I know that quantum momentum is inversely proportional to the wavelength of the probability or matter wave of a given particle, but I don't get how this relation of this abstract mathematical construct (the probability wave) relates to the actual observable property (momentum). I don't get how $mv = h/λ$ when mass times velocity is something very "real" and classical while wavelength times Planck's constant is not.
Basically, can someone please explain how the momentum of a probability wave (given by $p = h/λ$) is the same as the momentum of the particle that the probability wave describes (given by $p = mv$)? Please do not use too much math in your answer because I don't know too much of it.


3 Answers 3


Basically, can someone please explain how the momentum of a probability wave (given by p = h/λ) is the same as the momentum of the particle that the probability wave describes (given by p = mv)?

It's hard to explain this with a short answer since to even start moving towards an "answer" to this question requires explaining many of the basics of Quantum Mechanics. And these basics are somewhat inexorably stuck in a mathematics formalism that at least requires you to understand calculus... but anyways, let's try...

In single-particle quantum mechanics, one finds out that, unfortunately, one simply can not make fully deterministic predictions about the position of a "particle" (for this discussion a "particle" means something like an electron). Instead, one has to characterize a particle as being described by a probability-amplitude wave function (often called $\Psi(x,t)$, where $x$ is a spacial position argument).

The absolute square of the probability-amplitude gives the probability density that the particle is "at x."

For example, $\int_{x_1}^{x_2}|\Psi(x,t)|^2dx$ is the probability that the particle is between $x_1$ and $x_2$ at time t.

For example, $\int_{-\infty}^{\infty}|\Psi(x,t)|^2dx=1$ (since the integral of a probability density over all possible values is 1).

Quantum mechanics also give us a way to figure out how the wave function $\Psi(x,t)$ changes with time. The equation that determines how $\Psi(x,t)$ changes with time is called Schrodinger's equation.

For example, the "free" Schrodinger equation that determines how a "free" particle wave-function $\Psi_f$ changes with time is: $$ \frac{i\hbar\partial \Psi_f(x,t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi_f(x,t)}{\partial x^2}\;, $$ where $\hbar = h/(2\pi)$ and $m$ is the particle mass.

Given that $|\Psi(x,t)|^2$ is the probability density, we can (by using probabilty theory, as usual) calculate the expected value of "X", the position. This expected value is: $$ E[X] = \int_{-\infty}^{\infty}x|\Psi(x,t)|^2dx $$

We can also calculate the expected value of "P", the momentum. This expected value is: $$ E[P] = \int_{-\infty}^{\infty}\Psi^*(x,t)\frac{-i\partial\Psi(x)}{\partial x}dx\;, $$ where the "$*$" means "complex conjugate" (the absolute square of a complex function is $|\Psi(x,t)|^2=\Psi^*(x)\Psi(x)$).

This leads one to define the "momentum operator" as $$ \hat P = -i\frac{\partial{\;\;}}{\partial x} $$

In quantum mechanics, the "momentum of" a particle in a state $\Psi$ is determined by sandwiching the "momentum operator" in between $\Psi^*$ and $\Psi$ and integrating over all space. Note that even though I said the "momentum of" what I should have said was "the expected value of the momentum." The actual measured value of the momentum can actually be any value from negative infinity to infinity and the probability density for the momentum (it turns out) is actually the Fourier transform (with respect to x) of $\Psi(x,t)$. Similarly, the actual measured value of the position can be anything from negative infinity to infinity, but the "expected value" is the weighting of all these by the probability $|\Psi(x)|^2$.

People will often just call the "expected value of the momentum" the "momentum" and might write: $$ "p" = E[P] = \int_{-\infty}^{\infty}\Psi^*(x,t)\frac{-i\partial\Psi(x)}{\partial x}dx\;, $$ It's good to keep this in mind since the particle doesn't "have" any specific momentum, i.e., if we measure the momentum any value is possible, but depending on the wave function some possibilities may be more likely than others and also given the wave function the expected value can be determined.

