Does the Heisenberg uncertainty principle only allow location OR momentum to exist?

Question

To elaborate, if you know the exact location of a particle with perfect accuracy, does this mean that the momentum of it still exists but we just don't know what it is/ can't measure it? Or does this mean that the particle at this point literally does not have momentum at all (while having perfect 100% knowledge of the location)?

Answer 1

If you think of it as position and frequency, you might understand it better. If you have a wave with a single frequency, then the wave must be spread throughout all space, so it does not have a single point in space at which you can consider it to be located. On the other hand, if you have a wave that is more confined in space, it doesn't have a single frequency - if you perform a Fourier analysis of it you will find it has a large range of frequency components, so you cannot assign a single frequency to it.

A wave function that is localised at a single point in space has an infinite spread of frequency (i.e., momentum) components - it does not have a zero frequency!

Answer 2

It does not mean that a particle whose location is known with perfect accuracy has no momentum at all- that is, zero momentum. If it had zero momentum, then it would have (for a massive particle) zero velocity and then the uncertainty in both position and momentum would simultaneously be zero, which is explicitly forbidden by the uncertainty principle.

The standard way of visualizing this is as follows:

We imagine a massive quantum particle trapped in a one-dimensional potential well with some width w. We discover that it spends most of its time somewhere around the middle of the well and it jitters around in there with some average velocity v.

Now we squeeze the width down so the particle has less room in which to exist. Its positional uncertainty goes down but its average momentum goes up: it is jittering around faster than it did before. We squeeze it down some more, its positional uncertainty does down again and in response its random jittering becomes increasingly frantic.

At some point in the squeezing process the jittering velocity of the particle approaches the speed of light and its average momentum no longer tracks its mass: to us, it behaves as if its mass (which is actually invariant) were increasing. At some point it is energetic enough that it can pluck another particle out of the vacuum, and then not only do we not know how fast it is moving, but we can no longer say how many particles are present.

This whole process is explained very clearly in the introduction of Peter Woit's book, Not Even Wrong.

Answer 3

The assumption in the standard formulation of QM is that regarding precise values: when the position of a particle along $x$ is defined (= it exists), then its momentum along $x$ is not defined (= it does not exist).

In common situations both observables are unsharply defined.

That is because quantum objects like particles are theoretically different from classical ones. There are (also very important) attempts to construct a deeper theory of classical type (where a particle may be described as we should expect in classical mechanics) capable to explain the whole quantum phenomenology.

All these attempts face a number of no-go results which prove that something physically important (related to causality, realism, non contextuality) should be however paid. These sort of deeper classical explanations, if any, cannot exist in the form one naively expects.

Finally, none of these attempts, with the important exception of the so called Bohm de Broglie formulation in the non relativistic regime, can be considered a true competitor of the standard non-classical like formulation(s).

At the end of the day the overall problem concerns the interpretation of uncertainty presented by QM: is it ontic or epistemic? Something epistemic remains (some intepretations of mixed states can be considered epistemic to some extent), but it is very difficult (personally I think it is impossible) to assert that the quantum uncertainty is fully epistemic.

Answer 4

When trying to gain intuition into a problem, going to extremes can be helpful, e.g. those questions about the water level when you throw something overboard: imagine it's something extreme, like Thor's hammer (Neil deGrasse Tyson says it's white dwarf matter), and it helps.

Other times it doesn't help. How many questions on this site have "A massive object moving at the speed of light" at which point nothing can be done. I kind of think you did that with:

"the exact location of a particle with perfect accuracy"

From that I can make a coordinate system where the exact location is $x_0$, so:

$$ \langle x \rangle = x_0 $$

and without starting an accuracy vs precision war:

$$ \langle (x-x_0)^2 \rangle = 0 $$

That leaves me with

$$ \psi(x) = A\delta(x-x_0)e^{i\phi} $$

I'm never sure about normalization, but I am going to set $A=1$ and get back to it if it causes problems. Global phase is not an observable, so I'll set $\phi=0$. Finally, I'm doing a coordinate translation: $ x \rightarrow x + x_0 $, so your question can now be rephrased as:

If I have a particle in a pure state:

$$ \psi(x) = \delta(x) $$

what can I know about the momentum?

Well QM tells us the average momentum is:

$$ \bar p = \langle \psi|-i\hbar\frac d{dx}|\psi \rangle = 0$$

and the uncertainty is:

$$ \sigma_p = \langle \psi|(-i\hbar\frac d{dx}-\bar p)^2|\psi \rangle$$

$$ \sigma_p = \langle \psi|-\hbar^2\frac {d^2}{dx^2}|\psi \rangle=\infty $$

Now I didn't actually evaluate those expressions. I just put in $0$ and $\infty$, respectively, to set the mood.

Since the original question is:

"What do we know about the momentum?"

The answer is:

EVERYTHING

We can compute the (unnormalized) wave function in momentum space with the general expression:

$$ \phi(p) = \frac 1 {\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} \psi(x)e^{-ipx/\hbar} dx $$

which for us is:

$$ \phi(p) = \frac 1 {\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} \delta(x-x_0)e^{-ipx/\hbar} dx = e^{2\pi ipx_0/\hbar} = 1 $$

Note: I put $x_0$ back into $\psi$ just to show how it affects the phases of the momentum eigenstates.

Being in a momentum eigenstate means a spatial translation is equivalent to a phase rotation, with the rate of rotation proportional to momentum.

