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LarryS
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Apparently, in QM, the photon does not have a position operator. Why is this so?
As usual, thanks in advance.
As usual, thanks in advance.
So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?MrRobotoToo said: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.
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 16:18referframe said: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).referframe said:So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?
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.jostpuur said: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?
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.referframe said:So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?
MrRobotoToo said: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.
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.
jostpuur said:That almost sounds like that humans' decision about the meaning of ##|x\rangle## would affect the behavior of Schrödinger's equation.
jostpuur said: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.
PeterDonis said: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 said: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.
jostpuur said: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?
PeterDonis said: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.
referframe said:at some point, especially when we do an experiment, we do choose a specific reference frame - the reference frame of the lab
referframe said:can we not also consider the time evolution (within Minkowski spacetime), of the system within that frame?
PeterDonis said: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##.
unusualname said:the Schrodinger Equation is a postulate of quantum mechanics
unusualname said: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.
jostpuur said:Perhaps if we replace the differential equation by
[tex]
|\psi(t)\rangle = e^{-\frac{it}{\hbar} H}|\psi(0)\rangle
[/tex]
it will look more familiar for the QFT contex?
PeterDonis said: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 the Heisenberg picture, we make the operators [itex]\phi[/itex] and [itex]\pi[/itex] time-dependent in the usual way:
[tex]
\phi(x) = \phi(\textbf{x},t) = e^{iHt}\phi(\textbf{x})e^{-iHt}\quad\quad\quad (2.43)
[/tex]
PeterDonis said:In QFT, there are no wave functions ##\psi(t)##.
jostpuur said:You should not deny this
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.PeterDonis said: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.
vanhees71 said:by definition the Hamiltonian defines the time evolution of observable quantities
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.Fred Wright said:The photon has no rest frame. Computing an expectation of position for such an object is nonsense. Speedy Gonzalez will laugh at you.
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 said:This requires choosing a particular coordinate chart, whose coordinate time the time evolution is with respect to, correct?
vanhees71 said:so it's with quantum theory, which is based on the Hamiltonian formalism
True, but where in your calculation do quantum field expectations come into play? The photon by definition is a quantum object.Nugatory said: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.
There are no violations of causality in QFT. QFT is formulated explicitly such that there are none!referframe said:So, the violations of causality in QFT also apply to massive particles that are moving fast but still less than c?
PeterDonis said:This requires choosing a particular coordinate chart, whose coordinate time the time evolution is with respect to, correct?
HomogenousCow said:Choosing a particular frame does not break lorentz invariance
The position operator is a mathematical tool used in quantum mechanics to describe the position of a particle. However, photons are massless particles that do not have a defined position in space. As a result, the concept of a position operator does not apply to photons.
While we can detect the presence of a photon using a detector, the position of a photon cannot be precisely measured due to the uncertainty principle. This principle states that the more precisely we know the position of a particle, the less we know about its momentum, and vice versa. As photons have no mass, their momentum is always uncertain, making it impossible to accurately measure their position.
In experiments involving photons, their position is usually determined indirectly by measuring the position of the detector where they are detected. The position of the detector can then be used to infer the position of the photon at that moment. However, this does not give us a precise measurement of the photon's position, only an estimation.
While photons do not have a defined position in space, they do have a defined direction of travel. This direction is determined by the momentum of the photon, which is always in the direction of its propagation. As photons have no mass, they are not affected by external forces and can travel in a straight line.
No, there are no exceptions where a position operator can be applied to photons. The concept of a position operator only applies to particles with mass, and photons are massless particles. Therefore, the position of a photon cannot be described using a position operator.