Why no position operator for photon?

[LarryS]

Apparently, in QM, the photon does not have a position operator. Why is this so?

As usual, thanks in advance.

 

[MrRobotoToo]

It can be shown that in the context of relativistic quantum theory the position operator leads to violations of causality. The demonstration can be found, for example, in the first chapter of Ticciati's "Quantum Field Theory for Mathematicians". This is why QFT dispenses with the position operator altogether, relegating position to the role of a parameter.

 

[vanhees71]

It's also in Arnold Neumaiers FAQ (who is a member of PF):

http://arnold-neumaier.at/physfaq/topics/position.html
http://arnold-neumaier.at/physfaq/topics/localization

 

[LarryS]

It can be shown that in the context of relativistic quantum theory the position operator leads to violations of causality. The demonstration can be found, for example, in the first chapter of Ticciati's "Quantum Field Theory for Mathematicians". This is why QFT dispenses with the position operator altogether, relegating position to the role of a parameter.

So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?

 

[MrRobotoToo]

So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?

QFT is what's needed to avoid the causality violations that result from applying QM and SR in a simple-minded way to particles. But yes, they're present for massive particles as well. I had forgotten that Markus Luty, a physicist at UC Davis, had provided a demonstration of causality violation in a video lecture of his. It start at 1:40 and ends around 16:18

 

[vanhees71]

So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?

There are no violations of causality in QFT, because QFT is precisely constructed such that it is causal (see e.g., Weinberg Quantum Theory of Fields, vol. 1).

 

[jostpuur]

I once complained to some physicists that isn't it amazing that there is no spatial representation for the relativistic quantum mechanical states. They responded to me that if you define the excitations of the electromagnetic field the usual way with lots of operators, then it will be equivalent to the spatial wave functions of photons obeying the Maxwell's equations. To me it sounds like that if there exists a spatial wave function for photon, then there also exists a position operator, so is this answer that I got back then contradicting the answer you guys are putting forward in this thread?

 

[MrRobotoToo]

I once complained to some physicists that isn't it amazing that there is no spatial representation for the relativistic quantum mechanical states. They responded to me that if you define the excitations of the electromagnetic field the usual way with lots of operators, then it will be equivalent to the spatial wave functions of photons obeying the Maxwell's equations. To me it sounds like that if there exists a spatial wave function for photon, then there also exists a position operator, so is this answer that I got back then contradicting the answer you guys are putting forward in this thread?

If you try to interpret the $A_{\mu }$ of Maxwell's equations as the wave function of a single photon, then you'll run into problems (negative energies, causality violation, etc.). But the equation can be salvaged by instead viewing $A_{\mu }$ as a quantum field that creates/anninilates photons at different points of space and time. That's the basic trick of QFT: the relativistic wave equations don't describe the time evolution of wave functions, but of quantum fields.

 

[kith]

So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?

For the phrasing of your question see vanhees71's reply. If your question is whether a position operator can be defined for massive relativistic particles, the answer is yes, see https://en.wikipedia.org/wiki/Newton–Wigner_localization.

 

[jostpuur]

It can be shown that in the context of relativistic quantum theory the position operator leads to violations of causality. The demonstration can be found, for example, in the first chapter of Ticciati's "Quantum Field Theory for Mathematicians". This is why QFT dispenses with the position operator altogether, relegating position to the role of a parameter.

On the first page of the first chapter the author says

Section 1.6 shows that, if we adjoin eigenstates of the position operator to our state space, then there is a probability for signals to travel faster than light.

That almost sounds like that humans' decision about the meaning of $|x\rangle$ would affect the behavior of Schrödinger's equation. Like if humans decide to give $|x\rangle$ a meaning, then the solutions of $i\hbar\partial_t|\psi(t)\rangle = H|\psi(t)\rangle$ will become such that they violate causality. If humans instead decide to leave the notation $|x\rangle$ without any meaning, then the solutions of $i\hbar\partial_t|\psi(t)\rangle = H|\psi(t)\rangle$ will remain such that causality doesn't get violated. We of course don't won't the solutions of $i\hbar\partial_t|\psi(t)\rangle = H|\psi(t)\rangle$ to become such that they would start violating causality, so it is very important that we don't assign any meaning to the notation $|x\rangle$.

