Neutrons don't decay in nuclei because no available states incorrect?

[nonequilibrium]

"Neutrons don't decay in nuclei because no available states" incorrect?

Hello,

If I understand correctly, the argument for a neutron (usually) not decaying when in a nucleus, is that the resulting proton would then have to occupy a high energy level, the lower levels already being occupied by the protons that are already there.

But that argument presupposes that a particle has to be in an energy eigenstate (or at least immediately after decaying). Is there any argument for this? There are an infinite number of states with average energies lower than that "high energy state" it would be obliged--according to the traditional argument--to occupy.

 

[fzero]

Hello,

If I understand correctly, the argument for a neutron (usually) not decaying when in a nucleus, is that the resulting proton would then have to occupy a high energy level, the lower levels already being occupied by the protons that are already there.

But that argument presupposes that a particle has to be in an energy eigenstate (or at least immediately after decaying). Is there any argument for this? There are an infinite number of states with average energies lower than that "high energy state" it would be obliged--according to the traditional argument--to occupy.

The quantum mechanical bound state problem has a discrete set of bound states with energies below the configuration of free particles. The continuous spectrum is the free particle spectrum.

Stable nuclei and molecules are in the ground state of the bound state problem. Furthermore, stability requires that there are no configurations whatsoever that have lower energy, no matter what the individual particles might be able to decay into were they free. Any intermediate state should be well approximated by a linear combination of energy eigenstates, but there are none with an energy lower than that of the stable nucleus.

The "traditional argument" is nothing more than conservation of energy. It doesn't rely solely on the spectrum of intermediate states, but mainly on the energy of the ground state.

 

[nonequilibrium]

Thanks for trying to help, but I don't think you're getting my point. For example

The quantum mechanical bound state problem has a discrete set of bound states with energies below the configuration of free particles. The continuous spectrum is the free particle spectrum.

doesn't answer my question, it's only relevant if you presuppose the particles have to be in an energy eigenstate. So maybe I should rephrase my question like that, shortly put: why is it assumed that the nucleons are in energy eigenstates?

 

[Vanadium 50]

It isn't assumed. A nucleus has a total energy, therefore it is in a total energy eigenstate. It's a conclusion, not an assumption.

 

[nonequilibrium]

Okay, why does it have a total energy then?

 

[kurros]

...doesn't answer my question, it's only relevant if you presuppose the particles have to be in an energy eigenstate.

You seem to have missed this crucial sentence:

Any intermediate state should be well approximated by a linear combination of energy eigenstates, but there are none with an energy lower than that of the stable nucleus.

So the argument does not rely on the nucleus actually being in an energy eigenstate.

 

[nonequilibrium]

Oh I indeed seem to have missed that part, my apologies. But either I'm misinterpreting it, or I don't understand why it's true: to me the quote

Any intermediate state should be well approximated by a linear combination of energy eigenstates, but there are none with an energy lower than that of the stable nucleus.

means that, for example, any superposition of the 1st and 2nd lowest energy eigenstates have a higher (average) energy than, for example, the 2nd lowest energy eigenstate itself.

 

[kith]

With "stable nucleus", he means the ground state of the nucleus. And you agree that you can't find a linear combination with an average energy below the ground state? Sure, you can have all kinds of excited states, but eventually, they will decay into the ground state.

btw: this is nothing specific to nuclear physics. If your argument was true, it would also apply to electrons in atoms.