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I've asked this on different websites and never gotten an answer that a layperson can understand. Most people just say that light does not have a trajectory and then they do some hand waving. If light does not have a trajectory, then how do we receive coherent streams of photons from billions of lightyears away all lining up to produce an image?

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    $\begingroup$ Have you read Feynman's QED: The Strange Theory of Light and Matter? IMHO, his explanation is probably the best that's possible without using advanced mathematics. $\endgroup$
    – PM 2Ring
    Feb 16, 2023 at 5:53
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    $\begingroup$ The uncertainty principle has nothing to do with how photons travel. You have it correct in your first paragraph. $\endgroup$ Feb 16, 2023 at 6:23
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    $\begingroup$ Why doesn't light have a trajectory? A reference would be useful for context. "Trajectory" is a particle-dynamics concept, but is only vaguely mapped to wave dynamics. "Light" could mean either a particle or a wave. Heisenberg Uncertainty Principle is focused on particle approximations. Astronomy is typically in the "ray optics" subdomain of wave dynamics. $\endgroup$ Feb 16, 2023 at 17:00
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    $\begingroup$ Diffraction is the uncertainty principle for light. Passing through a smaller pinhole (tighter position) will cause more diffraction (looser momentum). The classical wave equation as well as Maxwells equations also have this property, so "uncertainty principle" is related to waves in general and isn't irrelevant for non-quantum systems. $\endgroup$ Feb 16, 2023 at 21:00
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    $\begingroup$ For an argument about how a spherical wavefunction can give rise to interactions along a straight line, see the paper by Mott described in this answer. $\endgroup$
    – rob
    Feb 17, 2023 at 17:13

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If you think about light as a particle, a photon, you can understand that it will continue on its path through space at velocity $c$ unless it scatters off another particle, is absorbed by an atom, etc. For other particles like electrons, which are charged, we can note that their momenta will remain unchanged unless they encounter an external force from an electromagnetic field, or they scatter off some particle, etc. Thus, particles travel "straight" (or more technically at a fixed momentum) unless they experience a force or "hit something".

Essentially, a quantum mechanical viewpoint is not necessary to answer this question, because momentum conservation ensures that all three components the the momentum are unchanged unless the particle experiences a force/an interaction with another particle. We may not know the momentum components to arbitrary precision due to the uncertainty principle, but they won't change unless subjected to an external force. Note that the uncertainty principle holds independently for each spatial axis because $[x_i, p_j] = 0 \ \ \forall i \neq j$.

A quantum mechanical perspective will rely on the idea that particles are mathematically described by a wavefunction. Depending on the basis you're working in, this wavefunction determines the probability of finding a particle at a certain position or momentum. The time evolution of a wavefunction is governed by Schrödinger's equation. It is not meaningful to think about the exact momentum of a wave function until you make a measurement, however, due to the fact that momentum is conserved and that the Heisenberg uncertainty principle applies independently to each axis, you can with certainty state that: if a photon is emitted from a star thousands of light-years away, and it's momentum is 100% along the axis pointing at the earth (let's call it the x axis), that photon can never spontaneously acquire momentum along another axis without violating conservation of momentum. We will not know the precise value of p along the x axis, but we can know its position along the x axis to arbitrary precision and that it has zero momentum along the y and z axes without violating the HUP.

