No uncertainty principle for scalar particles?

scope
Messages
61
Reaction score
0
hi,

I do wonder if the Higgs boson is a quantum object because since it is the (only) particle with spin 0, then it should not behave like a wave(since the wave aspect is connected to the fact that it is spinning) and therefore not experience the uncertainty principle.

Or am I wrong?
 
Physics news on Phys.org
Yes, this is wrong. The uncertainty relations hold in all quantum theories, because they all use non-commuting operators to represent observables.

The quantum theory that's taught in all the introductory courses on quantum mechanics is the theory of a single spin-0 particle that interacts with the rest of the world only via a potential. If you have taken such a course, you may recall that the uncertainty relation for position and momentum is derived from the commutator [x,p]=i. Other properties of the particle don't enter into it.

I posted a proof of the uncertainty theorem here.
 
but this has never been observed. the non commutating relations can be explained because of the Fourrier transforms, because it would behave as a wave. but if that particle does not behave like a wave(because of spin 0) then there is no uncertainty. What can be the cause of the wave behavior if the particle does not spin?
 
First of all, even particles with spin do not spin in the classical sense. Second, as I said, the theory you seem to be somewhat familiar with (since you mention Fourier transforms) is the quantum theory of a single spin-0 particle. Are you saying that you think that theory (the theory of wavefunctions, the Schrödinger equation, etc.) doesn't include any wave-like behavior?

I don't know what you're referring to when you say that it "has never been observed".
 
thank you,
elementary spin-0 =scalar fields have not never been observed so far (the Higgs boson would be the first particle), that's why I do wonder if its quantum nature will be the same, that's what i meant. spin -0 particles do not follow the Schrödinger equation but rather the Klein Gordon equation, but there are many theoretical problems with it(total probability for example), or i am wrong?
 
It doesn't matter if there are any elementary particles with spin 0. The point is that all quantum theories involve a non-commutative algebra of operators. You can even take that as the definition of a quantum theory. And where there are operators that don't commute, there's an uncertainty relation.

The problems you're referring to are only problems if you try to interpret the Klein-Gordon field as a wavefunction. That was just one of the first approaches to relativistic QM that was attempted, and quickly dismissed, for this very reason. The modern approach can't be explained without a lot of mathematics. (Representation theory of groups, etc.) The Klein-Gordon equation is satisfied by quantum fields, not by state vectors. The time evolution of the state vectors is described by the Schrödinger equation in the relativistic case too (but the Hamiltonian that appears in it can't be expressed as p2/2m+V).
 
I would ask the OP where he got the idea he wrote in the brackets: "since the wave aspect is connected to the fact that it is spinning". It's interesting to know this part as well.
 
Back
Top