What is the physical significance of the value of wave amplitude being $1$?

Question

In Feynman Lectures Vol.1, it is written that:

First of all, we know that the new way of representing the world in quantum mechanics - the new framework - is to give an amplitude for every event that can occur, and if the event involves the reception of one particle, then we can give the amplitude to find that one particle at different places and at different times. The probability of finding the particle is then proportional to the absolute square of the amplitude. In general , the amplitude to find a particle in different places at different times varies with position and time.

In some special case it can be that the amplitude varies sinusoidally in space and time like $e^{i(\omega t-\vec k\cdot r)},$ where $\vec r$ is the vector position from the origin. (Do not forget that these amplitudes are complex numbers, not real numbers.) Such an amplitude varies according to a definite frequency $\omega$ and wave number $\vec k$...

But when $\omega t=k.r$, the value of the amplitude becomes $1$ which is a real number . What does this mean? What is the physical significance of value of wave amplitude being $1$? Does this mean that $\omega t$ can not be equal to $k.r$?

Answer 1

There's no observable physical significance. Since you can only ever observe a probability density given by a wavefunction, the exact phase of the wave is not observable. Physically, it makes no difference whether at time $t$ and location $r$ the amplitude is $1$, $i$, or $\frac{1-i}{\sqrt{2}}$.

What you can, however, measure, is the relative phase between two waves, via an interference pattern experiment. But even then, this doesn't tell you whether the amplitude of one wave is $1$ or $i$ or whatever.

Actually, the fact that physical observables of wavefunctions don't depend on the absolute phase of the wave is an extremely important symmetry; according to Noether's theorem applied to the Schrodinger equation, this symmetry gives rise to conservation of a "probability current" (which states that probability density obeys a continuity equation, just like electrical current and charge does).

Answer 2

There is no physical significance of the complex amplitude being exactly 1 in the above example. It just means it will constructively interfere with an amplitude of 1 and destructively interfere with an amplitude of -1. Just as an amplitude of i will constructively interfere with an amplitude of i and negatively with an amplitude of -i.

1 is in no way exceptional here. Also there is a global phase invariance in QM, which means that if you shift the phase of the whole wavefunction by a constant the resulting wavefunction will be physically indistinguishable from the original wf. Using this you can set any one point of your wf to 1 if you want, only the relative phases will matter.

Answer 3

This should be a comment, but it is too long. The amplitude, $Ψ$ as :

$e^{i(\omega t-\vec k\cdot r)},$

The observable is the complex conjugate squared $Ψ$ , which gives the probability, the only measurable quantity.

When

$\omega t=k.r$

$Ψ$ becomes $e^{i(0)}$, a complex number .

It is $Ψ^*Ψ$ that becomes equal to 1, a real number. When the probability becomes 1 it means that you have a fixed (not time or space depended) measurement at that value of the variables.

The expression $e^{i(\omega t-\vec k\cdot r)},$ defines a plane wave, the $t$ and $r$ are independent variables in the expression.I cannot see how the equality is physically relevant.