Particle number lorentz invariant?

1. metroplex021

146
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!

2. bcrowell

6,397
Staff Emeritus
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. Einj

451
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. Bill_K

4,157
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. Einj

451
Why? Could you explain better?

6. bcrowell

6,397
Staff Emeritus
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. samalkhaiat

1,196
The number operator is a generator of $U(1)$ symmetry, like the electric charge, the baryon number and the lepton number. These are Lorentz invariant scalars and one can prove this.

18,498
Staff Emeritus
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. Bill_K

4,157
The particle number is not a constant in a QFT with interactions. Such as φ4 theory.

10. Einj

451
Could you explain that better? As far as I know the action of the representation of the Lorentz group, $U(\Lambda)$, 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. Bill_K

4,157
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. Einj

451
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. bcrowell

6,397
Staff Emeritus
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. Bill_K

4,157
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. Einj

451
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?

1 person likes this.
16. samalkhaiat

1,196
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
$$Q = \int d^{ 3 } x \ \psi^{ \dagger } ( x ) \psi ( x ) .$$
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 $U(1)$ 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 $SU(2)$ doublet of fermions and triplet of bosons (neglecting em-interaction)
$$\mathcal{L} = i \bar{ \psi } \gamma^{ \nu } \ \partial_{ \mu } \psi + \frac{ 1 }{ 2 } \partial_{ \mu } \phi^{ a } \partial^{ \mu } \phi^{ a } + i g \bar{ \psi } \gamma^{ 5 } \frac{ \sigma^{ a } }{ 2 } \ \psi \ \phi^{ a } .$$
This theory has two global internal symmetries:

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

2) $SU(2)$:
$$\phi^{ a } \rightarrow \phi^{ a } + \epsilon^{ a b c } \phi^{ b } \beta^{ c } ,$$
$$\psi \rightarrow \psi + i \frac{ \sigma^{ a } \beta^{ a } }{ 2 } \psi .$$
The invariance under these transformations leads to the following time-independent Lorentz invariant (total) iso-spin number operators
$$T^{ a } = \int d^{ 3 } x \left( \psi^{ \dagger } \frac{ \sigma^{ a } }{ 2 } \psi + \epsilon^{ a b c } \phi^{ b } \partial_{ 0 } \phi^{ c } \right) .$$
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. strangerep

2,292
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.

Last edited: Apr 15, 2014
18. metroplex021

146
Strangerep, do you have the section ref for Weinberg? I can't seem to find thie discussion of this. Thanks mate!

19. strangerep

2,292
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. metroplex021

146
Thank you very much! So appreciated. :)