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It's been mentioned elsewhere on this site that one cannot define a position operator for the one-photon sector of the quantized electromagnetic field, if one requires the position operator have certain formal properties. This is a theorem that holds only for massless particles of helicity $|\lambda| \geq 1$, in particular it does not apply to massless scalars.

A lot of people, particularly mathematical physicists or older quantum field theory textbooks, seem to interpret this to mean that we should never speak of the position of anything in relativistic quantum field theory. But it still seems possible to say something about where a photon is. For example, if I have an ideal cavity and excite the lowest mode with one photon, I know that the photon is in that cavity. Furthermore, I can localize the photon arbitrarily well using smaller and smaller cavities.

When an optics experiment is done using a laser beam, it is perfectly meaningful to talk about photons being in the beam. We can also speak of a photon being emitted by an atom, in which case it is obviously localized near the atom when the emission occurs. Furthermore, in the usual analysis of the double slit experiment one has, at least implicitly, a wavefunction for the photon, which successfully recovers the high school result.

When one talks about scattering experiments, such as in photon-photon scattering, one has to talk about localized wavepackets in order to describe a real beam. Furthermore, unlike the massive case, where the Compton wavelength provides a characteristic length, there is no characteristic length for photons, suggesting that beams can be made arbitrarily narrow in principle: the complaint that you would start causing pair production below the Compton wavelength doesn't apply.

In other words, while the theorem is airtight, it doesn't seem to impose any practical limitations on things we would actually like to do experimentally. But you can find very strange-sounding descriptions of what this theorem is telling us online. For example, on PhysicsForums you can read many obviously wrong statements (e.g. here and here and here) such as:

The photon has no rest frame. Computing an expectation of position for such an object is nonsense.

One good reason is that photons are massless and move at the speed of light and have no rest frame! Then also they are bosons, so you can't tell which are which.

These are wrong because they also apply to massless scalars, for which there does exist a (Newton-Wigner) position operator. It also just doesn't make sense -- if you can't measure the position of something if you're not in its rest frame, then how can I catch a ball?

In relativistic quantum (field) theory there is no concept of single photons.

You cannot define "position" for an electromagnetic field or of photons, which are certain states of this field (namely single-photon Fock states). Nobody thinking about classical electromagnetic waves would ever come to the idea to ask, what the position of a field might be.

This is wrong because the one-particle sector of a quantum field theory is perfectly well-defined, and it is perfectly valid to define operators acting on it alone.

It can be shown that in the context of relativistic quantum theory the position operator leads to violations of causality.

This is rather vague because quantum field theory is causal, so it's unclear how "the position operator" overturns that.

It could just be that PhysicsForums is an exceptionally low-quality site, but I think the real problem is that interpreting this theorem is actually quite tricky. What nontrivial physical consequences does the nonexistence of a formal photon position operator have?

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  • $\begingroup$ I’m a long way from an expert but for QED photons I would start down two tracks: thinking in terms of cavity modes in a cavity that has been allowed to grow without bound, and noting that on-shell photons are eigenstates of momentum. Alas, I’m unsure of how to close the deal either way. $\endgroup$ – dmckee --- ex-moderator kitten Jul 20 '19 at 18:29
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    $\begingroup$ "Furthermore, I can localize the photon arbitrarily well using smaller and smaller cavities" - not particularly, or at least not 'that' photon (i.e. not without changing its frequency. You can try to localise tighter than the wavelength, and get a 'confined' mode, but at the price of having the majority of the power (/population) in evanescent tails. $\endgroup$ – Emilio Pisanty Jul 21 '19 at 10:21
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We could spend forever playing whac-a-mole with all of the confusing/confused statements that continue popping up on this subject, on PhysicsForums and elsewhere. Instead of doing that, I'll offer a general perspective that, for me at least, has been refreshingly clarifying.

I'll start by reviewing a general no-go result, which applies to all relativistic QFTs, not just to photons. Then I'll explain how the analogous question for electrons would be answered, and finally I'll extend the answer to photons. The reason for doing this in that order will probably be clear in hindsight.

A general no-go result

First, here's a review of the fundamental no-go result for relativistic QFT in flat spacetime:

  • In QFT, observables are associated with regions of spacetime (or just space, in the Schrödinger picture). This association is part of the definition of any given QFT.

  • In relativistic QFT, the Reeh-Schlieder theorem implies that an observable localized in a bounded region of spacetime cannot annihilate the vacuum state. Intuitively, this is because the vacuum state is entangled with respect to location.

  • Particles are defined relative to the vacuum state. By definition, the vacuum state has zero particles, so the Reeh-Schlieder theorem implies that an observable representing the number of particles in a given bounded region of spacetime cannot exist: if an observable is localized in a bounded region of spacetime, then it can't always register zero particles in the vacuum state.

That's the no-go result, and it's very general. It's not restricted to massless particles or to particles of helicity $\geq 1$. For example, it also applies to electrons. The no-go result says that we can't satisfy both requirements: in relativistic QFT, we can't have a detector that is both

  • perfectly reliable,

  • localized in a strictly bounded region.

But here's the important question: how close can we get to satisfying both of these requirements?

Warm-up: electrons

First consider the QFT of non-interacting electrons, with Lagrangian $L\sim \overline\psi(i\gamma\partial+m)\psi$. The question is about photons, and I'll get to that, but let's start with electrons because then we can use the electron mass $m$ to define a length scale $\hbar/mc$ to which other quantities can be compared.