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    $\begingroup$ The position of a travelling photon is a bit tricky. See physics.stackexchange.com/q/492711/123208 $\endgroup$
    – PM 2Ring
    Feb 16, 2023 at 6:39
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    $\begingroup$ I'd add a note that the momentum of photon is identified with the light's corresponding wavelength, so that in 1D ("x"), HUP says that you can't know both position and wavelength with arbitrary precision. $\endgroup$ Feb 16, 2023 at 16:53
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    $\begingroup$ The final sentence (v1) is just plain wrong. The only way to know that a photon has zero momentum along a $y$- or $z$-axis is to begin with a plane wave of infinite extent. $\endgroup$
    – rob
    Feb 17, 2023 at 3:55
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    $\begingroup$ @klippo That isn’t my complaint. Suppose you have starlight traveling along the x-direction and you detect it on Earth (instead of, say, on Jupiter). This constrains the location of the light in the y-z plane, which means the beam of starlight must have nonzero divergence by the uncertainty principle. $\endgroup$
    – rob
    Feb 17, 2023 at 10:55
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    $\begingroup$ @klippo A photon from a distant star is drawn from a spherical distribution. For a quantum-mechanical argument about how a straight track can emerge from a spherical wavefunction, see the paper by Mott described briefly in this answer. $\endgroup$
    – rob
    Feb 17, 2023 at 17:15
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The uncertainty principle guarantees that a beam of light will have some minimum divergence in the plane perpendicular to the motion of the beam. If your light beam passes through some aperture with width $\Delta x$, the aperture introduces a momentum uncertainty $\Delta p_x \gtrsim\hbar/ \Delta x$ in the $x$ direction. If the momentum of the beam is $p=h/\lambda$, the quantum-induced angular divergence goes like $\Delta p_x/p ≈ \lambda/\Delta x $. You might recognize this ratio from expressions for diffraction, which becomes unignorable when the size of an aperture is comparable to the wavelength of light.

For a classical example, imagine you make a beam of parallel light rays by placing a point source at the focus of a converging lens. If you believe the thin lens equation,

$$ \frac1f=\frac1{d_\text{object}}+\frac1{d_\text{image}} $$

then the image distance should approach infinity as the object distance approaches the focal length. But if you look more carefully, you’ll learn that

  1. the thin lens approximation is an approximation, and
  2. all “point” sources have finite size.

The best you can hope for with a telescope is to project a magnified image of your point source onto an image plane that is very, very far away. In this case the beam from your “point” source will pass through a waist, after which it diverges again. If the waist is far enough away, then its properties will depend more on the diffraction from the lens than on the details of the point source.

You ask how starlight remains in coherent streams over billions of light-years. This only happens if there are no intervening absorbers. The direction of the starlight which reaches your eye is maintained by constructive interference with the light from that same star which reaches my eye, and all of the other nearly-parallel rays which pass through the interval between you and me. In our shadows, the starlight undergoes diffraction.

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Why does light travel in a straight line if the uncertainty principle is true?

Many physics theories exist for modeling observations. By mathematical construction in the region of overlap in phase space, they coherently give the same results.

In this case you are confusing light (which is modeled with classical electromagnetism and optics) with photons. Photons are in the realm of quantum mechanics and quantum electrodynamics, which are necessary in modeling the behavior of photons. The Heisenberg uncertainty principle is in the realm of quantum mechanics and does not apply to light. Light emerges from many photons, but photons are not light. See my answer here.

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  • $\begingroup$ I have difficulty following your sentences here. If I use the edit function to propose changes, will it come up as a suggested edit for you to approve? I don't feel appropriate directly editing someone else's words. $\endgroup$
    – g s
    Feb 16, 2023 at 5:47
  • $\begingroup$ @gs let me try editing and if you still have difficulty , feel free to edit . $\endgroup$
    – anna v
    Feb 16, 2023 at 5:56
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First there is light, a thing that exists even if people don't think about it. Then there is physics, a mathematical description of how those things behave.

Physics also has a description in words that explain what the things in the universe are and how they work. This is a guide to thinking, and is generally seen as less the important part. The math does all the accurate prediction of how the universe behaves.

So physics is how people think about the universe. It is not the universe itself. This point needs to be said because it is easy to confuse the two. For example, physics talks about light, and light is a thing in the universe. The thing physicists talk about must therefore be real, right? No, it need not be.

There are several versions of physics, each of which explains a piece of how the universe behaves. We don't know the total behavior. People switch back and forth between different ways of thinking that are all simplifications of the universe.

