Point-like particles, Lorentz invariance and QM/QFT

[TrickyDicky]

As we know nonrelativistic quantum mechanics doesn't have the Lorentz invariance property and yet it makes a number of powerful predictions and gives rise to all the fundamental quantum properties (HUP, tunnelling effec, harmonic oscillator, superposition, wave-particle duality etc).
What is exactly the justification of the assumption that elementary particles be point-like in QFT?

 

[Dickfore]

Elementary particles are created and annihilated by field operators which are (operator) functions of the space-time coordinate. So, you really create and annihilate a particle at a particular point in space-time, and, not, for example, in a domain lying on a spacelike hypersufrace.

Let me comment on the examples that you enumerated:
*) HUP, harmonic oscillator - depend only on the operator character of non-commuting observables. The same immediately translates to QFT, where observables are still described by operators, albeit in second quantized formalism

**) tunneling effect, superposition, wave-particle duality - are most easily deduced from Wave Mechanics. In wave mechanics, the fundamental quantity of interest is the wave function of the system which evolves according to a linear partial differential equation. Since the equation is linear, the principle of superposition is obvious. The tunneling effect is characteristic for the propagation of any kind of (classical) wave, see Evanescent waves. Wave-particle duality is an outdated concept taught in Introductory textbooks of Modern Physics for non-scientist students.

A modern view of the wave-function and the wave equation is that the propagator for a quantum particle/field also obeys it. The equation is linear for free fields only, though and QFT allows for the essential complication of interacting fields.

An example of merging QM and SR is spin and explaining the spin/statistics correlation. This is not predicted in non-relativistic QM, but is only described phenomenologically as some kind of quantized intrinsic angular momentum. Relativistic QM correctly predicts that spin is a consequence of the transformation properties of the quantum field under Lorentz transformations. Furthermore, it predicts that fields with integer spin commute (bosons) and fields with half-integer spin anticommute (fermions).

 

[TrickyDicky]

Elementary particles are created and annihilated by field operators which are (operator) functions of the space-time coordinate. So, you really create and annihilate a particle at a particular point in space-time, and, not, for example, in a domain lying on a spacelike hypersufrace.

Thanks, Dickfore
So I guess the questions is why did they choose that kind of field operators? was there some theoretical reason to impose that in the theory besides demanding the Lorentz invariance of SR? was it not hard to reconcile with the fact that for instance GR is a generally covariant theory and this does not seem very compatible with field operators that are coordinate dependent?

 

[Dickfore]

I'm afraid I don't understand your questions. Who is they? What kind of field operators? Impose what?

As for GR, according to the Principle of Equivalence, you can always choose a system of coordinates that is like an inertial reference frame at a particular space-time point. Then, you can apply the principle of local Gallilean covariance. But, then, derivatives do not form covariant objects and you need to include a connection, which is the Christoffel symbol. Then, it is fairly straightforward to formulate a QFT in curved space-time. Actually Hawking Radiation is one prediction coming from such a generalization.

The problem comes when you start treating the gravitational field (which is the metric of space-time itself according to Einstein's GR) as a quantum object itself. As long as you treat it as a classical object that does not evolve according to its own action, you do not encounter any problems. But, as soon as you try to quantize the Einstein-Hilbert action for the gravitational field, the theory becomes non-renormalizable. This is one of the unsolved problems of Modern Physics. As overly speculative posts are not allowed according to the rules of these forums, I will refrain myself from discussing it further.

 

[TrickyDicky]

I'm afraid I don't understand your questions. Who is they? What kind of field operators? Impose what?

They would be the physical theorists that formulated QFT in the 30's and 40's. The field operators are the coordinate dependent ones that create and annihilate particles. I was referring to the theoretical reasons for introducing this operators when I used the word impose.

 

[TrickyDicky]

An example of merging QM and SR is spin and explaining the spin/statistics correlation. This is not predicted in non-relativistic QM, but is only described phenomenologically as some kind of quantized intrinsic angular momentum. Relativistic QM correctly predicts that spin is a consequence of the transformation properties of the quantum field under Lorentz transformations.

Are you sure? https://www.physicsforums.com/showpost.php?p=3532404&postcount=23

 

[Dickfore]

Yes, I am most definitely sure. I don't care who he is, the fathers of Quantum Mechanics tell otherwise.

I think the confusion arises first because he is talking about intrinsic magnetic moment of the electron. According to the Dirac theory (which is linearized in energy, but still relativistic!), the g-factor for the electron is exactly 2. This was one of the great triumphs of Dirac's theory.