To construct observables that count electrons, we can use the creation/annihilation operators. We know from QFT $101$ how to construct creation/annihilation operators from the Dirac field operators $\psi(x)$, and we know that this relationship is non-local (and non-localizable) because of the function $\omega(\vec p) = (\vec p^2+m^2)^{1/2}$ in the integrand, as promised by Reeh-Schlieder.

However, for electrons with sufficiently low momentum, this function might as well be $\omega\approx m$. If we replace $\omega\to m$ in the integrand, then the relationship between the creation/annihilation operators becomes local. Making this replacement changes the model from relativistic to non-relativistic, so the Reeh-Schlieder theorem no longer applies. That's why we can have electron-counting observables that satisfy both of the above requirements in the non-relativistic approximation.

Said another way: Observables associated with mutually spacelike regions are required to commute with each other (the microcausality requirement). The length scale $\hbar/mc$ is the scale over which commutators of our quasi-local detector-observables fall off with increasing spacelike separation. Since the non-zero tails of those commutators fall off exponentially with characteristic length $\hbar/mc$, we won't notice them in experiments that have low energy/low resolution compared to $\hbar/mc$.

Instead of compromising strict localization, we can compromise strict reliability instead: we can construct observables that are localized in a strictly bounded region and that almost annihilate the vacuum state. Such an observable represents a detector that is slightly noisy. The noise is again negligible for low-resolution detectors — that is, for detector-observables whose localization region is much larger than the scale $\hbar/mc$.

This is why non-relativistic few-particle quantum mechanics works — for electrons.

Photons

Now consider the QFT of the elelctromagnetic field by itself, which I'll call QEM. All of the observables in this model can be expressed in terms of the electric and magnetic field operators, and again we know from QFT $101$ how to construct creation/annihilation operators that define what "photon" means in this model: they are the positive/negative frequency parts of the field operators. This relationship is manifestly non-local. We can see this from the explicit expression, but we can also anticipate it more generally: the definition of positive/negative frequency involves the infinite past/future, and thanks to the time-slice principle, this implies access to arbitrarily large spacelike regions.

In QEM, there is no characteristic scale analogous to $\hbar/mc$, because $m=0$. The ideas used above for electrons still work, except that the deviations from localization and/or reliability don't fall off exponentially with any characteristic scale. They fall of like a power of the distance instead.

As far as this question is concerned, that's really the only difference between the electron case and the photon case. That's enough of a difference to prevent us from constructing a model for photons that is analogous to non-relativistic quantum mechanics for electrons, but it's not enough of a difference to prevent photon-detection observables from being both localized and reliable for most practical purposes. The larger we allow its localization region to be, the more reliable (less noisy) a photon detector can be. Our definition of how-good-is-good-enough needs to be based on something else besides QEM itself, because QEM doesn't have any characteristic length-scale of its own. That's not an obstacle to having relatively well-localized photon-observables in practice, because there's more to the real world than QEM.

Position operators

What is a position operator? Nothing that I said above refers to such a thing. Instead, everything I said above was expressed in terms of observables that represent particle detectors (or counters). I did that because the starting point was relativistic QFT, and QFT is expressed in terms of observables that are localized in bounded regions.

Actually, non-relativistic QM can also be expressed that way. Start with the traditional formulation in terms of the position operator $X$. (I'll consider only one dimension for simplicity.) This single operator $X$ is really just a convenient way of packaging-and-labeling a bunch of mutually-commuting projection operators, namely the operators $P(R)$ that project a wavefunction $\Psi(x)$ onto the part with $x\in R$, cutting off the parts with $x\notin R$. In fancy language, the commutative von Neumann algebra generated by $X$ is the same as the commutative von Neumann algebra generated by all of the $P(R)$s, so aside from how things are labeled with "eigenvalues," they both represent the same observable as far as Born's rule is concerned. If we look at how non-relativistic QM is derived from its relativistic roots, we see that the $P(R)$s are localized within the region $R$ by QFT's definition of "localized" — at least insofar as the non-relativistic approximation is valid. In this sense, non-relativistic single-particle QM is, like QFT, expressed in terms of observables associated with bounded regions of space. The traditional formulation of single-particle QM obscures this.

Here's the point: when we talk about a position operator for an electron in a non-relativistic model, we're implicitly talking about the projection operators $P(R)$, which are associated with bounded regions of space. The position operator $X$ is a neat way of packaging all of those projection operators and labeling them with a convenient spatial coordinate, so that we can use concise statistics like means and standard deviations, but you can't have $X$ without also having the projection operators $P(R)$, because the existence of the former implies the existence of the latter (through the spectral theorem or, through the von-Neumann-algebra fanciness that I mentioned above).

So... can a photon have a position operator? If by position operator we mean something like the projection operators $P(R)$, which are both (1) localized in a strictly bounded region and (2) strictly reliable as "detectors" of things in that region, then the answer is no. A photon can't have a position operator for the same reason that a photon can't have a non-relativistic approximation: for a photon, there is no characteristic length scale analogous to $\hbar/mc$ to which the size of a localization region can be compared, without referring to something other than the electromagnetic field itself. What we can do is use the usual photon creation/annihilation operators to construct photon-detecting/counting observables that are not strictly localized in any bounded region but whose "tails" are negligible compared to anything else that we care about (outside of QEM), if the quasi-localization region is large enough.

What is a physical consequence?