 

[PeterDonis]

That almost sounds like that humans' decision about the meaning of $|x\rangle$ would affect the behavior of Schrödinger's equation.

All it's saying is that including position eigenstates in the model means the model predicts violation of causality. You can't change reality by changing your model; all you can do is make the model match or not match reality.

 

[jostpuur]

Even the model's predictions should not depend on a choice of assigning a meaning to $|x\rangle$, because the model's prediction should come from the time evolution giving Schrödinger's equation.

What the author is really saying is that the behavior of relativistic quantum systems is so strange that it looks paradoxical (which is typical with special relativity, of course), and then the author is trying to convince the reader that the paradox can be nicely dealt with by refusing to think about it (which is a very typical way of "solving" paradoxes, when you don't know the real solutions).

"Never use $|x\rangle$, and then you are not going to see the difficult paradox (related to special relativity)".

 

[MrRobotoToo]

To expand on PeterDonis’ point, when developing a model of a system one has to decide which Hermitian operators will be elevated to the status of observables. The usual way of approaching this task is to identify the symmetries that the system obeys, which gives one the system’s symmetry group(s). The generators of the symmetry group(s) are then identified as observables of the system. This approach leads to many of the usual suspects: energy, momentum, angular momentum, charge, etc. In non-relativistic quantum mechanics, in which the relevant symmetries are encapsulated by the Galilean group, if one elevates the position operator to the status of an observable and then calculates its commutation relations with the generators of the Galilean group, one finds that the generators can be expressed completely in terms of the position operator and a naturally defined velocity operator (this is in fact how one derives the position representation of the Schrödinger equation, as is shown, for example, in the third chapter of Ballentine’s “Quantum Mechanics: A Modern Development”). However, if one attempts something similar with the relativistic Poincaré group, one ends up with some funky commutation relations that can’t be easily made sense of, as is explained in the Wikipedia article linked to by kith.

 

[PeterDonis]

Even the model's predictions should not depend on a choice of assigning a meaning to $|x\rangle$, because the model's prediction should come from the time evolution giving Schrödinger's equation.

First, Schrodinger's equation is non-relativistic, so if we're talking about relativistic quantum theory, it's not the right equation.

Second, "time evolution" is problematic in any relativistic (more precisely, Lorentz invariant) theory, because there is no preferred frame and therefore no preferred "time". So that's not really a good way of thinking about the model's predictions.

That said, the model's predictions certainly do depend on what states you include in the state space, because the states in the state space are the possible states of the system (in the model). So including position eigenstates in the state space is saying they are possible states of the system, and what the author appears to be saying (at least based on what you quoted--I do not have the book itself) is that if position eigenstates are possible states of the system, you can get violations of causality.

 

[jostpuur]

First, Schrodinger's equation is non-relativistic, so if we're talking about relativistic quantum theory, it's not the right equation.

When I wrote down Schrödinger's equation, by $|\psi(t)\rangle$ I meant something very general, like a wave functional, that contains the information about all the excitation states of the quantum fields, and so on. QFT texts usually speak about things like Heisenberg picture or interaction picture, but you should keep in mind that they are defined in such way that ultimately they should be equivalent with some Schrödinger picture. Perhaps if we replace the differential equation by

$$|\psi(t)\rangle = e^{-\frac{it}{\hbar} H}|\psi(0)\rangle$$

it will look more familiar for the QFT contex?

Second, "time evolution" is problematic in any relativistic (more precisely, Lorentz invariant) theory, because there is no preferred frame and therefore no preferred "time". So that's not really a good way of thinking about the model's predictions.

I am confident that if a theory is really good, it will only look like that you could try to generate many time evolutions by using different frames of references, but eventually a closer look will reveal that the different time evolutions are actually equivalent. That is how Lorentz invariance often works out.

That said, the model's predictions certainly do depend on what states you include in the state space, because the states in the state space are the possible states of the system (in the model). So including position eigenstates in the state space is saying they are possible states of the system, and what the author appears to be saying (at least based on what you quoted--I do not have the book itself) is that if position eigenstates are possible states of the system, you can get violations of causality.

