Is decay spectrum continuous or discrete?

[zhangyang]

For a definite particle,the decay mode is determinant,finite kind,which embody the characteristic of quantum mechanics.

But for a specific mode of a definite particle's decay,the decay spectrum,ie,energy of products,continuous,or discrete?

Decay is a process which has unique initial condition,a static parent particle.So the result is very interesting.

If it is continuous,why the gamma decay of nucleon has some characteristic energy?

If it is discrete,the amplitude function on energy and momentum would not be continuous,so how to use the analytic way to calculate?

 

[Orodruin]

But for a specific mode of a definite particle's decay,the decay spectrum,ie,energy of products,continuous,or discrete?

This depends on the final state, i.e., what particles are created in the decay. In a two-body decay, the energy spectrum is discrete (in the rest frame of the decaying particle). With more particles in the final state, the spectra are generally continuous.

If it is continuous,why the gamma decay of nucleon has some characteristic energy?

If you mean when an excited nucleus (nucleons do not decay by gamma emission) goes to a ground state, this is a two-body decay. The spectrum is therefore discrete.

 

[mfb]

For a definite particle,the decay mode is determinant

No it is not, if multiple decay modes are available you cannot tell in advance which one will be measured. Only branching fractions (probabilities) can be given.

See Orodruin's post for the difference between two and more decay products.

If it is discrete,the amplitude function on energy and momentum would not be continuous,so how to use the analytic way to calculate?

To calculate what, how?

 

[zhangyang]

This depends on the final state, i.e., what particles are created in the decay. In a two-body decay, the energy spectrum is discrete (in the rest frame of the decaying particle). With more particles in the final state, the spectra are generally continuous.

If you mean when an excited nucleus (nucleons do not decay by gamma emission) goes to a ground state, this is a two-body decay. The spectrum is therefore discrete.

I agree.The reason is the conservation of momentum and the relation between momentum and energy which make the two-body-product decay's energy spectrum discrete.
And the gamma decay is always after α or β decay,which make a nucleus excited.So gamma decay is very particular.

 

[zhangyang]

No it is not, if multiple decay modes are available you cannot tell in advance which one will be measured. Only branching fractions (probabilities) can be given.

See Orodruin's post for the difference between two and more decay products.

To calculate what, how?

Is there some contradiction in application of Feynman rule on the two-body-product decay?Because the probability amplitude is a function on momentum,if the momentum is discrete,or even unique,what is the meaning of it?

Or,QFT is only about elementary particle,whose decay are all many-bodied?You can say τ which decay into ρ and ν(τ),but ρ is not elementary,and make this decay many-bodied.And the contradiction solved?

 

[mfb]

Which Feynman rule?
There is no contradiction.

There are two-body decays of elementary particles. W and Z bosons, the Higgs and the top quark usually decay to two other particles. In addition, it does not matter, hadrons have a well-defined mass in the same way elementary particles have.

 

[ChrisVer]

the tau decays are not 2 bodies...
The tau is in general going to decay into a tau-neutrino + some sets of products = hadrons and leptons...

The hadronic decay will generally form a jet (in general with small R). There your main problem is identifying the jets coming from taus, which need a good algorithm for tau-identifications+ many bodies problem...

The leptons that you will get will be 2 (lepton+ neutrino)

 

[Orodruin]

The leptons that you will get will be 2 (lepton+ neutrino)

Just for clarification, the lepton+neutrino are in addition to the neutrino already mentioned, making the leptonic tau decays three-body decays. I think this is what you meant to say, but it was not obvious to me on first read.

 

[Hepth]

I think I understand what he's asking. Assume you have a theoretical particle $h$ that decays to two other particles $(b \bar{b})$, I think he's trying to ask if there is some problem with $\frac{d\Gamma}{d\Omega}$ being "discrete" in that its proportional to a delta function over the angle, thus giving a "static" result, in that it is not technically continuous in any given energy spectrum or over any angular spectrum.

I don't think there is any problem with this. Most decays are not continuous in that sense, even in 3-body decays. Remember the final partial width $\Gamma$ is just a number. The mathematical representation of the amplitudes commonly contain delta functions and their derivatives, theta functions if there are kinematic cuts, as well as some cusps/discontinuities due to crossing kinematic thresholds where amplitudes develop an imaginary part. The spectrum has no requirement to be a smooth function I believe.

 

[ChrisVer]

Just for clarification, the lepton+neutrino are in addition to the neutrino already mentioned, making the leptonic tau decays three-body decays. I think this is what you meant to say, but it was not obvious to me on first read.

yes that was what I meant...
I said that tau gives tau neutrino + other products , and then I wrote about those products ^^