# Green's function for the inhomogenous free particle time-dependent Schrodinger equation in 3D

I need to solve the equation $$\mathrm{i}\hbar \frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\left(\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}\right)+ f(x,y,z,t)$$ in the domain $$t\ge 0$$ with boundary condition $$\Psi=0$$ at $$t=0$$. Note that there is no $$\Psi$$ multiplying the $$f$$, it is not a potential problem. The function $$f$$ can be assumed to be square integrable and arbitrarily nice. The obvious thing to try is a Green's function, but when I try to write a Green's function for this in terms of functions of radius and time things get complicated. I cannot find a Green's function for this case anywhere, but I am sure that it must have been done. I would be very grateful if anyone could offer some advice or give a place to look.

This came up in study of the Schrödinger equation in noncommutative space time, and I specified the exact equation to be solved to minimize misunderstandings — it is a free particle equation (no potential) with inhomogenous forcing term.

• I think you could go to the momentum frequency representation and find something like: $\Psi(\omega, k) = f(\omega,k) \left(\hbar \omega - \hbar^2 k^2/2m\right)^{-1}$ Jan 23 at 15:30
• Jan 24 at 9:42
• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. Feb 2 at 3:06

If there were no "$$i$$" this would be the inhomogeneous heat equation. In one space and one time dimension it would be $$\frac{\partial \psi}{\partial t}= \frac{\partial^2 \psi}{\partial x^2}+f(x,t).$$ The solution to this involves the heat kernel $$G(x,t,\xi,\tau) =\theta(t-\tau)\frac{1}{\sqrt{4\pi (t-\tau)}} \exp\left\{-\frac{1}{4(t-\tau)}(x-\xi)^2\right\}$$ were $$\theta(t)$$ is the Heaviside step function.

With initial conditions $$\psi(x,0)$$ at $$t=0$$, the solution for $$t>0$$ is
$$\psi(x,t)= \int_{-\infty}^\infty G(x,t,\xi,0)\psi(\xi,0) \,d\xi + \int_{-\infty}^\infty \left( \int_0^t G(x,t,\xi,\tau)f(\xi,\tau)\,d\tau\right)d\xi.$$ Your problem is similar but the heat kernel needs to be replaced by the Schroedinger propagator. Hope this hint helps!

Green's function definition
The Green's function is defined by equation $$\left[i\frac{\partial }{\partial t}+\frac{\hbar^2\nabla^2}{2m}\right]G(\mathbf{x},t;\mathbf{x}',t')=\delta(\mathbf{x}-\mathbf{x}')\delta(t-t').$$ I also assume that we are looking here for a retarded Green's function, i.e., $$G(\mathbf{x},t;\mathbf{x}',t')=0\text{ for } t (This bit is not crucial for the derivation below, but may be a of key importance, e.g., if using Fourier transform).

Green's function of a Schrödinger operator
There are exist many methods for solving the equationf or the Green's function. One method, that is useful to know, is the general expansion for the Green's function of a Schrödinger operator in terms of its eigenvunstions:
$$\left[i\frac{\partial }{\partial t}+H\right]G(\mathbf{x},t;\mathbf{x}',t')=\delta(\mathbf{x}-\mathbf{x}')\delta(t-t'),H\varphi_n(\mathbf{x})=\epsilon_n\varphi_n(\mathbf{x})\Rightarrow\\ G(\mathbf{x},t;\mathbf{x}',t')=\sum_n\varphi_n(\mathbf{x})\varphi_n(\mathbf{x}')^*e^{-i\epsilon_n(t-t')/\hbar}\Theta(t-t'),$$ where $$\theta(t-t')$$ is the Heavyside step function: $$\Theta(t-t') \begin{cases}1, \text{ if } t>t',\\ 0,\text{ otherwise.}\end{cases}$$ One can check that this expression is indeed the Green's function by substituting it to the defining equation.

Solution
In our case $$\varphi_\mathbf{k}(\mathbf{x})\propto e^{-i\mathbf{k}\mathbf{x}},\epsilon_\mathbf{k}=\hbar\omega_\mathbf{k}=\frac{\hbar^2\mathbf{k}^2}{2m},$$ and the sum is replaced by an integral, so that we have $$G(\mathbf{x},t;\mathbf{x}',t')=\Theta(t-t')\int \frac{d^3\mathbf{k}}{(2\pi)^3}e^{i\mathbf{k}(\mathbf{x}-\mathbf{x}')-i\omega_\mathbf{k}(t-t')}$$ The rest is math. For one, it is easier to carry out integration in polar coordinates, assuming that the z-axis is along $$\mathbf{x}-\mathbf{x}'$$. The expression then becomes $$G(\mathbf{x},t;\mathbf{x}',t')=\Theta(t-t')\frac{-1}{(2\pi)^3}\int_0^{2\pi}d\phi\int_0^\pi d\theta\int_0^{+\infty}dk k^2\sin\theta e^{ik|\mathbf{x}-\mathbf{x}'|\cos\theta-i\frac{\hbar k^2(t-t')}{2m}}$$ Carrying integration over $$\phi$$ and $$\theta$$ gives $$G(\mathbf{x},t;\mathbf{x}',t')=\Theta(t-t')\frac{2}{(2\pi)^2|\mathbf{x}-\mathbf{x}'|}\int_0^{+\infty}dk k\sin\left(k|\mathbf{x}-\mathbf{x}'|\right)e^{-i\frac{\hbar k^2(t-t')}{2m}}=\\ \frac{1}{(2\pi)^2|\mathbf{x}-\mathbf{x}'|}\int_{-\infty}^{+\infty}dk ke^{ik|\mathbf{x}-\mathbf{x}'|-i\frac{\hbar k^2(t-t')}{2m}}$$ (In the second line I used the fact that the real part of $$k\cos\left(k|\mathbf{x}-\mathbf{x}'|\right)e^{-i\frac{\hbar k^2(t-t')}{2m}}$$ is an odd functiona nd gives zero, when integrated in symmetric limits $$(-\infty,+\infty)$$.) The last integral is a Gaussian-like integral, and is easily calculated - if you are not at ease with Gaussian integrals, Gradshtein&Ryzhik give a formula (GR3.462-6): $$\int_{-\infty}^{+\infty}xe^{-px^2+2qx}dx=\frac{q}{p}\sqrt{\frac{\pi}{p}}ee\frac{q^2}{p}$$ The final answer is therefore: $$G(\mathbf{x},t;\mathbf{x}',t')=i\Theta(t-t')\left(\frac{m}{2\pi i\hbar (t-t')}\right)^{\frac{3}{2}}\exp\left[-\frac{m(\mathbf{x}-\mathbf{x}')^2}{2 i\hbar (t-t')}\right]$$

Remark: I urge you to go through the derivation, as I cannot guarantee that there have not been minor errors in my derivation, as well as typos.