What is a physical consequence of the non-existence of a strict position operator? Real localized detectors are necessarily noisy. The more localized they are, the noisier they must be. Reeh-Schlieder guarantees this, both for electrons and for photons, the main difference being that for electrons, the effect decreases exponentially as the size of the localization region is increased. For photons, it decreases only like a power of the size.

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    $\begingroup$ Thanks for the answer! This actually seems to contradict some other stuff I heard about -- I hear that you can define a Newton-Wigner position operator for all massive particles and massless particles of low helicity, it's just the photon that's the problem. (Specifically, it is defined as $i \partial_k$ in the one-particle sector, where the derivative is in momentum space. For the photon this fails because applying it produces longitudinal photons.) The eigenkets are perfectly localized states. $\endgroup$ – knzhou Jul 21 '19 at 10:40
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    $\begingroup$ What would you say about the construction of such position operators? I don't doubt the validity of what you're saying, but it means there must be slightly different notions of "position operator" at play here. $\endgroup$ – knzhou Jul 21 '19 at 10:41
  • $\begingroup$ @knzhou Newton-Wigner defined their position operator only for 1-particle states. They don't put it in the context of QFT, but from the form of their states, if we put their 1-particle theory into QFT in what seems like the obvious way, they aren't strictly localized according to QFT's local observables: they have tails. This is consistent with arxiv.org/abs/quant-ph/0007060, which critiques variants of the NW idea and concludes that their "local" observables don't commute at spacelike separation. So you're right: NW achieved localized states by changing the meaning of "localized." $\endgroup$ – Chiral Anomaly Jul 22 '19 at 0:19
  • $\begingroup$ I just want to say, I'd think that not having a local operator that kills vacuum is actually quite intuitive and sensible. Think about this way: the vacuum state $|0\rangle$ means the entire Universe is empty, right now. In order to verify such a fact, you would need to sense all of space to make sure it is free of particles "hiding" somewhere. Killing vacuum with a local operation means that you have somehow been able to find out what is going on even in the most ungodly-far-out reaches of space with that operation alone - something that clearly contradicts the information transmission $\endgroup$ – The_Sympathizer Jul 23 '19 at 7:05
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    $\begingroup$ @The_Sympathizer I'm not sure where the pop-sci phrase actually comes from. It might be an allusion to the structure of perturbation theory (Feynman diagrams), whereas the phenomenon highlighted in this post makes no reference to perturbation theory. Or maybe it's alluding to the fact that local observables necessarily have non-zero variance in the vacuum state, which is more directly related to the non-perturbative phenomenon highlighted here -- but the only way to actually measure an observable is to introduce a physical measuring device, and then the state isn't the vacuum state anymore. $\endgroup$ – Chiral Anomaly Jan 29 at 3:50
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The idea "photons do not have position operator" may have more meanings depending on who you ask.

To me, this statement means something very specific: EM radiation does not consist of particles that could be observed at some point of space and could be described by $\psi(r_1,r_2,...r_N)$ function in the sense of Born's interpretation. Instead, EM radiation itself is everywhere, and properly described by a function of 3 spatial coordinates - the thing to be studied is the EM field, not some particles of light. The field can be c number or q number, but the point is that the entity to be described is a field, not any set of particles. This view means there are no actual "particles of radiation" flying in hydrogen molecules, in contrast to electrons, which there are two in every neutral hydrogen molecule.

"Particles of light" or "photons" is a somewhat problematic word, because it doesn't have clear universally adopted concept behind it. The originator of the word meant something very different from what we use this term for after end of 1920's. Today, often it is meant as a short hand for "chunk of energy $hf$ transferred between matter and radiation of frequency $f$"; it may be distributed in some region of space but it is not localized at any single point of space.

Of course, one can go to the simple examples and talk about things such as "1 photon in mode (1,1,1,1), 2 photons in mode (2,2,2,2)" as a state of EM field in a box, but these states are of the whole system, one cannot go and find some real things at some point of space within the box more precisely than "in the box".

When an optics experiment is done using a laser beam, it is perfectly meaningful to talk about photons being in the beam.

Usual laser light is well described by a classical EM wave with definite electric strength vector and wave vector. This means it does not have any definite number of photons in it, it is better described (if needed) as coherent state. One can talk about photons in superposition, but then there is not definite number of photons of any definite kind there. The photons there are a mathematical fiction, spread from minus infinity to plus infinity.

We can also speak of a photon being emitted by an atom, in which case it is obviously localized near the atom when the emission occurs.

Yes, but this region is huge, its size is greater than wavelength of the emitted radiation. The claim is that it makes no sense to assign position to that emitted radiation within this region.

Furthermore, in the usual analysis of the double slit experiment one has, at least implicitly, a wavefunction for the photon, which successfully recovers the high school result.

Yes, this is because diffraction on slit can be roughly analyzed with simplified models such as diffraction of scalar field. This does not necessarily mean wave function of photons is a useful concept in general problems of interaction of light and matter. Try to describe spontaneous emission in terms of "wave function of photon".

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    $\begingroup$ In my mind a photon is an exchange of energy requiring 2 atoms whether they are an inch or a million light years apart. No photon is ever generated until atom 1 finds atom 2 and then the EM field exists for a period of time depending on this distance, this is the wave function. We can alter this wave function before transfer/collapse occurs. The wave function prefers with high probability a path that is an integer times its wavelength (i.e. laser cavity). We observe the double slit pattern (which is not an interference) as a pattern of allowed paths, bright; and dark areas of no paths. $\endgroup$ – PhysicsDave Jul 21 '19 at 0:41
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As other answers have noted, the first task is to define what is meant by position operator. It helps to start from something more basic than QFT.

