Undergrad Relativistic Quantum Mechanics & Localized Particles

[Mentz114]

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.

 

[HomogenousCow]

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.

 

[WWCY]

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:

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.

 

[Michael Price]

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.

 

[Demystifier]

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.

 

[Demystifier]

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 $|n_{\bf P_1}, n_{\bf P_2}, n_{\bf P_3}... \rangle$?

Since your state above is not written in the position basis, it sweeps the issue of position under the carpet so you don't need to worry about Lorentz covariance of the position (operator).

 

[Demystifier]

How does one show that normalisation (to Dirac Delta) for position states is not Lorentz invariant?

A simple way to see this is as follows. One has $\int d^3x\,\delta^3({\bf x})=1$. The right-hand side is Lorentz invariant, so the left-hand side is also Lorentz-invariant. But the measure $d^3x$ is not Lorentz invariant (do you know how to show that?), and the product of a Lorentz-noninvariant quantity with a Lorentz-invariant one cannot be Lorentz invariant. Therefore the other factor, $\delta^3({\bf x})$, cannot be Lorentz invariant.

And similarly, how would we show that normalisation for momentum states doesn't suffer from the same problem?

The momentum Dirac delta $\delta^3({\bf p})$ is not Lorentz invariant for a similar reason. But there is a Lorentz invariant version of Dirac delta in the momentum space, namely
$$\delta^3({\bf p}) 2\omega({\bf p})$$
The Lorentz invariance of this can be seen easily by noting that there is a Lorentz invariant measure in the momentum space
$$\frac{d^3p} {2\omega({\bf p})}$$
Finally, note that one cannot do something similar in position space because there is no such thing as $\omega({\bf x})$.

 

[Michael Price]

Yes, but what do you mean by beam width? There is no quantum width without a quantum concept of position.

Can we use the field operators to define position? So φ(x)|0> creates a particle at x, by definition.

 

[Demystifier]

Can we use the field operators to define position? So φ(x)|0> creates a particle at x, by definition.

You can define it so, but it will not obey the desired properties. The state $|{\bf x}\rangle=\varphi({\bf x})|0\rangle$ will not satisfy $\langle {\bf x}'|{\bf x}\rangle\propto \delta^3({\bf x}'-{\bf x})$.

 

[Michael Price]

You can define it so, but it will not obey the desired properties. The state $|{\bf x}\rangle=\varphi({\bf x})|0\rangle$ will not satisfy $\langle {\bf x}'|{\bf x}\rangle\propto \delta^3({\bf x}'-{\bf x})$.

Agreed, we have to give up that property. But it is the position as appears in Feynman diagrams, and space-like separated positions are causally disconnected. It seems to be crying out to called "position".

 

[Demystifier]

Agreed, we have to give up that property. But it is the position as appears in Feynman diagrams, and space-like separated positions are causally disconnected. It seems to be crying out to called "position".

Fine, in that sense massive particles are exponentially localized within a Compton wavelength, as I already explained.

 

[Demystifier]

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.

For that to make sense you must use an extended Hilbert space. See e.g. my http://de.arxiv.org/abs/0811.1905

 

[HomogenousCow]

You can define it so, but it will not obey the desired properties. The state $|{\bf x}\rangle=\varphi({\bf x})|0\rangle$ will not satisfy $\langle {\bf x}'|{\bf x}\rangle\propto \delta^3({\bf x}'-{\bf x})$.

What exactly does $|\bf{x}\rangle$ mean in QFT? If we Lorentz transform this state, what do we get? What is $U(\Lambda)|\bf{x}\rangle$? Surely the transformation can't evolve the state forward in time or anything like that.

 

[Demystifier]

What exactly does $|\bf{x}\rangle$ mean in QFT? If we Lorentz transform this state, what do we get? What is $U(\Lambda)|\bf{x}\rangle$?

It depends on how one defines $|\bf{x}\rangle$. Various inequivalent definitions can be found in the literature.

 

[vanhees71]

Well, all you want to describe with QFT is with which probability do you measure particles with some properties in a detector. The detector defines the position of the corresponding detection events. For massive particles you have a position observable (defined, e.g., by Newton+Wigner) for massless quanta (in practice practically only photons) you don't have a position observable, but still you can define the probability with which a detector located at a given place with some finite extent registers a photon (and without foundational problems since massive objects have well defined position observables). As in classical physics the electromagnetic field it's basically a distribution given by the energy density of the em. field. See any textbook on quantum optics (my favorite at the time is Garrison and Chiao) about various detection measures (single-photon and two-photon, which are, roughly speaking, simply autocorrelation functions of the electric field).

