# Are particles frame independant?

1. May 17, 2010

### TCS

Could an observer in a gravity well perceive a partical to be a differnt type of particle than an obsever who is outside of the gravity well?

Can an observer in one frame see a hydrogen atom and an observer in another frame see a neutron?

Last edited: May 17, 2010
2. May 17, 2010

Staff Emeritus
No.
.

3. May 18, 2010

### tom.stoer

Partices are mathematically elements of unitary reps. of the Poincare algebra. According to this algebra only kinematical variables (energy, momentum, orbital angular momentum) are frame dependent; inner variables like isospin etc. commute with the Poincare algebra, therefore particles types are not affected by choosing different reference frames.

4. May 18, 2010

### meopemuk

This argument is correct when applied to stable particles, such as electron or proton. However, unstable particles are not described by irreducible representations of the Poincare group. There is no law that forbids seeing different particle content in different frames. For example, one observer can see a single neutron, while another (moving) observer can see the decay products (proton + electron + antineutrino). I think that TCS has a valid point.

Eugene.

5. May 18, 2010

### ansgar

if the particles are not related by a physical law then?

6. May 18, 2010

### tom.stoer

Can you show this in a simple example, e.g. for the weakly interacting Z boson?
How does the state |Z°> look like and how is it transformed by a boost?

7. May 18, 2010

### meopemuk

There are dozens of decay modes for the Z boson. This means that if a pure Z-boson state is prepared at time 0, then at later time (e.g., with respect to the time-translated oberver) this state will look like a linear combination of the original state plus all possible decay products. So, the time-translated Z-boson state is

$$exp(iHt) |Z^0 \rangle = a_0(t) |Z^0 \rangle + a_1(t) |e^+ e^- \rangle + a_2(t) |\mu^+ \mu^- \rangle + \ldots$$

where H is the Hamiltonian, $$a_0(0) = 1, a_1(0) = 0, \ldots$$ and the sum of squares of all coefficients $$a_i(t)$$ is equal to 1 at all times.

The same arguments can be applied to the transformation of the Z-boson state with respect to boosts. If a pure Z-boson is prepared in one frame, then in the moving frame we can expect seeing the state

$$exp(iKv) |Z^0 \rangle = b_0(v) |Z^0 \rangle + b_1(v) |e^+ e^- \rangle + b_2(v) |\mu^+ \mu^- \rangle + \ldots$$

where v is the rapidity of the boost, K is the boost operator, $$b_0(0) = 1, b_1(0) = 0, \ldots$$, and the sum of squares of all coefficients $$b_i(v)$$ is equal to 1.

Eugene.

8. May 18, 2010

### tom.stoer

You can define the original Z° state via a Z° creation operator acting on the vacuum; then one can understand new particle states via the commutation of the creation operator with the exponentiated boost. It seems that this commutation leaves behind all these creation operators from K acting on the vacuum.

=> Questions:
1) doesn't this mean that we have a problem to define a vacuum which is invariant under the (normal ordered) boost?
2) why is this restricted to unstable particles; it seems that this argument applies to all "interaction" terms in K, i.e. third or higher order terms in K
3) if this is true then this does not only generate states which are decay channels of Z°, but all states to which Z° can couple in principle, including any virtual intermediate state

9. May 18, 2010

### meopemuk

I think that vacuum invariance is preserved in this case. In terms of creation $$(a^{\dag})$$ and annihilation $$(a)$$ operators the interacting boost generator K can be written as

$$K = K_0 + a^{\dag}(e^+) a^{\dag}(e^-)a(Z^0) + a^{\dag}(Z^0) a(e^+) a(e^-)+ \ldots$$

where $$K_0$$ is the non-interacting boost generator. The operator K yields zero when applied to the vacuum vector. So, the exponent of K leaves the vacuum vector invariant.

Yes, the structure of K should be just as complicated as the structure of the full interacting Hamiltonian H. So, terms of higher perturbation order can be present in K as well.

Do you mean that we can add terms of the type $$a^{\dag}(Z^0) a^{\dag}(e^+)a^{\dag}(e^-) a(Z^0)$$ which violate the energy conservation condition? I don't think so. I think that such terms should not be present neither in the Hamiltonian nor in the boost operator.