The notion of position operator in QM derives from the notion of position in classical physics. In classical physics this notion is obviously well-defined: you can tell where an apple is by simply looking at it. This position has a well-defined evolution and does not depend on the way you measure it.

In QM we know that position operator does not have to have a definite value in a state. In principle, one could anticipate something like this: as things that you measure get smaller, it gets harder to measure the position without disturbing it. If you can't measure something without disturbing its value, how can you say that it is well-defined? However, this anticipation is not what happens. In QM the lack of a definite value of position in some (most) states is not due to the disturbance from the measurement, but is instead a fundamental property of our quantum world. QM is very interesting because this property kicks in before measurements start being too invasive. Let's consider a concrete example: measuring the position of an non-relativistic electron. We can do so by scattering a photon off it, and detecting where this photon goes. If we use photons of energy $h\nu$, we can localize the electron to within $\Delta x= c/\nu$. Suppose the electron does not receive a relativistic kick from the photon, so that we stay in non-relativistic realm. This requires $h\nu\ll mc^2$. The during the measurement time $1/\nu$ the electron will travel at most $c/\nu$, and so our estimate of measurement error is $\Delta x$ is valid. This error is $\Delta x= c/\nu\gg \frac{h}{mc}$, where the right hand side is arbitrarily small in the non-relativistic limit $c\to \infty$, and thus $\Delta x$ can be made arbitrarily small as well.

So in non-relativistic QM position operator is of quantum-mechanical nature, but there is no practical problem with measuring it experimentally. The important point is that there is a universality in measurements: we can perform different measurements of the position, but all these measurements can be mathematically described by measuring the position operator.

In relativistic QM, aka QFT, we now have both problems: the system is quantum mechanical and there are practical problems with measuring the position experimentally. In the above discussion, we can use photons of energies $h\nu\sim mc^2$ to localize electron in $\Delta x\sim\frac{h}{mc}$, but if we go to higher $h\nu$, we will strart creating electron-positron pairs, and not it is not clear what we are measuring anymore: say if we spawned an electron-positron pair, position of which electron are we measuring?

Here let me step back and discuss the formal problem of defining position in classical relativistic theory with indistinguishable particles. Because particles are indistinguishable, we cannot ask for space position of a single particle as a function of time. Instead, the only sensible question to ask is "how many worldlines intersect a given spacelike surface element?" In other words, we want to define a conserved particle number current $J_N^\mu(x)$ and measure its flux through a spacelike surface $S$ ($S$ can have a boundary and be small), $$ N_S = \int_S J_N^\mu(x) dS_\mu. $$

Getting back to QFT, the problem is that there is no particle number current, since the particle number is not conserved by interactions. One can define something that, to one's taste, "looks like" a particle number current, but it will not have the property of being the universal quantity measured by different experiments. Instead, different experiments will each measure its own observable, with these observables being hopefully equivalent in non-relativistic limit.

One can ask what happens in free theories, where one can imagine defining particle number operator. The answer is that you can't measure anything in a free theory, since there are no interactions. You can write any observable and declare it to be the position operator, but it won't be related to any experiment. As soon as you imagine doing an experiment, you introduce interactions which break particle number conservation. (I am ignoring here 2d integrable QFTs with no particle production, which perhaps merit their own discussion.)

That said, there are conserved currents in QFT, for example the electric current, and it is possible to measure those. In particular, for a conserved current $J$ one can consider observables of the form $$ Q_S = \int_S J^\mu(x) dS_\mu. $$ These observables are sufficiently universal because gauge fields couple to conserved currents, and you can design experiments which interact with your system via these gauge fields. For example, in deep inelastic scattering one, to a good approximation, measures matrix elements $$ \langle H|J^\mu(x)|X\rangle $$ where $H$ is a hadron state and $X$ are various final states, and $J$ is the electric current of QCD. This comes from scattering an electron off $H$. To the leading order in fine structure constant, the electron emits a single virtual photon, which in turn couples to $J$ of QCD.

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  • $\begingroup$ Note that these observables $Q_S$ do not go against what is said in Chiral Anomaly's answer. In particular, unless $S$ is a complete spacelike slice, so that $Q_S$ is the total charge, $Q_S$ will not have definite value in the vacuum state. In particular, it will not annihilate the vacuum state. This is because any experiment which measures $Q_S$ will interact with the system sufficiently strongly as to have an amplitude of creating particle-antiparticle pairs. The formal proof of this is in Chiral Anomaly's answer. $\endgroup$ – Peter Kravchuk Jul 21 '19 at 18:19
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Introduction

What is meant by this is really that, unlike in non-relativistic quantum mechanics, in Relativistic Quantum Field Theories (RQFTs) - such as those which describe photons - the position of a particle, any one, including massive particles like electrons, cannot ever be arbitrarily high-information. It does not mean that there is no sense in talking about position whatsoever, contrary to how this often gets put across, but it does have consequences for how to describe it mathematically.

And I think part of the problem is that the existing formalism often handed down unquestioningly is pretty dated conceptually and we have much better ways to talk about these things in the modern age. This post, for better or worse, tries to cut through some of that legacy cruft and ends up as a whirlwind "tour de force" of classical to modern physics basically because we have to connect back to so many other concepts to really get at what is going on here and put it on a sound conceptual footing. And I think it's a shame because much of the real beauty of these theories goes unappreciated with the treatments they so often get.