 

[PeterDonis]

it is the position as appears in Feynman diagrams

Not if you draw the diagrams in momentum space, which is the most common way of doing it, at least in particle physics.

 

[Michael Price]

Not if you draw the diagrams in momentum space, which is the most common way of doing it, at least in particle physics.

Obviously I am referring to diagrams drawn in configuration space. They are equivalent, of course.

 

[Michael Price]

What exactly does $|\bf{x}\rangle$ mean in QFT? If we Lorentz transform this state, what do we get? What is $U(\Lambda)|\bf{x}\rangle$? Surely the transformation can't evolve the state forward in time or anything like that.

For a scalar field the Lorentz transform is just the usual Lorentz transform. For fermions you would also have the spinor indices transforming, which are also well defined.

 

[A. Neumaier]

Obviously I am referring to diagrams drawn in configuration space. They are equivalent, of course.

Only in nonrelativistic quantum field theory.

 

[vanhees71]

Feynman diagrams depicting S-matrix elements lead to the same result in both the space-time and the energy-momentum representation. The corresponding (amputated) time-ordered $N$-point Green's functions are Fourier transforms of each other.

 

[Michael Price]

Only in nonrelativistic quantum field theory.

The equivalence is shown using 4-D delta functions, so it works in relativistic QFT.

 

[A. Neumaier]

Feynman diagrams depicting S-matrix elements lead to the same result in both the space-time and the energy-momentum representation. The corresponding (amputated) time-ordered $N$-point Green's functions are Fourier transforms of each other.

But if you take 4D Fourier transforms then the N-point Green's functions get N times, and can for N>2 no longer be interpreted in terms of inner products of multiparticle states (where there is one time only but multiple space coordinates).

 

[Michael Price]

But if you take 4D Fourier transforms then the N-point Green's functions get N times, and can for N>2 no longer be interpreted in terms of inner products of multiparticle states (where there is one time only but multiple space coordinates).

Expand it all out in terms of propagators and point vertices and the equivalence becomes trivial.

 

[A. Neumaier]

Expand it all out in terms of propagators and point vertices and the equivalence becomes trivial.

I don't complain about the equivalence but about the interpretation that you give in post #38.
Applying two field operators gives a state with two time variables.

 

[Michael Price]

I don't complain about the equivalence but about the interpretation that you give in post #38.
Applying two field operators gives a state with two time variables.

Representing creation at X and destruction at Y. Or vice versa for the antiparticle.

 

[A. Neumaier]

Representing creation at X and destruction at Y. Or vice versa for the antiparticle.

And what does $\phi(x)\phi(y)\phi(z)|vac\rangle$ with 4D $x,y,z$ represent? Surely not an ordinary wave function ket, which would be $|x,y,z\rangle$ with 3D position vectors $x,y,z$.

 

[HomogenousCow]

And what does $\phi(x)\phi(y)\phi(z)|vac\rangle$ with 4D $x,y,z$ represent? Surely not an ordinary wave function ket, which would be $|x,y,z\rangle$ with 3D position vectors $x,y,z$.

Is the whole idea of particle positions muddled by the lost simultaneity in relativistic quantum mechanics? As in, how do we make sense of unitary time evolution when particle positions defined at the same times are boosted to different times?

 

[A. Neumaier]

Is the whole idea of particle positions muddled by the lost simultaneity in relativistic quantum mechanics?

Not only that of particle positions but that of more than one particle.

Even in classical relativity, there is no good relativistic multiparticle theory: see

• Currie, Jordan and Sudarshan, Relativistic Invariance and Hamiltonian Theories of Interacting Particles, Reviews of Modern Physics 35 (1963), 350.

 

[HomogenousCow]

Not only that of particle positions but that of more than one particle.

Even in classical relativity, there is no good relativistic multiparticle theory: see

• Currie, Jordan and Sudarshan, Relativistic Invariance and Hamiltonian Theories of Interacting Particles, Reviews of Modern Physics 35 (1963), 350.

Wasn’t there a recent textbook on classical electrodynamics where the problem of charged point particles interacting via dynamical fields was given a satisfactory treatment?

 

[A. Neumaier]

Wasn’t there a recent textbook on classical electrodynamics where the problem of charged point particles interacting via dynamical fields was given a satisfactory treatment?

To my knowledge only the problem of a single point particle in a field is tractable, not that of several...