# How is the quantum propagator related with Huygens principle?

Usually in quantum mechanics the wave function can be propagated via the so-called Kernel or Amplitude: $$\Psi(x,t) = \int K(x,t;x',t')\Psi(x',t')dx'$$. I have read in some paper that this comes from Huygen's principle, and it makes somehow sense, since we are calculating the wave function at some point and time from the wave function at all the other points and times. However, I have two questions:

First of all, Huygens' principle is referred to waves, i.e., to solutions of the wave equation

$$\frac{\partial^2u}{\partial t^2} = v^2\frac{\partial^2u}{\partial x^2}.$$

However, the Schrödinger equation is not a wave equation since the time and position derivatives are not of the same order.

Second, if $$\Psi(x,t) = \int K(x,t;x',t')\Psi(x',t')dx'$$ is true because of Huygens' principle, could we just say that for every wave function $$u(x,t)$$ that satisfies the above wave equation (which is not the Schrödinger equation) something like

$$u(x,t) = \int G(x,t;x',t')u(x',t')dx'$$

always holds? Could we say that this is the mathematical formulation of Huygens' principle?

• Aug 5, 2019 at 17:04
• Aug 5, 2019 at 18:45
• Feynman, 1948, section 7 reminds you Schroedinger's eqn is a dispersive wave equation albeit different than d' Alembert's, and more generally compatible with Huygen's principle, not less! Aug 5, 2019 at 19:29
• Leaving Huygens' principle aside, the weird propagation equation you posit for some G of the 1-D d'Alembert, integrating over just x, looks unlikely (it is not the Green's function definition). You do not get the miracles operative in the complex diffusion equation (Schroedinger's). Aug 6, 2019 at 0:33
• To see this, note the general solution of d'Alembert is $u(x,t)=R(x-vt)L(x+vt)$, for arbitrary R,L for the right and left movers. So G should be $\delta(x-x')\delta (t-t')$ and you are missing an integration in t'. Aug 6, 2019 at 13:46

Huygens' principle really just says the space a wave travels through is homogeneous. The wave equation $$\partial_t^2u(x,t)-$$ $$c^2\partial_x^2u(x,t)=0$$ is from Maxwell's equations. We can just we well consider the Schroedinger equation for a wave $$\psi(x,t)=\psi(x)e^{-iEt/\hbar}$$ with $$i\hbar\partial_t\psi(x,t)=$$ $$E\psi$$. The Schroedinger wave equation $$i\hbar\partial_t\psi(x,t)=$$ $$H\psi(x,t)$$ then reduces to $$E\psi(x,t)=H\psi(x,t)$$.
A typical propagator for Huygens' principle is found from the case of a wave passing through a slit with $$\frac{e^{i\vec k\cdot\vec r}}{r}K(r)$$ and the propagator would be $$U(r)=\int\frac{e^{i\vec k\cdot\vec r'-iEt/\hbar}}{r'}K(r')dS.$$ This is an integration over the hemispherical wave front so $$dS=2\pi r'dr'$$ and we have $$U(r)=2\pi\int e^{i\vec k\cdot\vec r'-iEt/\hbar}K(r')dr'.$$ The Hamiltonian operator for a free particle $$\hat H=-\frac{\hbar}{2m}\nabla^2$$ gives the Schroedinger equation $$U(r)=\int e^{ikr'-iEt/\hbar}\left(\frac{\hbar}{2m}(k^2K(r')+2i\vec k\cdot\nabla K(r')+\nabla^2K(r')\right)=2\pi\int e^{i\vec k\cdot\vec r'-iEt/\hbar}E.$$ The energy can be read directly off.
The function $$K(r')$$ can be a Bessel function, or it can just be for the simple Huygen's-Fresnel case just $$K(r')=\frac{1}{2}(1+cos\theta)$$ for the angle $$\theta$$ the angle between a normal on the wave front and a line to the point $$r$$. This can be found with elementary analytic geometry.