To understand this requires us to be careful - to exercise Discernment - about a number of things:

  1. what constitutes a "particle",
  2. what is "position",
  3. what does it mean to have "information about" something like a particle's position,
  4. what is a "quantum field", and
  5. how do we describe "particles" in terms of such a thing, and how does a description in terms of such affect 1-3 above.

Without being precise as to what each of these means, we cannot properly understand this statement, nor tease out what is wrong with the various jabbings given at it from many admittedly not-so-great quality sources. Hence,

What is a "particle"?

For the first point, we will say that we cannot, actually, define this kind of concept from a viewpoint of formal mathematics, and we shouldn't. It's just like how that in theoretical mathematics alone, we have certain "primitive concepts" like in axiomatic Euclidean geometry, we have straight lines or points, or otherwise, in set theory, sets are considered as such. They are not necessarily "meaningless" although it's often, and I think very unhelpfully, claimed that that is how they should be dealt with when really we need to exercise our Discernment in separating the "meaning" from the usage in the mathematical formalism. It is, rather, that to describe their meaning goes outside the realm of mathematics - from within mathematical formal language alone ("formal language" is roughly the language of mathematical and logical symbols, here), there isn't a "meaning" in the sense that we cannot write another formal language statement saying what it is. Yet to say it has "no meaning" as an absolute, without due attention to this qualifier, is wrong - the meaning is to us, not the symbols. It would be like saying the words on this paper have no meaning, when clearly, they do, or the individual letters.

So a "particle" here, has a meaning. It is an imagined entity that we are using in our model - we don't know if any "really exist", but they exist in the mental model of reality we're trying to make. A particle is a very tiny object - so small that we would mathematically assign it a size of zero: it occupies an amount of space equal to a point.

What is "position"?

"Position" is a bit more complicated to deal with - since it seems that, again, very often, the conflation seems to occur here that phenomena we will be discussing regarding position end up as somehow having bearing on size, which is not correct. To understand it, having experience with computer graphics and computer game design and modification, I think, really helps. In computer games, you have "avatars" or "objects" that are abstract geometric objects. They are specified by a geometry file that is independent of them being used within a game world. When put into such, they are given a parameter called the position, which effectively references a point int the space of the game world, and which nails a copy of the object described by the geometry in the geometry file to that point. The important point here is that while the position references a single point, its fact of being such is not the same thing as the object being of point-like nature in size: the size of the object is defined by the geometry in the avatar - what is its width if you take a (virtual) tape measure from one end to the other. Instead, what happens is we have some reference point on the avatar and we move that to coincide with the position point.

In the case of "particle" and "position" taken together, the particle is an "avatar" consisting of a single geometric point only. Position is then a parameter we are going to affix to that avatar that tells us where it appears in our model of the world we have in our head (which could be translated into an actual computer model, though QM and especially RQFT are notoriously intractable to actually do in practice). Note that whatever happens to position has no bearing on the "size" of the particle: that is defined by the geometry in the avatar, and that doesn't change even if we were to delete the parameter "position" altogether.

(If you want mathematics, an avatar is a set of points taken out of a Euclidean space with their metrical interrelations preserved, plus a designated center or pivot point. Using the avatar concept also helps greatly, I think, when dealing with, say, classical rigid body dynamics and the position and orientation coordinates. "Positioning" the avatar can be thought of as dropping it into the space and then applying geometric transformations, e.g. translations and rotations, to align the pivot to the given coordinates. The usual physics formalism is really, I think, rather dated, as said.)

In classical mechanics, position is defined by a triple of real numbers, e.g. the Cartesian coordinates: $(x, y, z)$. For extended avatars, we also have the orientative coordinates, e.g. $(\theta_R, \theta_P, \theta_Y)$ (yes, I'm partial to the Tait-Bryan angles; sue me but they're more intuitive, I find, than the Euler angles.). For a particle, there are no orientative coordinates, or they are irrelevant, as it's a single point.

Such a specification of position, we say, takes infinite information, because as these are real numbers, they require an infinite number of digits to write them down exactly in a truly arbitrary, general case. Classical mechanics is thus a "theory with infinite information".

What does "information about" mean and what does QM do?

In quantum mechanics, what happens now is that we change two things - one is that we have to go from an "objective" to "subjective" view: we aren't going to be talking any more about what position a particle "really" has without perhaps a few qualified exceptions, but instead about what information an agent - some entity capable of interacting with and getting information about an external system - has about the position of that particle. Thus the Universe always has at least two elements in it: object and agent. We can't take an inconsequential "view from nowhere" or "God's eye trick", to use terminology echoing the feminist philosopher Donna Haraway, and perhaps others in a similar vein. Our "view" is from "somewhere", and we have to take account of the viewing agent's interactions with its world.

Hence we talk less of the position of the particle and instead more of agent's knowledge of said position.

When we do this, we actually gain descriptive flexibility in that we can then talk about varying levels of knowledge through the machinery of Bayesian probability and information theory, "probability as information", "it from bit" (John Archibald Wheeler), sue my socks, it works.

