Relativistic Quantum Mechanics & Localized Particles

In summary, a lecturer discussed the flaws of relativistic QM for single particles by showing that for a state centered at the origin, it was possible to have a nonzero probability of ##Pr(\vec{x}>ct)>0##. He explained that this is due to the fact that multi-particle states should be considered in relativistic situations before introducing Fock-space states. Additionally, he mentioned that in relativistic QM, it is not the particles that are localized, but rather the measurements given by ##\hat{\phi}(x)##. He used the analogy of calorimeters in the LHC to illustrate this concept. The conversation also touched on the ability to predict the trajectories of particles using spatial distributions in QFT, and the possibility
  • #1
WWCY
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TL;DR Summary
Are we able to think about positions of particles in QFT?
A lecturer today told the class that relativistic QM for single particles is flawed by showing us that for a state centered at the origin, it was possible that ##Pr(\vec{x}>ct)>0##.

He said that this was down to the fact that we should be considering multi-particle states in relativistic situation, before introducing Fock-space states. He then mentioned that
a) We have evaded the ##\vec{x}>ct## problem as "we can't"/"it doesn't make sense to" think about localised particles in such situations.
b) In relativistic QM, it is not the particles that are localised, but the measurements given by ##\hat{\phi}(x)##. He then made an analogy to the calorimeters found in the LHC being localised, rather than the particles(?), which I couldn't really understand.

I can't make out the specifics of what he was trying to bring across;
For a), why can't we / doesn't it make sense to think about localised, relativistic particles? Does this have anything to do the the Fock-space states? As for b), suppose I take the mean value ##\langle n_1 , n_2, ...|\hat{\phi}(x,t)| n_1, n_2 ... \rangle##. What are the measurements given by ##\hat{\phi}(x,t)## made on, if not particles? And doesn't the calorimeter measure the energy deposited by particles that are in its locality (ie localised)?

Thanks in advance.
 
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  • #2
WWCY said:
Summary: Are we able to think about positions of particles in QFT?

[]
He then made an analogy to the calorimeters found in the LHC being localised, rather than the particles(?), which I couldn't really understand.

[]
And doesn't the calorimeter measure the energy deposited by particles that are in its locality (ie localised)?

Thanks in advance.
QFT allows one to calculate the probability of energy E to be detected at a certain place and time in a given volume. Which is where the detector is.

That says nothing at all about positions of particles.
 
  • #3
Thanks for your response.

Mentz114 said:
QFT allows one to calculate the probability of energy E to be detected at a certain place and time in a given volume. Which is where the detector is.

That says nothing at all about positions of particles.

i'm not quite sure I follow. Doesn't the particle have to hit a calorimeter cell to deposit energy? And what if we think of muon detectors, don't those provide information about where a muon is at some point in time?
 
  • #4
WWCY said:
Thanks for your response.
i'm not quite sure I follow. Doesn't the particle have to hit a calorimeter cell to deposit energy? And what if we think of muon detectors, don't those provide information about where a muon is at some point in time?
I think by position you mean 'trajectory', that is being able to predict a position of an individual indistinguishable thing at any time.

Obviously if a bunch of particles activate a detector then then one might make the entirely redundant deduction that they were briefly somewhere near the detector !
 
  • #5
Mentz114 said:
I think by position you mean 'trajectory', that is being able to predict a position of an individual indistinguishable thing at any time.

Obviously if a bunch of particles activate a detector then then one might make the entirely redundant deduction that they were briefly somewhere near the detector !

So would I be right to say that if we were able to come up with some spatial distribution, we could ##\textit{predict}## the trajectories of these particles, but since we can't come up with such distributions, we can only rely on local measurements ##\hat{\Phi}(x)## to tell us whether or not a particle is in some vicinity?
 
  • #6
WWCY said:
So would I be right to say that if we were able to come up with some spatial distribution, we could ##\textit{predict}## the trajectories of these particles, but since we can't come up with such distributions, we can only rely on local measurements ##\hat{\Phi}(x)## to tell us whether or not a particle is in some vicinity?
Yes, I believe that is true. But I am not an expert and I'm sure that one will be along eventually to tell us.

