Particle number lorentz invariant?

In summary: No, I'm saying they're two different things. Lorentz transformations are coordinate changes. Time translation means you replace t by t + t0. A system is time-translation invariant if it...If you define particle number as the number of particles in a box, this is invariant, although the shape of the box is not. If you define particle number as the number of particles at some time t, it is not invariant.
  • #1
metroplex021
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Hi folks -- does anyone know of a proof that particle (quanta) number in QFT is / is not Lorentz invariant? I'd be happy to hear of it -- so thanks!
 
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  • #2
If the particle carries a conserved quantum number such as charge, then clearly the number of quanta has to be Lorentz invariant. If there is a more general proof of this fact, then there must be something in the proof that restricts it to Lorentz boosts between inertial frames, because otherwise the Unruh effect would be a counterexample.
 
  • #3
I don't think there is any actual proof. It's actually the relativity principle itself. The laws of Physics must be the same in all inertial frames. This means that, if an observer sees N particles, another observed must see the same set of particles, otherwise you would allow different kind of interactions in one system which are not present in the other.
 
  • #4
Particle number is NOT Lorentz invariant. Particle number is defined with respect to a particular spacelike hypersurface, and specifying its value includes specifying the hypersurface. Particle number at zero time in one reference frame does not necessarily equal the particle number at zero time in another reference frame.
 
  • #5
Why? Could you explain better?
 
  • #6
Bill_K said:
Particle number is NOT Lorentz invariant. Particle number is defined with respect to a particular spacelike hypersurface, and specifying its value includes specifying the hypersurface. Particle number at zero time in one reference frame does not necessarily equal the particle number at zero time in another reference frame.

That makes sense. But suppose I define particle number as the number of particles minus the number of antiparticles. E.g., this could be the number of electrons minus the number of antielectrons, or more generally it could be lepton number. Isn't this the same on all spacelike surfaces?
 
  • #7
metroplex021 said:
Hi folks -- does anyone know of a proof that particle (quanta) number in QFT is / is not Lorentz invariant? I'd be happy to hear of it -- so thanks!

The number operator is a generator of [itex]U(1)[/itex] symmetry, like the electric charge, the baryon number and the lepton number. These are Lorentz invariant scalars and one can prove this.
 
  • #8
The confusion is caused by the ill-defined "particle number". If you define particle number as the number of particles in a box, this is invariant, although the shape of the box is not. If you define particle number as the number of particles at some time t, it is not invariant.
 
  • #9
samalkhaiat said:
metroplex021 said:
Hi folks -- does anyone know of a proof that particle (quanta) number in QFT is / is not Lorentz invariant? I'd be happy to hear of it -- so thanks!
The number operator is a generator of [itex]U(1)[/itex] symmetry, like the electric charge, the baryon number and the lepton number. These are Lorentz invariant scalars and one can prove this.
The particle number is not a constant in a QFT with interactions. Such as φ4 theory.
 
  • #10
Bill_K said:
The particle number is not a constant in a QFT with interactions. Such as φ4 theory.

Could you explain that better? As far as I know the action of the representation of the Lorentz group, [itex]U(\Lambda)[/itex], on a N-particle state is:
$$
U(\Lambda)|p_1,p_2,\cdots,p_N\rangle=|\Lambda p_1,\Lambda p_2,\cdots,\Lambda p_N\rangle.
$$
I don't see how this can create/destroy particles. What am I doing wrong?
 
  • #11
Einj said:
Could you explain that better? As far as I know the action of the representation of the Lorentz group, [itex]U(\Lambda)[/itex], on a N-particle state is:
$$
U(\Lambda)|p_1,p_2,\cdots,p_N\rangle=|\Lambda p_1,\Lambda p_2,\cdots,\Lambda p_N\rangle.
$$
I don't see how this can create/destroy particles. What am I doing wrong?
You're confusing time translation with time evolution. The way to get the state at a later time t is to use the evolution operator, U(t) = exp(iHt), where H is the Hamiltonian. And in the general case, H contains terms like φ4 which don't conserve particle number.
 
  • #12
So this is probably my confusion on the definition. Shouldn't all the possible Lorentz transformations be represented by the representation U?
From what I understand you are saying that we also must consider the evolution operator as the action of a Lorentz transformation. Is that right?
 
  • #13
Vanadium 50 said:
The confusion is caused by the ill-defined "particle number". If you define particle number as the number of particles in a box, this is invariant, although the shape of the box is not. If you define particle number as the number of particles at some time t, it is not invariant.

This doesn't sound right to me. The box is a spacelike 3-surface, right? The only difference is that the box is finite, rather than being an infinite 3-surface.
 
  • #14
Einj said:
From what I understand you are saying that we also must consider the evolution operator as the action of a Lorentz transformation. Is that right?
No, I'm saying they're two different things. Lorentz transformations are coordinate changes. Time translation means you replace t by t + t0. A system is time-translation invariant if it does not depend explicitly on the value of t. You pick the system up and put it down unchanged. The switch from standard time to daylight sayings time is a time translation.

Even a system which is time-translation invariant will evolve, and the properties of the system will also evolve, the ones whose operators don't commute with H, including perhaps the particle number.
 
  • #15
Ok, then the answer to the initial question is that particle number is invariant under Lorentz transformation but not under the evolution of the system right?
 
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  • #16
Bill_K said:
The particle number is not a constant in a QFT with interactions. Such as φ4 theory.