Glossing details, the result is that we jettison the usual coordinate assignment $(x, y, z)$ in favor of a probability distribution function

$$\psi(x, y, z)$$

instead. Moreover, due to other reasons which aren't immediately relevant to this discussion, we have to make this function a complex-valued, not real-valued, probability function. Such a distribution function can give "bad information" about the position, or "restricted information". Now you may be wondering how we can call this limited - I said it was real-valued, didn't I? Doesn't it still take infinite information thus to describe $\psi$, if not perhaps in a sense "even more"?

Sure, but then we should again make a distinction between "reality" and our model thereof. $\psi$ is not information we can reify as literally possessed by anything any more than it makes sense to reify it as a real extant wavefield as some do. It is a model for the agent's information, one that has a lot of verbiage to talk about little, so to speak, a lot of "ado", because that extra verbiage makes it very useful in constructing an accurate, predictive theory. But why probability specifically to capture this notion of "lesser information"? Well, probability tells us more about less because it says instead of a single alternative, there are a number of differently-weighted "possible" alternatives. If I say I am only 75% certain of something, that is "less informative" to you then if I say I am 100% certain. Likewise, for a probability distribution, the "broader" it is, encompassing more possibilities, the less informative it is, and the "tighter", the more informative. (The exact "content of information" or, better, "degree of privation of information" in a PD can be quantified by its Shannon entropy, $H$.)

Quantum fields

Now I'm admittedly gonna pick up the pace as I don't want to recapitulate all of physics in one post, but the next step is to go to quantum fields as quick as I can. You see, more generally we don't talk about solely functions of the form given above for a single particle. Instead, we talk about a mathematical object called quantum state vector that can be "decoded" to reveal probability distributions about many different parameters of that particle such as not just its position but also its velocity, orientation (if we have such), and so forth. These things are denoted with symbols like $|\psi\rangle$, called a "ket sign". "Decodings" of it into positions and velocities (better, momenta) are described by operators that act on these vectors - basically just functions, that eat a vector and make another.

In non-relativistic QM, that translates into having a positional operator $\hat{X}$ and a momental operator (also called impulsion operator) $\hat{P}$.

These operators "decode" the position and momentum by effectively "tagging" quantum state vectors as representing cases where we do have infinite information about the position and momentum, respectively. that is, the existence of positional operator $\hat{X}$ goes hand-in-hand with the existence of cases $|\mathbf{x}\rangle$ where the corresponding wave function $\psi$ is a delta function centered at $\mathbf{x}$. These are called "eigenstates" of position, and the decoding happens through expanding a state vector into components with these treated a linear algebra-style basis set.

Now, this formalism works all fine and good when we are considering a single particle, but it quickly gets bad for dealing with multiple particles - again leaving out details as to why, I wanna get there, PLEASE... And because of that, quantum field theory is, effectively, a way to deal with those multiple particles much more cleanly, through the use of a mathematical device called a "quantum field".

Basically, what that means is that we'll talk about a state vector (information datum) not of just one particle or a set number of particles, but rather for a system which can contain any number of particles, and moreover to which particles can be added or removed. Here's how that works. We start with a vacuum state vector $|0\rangle$, which is said to contain no particles, that occupies a suitably-rich vector space to make all of what we're going to be doing with it feasible. We then proclaim the existence of a creation and destruction operator (vector-to-vector function, remember?) $a^{\dagger}$ and $a$. There is one such operator for each position vector $\mathbf{x}$, e.g. $a^{\dagger}(\mathbf{x})$. (Alternatively, we can write $a^{\dagger}(x, y, z)$ to make the position coordinates explicit.)

Now, this $a^{\dagger}$ effectively acts as a "paintbrush" we can use to "paint" particles on the quantum field. If I apply $a^{\dagger}(\mathbf{x})$ to $|0\rangle$, it creates a vector with a particle with exact position (i.e. like the delta function) $\mathbf{x}$. That is, the vector $|\phi_\mbox{1 particle}\rangle := a^{\dagger}(\mathbf{x}) |0\rangle$, represents (information saying that) the quantum field is holding a single particle with exact position $\mathbf{x}$, i.e. a particle whose wave function

$$\psi(x, y, z)$$

is a delta-spike at $\mathbf{x}$. If we were to apply $a^{\dagger}$ again, i.e. say $a^{\dagger}(\mathbf{x}_2) |\phi_\mbox{1 particle}\rangle$, we now instantiate a second particle in the quantum field with exact position $\mathbf{x}_2$. Note that what the particle is has not changed: the denotation of what $a^{\dagger}$ created is still the location to pin the point avatar, only the mathematics we're using to talk about it, and that's something to keep in mind for the last few bits here.

Hence, you should note that it is not proper, then, to try to repeatedly apply $a^{\dagger}$ to try and get a particle with underdetermined position. Instead, and to really make clear why I use the term "paintbrush", to represent a particle with underdetermined position, we must superpose a number of the one-particle states, obtained by operating with $a^{\dagger}$ only once on the vacuum state but at each possible position, which we do with an integral:

$$|\phi_\mbox{1 fuzzily-posed particle}\rangle := \int_{\mathbb{R}^3} [\psi(x, y, z)\ dV]\ a^{\dagger}(\mathbf{x}) |0\rangle$$

This is just as how we would express the $\psi$ function in terms of superposing eigenstates of position in ordinary quantum mechanics to build the wave function, except now we're superposing states of the quantum field.