The cloud chamber is interesting and a search brings many hits like
https://en.wikipedia.org/wiki/Cloud_chamber
 
  • #7
I think you guys have gotten to the crux of what WWCY was trying to talk about, but I would like to add my two cents.

QFT does actually allow for the measurement of the exact position of a particle, and it is in such a way that does not violate the Uncertainty Principle. Two good theoretical descriptions of this are "Quantum Field Theory and the Standard Model" by Mathew D. Schwartz and "Quantum Field Theory for the Gifted Amateur" by Tom Lancaster and Stephen J. Blundell. The discuss in on page 22 of Schwartz's book and page 38 of Lancaster and Blundell. Basically how this is accomplished is by applying the field operator to the ground state $$\phi_0(\vec{x})|0>=\int\frac{d^3 p}{(2\pi)^3}(\hat{a}_{p}e^{i\vec{p}\cdot\vec{x}}+\hat{a}_{p}^{\dagger}e^{-i\vec{p}\cdot\vec{x}})|0>$$ This results in $$\int\frac{d^3 p}{(2\pi)^3}\hat{a}_{p}^{\dagger}e^{-i\vec{p}\cdot\vec{x}}|0>=\int\frac{d^3 p}{(2\pi)^3}e^{-i\vec{p}\cdot\vec{x}}|\vec{p}>$$ This is then ##|\vec{x}>##. Yes this is the non-relativistic case, but you get the same result even when you transition to four space. Now when we apply the state ##<y|## to this we get the result ##<\vec{y}|\vec{x}>=\delta^{(3)}(\vec{y}-\vec{x})##. Which shows us the projection to a single location in space. Futher the fact that ##[\phi(\vec{x}),\phi(\vec{y})]=0## tells us that we can measure the positions of the field operators at the same time.

Now for why this is not violating the Heisenberg uncertainty. Basically what the calculations are doing is putting infinite uncertainty into the momentum state of the particle. Thus, allowing us to know the exact location of the particle.

There is also research being done in Quantum Metrology where they are using a method known as quantum squeezing to take advantage of this very property. They are using this property to measure the displacement of particles. Burd, et. al. in their recent Science article https://science.sciencemag.org/content/364/6446/1163 showed that they have gotten down to "magnetudes well belowe these zero-point fluctuations". They performed the experiment with an Mg ion held above a surface electrode. Very fascinating read!
 
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  • #8
SisypheanZealot said:
Yes this is the non-relativistic case, but you get the same result even when you transition to four space. Now when we apply the state <y|<y|<→y|→x>=δ(3)(→y−→x)<y→|x→>=δ(3)(y→−x→)=\delta^{(3)}(\vec{y}-\vec{x}). Which shows us the projection to a single location in space.

This is not true in the relativistic case because of the lorentz-invariant measure $$\int \frac{d^3 p}{2 \omega_p}$$.
 
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  • #9
HomogenousCow said:
This is not true in the relativistic case because of the lorentz-invariant measure $$\int \frac{d^3 p}{2 \omega_p}$$.
So if I make the appropriate changes to account for the relativistic realm. We get $$|p>=\frac{1}{\sqrt{2\omega_p}}\hat{a}^{\dagger}_{p}|0>$$ and $$\phi(\vec{x},t)=\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(\hat{a}_{p}e^{-ipx}+\hat{a}_{p}^{\dagger}e^{ipx})$$ Applying the field operator to the ground state yields $$\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\hat{a}_{p}^{\dagger}e^{ipx}|0>=\int\frac{d^3 p}{(2\pi)^3}|p>e^{ipx}$$ Which still results in |x>. Where am I violating Lorentz-invariance, because I do not see it.

[Edit: typo in finaly equation]
 
  • #10
SisypheanZealot said:
So if I make the appropriate changes to account for the relativistic realm. We get $$|p>=\frac{1}{\sqrt{2\omega_p}}\hat{a}^{\dagger}_{p}|0>$$ and $$\phi(\vec{x},t)=\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(\hat{a}_{p}e^{-ipx}+\hat{a}_{p}^{\dagger}e^{ipx})$$ Applying the field operator to the ground state yields $$\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\hat{a}_{p}^{\dagger}e^{ipx}|0>=\int\frac{d^3 p}{(2\pi)^3}|p>e^{ipx}$$ Which still results in |x>. Where am I violating Lorentz-invariance, because I do not see it.