Nevertheless, the term linearized might mean two different context and I cannot reach a conclusion from that quote alone. It could mean, as I said, that you "linearize" the energy momentum relation:
[tex]
(E - e \Phi)^{2} = c^{2} (\mathbf{p} - \frac{e}{c} \mathbf{A})^{2} + m^{2} c^{4}
[/tex]
to
[tex]
\left[ \hat{\beta} \, \left(i \hbar \frac{\partial}{\partial t} - e \, \Phi \right) - c (\hat{\beta} \cdot \hat{\alpha}_{k}) \left( - i \hbar \frac{\partial}{\partial x^{k}} - \frac{e}{c} \, A_{k} \right) - m c^{2} \, \hat{1} \right] \Psi = \hat{0}
[/tex]
as was first done by Dirac (who showed that the matrices have to anticommute!).

Alternatively, it could mean that the full theory, according to QED, is an interacting theory of a Dirac field with a U(1) vector gauge field (the Electromagnetic field). If we treat the electromagnetic field as a classical field (with no dynamics on its own), then we would get the same equation as before for the Dirac spinor field. Nevertheless, if we recalculate the vertex at one-loop, we would get an anomalous magnetic moment as predicted by QED and not by Dirac theory and measured experimentally.

Nevertheless, for the purposes of our discussion, it is essential that the Dirac equation is fully relativistically covariant equation and it predicts an intrinsic magnetic moment for the electron. It also predicts an intrinsic spin. But, it is not non-relativistic!

 

[TrickyDicky]

Well actually by saying that spin can be predicted also fron the nonrelativistic equation doesn't mean that the relativistic one doesn't predict it, I think what is suggested is that you can obtain the spin both ways, that it is not something exclusive of relativistic QM but that traditionally that was the idea that remained.

 

[Dickfore]

what non-relativstic equation?

 

[TrickyDicky]

what non-relativstic equation?

according to the quote "linearization of Schrodinger equation"

 

[Dickfore]

But Schroedinger equation is already linear. Can you write down that equation?

 

[TrickyDicky]

But Schroedinger equation is already linear. Can you write down that equation?

Sorry,I can't transcribe it. It's in Google books "Quantum mechanics: an introduction" by Greiner in page 355.
It says he obtains it by symmetrizing wrt time.

 

[George Jones]

Wigner's work (1939 and later) showed that irreducible unitary representations of the Poincare group lead to the free-particle Dirac equation. In 1967, Levi-Lebond showed that irreducible unitary representations of the Galilean group leads not the free-particle Schrodinger equation, but to the Schrodinger-Pauli equation. In this type of treatment, the Schrodinger equation corresponds to particular representation with s = 0.

 

[strangerep]

Details of the paper George mentioned...

J.M. Levy-Leblond,
"Nonrelativistic Particles and Wave Equations",
Comm. Math. Phys., 6, 286-311 (1967)
Available from
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1103840281

----------------

Greiner's treatment of the linearized Schrodinger equation is in the chapter titled "A Nonrelativistic Wave Equation with Spin". (The page number differs considerably between successive editions of the book).

Greiner does some things following Levy-Leblond's ideas, but brings in some other techniques to try and make it clearer. Basically, he uses gamma-like matrices to perform the linearization (though of course they satisfy a slightly different algebra compared to the relativistic Dirac case).

Spin-1/2 arises merely by insisting that SO(3) or SU(2) be represented unitarily (i.e., as operators on a Hilbert space). So as soon as one has an SO(3) little group (which happens in both rel and nonrel cases), half-integral spin becomes unavoidable. Hendrik van Hees gave a more detailed post related to this in a parallel thread:
https://www.physicsforums.com/showthread.php?t=534819&page=2

But Schroedinger equation is already linear. [...]

They probably meant: make the Schroedinger equation linear in the derivatives.

 

[TrickyDicky]

Thanks a lot, really interesting the Levy-Leblond paper.

 

[Demystifier]

Are you sure? https://www.physicsforums.com/showpost.php?p=3532404&postcount=23

Unusualname is making a good point that spin is not really a relativistic effect. After all, it survives even in the limit v/c->0. However, 2 points should be stressed:

1. It is hard to find a physical motivation for linearization of the nonrelativistic Schrodinger equation.

2. No derivation of the spin-statistics theorem is known which does not include the assumption of relativity.

 

[Demystifier]

But Schroedinger equation is already linear.

In this context, "linear" means that only first derivatives appear. Schrodinger equation contains second space derivatives, so is not "linear" in this sense.

 

[Demystifier]

Yes, I am most definitely sure. I don't care who he is, the fathers of Quantum Mechanics tell otherwise.