Eugene.

10. May 18, 2010

### tom.stoer

No.

Regarding 3) I only say that based on 2) your argument applies not only to instable particles like the Z° but to all interacting particles.

Our discussion started with the question whether particles can change when a boost is applied. You said that acting with a boost on an instable particle will create states of its decay channels. My argument was that this is not limited to decay channels but one must include all particles to which the original particle couples. You agreed with that.

Now this is my point: if this is true (and we already agreed that it is true) this means that a boost acting on an electron (which is a stable but interacting particle) will "create" other particles. So two observers will in general not agree if they observe an electron or not.

But if this is were true, then the electron would no longer be in an irred. rep. of the Poincare group. This contradicts our starting point!

11. May 18, 2010

### meopemuk

I disagree with that. Since electron is a stable particle, there should be no interaction terms in the Hamiltonian or boost operator, which act on the one-electron state and produce something different. For example, there should be no terms like $$a^{\dag}(e^-) a^{\dag}(\gamma)a(e^-)$$. In other words, even in the interacting theory, 1-electron states transform as irreducible representations of the Poincare group.

However, this does not contradict the fact that electrons interact with themselves and with other particles. For example the term $$a^{\dag}(e^-) a^{\dag}(e^-) a(e^-) a (e^-)$$ is responsible for the electron-electron interaction (in both the Hamiltonian and the boost operator). Such a term acts only on states with two or more electrons.

Eugene.

12. May 18, 2010

### tom.stoer

It will be hard to check these claims for all interacting theories. The problem becomes worse in a fully gauge fixed version of the theory; in a non-perturbative formulation of QCD you have terms like $$\rho((D^2)^{-1})\rho$$ where $$\rho$$ is bilinear in the quarks and $$D$$ is a differential operator depending on the gauge fields.

You say that 1-electron states form irred. reps. because the electron is stable, whereas the Z° is not stable - and should therefore not belong to an ired. rep.? What about the photon? Where is the fundamental difference between the photon and the Z° in the el.-weak theory?

Another idea: the photon cannot decay in a physical electron-positron pair (only into a virtual pair), but this is not due to the algebraic structure of interaction term but due to phase space.

13. May 18, 2010

### meopemuk

I am not sure about electro-weak theory, but from experiment it is pretty obvious that the photon is massless and stable while the Z-particle is massive and decays.

I am pretty sure that "virtual" electron-positron pairs have not been directly observed in experiments. The same is true for "virtual" photon decays. These ideas are purely speculative. So, it seems reasonable to assume that a consistent theory should not contain interaction vertices like "photon -> electron + positron."

Eugene.

14. May 19, 2010

### tom.stoer

I am sorry, but you have to prove that from the Hamiltonian. Everything is clear as long as you look at the Hamiltonian of a free theory, that means a Hamiltonian that is bilinear in the creation and annihilation operators. It becomes difficult in an interacting theory. I use the el.-weak model as there are two nearly identical particles, the photon and the Z°, no "bound states" like nucleons in QCD. The main difference between photon and Z° is not their mass but their coupling to the Higgs boson (the mass is a trick due to the vev of the Higgs, but the formal structure of the interaction term simply says that the Z° couples to the Higgs whereas the photon does not; this applies even in the symmetric phase w/o vev)

Regarding virtual particles: forget about them if you don't like them.

I only say that from the algebraic structure of the Hamiltonian the coupling of the photon to leptons is rather similar to the coupling of the Z° to leptons. As the interaction terms are similar one expects that you above argument applies to photon states as well; that means a "boosted photon states" should contain lepton-antilepton pairs just as for the Z°.

Your argument regarding stability relies on the phase space and physical masses which you cannot read of from the Hamiltonian directly. So from the fundamental theory I do not see why you should get the series of complicated states when you act with a boost on the Z° only; I expect (simply based on the structure of the Hamiltonian) that a similar series of states occures for the photon - from which we know that it's stable.