RQFTs

So what does relativistic quantum theory do? Well, introducing relativity causes something funny to happen. Effectively, intuitively, our "sharp" paintbrush $a^{\dagger}$ that is probably more aptly thought of as a pen, becomes a thick, frizzy one, a "true" paintbrush indeed: it itself can only paint states that are missing position information in the sense above, that they have nontrivial spread (and actually infinite support, i.e. they never entirely go to zero). Even worse, states with unlimited position information do not even exist, to begin with! The same painting technique will work, but it becomes a sort of "fuzz of fuzz" and the weight function $\psi$ in the integral loses some of its original significance. The Universe, effectively, has a strong upper limit as to how much information can ever exist to define a particle's position, not just a limit on the joint information of position and momentum together as per Heisenberg's principle.

This does not mean that position is non-existent or nonsense to even talk about, any more than that the fact that position being "fuzzy" (missing information) in ordinary quantum mechanics would. Nor does it mean the particle is not point-sized - remember, that question pertains to the "avatar" we separated out earlier, not whatever we're using to position it in space, and there are experiments to this effect that set the "size" of particles as being very small indeed (these work not by locating, but rather by scattering off particles, in a far-matured version of the techniques pioneered by Rutherford to study the atomic nucleus.).

Yet it does necessitate a change in the mathematical description of such "position" - remember that I just said that before we were describing one-particle positions with operators that "tagged" exact position states? Well, we don't have those anymore (if we did, then we could use them to make a sharp $a^{\dagger}$ brush, but we can't), so the original idea of trying to figure what $\hat{X}$ meant in terms of "eigenstates", is gone! The operator formalism we had been using before, no longer works to talk about the position of particles! (It still does work in other ways as, above, we just used the "painting operator" $a^\dagger$, just not for this way!) Instead, we must use other tools to describe the situation of "what is going on in space", which some of the other posts here have covered, and while I could get into that, I'm getting a little suppressed now and moreover I think this is far enough to nail the statement in question and what it means.

(Moreover, perhaps, this suggests we should call quantum field theory better as "paintbrush quantum mechanics" or "painter's physics" :) )

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I will answer as an experimental particle physicist. This is the double slit experiment single photon at a time:

singlphot

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

It is experimentally evident that single photons do hit the screen of the camera, so the problem of a position operator is a mathematical theory problem, not an experimental observation problem.

My impression of quantum field theory is that in general the fields representing all the particles on which operators act to create or annihilate particles, are plane wave solutions of the corresponding quantum mechanical equation: Dirac for fermions, Klein Gordon for bosons , quantized Maxwell for photons. As is well known plane waves cover the whole space time so I do not think there is a meaning in position operators acting on the fields themselves. The theory has to use a wavepacket to define a real particle localized in space and time, afaik.

In general , I find that most people theoretically inclined tend to view the mathematics of the theory as generating the real world, and not the other way, the mathematics modeling the real world. The useful QFT models use wavepackets for real particles to be measured in experiments, if it is necessary to define the probable location of the particle, as in the double slit experiment, again AFAIK.

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After the correct comments, I agree with the other answers on classical QM, and the position operator. My answer is about the proper time, and relativistic QM (QFT) and that photons do not have a proper time. Though, you can define a λ affine parameter that increases monotonically along the lightlike worldline. But still, photons do not have a proper time and in relativistic QM (QFT) this will (defining a position eigenstate) violate casuality.

How to define the proper time of a photon?

A point in Minkowski space is a time and spatial position, called an "event", or sometimes the position four-vector or four-position or 4-position, described in some reference frame by a set of four coordinates: When considering physical phenomena, differential equations arise naturally; however, when considering space and time derivatives of functions, it is unclear which reference frame these derivatives are taken with respect to. It is agreed that time derivatives are taken with respect to the proper time {\displaystyle \tau } \tau . As proper time is an invariant, this guarantees that the proper-time-derivative of any four-vector is itself a four-vector.

https://en.wikipedia.org/wiki/Four-vector

IF you don't have proper time for photons, you cannot use the differential of the four-position vector. Thus, you cannot define a position operator for the photon.

A photon is an elementary particle in the SM, the quantum of EM field and radiation, point-like, with no spatial extension or substructure, the force carrier of the EM force, and is massless.

It is possible to emit single photons, so I will assume you are asking about that.

Like all elementary particles, photons are currently best explained by quantum mechanics and exhibit wave–particle duality, exhibiting properties of both waves and particles. For example, a single photon may be refracted by a lens and exhibit wave interference with itself, and it can behave as a particle with definite and finite measurable position or momentum, though not both at the same time as per Heisenberg's uncertainty principle. The photon's wave and quantum qualities are two observable aspects of a single phenomenon—they cannot be described by any mechanical model;[2] a representation of this dual property of light that assumes certain points on the wavefront to be the seat of the energy is not possible. The quanta in a light wave are not spatially localized.

https://en.wikipedia.org/wiki/Photon

So let's say there is consensus on what a photon is.

Let's start understanding the problem, why photon's do not have positions, by understanding how photons can be created and annihilated (because they can only have a position while they exist).

Photons are emitted in many natural processes. For example, when a charge is accelerated it emits synchrotron radiation. During a molecular, atomic or nuclear transition to a lower energy level, photons of various energy will be emitted, ranging from radio waves to gamma rays. Photons can also be emitted when a particle and its corresponding antiparticle are annihilated (for example, electron–positron annihilation).

So basically there is consensus on how photons can be created and annihilated:

  1. emission (creation)

  2. absorption (annihilation)

So basically we can say that there is at least consensus on the time of existence of a photon.