OK. So. Switching back to our free particle Schrodinger equation: We can solve it the usual way by simply guessing the right answer (sorry). One solution happens to be: $$ \Psi_f(x,t) = e^{ip_0x/\hbar-ip_0^2t/(2m\hbar)}\;, $$ where $p_0$ is a parameter with units of momentum.

Unfortunately, this solution is not really acceptable since it cannot be normalized (it's squared integral cannot be made to equal one as must be done for real probability amplitudes). Nevertheless, let's press on.

According to our machinery, in order to calculate the expected value of the momentum. We should do: $$ "p" = E[P] = \int_{-\infty}^{\infty}\Psi_f^*(x,t)\frac{-i\partial\Psi_f(x,t)}{\partial x}dx = p_0\int|\Psi(x,t)|^2\;, $$ which would just equal "$p_0$" except for the unfortunate fact that our poor wavefunction is unnormalizable. But nevertheless seems to somewhat justify the statement that $\frac{-i\partial}{\partial x}$ is a reasonable "momentum operator."

The spatially-dependent part of our $\Psi_f(x,t)$ is just $$ e^{ip_0x/\hbar}\;, $$ which is a (complex) wave in space.

The wave length of this wave is given by how far in x the wave has to go until the argument goes from zero to $2\pi$ (since a wave repeats every time the phase (the part multiplying the $i$) goes through another $2\pi$ multiple). That is, when $ip_0 x$ is $i2\pi$: $$ 2\pi = p_0 (wavelength)/\hbar\;, $$ which means that: $$ (wavelength) = \frac{2\pi\hbar}{p_0}\;. $$

So in our unacceptable (unnormalizable) case it looks like the "momentum" of the "free particle" is: $$ "p_f" = \frac{2\pi\hbar}{(wavelength)}\;. $$

Or, switching back from $\hbar$ to $h$, this is: $$ "p_f" = \frac{h}{(wavelength)} $$

  • 1
    $\begingroup$ +1: Great answer! I really appreciate it that you took the pains to go through the details and explained the results with minimal assumptions rather than with minimal work. :D ;-) $\endgroup$
    – user87745
    Commented May 14, 2019 at 8:35
  • $\begingroup$ On a tangential note, the OP should notice that the momentum operator in QM is constructed in such a way so as to follow the commutation relations which mirror the Poisson bracket relations of Classical Mechanics. Thus, the classical momentum that corresponds to the momentum operator of QM (and thus, the wavelength) is the canonical momentum $\frac{\partial\mathcal{L}}{\partial{\dot{q}}}$ and not necessarily the kinetic momentum $m\dot{q}$. The two are different in the cases where there is, for example, a vector potential. $\endgroup$
    – user87745
    Commented May 14, 2019 at 8:43
  • $\begingroup$ Thanks for the answer! But let's say I have a really localized position probability wave, which means that I have a high certainty of where it is, no matter if it's being observed or not. If I try to find the momentum from that wave, I'll get a definite "expected" value, but Heisenberg's uncertainty principle tells us that I can't be certain about position and momentum at the same time. Does this mean that the expected value I get isn't actually very accurate? $\endgroup$
    – Wanf
    Commented May 14, 2019 at 12:04
  • $\begingroup$ If you have a highly localized spacial probability amplitude then the momentum probability amplitude will be very flat because the two amplitudes are Fourier transforms of each other. This doesn't mean that the expected value in either case is inaccurate. Heisenberg's uncertainty principle has to do with measurement. Suppose you were to exactly measure the position. The wave function would thus collapse to be a "delta function" at the measured position. But the fourier transformation of a delta function is completely flat (in momentum-space). So any momentum is equally likely now. $\endgroup$
    – hft
    Commented May 14, 2019 at 16:43
  • $\begingroup$ (BTW, wave function "collapse" is another of those fundamental tenets of QM that are hard to explain in a short answer. But, basically, measurements correspond to operators that can be thought of as matrices in a certain set of basis vectors. A measurement can force the wavefunction to "collapse" down to a subset of the eigenvectors of the measurement operator.) $\endgroup$
    – hft
    Commented May 14, 2019 at 16:47

It is a hypothesis that fitted experimental data. It is worth reading how the quantum theory was slowly evolved from experimental observations.