Thank you for that valuable piece of information :) . I said the NUMBER OPERATOR, i.e. the number of FERMIONS minus the number of ANTI-FERMIONS which has the following fields-operator expression
[tex]Q = \int d^{ 3 } x \ \psi^{ \dagger } ( x ) \psi ( x ) .[/tex]
This (number) operator is (as one can show) time-independent and Lorentz-invariant scalar. This is true as long as your INTERACTING theory has a global [itex]U(1)[/itex] symmetry. In general, PARTICLE number is ill-defined concept in the interacting QFT specially when there are bosons around. But we can always deal with conserved and Lorentz-invariant NUMBER operators of the internal symmetries of our interacting theory. To make this point clear, consider the following interacting theory of [itex]SU(2)[/itex] doublet of fermions and triplet of bosons (neglecting em-interaction)
[tex]\mathcal{L} = i \bar{ \psi } \gamma^{ \nu } \ \partial_{ \mu } \psi + \frac{ 1 }{ 2 } \partial_{ \mu } \phi^{ a } \partial^{ \mu } \phi^{ a } + i g \bar{ \psi } \gamma^{ 5 } \frac{ \sigma^{ a } }{ 2 } \ \psi \ \phi^{ a } .[/tex]
This theory has two global internal symmetries:

1) [itex]U(1)[/itex]:
[tex]\phi^{ a } \rightarrow \phi^{ a }, \ \ \ \psi \rightarrow \psi + i \alpha \psi .[/tex]
This leads to the following time-independent Lorentz-invariant Number operator
[tex]B = \int d^{ 3 } x \ \psi^{ \dagger } ( x ) \psi ( x ) .[/tex]
This is nothing but the conserved Baryon number.

2) [itex]SU(2)[/itex]:
[tex]\phi^{ a } \rightarrow \phi^{ a } + \epsilon^{ a b c } \phi^{ b } \beta^{ c } ,[/tex]
[tex]\psi \rightarrow \psi + i \frac{ \sigma^{ a } \beta^{ a } }{ 2 } \psi .[/tex]
The invariance under these transformations leads to the following time-independent Lorentz invariant (total) iso-spin number operators
[tex]T^{ a } = \int d^{ 3 } x \left( \psi^{ \dagger } \frac{ \sigma^{ a } }{ 2 } \psi + \epsilon^{ a b c } \phi^{ b } \partial_{ 0 } \phi^{ c } \right) .[/tex]
This means that the iso-spin of the interacting fermions by themselves is not conserved. This is, of course, reasonable since the bosons in the theory also carry iso-spin.

This, I hope, clarify what I meant by NUMBER OPERATORS.

Sam
 
  • #17
Einj said:
Shouldn't all the possible Lorentz transformations be represented by the representation U?
In an interacting theory (constructed according to the instant form of dynamics), the Lorentz boost operator contains another term determined by the interaction (just as the Hamiltonian has another term compared to the free theory).

Although one constructs interactions in terms of the free fields, the notion of "particle" becomes problematic, except (hopefully) at asymptotic times -- provided the interaction vanishes sufficiently fast in that limit.

For more detail: Weinberg vol 1. :biggrin:
 
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  • #18
Strangerep, do you have the section ref for Weinberg? I can't seem to find thie discussion of this. Thanks mate!
 
  • #19
metroplex021 said:
Strangerep, do you have the section ref for Weinberg? I can't seem to find thie discussion of this.
Yeah -- he "hides" it under "Symmetries of the S-Matrix", section 3.3. (That's where you can find stuff about modifications to the Lorentz boost operator for interacting theories.)

I have a love-hate relationship with Weinberg. His subsection numbering and indexing are very poor -- all the more so since his textbook is supposed to be a research reference tome (sigh). :grumpy:

You might also be interested in some of the other references in this old thread -- they might fill in more of the gaps.

(Btw, these don't cover my oblique mention of "interactions vanishing sufficiently fast at asymptotic times". If you want to know about that, I'll have to dig out some other references.)
 
  • #20
Thank you very much! So appreciated. :)
 
  • #21
you may find this reference on fields handy, I often refer to it in my self studies. Tends to help fill in the blanks on numerous articles and textbooks I read with regards to QFT


"Fields" by W.Siegel
http://arxiv.org/abs/hepth/9912205
 

What is the definition of a particle number Lorentz invariant?

A particle number Lorentz invariant is a physical quantity that remains constant regardless of changes in the frame of reference. It is a fundamental property of particles in the theory of special relativity.

What is the importance of particle number Lorentz invariance in physics?

Particle number Lorentz invariance is important because it allows for a consistent and universal way to describe the behavior of particles in different frames of reference. It also plays a crucial role in the development of quantum field theories.

How is particle number Lorentz invariance related to the conservation of particle number?

Particle number Lorentz invariance is closely related to the conservation of particle number, as it ensures that the total number of particles in a system remains constant regardless of the observer's frame of reference.

Can particle number Lorentz invariance be violated?

In certain physical systems, such as those involving strong interactions, the violation of particle number Lorentz invariance has been observed. However, it is still a fundamental principle in most areas of physics and is considered a key aspect of our understanding of the universe.

Are there any experimental tests for particle number Lorentz invariance?

Yes, there are several experimental tests for particle number Lorentz invariance, including high-energy particle collisions and measurements of particle lifetimes. These tests have shown that particle number Lorentz invariance holds to a high degree of precision in most physical systems.

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