[Edit: typo in finaly equation]
First, the invariant measure is ##d^3p/2\omega##, not ##d^3p/\sqrt{2\omega}##. Second, I think that your normalization of ##|p\rangle## makes ##\langle p'|p\rangle## Lorentz non-invariant (check it!).
 
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  • #11
It's not that the position operator does not exist. It can be constructed in the same way as in non-relativistic quantum theory, as SisypheanZealot outlined above. It's that a Lorentz covariant position operator (on the usual Hilbert space) does not exist. But Lorentz non-covariance is not a problem if one thinks of position not as an intrinsic property of the particle, but as a quantity that only makes sense when measured with a measuring apparatus at rest in some particular Lorentz frame.
 
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  • #12
Demystifier said:
First, the invariant measure is ##d^3p/2\omega##, not ##d^3p/\sqrt{2\omega}##. Second, I think that your normalization of ##|p\rangle## makes ##\langle p'|p\rangle## Lorentz non-invariant (check it!).
Thank you, I had written things down incorrectly in my notes. The proper relationship was ##a^{\dagger}_{p}|0>=\frac{1}{\sqrt{2\omega_p}}|p>## which when applied does reconstruct the proper measure of ##\int\frac{d{3}p}{2\omega_p}##.
 
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  • #13
SisypheanZealot said:
Thank you, I had written things down incorrectly in my notes. The proper relationship was ##a^{\dagger}_{p}|0>=\frac{1}{\sqrt{2\omega_p}}|p>## which when applied does reconstruct the proper measure of ##\int\frac{d{3}p}{2\omega_p}##.
What you need to compute with your normalizations is ##\langle {\bf x}|{\bf x}'\rangle##. It is either proportional to ##\delta^3({\bf x}-{\bf x}')##, in which case it is not Lorentz invariant, or not proportional to ##\delta^3({\bf x}-{\bf x}')##, in which case the states ##|{\bf x}\rangle## do not define a localized position operator. There is no way to escape from that conclusion.
 
  • #14
Demystifier said:
What you need to compute with your normalizations is ##\langle {\bf x}|{\bf x}'\rangle##. It is either proportional to ##\delta^3({\bf x}-{\bf x}')##, in which case it is not Lorentz invariant, or not proportional to ##\delta^3({\bf x}-{\bf x}')##, in which case the states ##|{\bf x}\rangle## do not define a localized position operator. There is no way to escape from that conclusion.
Assuming we have to retain Lorentz invariance, does that mean in QFT that particles can not be localised? To what, inside their Compton wavelength?
 
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  • #15
Michael Price said:
Assuming we have to retain Lorentz invariance, does that mean in QFT that particles can not be localised? To what, inside their Compton wavelength?
Yes, that's right. For more quantitative analysis see e.g. my http://de.arxiv.org/abs/hep-th/0202204 Eqs. (60), (59), (47), (48), (49).
 
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  • #16
Demystifier said:
It's not that the position operator does not exist. It can be constructed in the same way as in non-relativistic quantum theory,
No. It can be constructed only on the subspace of positive energy solutions (and after choosing a particular direction of time, breaking Lorentz invariance). These are the ones that are being used in quantum field theory.

WWCY said:
relativistic QM for single particles is flawed by showing us that for a state centered at the origin, it was possible that Pr(x>ct)>0
This is because the relativistic single-particle equations also have negative energy solutions (and superpositions of both kinds), which are unphysical. This makes single-particle quantum mechanics meaningless. (See also this thread.) Quantum fields do not have this drawback.
Michael Price said:
does that mean in QFT that particles can not be localised?
It means that the particle concept is not an exact concept in QFT but only an approximate one for asymptotic (nearly free) states.
 
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  • #17
SisypheanZealot said:
I think you guys have gotten to the crux of what WWCY was trying to talk about, but I would like to add my two cents.

