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In this Veritasium video "Infinite Slit Paradox: Something Strange Happens When You Trust QM", the Feynman integral is built up from just one formula (that is completely brushed under the rug by the video): that for $\phi$ the phase of a particle wave going along a path, on a very small almost-linear segment of the path,

$$\Delta \phi = \frac{2\pi}{\lambda}\Delta x - 2\pi f \Delta t,$$

where $\lambda, f$ I think are the de Broglie wavelength and frequency for that particle.

Can this formula be (rigorously?) derived from some simple/natural assumptions, at least in some simple/natural setting? I would appreciate if people could state these assumptions/settings explicitly.

If so, would that be a complete and satisfactory answer to this Phys.SE question on why exactly the formula in the Feynman path integral is the way it is?

Qmechanic
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D.R
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3 Answers3

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  1. Inspired by (among other things)

    the Veritasium video is postulating at 20:32 that the amplitude $K(b,a)$ is given by a sum of histories via the Feynman stopwatch hand/arrow picture $$ K(b,a) ~=~ \sum_{\text{paths $\gamma$ from $a$ to $b$}} e^{i\int_{\gamma}\mathrm{d}\phi},\tag{1}$$ with infinitesimal phase $$ \mathrm{d}\phi~=~k\mathrm{d}x-\omega\mathrm{d}t, \qquad k~\equiv~\frac{2\pi}{\lambda}, \qquad \omega~\equiv~ 2\pi f.\tag{2}$$

  2. Next recall the de Broglie relations for momentum $$\frac{\partial L}{\partial\dot{x}}~\equiv~p~=~\hbar k~\equiv~\frac{h}{\lambda},\tag{3}$$ and energy $$p\dot{x}-L~\equiv~E~=~\hbar\omega~\equiv~hf.\tag{4}$$

  3. Also recall the formula for an infinitesimal action $$\mathrm{d}S~=~L\mathrm{d}t~=~(p\dot{x}-E)\mathrm{d}t~=~p\mathrm{d}x-E\mathrm{d}t,\tag{5}$$ cf. the Lemma in my Phys.SE answer here.

  4. The Veritasium video is arguing at 22:33 that $$\mathrm{d}\phi~\stackrel{(2)+(3)+(4)+(5)}{=}~\frac{\mathrm{d}S}{\hbar},\tag{6}$$ so that we arrive at the Feynman path integral $$ K(b,a) ~\stackrel{(1)+(6)}{=}~ \sum_{\text{paths $\gamma$ from $a$ to $b$}} e^{i\int_{\gamma}\mathrm{d}S/\hbar},\tag{7}$$

Qmechanic
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This is related to any wave equation, the propagation of light follows this equation. You can solve it by expanding on the basis $e^{i\phi}=e^{i(kx-\omega t)}$ (in 1D) where $k=2\pi/\lambda$ and $\omega=2\pi f$. These solutions are called plane waves. Equivalently, you can write any solution as sum of these plane waves (Fourier decomposition).

For electrons, Schrödinger equation is a wave equation so it follows the same idea. For free electrons you find plane wave solutions.

Mauricio
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You might be familiar with the formula: $$\Delta \phi = \frac{2\pi}{\lambda}\Delta x$$ This is true for a snapshot of a wave or a standing wave. It gives you the phase difference between two points in space at a given instant. It can be easily proved considering the general wave equation and dropping the time dependent term.

The formula stated in the Veritasium video can be simplified as follows: $$\Delta \phi = \frac{2\pi}{\lambda}\Delta x - 2\pi f \Delta t$$ $$\Delta \phi = \frac{2\pi}{\lambda}\Delta x - 2\pi \frac{v}{\lambda}\ \Delta t$$ $$\Delta \phi = \frac{2\pi}{\lambda}(\Delta x - v \Delta t)$$ Clearly some term $v\Delta t$ has been subtracted. This terms accounts to the wave's progression over time.

Consider a plane wave travelling in the x direction: $$E(x,t) = A e^{i(kx - \omega t)}$$ The phase of a point on this wave is given by: $$\phi(x,t)=kx - \omega t $$
Consider a point at some distance $\Delta x$ and at after time $\Delta t$ The phase of this point will be $$\phi' = k(x + \Delta x) - \omega (t + \Delta t)$$

Therefore, the phase difference will be: $$\Delta\phi = \phi' - \phi$$ $$\Delta\phi = k(x + \Delta x) - \omega (t + \Delta t) - (kx - \omega t)$$ $$\Delta \phi = k\Delta x - \omega\Delta t$$ hence, $$\Delta \phi = \frac{2\pi}{\lambda} \Delta x - 2\pi f \Delta t$$

Kyathallous
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