The discussion centers on the interpretation of the Dirac equation and the concept of "free" particles in quantum mechanics. Participants clarify that a free particle can possess any momentum and that its energy includes both rest energy (mc²) and relativistic kinetic energy (p²/2m). The conversation highlights the necessity of a fixed Lorentz frame for defining momentum and energy, emphasizing that Newton's first law does not apply to quantum particles. The complexities of particle interactions and their implications for quantum field theory (QFT) are also addressed, particularly in relation to the Klein paradox and the Bethe-Salpeter equation.
PREREQUISITESPhysicists, quantum mechanics students, and researchers interested in the foundations of quantum field theory and the behavior of free particles in relativistic contexts.
No; this is only the rest energy. You also need to count the (relativistic) kinetic energy! A free nonrelativistic particle can also move with any momentum and then has an energy of ##p^2/2m##.ftr said:I mean the equation shows the particle could have any momentum, how did that came about. If it is truly free it should have only an energy of mc^2, shouldn't it.
A. Neumaier said:No; this is only the rest energy. You also need to count the (relativistic) kinetic energy! A free nonrelativistic particle can also move with any momentum and then has an energy of ##p^2/2m##.
It is against a fixed Lorentz frame in space-time. The frame is not a material entity but a way to describe space-time independent of its contents. It is needed even to describe the vacuum state - without it there is no meaning to the standard requirement that the vacuum state should be Lorentz invariant.ftr said:In both cases the momentum or speed should be against/relative to something else, but that cannot happen for "single" particle, correct or not.
At fundamental quantum level particles are subject to the "four" known forces, and these are relentless. Any particle subjected to them they will be under the influence no matter how small, but if they become so small then you are back to only mc^2. It seems that Newton's first law is ambiguous for quantum particles.Nugatory said:"Free" means not subject to any forces, so the momentum is not changing. It does not follow that the momentum and hence the kinetic energywill be zero.
A single electron alone in the universe will be free. Note that free particles are (like almost everything in physics) an abstraction form the all-too-complex reality. One simplifies and abstracts in order to be able to understand and classify.ftr said:At fundamental quantum level particles are subject to the "four" known forces
A. Neumaier said:A single electron alone in the universe will be free. Note that free particles are (like almost everything in physics) an abstraction form the all-too-complex reality. One simplifies and abstracts in order to be able to understand and classify.
That is simply not true, because both the momentum ##p## and the kinetic energy ##p^2/2m## are frame-dependent - you can make them take on any value you please simply by choosing coordinates that make them come out the way you want no matter what forces are present or not.ftr said:Any particle subjected to them they will be under the influence no matter how small, but if they become so small then you are back to only mc^2.
It would be better to say that it doesn't apply - it's a classical description that assumes that both position and momentum can be known simultaneously.It seems that Newton's first law is ambiguous for quantum particles.
What does the first Newton law say?ftr said:I mean the equation shows the particle could have any momentum, how did that came about. If it is truly free it should have only an energy of mc^2, shouldn't it.
Nugatory said:momentum ppp and the kinetic energy p2/2mp2/2mp^2/2m are frame-dependent
Demystifier said:What does the first Newton law say?
ftr said:That was my concern in the OP. That is why I said under realistic condition of interacting with another particle/s can give more realistic results.