No father is always right. You are making a classical appeal-to-authority fallacy
http://www.appealtoauthority.info/

 

[Dickfore]

Just as I suspected, the cited paper uses the projective representations of the Galilean group:

According to the point of view advocated in the Introduction, the
characterization of nonrelativistic particles is furnished by the theory
of unitary irreducible representations of the Galilei group. Let us briefly
review the results of this theory. The physical representations of the
Galilei group are nontrivial projective (ray) representations [4], the true
(vector) representations being devoid of physical content 3 because they
do not permit the existence of any sensible notion of localizability [5].

The point is that SU(2) forms a double cover of SO(3), but the half-integer representations are not true representations of SO(3). For example, a rotation by [itex]2 \pi[/itex] in a spin-1/2 representation:
[tex]
\binom{a}{b} \stackrel{ R_{z}(2\pi) } { \rightarrow } e^{-i \pi \sigma_{z}} \, \binom{a}{b} = -\binom{a}{b}
[/tex]
whereas a [itex]4\pi[/itex] rotation is the identity transformation. It is known that orbital angular momentum, for example, cannot attain half-integer values.

One might wonder, what principle would motivate us to consider the central extension of SO(3)? Furhtermore, the spin-statistics theorem, as had already been said, depends on relativity. Also, one cannot formulate consistent non-relativistic electrodynamics.

 

[Demystifier]

Also, one cannot formulate consistent non-relativistic electrodynamics.

Why not?

 

[Dickfore]

Why not?

Because, for example, Gauss's Law is not covariant w.r.t. to Galilean transformations.

 

[Dickfore]

No father is always right. You are making a classical appeal-to-authority fallacy
http://www.appealtoauthority.info/

And so is the op by quoting a book then.

 

[TrickyDicky]

And so is the op by quoting a book then.

lol, since when presenting external sources to enrich a debate is an appeal to authority fallacy?
I didn't pretend that just the fact that is in a book makes it right, you did that by mentioning that if the fathers of QM say so it must be right.

 

[Dickfore]

lol, since when presenting external sources to enrich a debate is an appeal to authority fallacy?
I didn't pretend that just the fact that is in a book makes it right, you did that by mentioning that if the fathers of QM say so it must be right.

No, I did not say that. I said I would rather believe the fathers of QM/QFT than a textbook author.

 

[Dickfore]

Regarding the issue of Galilean Electromagnetism, from the same author:

http://fisicavolta.unipv.it/percorsi/pdf/jmll.pdf

 

[Demystifier]

Gauss's Law is not covariant w.r.t. to Galilean transformations.

nabla E = rho
is covariant w.r.t. to Galilean transformations
t=t'
x=x'+vt'

 

[Demystifier]

And so is the op by quoting a book then.

lol, since when presenting external sources to enrich a debate is an appeal to authority fallacy?
I didn't pretend that just the fact that is in a book makes it right, you did that by mentioning that if the fathers of QM say so it must be right.

TrickyDicky is right and Dickfore is not. Presenting an argument by someone else is not appeal to authority, while presenting the conclusion by someone else without presenting his argument is appeal to authority.

 

[Demystifier]

I said I would rather believe the fathers of QM/QFT than a textbook author.

Perhaps that's what you meant, but that's not what you said.

 

[Dickfore]

nabla E = rho
is covariant w.r.t. to Galilean transformations
t=t'
x=x'+vt'

How does E transform w.r.t. Galilean transformations?

 

[Dickfore]

TrickyDicky is right and Dickfore is not. Presenting an argument by someone else is not appeal to authority, while presenting the conclusion by someone else without presenting his argument is appeal to authority.

Ok, then, here:
http://www.jstor.org/stable/94981"

 

[PhilDSP]

How does E transform w.r.t. Galilean transformations?

The Lorentz Transformation may be more highly underdetermined than is generally recognized. It does appear to support a bridge to the Galilean transformation by holding t=t'. The translation of the spatial basis in that case can be accommodated by applying the principle of convection.

 

[Demystifier]

How does E transform w.r.t. Galilean transformations?

It transforms like
partial x/partial x' =1
so E'=E.

 

[Demystifier]

Ok, then, here:
http://www.jstor.org/stable/94981"

Here spin is derived from relativity and QM. However, it does not prove that spin cannot be derived without assuming relativity.

 

[Dickfore]

But it says that the non-relativistic treatment gives a factor that is twice as big for the spin-orbit coupling term.

 

[Dickfore]

It transforms like
partial x/partial x' =1
so E'=E.

please give a reference for this conclusion.