I and my friend were disputing about some weird behaviour of the act of measuring some observables quantities e.g. Energy,position.

But I still don't think what he said is strictly true.

He said" each observable has its own Eigenstate, and when you measure it, the wave function will collapse to that eigenstate , giving its eigenvalue for that particle observable quantity. However ,I think some observables do share same eigenstate. i.e. a free particle with zero potential, you can prove this by Solving the TISE. When you know measure and obtained an exact momentum of a particle of a system, by uncertainty principle, the uncertainty of x will be infinite, then the wave function will be spread everywhere,therefore it has a well define wavelength, thus you have well defined k for wavenumber  , hence a well defined energy by p square / 2m.

To conclude, He said  :is it true that every time you measure a quantity, it only produces a eigenstate for that particular quantity?ie measure position gives position eigenstate, momentum for momentum eigenstate.

My argument shown above claimed the above is not strictly true: So my argument is energy and momentum do share same eigenfunction, when potential is zero, Who is correct?

I also mentioned something called "Conjugate variables" e.g. position and momentum, you can only know one at a time But I think energy and momentum are not, hence resulting in the reasoning i wrote above! Thanks

I and my friend were disputing about some weird behaviour of the act of measuring some observables quantities e.g. Energy, position.

But I still don't think what he said is strictly true.

He said" each observable has its own Eigenstate, and when you measure it, the wave function will collapse to that eigenstate , giving its eigenvalue for that particle observable quantity. However ,I think some observables do share same eigenstate. i.e. a free particle with zero potential, you can prove this by Solving the TISE. When you now measure and obtained an exact momentum of a particle of a system, by uncertainty principle, the uncertainty of $x$ will be infinite, then the wave function will be spread everywhere,therefore it has a well define wavelength, thus you have well defined $k$ for wavenumber, hence a well defined energy by $p^2 / 2m$.

To conclude, He said: is it true that every time you measure a quantity, it only produces a eigenstate for that particular quantity? I.e. measure position gives position eigenstate, momentum for momentum eigenstate.

My argument shown above claimed the above is not strictly true: So my argument is energy and momentum do share same eigenfunction, when potential is zero. Who is correct?

I also mentioned something called "Conjugate variables" e.g. position and momentum, you can only know one at a time. But I think energy and momentum are not, hence resulting in the reasoning i wrote above!

every one. I was disputing some weird behaviour of the act of measuring some observables quantities with my friend. But I still don't think what he said is strictly true. He said" each observable has its own eigenstate, and when you measure it, it will collapse to that eigenstate giving the corresponding eigenstate. I think some observables do share same eigenstate e.g. for a free particle with zero potential. They don't share same eigenstate if V is not zero. I think when you know its momentum then by uncertainty principle, the uncertainty of x will be infinite, then the wave function will be spreaded everywhere. Therefore it has a well define wavelength, and you have well defined k for wavenumber , hence a well defined energy by p square / 2m.

He said because the particle is not in a infinite square well, the energy can't be quantised, and hence can not have a well define energy, which I think is wrong.

I and my friend were disputing about some weird behaviour of the act of measuring some observables quantities e.g. Energy,position.

But I still don't think what he said is strictly true.

He said" each observable has its own Eigenstate, and when you measure it, the wave function will collapse to that eigenstate , giving its eigenvalue for that particle observable quantity. However ,I think some observables do share same eigenstate. i.e. a free particle with zero potential, you can prove this by Solving the TISE. When you know measure and obtained an exact momentum of a particle of a system, by uncertainty principle, the uncertainty of x will be infinite, then the wave function will be spread everywhere,therefore it has a well define wavelength, thus you have well defined k for wavenumber , hence a well defined energy by p square / 2m.

To conclude, He said :is it true that every time you measure a quantity, it only produces a eigenstate for that particular quantity?ie measure position gives position eigenstate, momentum for momentum eigenstate.

My argument shown above claimed the above is not strictly true: So my argument is energy and momentum do share same eigenfunction, when potential is zero, Who is correct?

I also mentioned something called "Conjugate variables" e.g. position and momentum, you can only know one at a time But I think energy and momentum are not, hence resulting in the reasoning i wrote above! Thanks