QFT does actually allow for the measurement of the exact position of a particle, and it is in such a way that does not violate the Uncertainty Principle. Two good theoretical descriptions of this are "Quantum Field Theory and the Standard Model" by Mathew D. Schwartz and "Quantum Field Theory for the Gifted Amateur" by Tom Lancaster and Stephen J. Blundell. The discuss in on page 22 of Schwartz's book and page 38 of Lancaster and Blundell. Basically how this is accomplished is by applying the field operator to the ground state $$\phi_0(\vec{x})|0>=\int\frac{d^3 p}{(2\pi)^3}(\hat{a}_{p}e^{i\vec{p}\cdot\vec{x}}+\hat{a}_{p}^{\dagger}e^{-i\vec{p}\cdot\vec{x}})|0>$$ This results in $$\int\frac{d^3 p}{(2\pi)^3}\hat{a}_{p}^{\dagger}e^{-i\vec{p}\cdot\vec{x}}|0>=\int\frac{d^3 p}{(2\pi)^3}e^{-i\vec{p}\cdot\vec{x}}|\vec{p}>$$ This is then ##|\vec{x}>##. Yes this is the non-relativistic case, but you get the same result even when you transition to four space. Now when we apply the state ##<y|## to this we get the result ##<\vec{y}|\vec{x}>=\delta^{(3)}(\vec{y}-\vec{x})##. Which shows us the projection to a single location in space. Futher the fact that ##[\phi(\vec{x}),\phi(\vec{y})]=0## tells us that we can measure the positions of the field operators at the same time.

Now for why this is not violating the Heisenberg uncertainty. Basically what the calculations are doing is putting infinite uncertainty into the momentum state of the particle. Thus, allowing us to know the exact location of the particle.

There is also research being done in Quantum Metrology where they are using a method known as quantum squeezing to take advantage of this very property. They are using this property to measure the displacement of particles. Burd, et. al. in their recent Science article https://science.sciencemag.org/content/364/6446/1163 showed that they have gotten down to "magnetudes well belowe these zero-point fluctuations". They performed the experiment with an Mg ion held above a surface electrode. Very fascinating read!
This is already a quite good ansatz, but you missed the crucial point! A true state must be normalizable to 1, and what' you've written down is not. You have to construct wave packets (written down for non-relativistic particles)
$$|\psi(\vec{x}) \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \tilde{\psi}_{\vec{x}}(\vec{p}) \hat{a}^{\dagger}(\vec{p}) |\Omega \rangle,$$
where you have something like
$$\tilde{\psi}(\vec{p}) = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' N \exp \left [-\frac{(\vec{x}'-\vec{x})^2}{4 \sigma_x^2} \right] \exp(-\mathrm{i} \vec{p} \cdot \vec{x}).$$
This describes a particle which is pretty well localized at the place ##\vec{x}## (with the standard deviation ##\Delta x_i = \sigma_x## for ##i \in \{1,2,3 \}##).
 
  • #18
A. Neumaier said:
No. It can be constructed only on the subspace of positive energy solutions (and after choosing a particular direction of time, breaking Lorentz invariance). These are the ones that are being used in quantum field theory.
I agree with that, so I don't know why did you start it with a "No".
 
  • #19
Demystifier said:
Yes, that's right. For more quantitative analysis see e.g. my http://de.arxiv.org/abs/hep-th/0202204 Eqs. (60), (59), (47), (48), (49).
Thanks for the article. Can I just double-check? In a flat space time in QFT the particle can not be localised to within its Compton wavelength? I asking again because, for the photon, the Compton wavelength is infinite.
 
  • #20
Michael Price said:
Thanks for the article. Can I just double-check? In a flat space time in QFT the particle can not be localised to within its Compton wavelength? I asking again because, for the photon, the Compton wavelength is infinite.
Yes, provided that by localization one means exponential Lorentz invariant localization. To understand where does the "exponential" clause come from, compare Eqs. (48) and (49); a massless particle obeys a power-law Lorentz invariant localization, but not an exponential Lorentz invariant localization. And if you give up Lorentz invariance, then you can even have delta-function localization, which doesn't depend on mass and is much stronger than exponential localization.
 