There are two possibilities concerning the time evolution giving operator $H$. Either it gives a time evolution that is consistent with the special relativity, or then it does not give that. I am not convinced that a decision to use or to avoid using $|x\rangle$ would have such effect on the behavior of $H$, because it does not look like that $H$ could depend on such thing. The decision to use or to avoid using $|x\rangle$ will obviously affect how much we can learn about the behavior of $H$ though, so the decision to avoid $|x\rangle$ looks like a decision to deliberately not study the behavior of $H$.

 

[PeterDonis]

When I wrote down Schrödinger's equation, by $|\psi(t)\rangle$ I meant something very general, like a wave functional, that contains the information about all the excitation states of the quantum fields, and so on.

That doesn't change the fact that Schrodinger's Equation is not relativistically correct. You can't make it relativistically correct by redefining what you mean by $\psi$.

Perhaps if we replace the differential equation by
$$
|\psi(t)\rangle = e^{-\frac{it}{\hbar} H}|\psi(0)\rangle
$$
it will look more familiar for the QFT contex?

No. Have you read any textbooks on QFT? I strongly suspect that you haven't, which means you don't have the requisite background for this discussion. In QFT, there are no wave functions $\psi(t)$. There are quantum fields, which are operators defined at particular points in spacetime.

 

[LarryS]

Second, "time evolution" is problematic in any relativistic (more precisely, Lorentz invariant) theory, because there is no preferred frame and therefore no preferred "time". So that's not really a good way of thinking about the model's predictions.

I've always been a little confused regarding this line of reasoning. It's true, especially in SR and GR, that we require the laws of physics to be independent of reference frame. But, at some point, especially when we do an experiment, we do choose a specific reference frame - the reference frame of the lab. In that experiment, we can measure energy and momentum separately, we are not forced to measure just four-momentum. So, can we not also consider the time evolution (within Minkowski spacetime), of the system within that frame?

 

[PeterDonis]

at some point, especially when we do an experiment, we do choose a specific reference frame - the reference frame of the lab

We can do this, yes, and we usually do, for reasons of convenience. But that doesn't mean we are forced to by the laws of physics. We're not.

can we not also consider the time evolution (within Minkowski spacetime), of the system within that frame?

Yes. But you will find that anything you want to express as a physical law or an actual observable, can be expressed in any frame, and cannot depend on the "time evolution" in your particular chosen frame.

Another way of putting this is: "time evolution in a particular frame" must be taken to mean looking at the coordinate times of various events, as assigned in that frame. But the only kind of "time" that has any invariant physical meaning is proper time along a particular worldline. And for a general physical system, there will be no coordinate chart in which coordinate time is the same as proper time along all worldlines belonging to the system: the best you can do, in general, is to construct a chart in which coordinate time equals proper time along one worldline. (The only exception I'm aware of is a system in which all particles move inertially and are always at rest relative to each other, in flat spacetime--in that case you can just use inertial coordinates in which the particles are all at rest.)

 

[unusualname]

That doesn't change the fact that Schrodinger's Equation is not relativistically correct. You can't make it relativistically correct by redefining what you mean by $\psi$.

Er, you make it relativistically correct by redefining what you mean by the Hamiltonian Evolution operator - the Schrodinger Equation is a postulate of quantum mechanics for a reason.

Just because people haven't worked out the correct evolution operator doesn't mean it's not possible to get relativistically consistent evolution operators in the Schrodinger Equation.

The actual, correct answer to the op, is that we haven't worked out a Hamiltonian operator for QED yet, so we can't have a totally correct position operator for a photon, there are suggested approximations in the literature.

 

[PeterDonis]

the Schrodinger Equation is a postulate of quantum mechanics

Not of quantum field theory. We are talking about quantum field theory, right?

Just because people haven't worked out the correct evolution operator doesn't mean it's not possible to get relativistically consistent evolution operators in the Schrodinger Equation.

In other words, you are not talking about actual mainstream physics but about your personal theory. Please review the PF rules on personal theories.

The same comment applies to the rest of your post.

