Let’s solve the free particle

I guess

Newton’s Law $ F = ma$

$ d^2 x / dt^2=0$

Hence $ x = x_0 + vt$

At least that works. Not sure I derived it particularly. Or proved it unique.

Whatever. Lagrangian version

$ L = T-V = \frac{1}{2}mv^2$

Euler Lagrange Equations

$ \frac{d}{dt} \partial L / \partial \dot{q} = \partial_q L$

How do you get that? By varying the action with fixed endpoints it’s the one that minimizes the path.

$ S = \int L dt = \int \partial/\partial\dot{q} L \delta \dot{q} + \partial_q L \delta q$


$ H = \frac{p^2}{2m}$

$ \dot{p}=-\partial_x H=0$

$ p = Const$

$ \dot{x}=\partial_p H=\frac{p}{m}$

Okay. What about the quantum version?

Well $ p = \frac{\hbar}{i}\partial_x$

How do I know that? In particular it’s hard to remember where the i goes. Well, I memorized it at some point. It follows that

$ [x,p]=-\hbar/i$

But what is

$ i \partial_t \psi = -\frac{\hbar^2}{2m}\nabla^2 \psi$

$ E\psi = $

Whatever. I’m bored.

Maybe I’ll do the path integral some other day