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  • #21
Demystifier said:
Yes, provided that by localization one means exponential Lorentz invariant localization. To understand where does the "exponential" clause come from, compare Eqs. (48) and (49); a massless particle obeys a power-law Lorentz invariant localization, but not an exponential Lorentz invariant localization. And if you give up Lorentz invariance, then you can even have delta-function localization, which doesn't depend on mass and is much stronger than exponential localization.
Thanks very much for the answer. Nice to see localisation is sorted - it was the one issue in QFT that always bothered me!
 
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  • #22
Thank you all for your replies.

Am I right to say then, that whether or not we "have" a position operator/position states depends on whether or not you want Lorentz covariance? A non-covariant position operator/state would mean that normalisation of position states is not Lorentz invariant, which we don't really want. Conversely Lorentz covariant position operator/states won't satisfy ##\langle \vec{x}|\vec{x}'\rangle \propto \delta(\vec{x} - \vec{x}')## in which case the idea of localisation falls apart.

But if we sacrifice the notion of localisation for Lorentz covariance, how would we know for example (purely from quantum theory and ignoring classical mechanics), that charged, relativistic quantum particles in electromagnetic fields will have the trajectories predicted by the Lorentz Force law? I ask as articles on the LHC always seem to mention the Lorentz Force, and I have always thought it was a classical equation of motion, rather than a quantum mechanical one.

Cheers.
 
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  • #23
Well when you think about it the fact that we have an arbitrary number of identical particles in any given state causes issues for the concept of particle position. We really shouldn’t be asking “where is this particle at time ##t##”, but rather what the probability is for finding some number of particles in a volume ##V## at some time ##t##.

Having said that what is the correct expression for a scalar particle density operator akin to ##\bar{\psi} \psi## for fermions?
 
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  • #24
Michael Price said:
Assuming we have to retain Lorentz invariance, does that mean in QFT that particles can not be localised? To what, inside their Compton wavelength?

Demystifier said:
Yes, that's right. For more quantitative analysis see e.g. my http://de.arxiv.org/abs/hep-th/0202204 Eqs. (60), (59), (47), (48), (49).

A. Neumaier said:
It means that the particle concept is not an exact concept in QFT but only an approximate one for asymptotic (nearly free) states

These sorts of points capture my understanding of (effectively localized) particle states in QFT. But recently I came across a line of research on neutrino wavepackets, which utilize wavepackets significantly below any reasonable estimate of the neutrino Compton length.

For example: https://arxiv.org/abs/1512.09068

"Given the special conditions with very high temperature and density in the neutrinosphere, where neutrinos are produced, a very short time scale for the microscopic production processes is realized. Consequently, the neutrino states are described by very short wave packets in configuration space. Indeed, previous estimates of the wave packet size were σx = 1.8 · 10−14cm for neutrinos produced in the core of the protoneutron star [17] and σx = 4.2 · 10−9cm for neutrinos emitted at a radius of 1000km"

How can we understand this as consistent with the standard notion of/limitations on particle states and localization discussed in this thread?
 
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  • #25
WWCY said:
But if we sacrifice the notion of localisation for Lorentz covariance, how would we know for example (purely from quantum theory and ignoring classical mechanics), that charged, relativistic quantum particles in electromagnetic fields will have the trajectories predicted by the Lorentz Force law? I ask as articles on the LHC always seem to mention the Lorentz Force, and I have always thought it was a classical equation of motion, rather than a quantum mechanical one.
That's a good question. To say that a quantum particle at LHC has a trajectory, you must either violate Lorentz covariance at some level or cheat a little bit by hiding the issue of Lorentz-covariant quantum position under the carpet.
 
  • #26
Demystifier said:
That's a good question. To say that a quantum particle at LHC has a trajectory, you must either violate Lorentz covariance at some level or cheat a little bit by hiding the issue of Lorentz-covariant quantum position under the carpet.
A proton's Compton wavelength is a about 10^-15 m, which is (I suspect) less than the beam width, so the beam trajectory is quite well defined?
 
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  • #27
Demystifier said:
That's a good question. To say that a quantum particle at LHC has a trajectory, you must either violate Lorentz covariance at some level or cheat a little bit by hiding the issue of Lorentz-covariant quantum position under the carpet.

Do you mind expanding a little on what you mean by sweeping the issue under the carpet? How would one do that given say, some ##|n_{\bf P_1}, n_{\bf P_2}, n_{\bf P_3}... \rangle##?