So your statement "... unstable particles are not described by irreducible representations of the Poincare group" is still not clear to me.

Last edited: May 19, 2010
15. May 19, 2010

### meopemuk

I see the contradiction. However, my resolution of this contradiction is different from yours. I think that the correct interacting Hamiltonian (and the boost operator) does not have terms like "photon <-> electron + positron." If such terms existed then there should be a non-zero probability for the photon<->pair conversion at finite times (such a temporary conversion is not forbidden by the energy conservation law). This has not been seen in experiments. So, perhaps the electro-weak Hamiltonian is not correct in this regard.

If a particle is described by an irreducible representation, then all Poincare transformations (including time translations and boost) do not change the particle's identity. This is definitely not true for unstable particles, because time translations destroy them.

Eugene.

16. May 19, 2010

### tom.stoer

I disagree!

Look at the simple interaction term

$$g\bar{\psi}\gamma^\mu A_\mu \psi \sim g j^\mu A_\mu$$

This term appears in the QED Lagrangian; it also appears in the QCD Lagrangian where an implicit trace over SU(3) color matrices is assumed and where A represents the gluon field; it also appears in the el.-weak theory with A to be replaced by the photon, W or Z-boson.

So all interaction terms in the standard model contain a term where one gauge boson couples to a lepton-antilepton pair. But some of these bosons or leptons are stable (photon, electron, electron-neutrino) whereas others are not (myon, W, Z, ...). Your conclusion based on the algebraic structure of the Lagrangian is not valid.

I think you have something in mind like a time evolution operator containing a complex part (which is used to model resonances). Look at a solution of QCD, e.g.

$$(H_\text{QCD} - E_p) |\text{neutron}\rangle = 0$$

Time evolution is simply

$$e^{-iH_\text{QCD}t}|\text{neutron}\rangle = e^{-iE_pt}|\text{neutron}\rangle$$

The problem arises once on includes the el.-weak interaction which couples to or mixes flavors; only these terms cause decay or mixing (CKM) of otherwise stable hadrons. So I agree that there must be something special in the el.-weak interaction.

Last edited: May 19, 2010
17. May 19, 2010

Staff Emeritus
It looks to me like this point isn't subtle. If I have a particle that decays at coordinates (x,t) by boosting my own reference frame, I can change back and forth between decayed and undecayed states. Nothing special with a decaying particle - I can make the same argument with an alarm clock or an egg timer.

This, though, doesn't really change the identity of an alarm clock, egg timer, or even elementary particle.

18. May 19, 2010

### tom.stoer

The state of a particle does not anything about a position where it decays. The question is very simple. In a non-interacting field theory I always have

$$e^{iK^0\eta}|Z^0, p=0\rangle = |Z^0, p=p(\eta)\rangle$$

with boost operator and boost parameter.

What happens in an interacting field theory? what is

$$|Z^0, p=0\rangle = \ldots$$

and what is

$$e^{iK\eta}|Z^0, p=0\rangle = \ldots$$

???

Last edited: May 19, 2010
19. May 19, 2010

### TCS

My initial thoughts were about accelerating with respect to a star, since increasing relative velocity would cause the diameter of the star to decrease and the mass of the star to increase. It seems that at some point the star would have to become a neutron star.

20. May 19, 2010

### meopemuk

This is exactly the major point of our disagreement. In my opinion, the interaction term that you wrote is problematic. (Yes, I know that this tri-linear term is present in all standard model theories, still I think it is not correct.) This interaction predicts processes like "photon <-> electron + positron" or "electron <-> electron + photon". It is true that such reactions are not present in the S-matrix, due to the energy-momentum conservation condition. But standard theory still allows such reactions to happen at finite times, which is in contradiction with observations. Moreover, these problematic interaction vertices are sources of ultraviolet divergences.

QED and other standard model theories can be re-formulated so that the problematic interaction vertices are not present. This is called the "dressed particle" approach. It has been published and peer-reviewed. See, for example

E. V. Stefanovich, "Quantum field theory without infinities", Ann. Phys. (NY), 292 (2001), 139

http://www.arxiv.org/abs/physics/0504062

Eugene.