 

[jostpuur]

Perhaps if we replace the differential equation by

$$|\psi(t)\rangle = e^{-\frac{it}{\hbar} H}|\psi(0)\rangle$$

it will look more familiar for the QFT contex?

No. Have you read any textbooks on QFT? I strongly suspect that you haven't, which means you don't have the requisite background for this discussion.

What I wrote should have looked more familiar to people who have read textbooks on QFT. For example Peskin and Schroeder write on their page 25

In the Heisenberg picture, we make the operators $\phi$ and $\pi$ time-dependent in the usual way:
$$\phi(x) = \phi(\textbf{x},t) = e^{iHt}\phi(\textbf{x})e^{-iHt}\quad\quad\quad (2.43)$$

Now recall carefully, that the Schrödinger and Heisenberg pictures are supposed to be equivalent, meaning that either they are both correct or both incorrect. It would make no sense to state that the Schrödinger picture would be incorrect, and that's why QFT would rely on the Heisenberg picture. According to our best knowledge, they are usually considered both correct.

In QFT, there are no wave functions $\psi(t)$.

The vector $|0\rangle$ appears everywhere in QFT, and lot of other vectors $|\psi\rangle$ can be generated out of it by adding all kinds of operators to its left side and taking linear combinations. In the Schrödinger picture ultimately there will be some time depending state vector $|\psi(t)\rangle$ that contains all the information about the state of the system, and its time evolution will be given by the Schrödinger equation. You should not deny this, because such denial would imply also denying the Heisenberg picture related equation I just quoted from Peskin and Schroeder.

 

[weirdoguy]

You should not deny this

But those are not wave functions. Read your QFT book carefully again.

 

[vanhees71]

First, Schrodinger's equation is non-relativistic, so if we're talking about relativistic quantum theory, it's not the right equation.

Second, "time evolution" is problematic in any relativistic (more precisely, Lorentz invariant) theory, because there is no preferred frame and therefore no preferred "time". So that's not really a good way of thinking about the model's predictions.

That said, the model's predictions certainly do depend on what states you include in the state space, because the states in the state space are the possible states of the system (in the model). So including position eigenstates in the state space is saying they are possible states of the system, and what the author appears to be saying (at least based on what you quoted--I do not have the book itself) is that if position eigenstates are possible states of the system, you can get violations of causality.

Position eigenstates are not in the state space, even not in non-relativistic quantum theory, where you always can define a position operator via the analysis of the Galilei group (which is a quite complicated topic, involving a lot of Lie-group and Lie-algebra theory as well as some topology; see Ballentine for a concise treatment). The "position eigenvectors" belong to the dual of the space, where the position operator is defined. Since this space is a proper dense subspace of the Hilbert space, its dual is larger than the Hilbert space, and the position eigenvectors are thus distributions, not Hilbert-space vectors, and only Hilbert-space vectors can define proper rays, which are representing pure states of the system.

Time evolution is not problematic in relativistic theories. That's why any comprehensible treatment of relativistic QFT starts with a careful analysis of the proper orthochronous Lorentz group. It turns out that you can define the physical observables, encoded in the S-matrix, i.e., transition matrix elements between asymptotic free states, in a covariant way (for the usual massive and massless unitary representations of the Poincare group). Together with micro causality this makes a causal picture of time evolution, and is a big part of the fact that relativistic QFT (and the Standard Model) is as it is!

Then there's more confusion about time evolution itself. Indeed, in any QT, including relativistic QFT, by definition the Hamiltonian defines the time evolution of observable quantities. The realization of this time evolution in the mathematical formalism is determined only up to unitary transformations, i.e., you can more or less arbitrarily choose how you distribute the time dependence to states (statistcal operators) and the operators representing observables. This choice is called the "picture of time evolution". Common examples are the Schrödinger picture (all time dependence is shuffled to the states), the Heisenberg picture (all time dependence is shuffled to the observables), and the interaction picture (observables evolve with the "free Hamiltonian", states with the "interaction Hamiltonian"). All pictures are FAPP unitary equivalent. In QFT there's however a formal obstacle, called Haag's theorem, but that's another story.