Classically, light consists of rays. This is the simplest version. Professional lens designers use these all the time. Rays leave a source and travel in straight lines until they hit something. Then they bend or get absorbed or reflected.

This works very well, but do rays actually exist? No. Lens designers know very well that light is an electromagnetic wave. If you look closely, light doesn't quite travel in a straight line. Lens designers account for this as diffraction. If light passes through a pinhole, it spreads out because of its wave nature. A lens can be thought of as a very large pinhole. Light spreads out a very small amount because of it. If you design a lens well enough, diffraction is the biggest imperfection left. There is nothing you can do about it except make the lens diameter bigger. Such a high quality lens is said to be diffraction limited.

OK, are electromagnetic waves real? See my answer to In what medium are non-mechanical waves a disturbance? The aether?. But leaving that aside, they are a simplification. For example, they cannot explain the photoelectric effect.

Photons are a better description. They are a much harder idea to wrap your head around because they are very different from anything you see everyday. People use them when they need to explain things that are different from everyday life. But they also need to understand them. So they are described as like a classical particle, but also like a classical wave. This can be a problem. The reason they work is because they are different from both. See How can a red light photon be different from a blue light photon?

Photons and quantum mechanics can explain diffraction. The uncertainty principal says $\Delta x \Delta p \ge h/4\pi$. This means there is always an uncertainty in both position and momentum of a photon. Uncertainty here doesn't mean the same thing as when you measure a distance and get an answer that is close to the right answer but not perfect because your ruler is only so precise. See Does the collapse of the wave function happen immediately everywhere? for more on what it does mean.

In classical physics, it is common to approximate a particle as a point, with a single position. Its speed is a single value. There is no limit to how precisely these can be measured, except for how good a ruler and speedometer you can make.

Like everything, a photon is not a point particle. What it means to say it doesn't have a position or momentum is where things get so different from everyday that saying a photon is like a particle breaks down. It is common to say it spread out like a wave. This can help, but it isn't exactly right either. We will go with it up to a point.

If you pass a classical wave through a pinhole, it fills the pinhole. It doesn't have a point-like position, but the position is restricted in the sideways directions. Because of this, it is impossible to know ahead of time the precise value of the sideways component of the momentum. Past the pinhole, light will not travel in a perfectly collimated beam. There will be some sideways component to the direction. You can see this if the beam hits a screen. It has spread out in all directions. The beam lights up a bigger spot.

Even though photons don't have a single value for momentum, it isn't right to say they have no value for momentum. The spot isn't the whole screen. You can see the relationship between uncertainty in position and momentum. A small pinhole makes a large uncertainty in momentum and a big spot. For a pinhole the size of a lens, the uncertainty in momentum is very small. The beam hardly affected very little. You can see it better if the lens focuses the beam onto the screen. The uncertainty limits the size of the spot.

The quantum mechanical description predicts the same spot as the electromagnetic description.

Final note: a photon isn't the ultimate answer. People are still working out better theories, such as string theory.

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If you model light propagation using the Huygens-Fresnel principle, you assume that it doesn't (classically) follow straight lines. The straight lines of ray optics are the paths where the light that follows paths close to the line stays in phase. "Close" means that the phase match is good enough that the summed wave has a considerable enhancement in amplitude.

When you make this quantitative, you successfully model the phenomena of diffraction, including the "diffraction limit" of optics. And then, when you carry that over into the quantum domain, you get the uncertainty principle. The position/momentum uncertainty principle for photons is exactly the same, mathematically, as the diffraction limit for telescope resolution.

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The uncertainty principle doesn't say that you can't have a well-defined momentum (and a well-defined direction of travel for the light), only that if you have a well-defined momentum, you must have a large spread of positions. The answer by rob on diffraction is correct and really the best answer in my opinion, but in contrast to that answer I want to talk about the uncertainty principle applied to the direction parallel to the direction of travel of the light.