Einstein proposed that $E=hν$ in the black body radiation conundrum, and the momentum for this hypothesis is $p=E/c=h/λ$ for the zero mass photons.

De Broglie extended it to massive particles.

($h$ the Planck constant, $c$ the velocity of light $ν$ the frequency of light and $λ$ the wavelength).

It was a hypothesis that was confirmed:

De Broglie's formula was confirmed three years later for electrons with the observation of electron diffraction in two independent experiments.

In the single electron at a time double slit experiment and the single photon at a time experiment it is seen that massive and massless particles generate the predicted interference patterns in the probability distributions.

Basically, can someone please explain how the momentum of a probability wave (given by p = h/λ) is the same as the momentum of the particle that the probability wave describes (given by p = mv)?

It was a hypothesis driven by analogy to photons, the quanta of light, that was confirmed by experiment, and finally exists within the quantum mechanical theory as it developed with rigorous mathematics.

  • $\begingroup$ -1: Well, it is more than just a hypothesis that fits the data, it is a result derived from a much more basic and richer theory (which fits the data just as well). The history is now pretty much irrelevant. Your answer touches physics only in the last line, the rest is a statement of affirmation of historical experimental evidence of the statement that the OP asked to explain. $\endgroup$
    – user87745
    Commented May 14, 2019 at 8:31
  • 1
    $\begingroup$ On a lighter note, I would definitely bet (if a casino takes bets for this) that you can answer each "explain to me why" question to your satisfaction with saying something along the lines of "because it fits the data". :P ;) $\endgroup$
    – user87745
    Commented May 14, 2019 at 8:33
  • $\begingroup$ @DvijMankad Yours is the platonic view of physics: "mathematics defines reality". Mine is an experimentalists : "mathematics models reality". In addition the OP asks for a simple explanation and the simple explanation is the historical : an analogy between light quanta and particles. Yes, mathematics can show how from simple hypothesis one arrives at specific data predictions. Why those hypothesis ? because they fit the data. data trumps theory. $\endgroup$
    – anna v
    Commented May 14, 2019 at 11:05
  • $\begingroup$ No, I am in as much opposition to "math defines reality" as you would be. We can construct mathematically consistent theories that don't describe reality. End of mathematical idealism/platonism or whatever it might be called. I am totally in agreement with "mathematics models reality". But I think there are explanations. Not just formulae that fit the data. :) $\endgroup$
    – user87745
    Commented May 14, 2019 at 11:16

Quantum momentum is a "fuzzy" version of classical momentum: in particular, it can be considered an information-limited version thereof.

To understand what "information-limited" means, consider this illustrative example. In classical mechanics, your classical momentum is represented by a real number, e.g.

$$p := 0.57701249053\cdots$$

or something similar. An arbitrary real number requires an infinite amount of information to specify: to specify almost any real number, without constraint, there is (provably) no other, more compact way to do so than to simply list out all its digits, which are infinite in number.

If, however, we only kept around a finite number of digits, e.g.


, we would now have a finite amount of information. Of course, one might, at first, upon seeing that digit string, wonder how this is any better - after all, isn't "0.577" just a special real number that is easy to represent? Well it would be. if we took this string as representing the real number


going out to infinity. But we're not doing that. We're taking "0.577" as specifying a number that is no more precise than the nearest 1000th. It is unspecified to any finer resolution, and thus it does not specify a specific point in space, or here a specific strength of momentum. This momentum is, rather, some unspecified quantity between 0.577 and 0.578 - and for our real Universe, it seems that the further-specified quantity, in its way, simply doesn't exist. The Universe is economical in allotting information to describing the objects it contains, and is not profligate.

However, the proper notion of "limited" information for the real Universe is not quite so simplistic as this "chopping" the digits in that fashion. Instead, we need a bit more complex way of talking about information limitation that allows it to be limited in different ways. To see an example - just to open your mind to the possibility that there are other ways to lack information besides just chopping digits off the end - note that another way we could write "$p$" with "less information" would be


where we have kept a few more digits now, but now only giving their parity: here $\mathrm{O}$ means an unknown odd digit (i.e. 1, 3, 5, 7, 9) and $\mathrm{E}$ means it's an unknown even digit (i.e. 0, 2, 4, 6, 8). We have thus, some information about these new digits, but not complete information. There is more information than our "0.577" string, but still less information than the whole real number.