Cheers.
 
  • #28
Demystifier said:
What you need to compute with your normalizations is ##\langle {\bf x}|{\bf x}'\rangle##. It is either proportional to ##\delta^3({\bf x}-{\bf x}')##, in which case it is not Lorentz invariant, or not proportional to ##\delta^3({\bf x}-{\bf x}')##, in which case the states ##|{\bf x}\rangle## do not define a localized position operator. There is no way to escape from that conclusion.

How does one show that normalisation (to Dirac Delta) for position states is not Lorentz invariant? And similarly, how would we show that normalisation for momentum states doesn't suffer from the same problem?

Thanks!
 
  • #29
WWCY said:
How does one show that normalisation (to Dirac Delta) for position states is not Lorentz invariant? And similarly, how would we show that normalisation for momentum states doesn't suffer from the same problem?

Thanks!
As I understand the post above, the expression ##\delta^3({\bf x}-{\bf x}')## cannot be Lorentz invariant because time is omitted. However ##\delta^4({\bf x}-{\bf x}')## is Lorentz invariant. There is a proof of this in the link below which is from an MIT course. It is in section 2.1.

http://web.mit.edu/edbert/GR/gr2b.pdf
 
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  • #30
Mentz114 said:
As I understand the post above, the expression ##\delta^3({\bf x}-{\bf x}')## cannot be Lorentz invariant because time is omitted. However ##\delta^4({\bf x}-{\bf x}')## is Lorentz invariant. There is a proof of this in the link below which is from an MIT course. It is in section 2.1.

http://web.mit.edu/edbert/GR/gr2b.pdf

So the issue is that we don't have a set of basis 4-vector states ##\{|x\rangle\}## such that ##\langle x|x'\rangle = \delta^4(x- x')##?

Cheers
 
  • #31
WWCY said:
So the issue is that we don't have a set of basis 4-vector states ##\{|x\rangle\}## such that ##\langle x|x'\rangle = \delta^4(x- x')##?

Cheers
I can't answer your question. Let's hope someone else can.
I'm talking about the normalization of ## \langle x|x'\rangle \propto \delta^4(x- x')##, being Lorentz invariant.
 
  • #32
WWCY said:
So the issue is that we don't have a set of basis 4-vector states ##\{|x\rangle\}## such that ##\langle x|x'\rangle = \delta^4(x- x')##?

Cheers

Such a state would be quite disturbing and incompatible with the entire quantum formalism IMO.
 
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  • #33
Mentz114 said:
I can't answer your question. Let's hope someone else can.
I'm talking about the normalization of ## \langle x|x'\rangle \propto \delta^4(x- x')##, being Lorentz invariant.

Apologies, perhaps I should have been more explicit in my reply.

Would I be right to say: since we know ##\delta^4(x - x')## is Lorentz invariant, this would mean that if there exists a set of position 4-vectors ##\{|x \rangle\}## that obey said normalisation, we would be able to talk about relativistic particles localised in spacetime. However as @HomogenousCow mentioned:

HomogenousCow said:
Such a state would be quite disturbing and incompatible with the entire quantum formalism IMO.

we don't, and so localisation goes out the window...?

Sorry if I sound confused, it's just that I have been trying to understand how we can talk about manipulating trajectories of relativistic particles to great accuracy in accelerator experiments, while having to consider problematic localisation in QFT.
 
  • #34
WWCY said:
Sorry if I sound confused, it's just that I have been trying to understand how we can talk about manipulating trajectories of relativistic particles to great accuracy in accelerator experiments, while having to consider problematic localisation in QFT.
Can we avoid confusion by picking the frame (the laboratory frame) in which the LHC is at rest? If we all agree to use that frame then we can agree about localization and retain the Dirac delta property for orthogonal positions.

If we insist on Lorentz invariance for all observers then we have to use the definition that a field Φ(x) creates a particle at x and give up on position orthogonality.
 
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  • #35
Michael Price said:
A proton's Compton wavelength is a about 10^-15 m, which is (I suspect) less than the beam width, so the beam trajectory is quite well defined?
Yes, but what do you mean by beam width? There is no quantum width without a quantum concept of position.
 

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