Now it is very important to understand why this question was raised at all. When does the photon have a position? When it exists. When does the photon exist? In what timeframe?

This is the main point. The question only does have a merit in our (why have rest mass and experience time differently then a photon) timeframe.

Now photons are massless, as per SR, they do not have a reference frame. It does not make sense to talk about what a photon sees when it is traveling between emission and absorption.

The question, why don't photons have a position is raised because in our timeframe, we move along spacelike and timelike worldlines (the photons' is lightlike).

There are three cases:

1.spacelike, in this case, it is the physical distance between two points in space.

space-like curves falling outside the light cone. Such curves may describe, for example, the length of a physical object. The circumference of a cylinder and the length of a rod are space-like curves.

  1. timlike, these must fall within the cones defined by lightlike cones

time-like curves, with a speed less than the speed of light. These curves must fall within a cone defined by light-like curves. In our definition above: world lines are time-like curves in spacetime.

  1. lightlike, this is what photons do, they basically travel spacetime distances of 0 between emission and absorption

light-like curves, having at each point the speed of light. They form a cone in spacetime, dividing it into two parts. The cone is three-dimensional in spacetime, appears as a line in drawings with two dimensions suppressed, and as a cone in drawings with one spatial dimension suppressed.

Now your question is because we, who have rest mass, do live along 1. and 2. worldlines. As per SR, we do not know 3. lighlike worldlines, at least we do not know what that would look like, since it does not make sense to talk about a frame of a photon.

The world line (or worldline) of an object is the path that object traces in 4-dimensional spacetime. It is an important concept in modern physics, and particularly theoretical physics.

https://en.wikipedia.org/wiki/World_line

Now you are asking why we can't define the position of a photon in space (3D).

The Born rule gives the probability of a QM measurement to yield a given result.

In its simplest form it states that the probability density of finding the particle at a given point is proportional to the square of the magnitude of the particle's wavefunction at that point.

So there is consensus about what position should be. It should be a probability to find the particle at a certain position in space (3D).

Now there are the reasons why we cannot give a position for the photon:

  1. you are asking for a position for a photon (of course because we live in a massive world where we experience time between emission and absorption) on a space or time like worldline.

  2. you are asking for a position in space (3D), and space is continuous.

  3. mwasurement, that is observation, which means interaction with the photon, now you want the position of the photon on the fly, that is between emission and absorption, and that is only possible with elastic and inelastic scattering

The photon only exist on the lightlike worldline, where the spacetime distance between emission and absorpion is 0. This makes the problem with 1., where you are asking for a position for the photon between these two points, that are separated by a distance (3D) only in our world (where we have rest mass), but in the photon's world it does not have such a position (or distance) inbetween.

You are asking for a position in space (3D) in a fabric of spacetime, that is continuous, and you are trying to model it as discrete, but in reality it is continuous.

In quantum mechanics, the uncertainty principle (also known as Heisenberg's uncertainty principle) is any of a variety of mathematical inequalities[1] asserting a fundamental limit to the precision with which certain pairs of physical properties of a particle, known as complementary variables or canonically conjugate variables such as position x and momentum p, can be known.

https://en.wikipedia.org/wiki/Uncertainty_principle

This means, that we can try to restrict the photon to a space that is small. Relative to what? Our world? Our devices? Whatever measurement you would use, spacetime is continuous and you are asking for a discrete measurement.

The only way to measure the position of the photon is to interact with it. But you want to do it on the fly, between emission and absorption, and to do that you need:

  1. elastic scattering, the photon will keep its energy and phase and change angle

  2. inelastic scattering, the photon will keep part of its energy and phase and change angle

Whichever you choose, it will have to obey the HUP.

Although position of a massive particle is an observable in nonrelativistic quantum mechanics, photon position is a controversial concept. It has been argued that there is no photon number density, only energy density [1]. Solutions to the photon wave equation are electric and magnetic fields, but the relationship between these fields and the normalizable Laudau-Peierls (LP) photon number amplitude is nonlocal [2, 3]. Localization of a converging (or diverging) photon pulse cannot be exact since, according to the Paley-Weiner theorem, it must have subexponential tails [4].

https://www.researchgate.net/publication/45927278_Photon_position_measure

So basically the answer for your question is:

  1. there is no position inbetween the creation and annihilation of a photon, on a lightlike worldline

  2. spacetime is continuous, and you are asking for a discrete position

  3. the only way to measure the position of the photon on the fly between emission and absorption is with elastic and inelastic scattering, and that again obeys the HUP

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  • $\begingroup$ This doesn't really make sense to me. Nowhere in the question does it demand position to be discrete, and the logic by which you conclude that zero spacetime distance somehow means you cannot ask for the position of an object moving along a null curve is both unclear and clearly false: There is no contradiction in classical relativity to ask for the position of a massless object moving along a null curve. It is only in relativistic quantum mechanics that this becomes ill-founded, and this answer does nothing to explain that. $\endgroup$ – ACuriousMind Jul 21 '19 at 16:01
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    $\begingroup$ The purpose of the large number of quote for rather standard material is also unclear. $\endgroup$ – ACuriousMind Jul 21 '19 at 16:01
  • $\begingroup$ @ACuriousMind I agree with the other answers on classical QM. Though, I will edit my answer as I am talking about proper time and relativistic QM. $\endgroup$ – Árpád Szendrei Jul 22 '19 at 7:22

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