# Why the nonexistence of a “one sentence layman phrase” for the de Broglie relation $p=\hbar k$? [closed]

(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)

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

1. 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)

1. 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.

1. The relativistic equation for energy is $$E = m\, c^2\, \frac{1}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} m v^2 + \frac{3}{8} m \frac{v^4}{c^2} + \frac{5}{16} m \frac{v^6}{c^4} + \ldots$$ and momentum is $$E^2=m^2c^4+(pc)^2$$ 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.

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

2. 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?

• The first three of your this's aren't links. I recommend re-reading your entire post and thinking about whether or not it makes sense to a reader who isn't you. One thing that would help a lot is to include whatever material is relevant from the this's so that your poor readers don't have to sort through any number of other links. The question is already pretty long. – DanielSank Aug 4 '15 at 15:19
• Just some more advice: all of the bold type is excessive and more distracting than helpful. This is a site where expert level physicists do Q&A and you can trust them to understand a physics question without the help of bolding every meaningful sentence. I recommend going through again with this in mind. If you would like I can take a pass at it. – DanielSank Aug 4 '15 at 15:58
• You should not expect to be able to formulate quantum mechanics in classical analogies a layman can understand. – ACuriousMind Aug 4 '15 at 16:45
• It is probably better to understand momentum as the generator of translation than as a "measure of how much mass moves"; cf. this PSE post and this one), among others. – Kyle Kanos Aug 4 '15 at 19:12
• Please stop making trivial edits to your question. That sort of thing is not appropriate SE etiquette. When you edit a post, go through and fix everything that you can find to fix all at once; in most cases you shouldn't be editing any single post more than 2 or 3 times. – David Z Aug 16 '15 at 20:44

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.

• I understood the formalism, but say if you want to describe how a wave's momentum is related to its spatial periodicity to e.g. a taxi driver, how to phrase it?  I know for the temporal case it is kinda straightforward, if your wave oscillates faster then by common sense it is more energetic (this can be modelled analogously with the commutation relation $[H,p]=i\hbar$) But I am at lost on the spatial equivalent, which is the question you point out, since the spatial periodicity is not something that is moving and our intuition on momentum is something that is moving – Secret Aug 4 '15 at 16:19
• I think for the taxi driver it suffices to say that the eigenfunctions of the momentum operator have the form $e^{ikx}$, with corresponding eigenvalues proportional to $k$. I suspect he's unlikely to request further detail. – WillO Aug 4 '15 at 16:22
• Using what you just said, for frequencies in:  Time: The faster a wave oscillates, the more energetic it is  Space: The denser the periodic arrangement of the wave, then (effectively) the more rapidly you go up and down for the same amount of leap forward? – Secret Aug 4 '15 at 16:35
• I have no idea what "the more rapidly you go up and down for the same leap forward" means. – WillO Aug 4 '15 at 17:26
• Suppose you have a wooden stick of length L placed on top of a flexible sculpture that is shaped like a plane wave (with a sine curve as its cross section) and you can extend or squish it like a Piano accordion.  Now if the sculpture suddenly get squished so that the spacing gets smaller (hence increasing spatial frequency at least from the point of view of the stick.), then effectively the stick will "feel" an increasing no. of contact points sliding against it.  Now think of the stick as spatial dimensions of spacetime, then the sculpture is a wave with some properties. So as the – Secret Aug 4 '15 at 17:40

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 $[x,p]=i\hbar$

1. "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

2. "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

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."

• what I am originally looking for is something like "The faster a wave oscillates, the more energy it is storing"----(1) that describes the expected result that when you increase E then you increase $\nu$ because $E=h\nu$  What I then want is an analogous statement for $p=\frac{h}{\lambda}$ that can be understood by a layman  However, as shown in the discussion results, the wavefunction does not store energy, (1) is meaningless, thus I am now not expecting a momentum analogue for (1) will exist – Secret Aug 16 '15 at 17:57
• "The faster a particle moves, the smaller its wavelength" satisfies the "something like" that you are looking for, describing the expected result that when you increase $v$ you decrease $\lambda$ because $\lambda = h / (m v)$... – CR Drost Aug 16 '15 at 18:54