Let's assume we are near a uniformly lit flat surface (like someone built a big wall of lights or we are close enough to the surface of the sun that it appears flat). If the direction of travel is z, then the light field is completely uniform in x and y. For a single color of light (a laser wall?), the wave function has a very well defined momentum. The uncertainty principle then requires that the wave function has a poorly defined position along z. A plane wave solution to the Schrödinger equation is exactly this: it has a precise momentum (precise wavelength), but is of infinite extent along z: it is a sine wave of a single frequency that continues oscillating toward infinity.

If you want to make a localized packet of light, you have to add many plane waves of different frequencies together. You can see this from a discussion of adding Fourier components. So if you localize sheet of light along z, you must have a wide mixture of frequencies as given by the uncertainty principle.

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It doesn't. If uncertainty principle is true light should not travel in any particular trajectory let alone straight line. So why we say light travels in a st. line? Light(photon) has an probability amplitude assigned to each path/trajectory possible and has a particular 'phase' association with each path/trajectory.

We know Fermat's principle of least time which roughly states that out of all possible paths that it might take to go from one point to another, light takes the path which requires the shortest time. So when light travels in vaccum or within a particular homogeneous medium it takes the straight line, because if light doesn't change the medium it's velocity remains constant and the straight line is the path which requires shortest time.

But how light can choose a particular path? What actually happens is that: the path which represents the least time is also the path where the time for the nearby paths is nearly the same, so the 'phase differences' are really really small which allows the probability amplitudes to aligne in nearly the same direction and add up to a substantial magnitude where the probability amplitudes corresponding to numerous other paths cancels each other. That is why we can say light takes the path where time is least i.e it goes in straight line in vaccum or within a particular homogeneous medium. It doesn't break uncertainty principle which originates from the fact that microscopic particles can't have a trajectory (In diffraction experiments we can see light isn't going in a straight line cause we have restricted the possible paths)

Feynman has discussed about this in a very simple manner in his lectures on QED .

Edit:

After reading my answer once again I realised there can be a misunderstanding. "...That is why we can say light takes the path where time is least i.e it goes in straight line in vaccum or within a particular homogeneous medium" doesn't mean it really does.

What it means:

In vaccum or Inside a particular homogeneous medium, without any kind of restriction on possible paths (what happens in diffraction experiments where we reduce possible paths significantly) if we consider light is going in a straight line and make predictions out of it we will find no significant error,We will get extremely fine, accurate results in any experiment whatsoever

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  • $\begingroup$ I still don't really understand how we can not know a photon's exact next position (because of the uncertainty principle) even if we know that light will take the path with the shortest time. My understanding from other answers is that we basically do know its next position, so my understanding of the uncertainty principle is flawed I guess. $\endgroup$
    – aa bb
    Mar 28, 2023 at 3:41
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The uncertainty principle in quantum mechanics states that there is a fundamental limit to the precision with which certain pairs of physical properties, such as position and momentum, can be simultaneously measured. However, this does not mean that light cannot travel in a straight line.

Light is an electromagnetic wave, and it propagates through space by oscillating electric and magnetic fields that are perpendicular to each other and to the direction of propagation. The wave nature of light is well-established and has been extensively studied, both theoretically and experimentally. In vacuum or a uniform medium, light travels in straight lines at a constant speed, which is one of the fundamental principles of classical optics.

The uncertainty principle applies to the behavior of particles at the quantum level, where the behavior of individual particles can be described by wave functions that exhibit wave-particle duality. However, the wave-particle duality of light does not affect its ability to travel in a straight line, as this behavior can be described by the classical wave theory of light.

In summary, while the uncertainty principle is true and affects the behavior of particles at the quantum level, it does not prevent light from traveling in a straight line, which can be explained by the classical wave theory of light.

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Everything travels in a straight line unless a force acts on it. The uncertainty principle has nothing to do with it.

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