \left(\frac{m}{2\pi i (t-t^\prime)} \right)^{\! 1/2}e^{i m(x-x^\prime)^2/2(t-t^\prime)} \theta(t-t^\prime). \tag{12}$$

In short, the link between the two is Duhamel’s principle.

Green’s functions arise whenever you want to solve an inhomogeneous linear problem. In the case of the Schrödinger equation, there are two possible sources. The first one is in the RHS assuming a zero initial condition: $$G_+(t=0)=0\\ (\partial_t+iH)G_+=\delta(x,t)$$ However the problem you are rather interested in is the initial value problem: $$G(t=0)=\delta(x)\\ (H-i\partial_t)G=0$$ The two Green’s functions are the same for $$t>0$$ because the equation is first order in time. Explicitly: $$G_+(t)=H(t)G(t)$$ with $$H$$ the Heaviside function. The general relation between the two is given by Duhamel’s principle allowing you to convert a source into an initial value problem.

In fact, you can already see this using ODE’s. Physically, you can get an intuitive feel of this considering Newton’s second law. The impulse response: $$m\ddot x=\delta(t)$$ is typically solved by assuming a discontinuity in the velocity converting the inhomogeneous force to an initial value problem with no force.

Heuristically, integrating the equation of $$G_+$$ over a small neighborhood of $$t=0$$ gives: $$G_+(t=0^+)= G_+(t=0^+)-G_+(t=0^-)=\delta(x)$$ and for $$t>0$$, both $$G_+$$ and $$G$$ satisfy the same homogeneous Schrödinger equation. By unicity of the IVP, they are equal for $$t>0$$.

Hope this helps.