In our perfect state, we get all momenta with equal weight (amplitude), and the phases work out just right so that the position-space representation land smack dab at $x_0$, and since that's just a coordinate I can set it to $0$ and get:

$$ \phi(p) \propto 1 $$

All momenta appear equally, which is required to get a delta function in position space, and they all have equal phase, which is required to put that delta function at $x=0$.

In retrospect, I think the extreme question actually works here. We always think of momentum as velocity, as in:

$$ p = mv $$

but that is a classical result from the QM fact that the momentum operator is the generator of translations, and momentum eigenstates are translation eigenstates with eigenvalue...let's work it out explicitly:

$$ T(\delta x)[\psi_p(x)] = T(\delta x)[e^{ipx/\hbar}]$$

$$ T(\delta x)[\psi_p(x)]= e^{ip(x + \delta x)/\hbar} $$

$$ T(\delta x)[\psi_p(x)]= e^{ipx/\hbar}e^{ip\delta x/\hbar} $$

$$ T(\delta x)[\psi_p(x)]= e^{ip\delta x/\hbar}\psi_p(x) $$

$$ T(\delta x)[\psi_p(x)]= \lambda \psi_p(x) $$

...with eigenvalue:

$$ \lambda = e^{ip\delta x/\hbar}$$

Anyway, that's what make momentum important in quantum mechanics, and if you think about a complex$^1$ function that is both certain in position while simultaneously being an eigenstate of a translation, you will get some intuition about the HUP for $\sigma_x$ and $\sigma_p$.

[1] I emphasize complex because you'll see a lot of pop-sci (read: YouTube) content which tries to explain HUP with real wave functions. This will not give intuition, it will cause confusion. You need to consider complex wave functions, and look at the relation between translation and phase rotation.

Answer 5

Heisenberg's uncertainty principle (HUP) is a consequence of quantum theory. So if you want to understand it you have to look at the equations of motion of quantum theory and their consequences. There are many theories called interpretations of quantum theory that claim you should use the equations of quantum theory to predict the results of experiments but that they don't describe reality. Since those theories say quantum physics is false they aren't very relevant to working out the consequences of quantum theory. So I'm just going to assume that the equations of motion of quantum theory describe reality.

Newtonian physics would describe the evolution of position and momentum in terms of variables whose values represent the result you would get if you measured the position or momentum. For example, the position vector of a particle $\mathbf{r}(t)=(x(t),y(t),z(t))$ would be used to write down the equation of motion for the particle: $$\frac{d^2\mathbf{r}(t)}{dt^2}=\frac{\mathbf{F}}{m}$$ where $m$ is the particle's mass and $\mathbf{F}$ is the force acting on the particle and it would be a function of the particle's position and momentum.

In quantum physics, the evolution of a measurable quantity is described by a Hermitian operator called an observable. The possible results of measuring that observable are its eigenvalues and quantum theory gives the expectation value of the observable: a sum of the eigenvalues weighted by probabilities. In general, explaining the evolution of a quantum system involves considering what happens to all of the possible values. This is called quantum interference, see Section 2 of this paper for an example:

https://arxiv.org/abs/math/9911150

The HUP places limits on how sharp one observable can be depending on the sharpness of another observable. See this review:

https://arxiv.org/abs/1511.04857

You ask about knowing the position of a particle with perfect accuracy. A measurement is an interaction that produces a record of some information about the measured system. No real measurement interaction will produce a state with perfectly accurate position. A real measurement will produce a relative state that is narrowly peaked in position and momentum compared to the scales of everyday life:

https://arxiv.org/abs/quant-ph/0306072

But that state will still have a nonzero standard deviation for position and in general explaining its motion will require that the state has a non-zero width in position. On the scales of everyday life it will usually be a good enough approximation to say that any object you can see with the naked eye has a particular position and momentum. Quantum theory will describe that system as having a small but non-zero spread in both position and momentum.

Answer 6

...if you know the exact location of a particle with perfect accuracy...

What do you mean by perfect accuracy? Do you mean that the error is zero? If so, how do you propose to make this measurement? (I think this was Heisenberg's real point: don't try to sound more precise than you are).

Imagine that you have a a metre ruler, marked out in mm: the most accurate measure you can make will be within plus or minus 0.5mm. Clearly this isn't good enough, so we'll mark it out in microns: the most accurate measure you can make will be within plus or minus 0.5 microns (assuming your eyesight is a lot better than mine). Hmmm, I don't think we'll get down to zero error this way.

When I was in the 6th form our physics teacher explained the uncertainty principle by a thought experiment. Try to get a zero error by using light to observe the particle (whoops, photon moves the particle slightly...). Try as we might we never get either error, position or momentum, down to zero.

Answer 7

If we know the position of a particle with 100% accuracy, it means the particle has stopped. If it has stopped, its momentum is zero. However, as a conclusion derived from this principle, in practice, we can never know the position of a particle with 100% accuracy.

Answer 8

Uncertainty principle is all about the particles exact physical conditions like Uncertainty principle says that we cannot know the exact position of particle and also it's momentum which also includes velocity As the particle at quantum level show wave nature also. But if you imagine a experiment in which a electron have only three points to move now by using graph you may calculate the exact location of electron at each time period but still you can't disprove uncertainty principle. Because still when electron travel between two points many random energy exchanges occur which still cause uncertainty in its momentum and also in its velocity which also causes the uncertainty in its position per unit time.