Now, for talking about the actual Universe, it has many, many ways in which the information can be limited, and the best language that we have created for discussing how it works is the language of probabilities, and this is the language we write quantum theory in. To understand how/why that probabilities are a form of information, or more particularly, a form of talking about lacking information, just consider how you may use them in an ordinary setting. If you say, in casual conversation, that "I'm only 75% sure" of something's being or not being the case, you're saying that you aren't really as informed of whether or not it will be than if you actually knew for sure. That is really a (admittedly ad hoc) probability, used to represent a state of limited information in your mind.

And that's what we do in quantum mechanics: we assign now a probability value now to each possible momentum $p$ of the particle - something we call a probability density function, or pdf:


from which we can obtain, through the use of calculus, the actual probability that the momentum $p$ is in any specified range. If there is a very high probability of it being in a narrow range and very little anywhere else, we can say that, by the same reasoning, we have a lot of information as to what the momentum is, and if the range is broad, we can say that we have little information.

Now, as for the relation

$$p = \frac{h}{\lambda}$$

, this requires us to talk a little more closely of why we need to use these probabilistic descriptions. In particular, the Universe appears to have a limit on information content, in the same fashion that it also has a limit on information movement (the famous speed of light). And a consequence of this appears as that certain pairs of physical parameters of an object, such as its momentum and position, are limited insofar as the simultaneous information that can exist for both at any given time.

And it's not simply that the Universe "hides" information from us or that we just aren't clever enough - if we try to assume that, and go far enough with that line of theorizing, we actually end up in a bit of a contradiction against the limit on information movement we just mentioned. It really does seem, at least for simplest hypothesis, that the information is limited. It's not that it has no information either, so yes, the Moon "is still there when you're not looking" [or at least, we have no reason to assume it's not, not on this basis anyways], contrary to what you may have heard in some pop-cult ideas about this subject.

In any case though, this limit on simultaneous information results in there being a tradeoff effect in that, once you demand information for one half of the pair of parameters past a certain point, it starts to exclude information in the other half of the pair. If you try to acquire more information than is already present for that one half, you will destroy some of the information in the other half.

And the formula you mention is actually related to just that. To actually describe how to get it following these lines would require a fair bit more math to do it full justice, but a plain-language description is like this. Since we established that there is a tradeoff between momentum and position information, it turns out that the manner of tradeoff - remember we said information can be lacking in different ways - is in "just so" a way that high-information momenta (i.e. a sharp, narrow probability distribution), well above the tradeoff threshold so as to mandate extensive loss of information in the position, manifest such loss in the positional probability distribution as it acquiring (simplistically speaking) a wavelike character. And the wavelength of those waves is related to the high-precision momentum value by the above formula.

Insofar as why the Universe trades information that way and not some other way, well that's one of those metaphysical questions, not physical ones. "It's built/formed/whatever you believe that way", at least given the extent of our knowledge. We can say that, though, if it didn't, life as we understand it could not exist, because these information limits and tradeoffs add great structure to matter, permitting all the complexity of chemistry, and thus biology as well, to occur.

  • $\begingroup$ Thank you for the answer, but I'm still confused as to why momentum and position have this "tradeoff" effect. If I find the momentum from a position probability wave, no matter the wavelength of the wave, I'll still get a definite momentum, right? $\endgroup$
    – Wanf
    Commented May 14, 2019 at 12:08
  • $\begingroup$ @Want : Only in the extremal case where that the position waves effectively occupy all of space and provide no information at all as to the particle position. It would require more maths to really detail this, though. $\endgroup$ Commented May 14, 2019 at 19:00
  • $\begingroup$ Though, I might try. I'll add some more and you can tell me if you're comfortable with it or not. $\endgroup$ Commented May 14, 2019 at 19:01

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