Why the nonexistence of a "one sentence layman phrase" for the de Broglie relation $p=\hbar k$?

Question

(My question seems most likely will be considered a duplicate of OP (and possibly 1, 2, 3), but it turns out to be WAY TOO LONG as a comment in OP, and the system has deleted the corresponding chat room due to 15 days of no answers thus I asked here instead for a clarification)

Background (All important materials have been listed here, links provided here only for further reference)

1. The 1st link asks about why the momentum is a derivative of the wavefunction.

Summarising the answers there, it says momentum of a particle is a superposition and that translations are described by partial derivatives of x and they describe momentum as conserved due to Noether's Theorem. Alternately, the result can be obtained by the plane wave solution of the Scrodinger equation which demands that $\frac{h}{i}\nabla$ is the momentum operator

2. The 2nd link asks what momentum is, relation that govern photon momentum and how is spin different from (linear) momentum.

The answer says momentum "measures how much a particle moves" and also answered about spin and how photon momentum's existence is deduced from experiments (which is not relevant to this question)

3. The 3rd link asks whether 4 spacetime dimensional events can be interpreted as travelling at a combined speed of c, so in its rest frame it is basically travelling at c through time

The answer says the 4-velocity is interpreted as an event moving through spacetime and the 4 velocity of an event in the rest frame is the same as saying it moves in time but stationary in space.

4. The relativistic equation for energy is \begin{equation} E = m\, c^2\, \frac{1}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} m v^2 + \frac{3}{8} m \frac{v^4}{c^2} + \frac{5}{16} m \frac{v^6}{c^4} + \ldots \end{equation} and momentum is \begin{equation} E^2=m^2c^4+(pc)^2 \end{equation} which in the special case of photons, $(m=0)$ gives $E=pc$

Since 4-momentum is just $p^\mu=mU^\mu$, it can be interpreted as "how much a mass energy moves in spacetime" using the 2nd and 3rd link's answer.

5. Physics forums says the photon momentum is only related to its energy, which by Planck, is related to its (temporal) frequency as $$E=h\nu$$

6. Finally, the OP asks how is the concept of (quantum) momentum extends to a guitar string

The answer said that a guitar string has many particles and each particle described by De Broglie relations, thus to ensure the following $$\vec{p}=\hbar \vec{k}$$ is valid, the number need to be scaled up by many times (e.g. N=no. of atoms times) both sides thus $\vec{P}=N\vec{p}$

The Question

But since the wavevector $\vec{k}$ is spatial frequency (how frequently something repeats itself in a region of space) ,and wavefunctions being probability amplitudes (which is related to how likely is something measured to have a certain state after superposition)

then how exactly does at a fixed moment in time, the "number of times a wave of probability amplitude repeats itself in a certain region in space" is related to "how much it moves (since at an infintesimal moment in time (a $\frac{\partial}{\partial t}$) it is technically speaking not moving at all)"

Or in short, what is the physical intuition/geometric consideration that lead one to relate periodicity in space (spatial frequency) to how much it moves (momentum as defined in 2.) since having a periodic arrangement in space does not imply the object is moving at all relative to some frame of reference (i.e. is at rest wrt that frame). Can this intuition be extended to the 4-momentum of a wave (e.g. photons) to describe it geometrically?

EDIT

In light of Kyle Kanos's suggestion and WillO's answer, we are made clear on the more fundamental meaning of momentum as a generator of (spatial) translation (analogous to how the energy is a generator of time translation by Noether's Theorem).

Now for the case of energy, it is very easy to find a "one sentence phrase" that capture the property of energy in the De Broglie relation $E=h\nu$

"The faster a wave oscillates, the more energy it is storing"--------(1)

(1) describes the following expected scenario in $E=h\nu$: increase in $\nu$ lead to increase in $E$

The comments by ACuriousMind, WillO etc. suggest the spatial analogue of (1) does not exists for the De Broglie relation $p=\frac{h}{\lambda}=\hbar k$, and the best we can say is that

" the eigenfunctions of the momentum operator have the form $e^{ikx}$, with corresponding eigenvalues proportional to $k$"

Why the discrepancy in terms of the existence of a one sentence layman description for $p=\hbar k$, given that both momentum $p$ and total energy $E$ have mathematically similar commutation relations of

$$[x,p]=i\hbar$$ and $$[t,E]=-i\hbar$$

thus should be able to be described simialrly?

Answer 1

I understand the main question to be "What is the physical intuition/geometric consideration that led one to relate periodicity in space to momentum?".

Or, in other words, why do we identify the physical property "momentum" with the mathematical operator $-i\hbar \partial/\partial x$? (This is essentially the same question, because the eigenfunctions of $-i\hbar \partial/\partial x$ are periodic.)

The answer is in four steps:

1. The bilinearity of the classical Poisson bracket implies that for non-commuting observables, the "Poisson bracket" (whatever that is going to mean) should be proportional to the Lie bracket. To see this, expand $\{uv,wx\}$ two different ways and compare.

2. We want the quantum "Poisson bracket" of two observables to be an observable, which means that if $C$ is the constant of proportionality from point 1), then for $u$ and $v$ self-adjoint, $C[u,v]$ should be self-adjoint. This requires $C$ to be pure imaginary, so we write $C=i\hbar$ for some real number $\hbar$.

3. Classically, if $q$ and $p$ are position and momentum (in the same direction), we have $\{q,p\}=1$. So we want to define $q$ and $p$ in such a way that $[q,p]=i\hbar$.

4. Once you've defined $q$ to be multiplication by $x$, you can satisfy point 3) by setting $p=-i\hbar\partial/\partial x$. This isn't the only possible choice, but it's the simplest, and a little algebra shows that any other choice will differ from this one only by a phase factor.

Answer 2

After some in depth discussion with the users, we found that there are a lot of assumptions in the question that does not hold water:

1. "The commutation relation of $E$ and $t$ is not even a commutation relation. Time is not an operator in quantum mechanics." (thanks ACuriousMind for reminding again), it is a parameter

thus there is no (straightforward) analogue for the energy operator $E$ to the commutation relation $[x,p]=i\hbar$

2. "The quantum mechanical wave function is NOT a physical wave. it doesn't "store energy". It acts as a wave because it is a solution to a wave equation, but it is not a wave like a wave on a string or an electromagnetic wave" (Thanks again!), even if it might possibly be a physical entity as strengthen by this research

3. "Wave mechanics works for naive quantum mechanics, but it doesn't capture the essence of what is quantum about the mechanics. I heavily dislike the wave analogies because they do not prepare you to deal with systems that cannot be described by waves, e.g. finite-dimensional system like qubits or unmoving particle with spin." (Thanks!)

Therefore the claim "The faster a wave oscillates, the more energy it is storing" is invalid and the question was built on a wrong premise

hence it is not sensible to expect a "one sentence layman explanation" for $p=\hbar k$ to exist, let alone for the following also: "the relationship between periodicity in space (spatial frequency) to how much it moves (momentum)"

The conclusion? Be careful when trying to interpret quantum mechanics via De Broglie relations to educate a layman

Answer 3

I have no idea what you're on about, here.

The one-sentence laymen's phrase for $p = \hbar~k$ is "Every moving object behaves like a wave, with wavelength equal to Planck's constant divided by that object's momentum." Usually you then add a one-sentence caveat, "The Planck constant is so tiny relative to most macroscopic masses that this wavelength is generally even smaller than the size of a proton for everyday objects at everyday speeds: but at the atomic scale the super-tiny masses generate super-tiny momentums which can give bigger wavelengths that we can see, like in electron diffraction."