 

[Fred Wright]

The photon has no rest frame. Computing an expectation of position for such an object is nonsense. Speedy Gonzalez will laugh at you.

 

[PeterDonis]

by definition the Hamiltonian defines the time evolution of observable quantities

This requires choosing a particular coordinate chart, whose coordinate time the time evolution is with respect to, correct?

 

[Nugatory]

The photon has no rest frame. Computing an expectation of position for such an object is nonsense. Speedy Gonzalez will laugh at you.

But I have no problem computing the position of a flash of light in special relativity... And it doesn't have a rest frame either.

 

[vanhees71]

This requires choosing a particular coordinate chart, whose coordinate time the time evolution is with respect to, correct?

Sure, Hamiltonian mechanics always works in a fixed reference frame, and so it's with quantum theory, which is based on the Hamiltonian formalism. Nevertheless the observable quantities like S-matrix elements and cross sections are invariant.

 

[PeterDonis]

so it's with quantum theory, which is based on the Hamiltonian formalism

But you can do QFT without using the Hamiltonian formalism, using the Lagrangian and path integral.

 

[vanhees71]

Indeed, that's a way to avoid going through non-covariant formulations in intermediate steps. It's a great help to do calculations in manifestly covariant gauges in the Standard Model.

 

[Fred Wright]

But I have no problem computing the position of a flash of light in special relativity... And it doesn't have a rest frame either.

True, but where in your calculation do quantum field expectations come into play? The photon by definition is a quantum object.

 

[Elemental]

Position and momentum are definitely not on an equal footing in relativistic quantum mechanics. Momentum eigenstates are easily represented, but position representations are fraught with problems.

At least for massive particles one has Newton Wigner states which are strictly localized at one point in time in one reference frame. But there are at least two formidable problems within the context of relativistic quantum mechanics: 1) The states spread out faster than the speed of light as mentioned in Marcus Ludy's lecture (referred to as the Hegerfeldt paradox after, e.g., Hegerfeldt and Ruijsenaars, Physical Review D 22, 377 (1980)), and 2) They are localized in one reference frame only, so we ask why should a moving observe see a particle state to be spatially distributed which a stationary observer finds to be strictly localized?

For photons there is an additional obstacle. Any attempt (that I know of) to define a photon wave function ψ would require the transversality condition ψ.k = 0, where k is the wave vector. One can easily construct momentum eigensates ψ(k). But when one performs a Fourier transform of ψ(k), integrating over all k, the transversality condition introduces a dependence on k within the integrand which prevents the integral from forming a delta function.

Yes, the problems are overcome in quantum field theory. But there remains a lot of interest in understanding these issues in the context of quantum mechanics (evidenced by numerous papers over the years). To me it suggests that momentum is a more fundamental property than position, but I'm not sure why that would be the case. Any other insights welcome.

 

[vanhees71]

So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?

There are no violations of causality in QFT. QFT is formulated explicitly such that there are none!

 

[HomogenousCow]

This requires choosing a particular coordinate chart, whose coordinate time the time evolution is with respect to, correct?

Choosing a particular frame does not break lorentz invariance, all quantities still have the same transformation properties. We solve Maxwell’s equations in specific frames all the time and There’s no problem.

 

[PeterDonis]

Choosing a particular frame does not break lorentz invariance

I didn't say it did. I was just making the point that the Hamiltonian formalism hides the Lorentz invariance by expressing the dynamics in terms of the time coordinate of a particular frame. The Lagrangian formalism, by contrast, keeps the Lorentz invariance manifest by writing everything in terms of scalars, 4-vectors, and tensors. See the subsequent exchange between me and @vanhees71 .

 

[meopemuk]

Regardless of the causality issue, there seems to be purely mathematical obstacles for building a position operator with 3 commuting components in the photon Hilbert space:

T. F. Jordan, "Simple proof of no position operator for quanta with zero mass and nonzero helicity", J. Math. Phys. 19 (1978), 1382.

Margaret Hawton is working on this problem for many years. You can start, e.g., from this article

M. Hawton, "Photon position operator with commuting components", Phys. Rev. A 59 (1999), 954.

and check its citations at Google Scholar.

Eugene.