RE: The Higgs Boson: what are the odds?
September 12, 2011 at 10:34 am
(This post was last modified: September 12, 2011 at 10:38 am by little_monkey.)
(September 12, 2011 at 9:14 am)lilphil1989 Wrote: I would say the odds are reasonably high for at least one Higgs' boson existing.
Otherwise, massive gauge bosons in a non-Abelian gauge theory make no sense at all. And they certainly exist in nature, the W's and Z bosons associated with the weak interaction have huge masses.
Indeed, just to give a general rundown of the theory, here:
(1) in QM: x → operator
But to satisfy Relativity, in which time is on an equal footing with space, in QFT: x → parameter, and Φ(x) → operator. Now Φ(x), a function of x, is called the “field”.
(2) L = T – V. The Lagrangian plays an important role. From Noether’s theorem, we know that if the Lagragian is invariant under a symmetry, this symmetry points to a conservation law.
Corresponding to L there is a Hamiltonian, H = T + V. The Hamiltonian is known to measure the energy of a system.
(3)In classical mechanics, let v = dx/dt, then L = (1/2) mv
2 – V(x). The corresponding Hamiltonian is, H = (1/2) mv
2 + V(x). Quantizing this, (ℏ =1),we get the Schroedinger equation: i∂Ψ(x)/∂t =( -(1/2m)∆
2 + V(x)) Ψ(x).
(4) In Relativity, the energy equation is:
E
2= p
2c
2 + m
2c
4.
Quantizing this, (c =1) yields the K-G equation:
(1/2) (∂
μΦ)(∂
μΦ) + (1/2) mΦ
2 = 0.
From this, the Lagrangian can be deduced as: L = (1/2) (∂μΦ)
2 – (1/2) mΦ
2.
(5) In QFT, the general Lagrangian is:
L = (1/2) (∂μΦ)
2 – V(Φ).
(6) Comparing (5) and (4), if V(Φ) contains any terms with Φ
2, its coefficient is taken to be the mass of the field quanta (particles).
Gauge theory:
From electromagnetism, it was known that Maxwell’s equations were gauge invariant. In QM, gauge invariance of the Lagragian involves three important steps:
(7) the wave function is redefined as Ψ → Ψe
iqX
(8) the operator ∂
μ → ∂
μ + iqA
μ
(9) the electromagnetic field A
μ → A
μ - ∂
μX
(10) In QED, in equation (5), V(Φ) → - ¼ F
μν F
μν, where F
μν = ∂
μA
ν - ∂
νA
μ
If you apply, 7,8,9,10 to equation (5), you get the invariance of the Lagrangian under gauge transformation, in which the photon mediates the electromagnetic force. Note that the photon has no mass.
In the weak force, the bosons involved have mass, and one had to figure out how to include a mass term, keeping the Lagrangian gauge invariant.
There is where number (6) comes into play under the notion of SPONTANEOUS SYMMETRY BREAKING.
In another post, I will draw the skeleton of the Higgs Mechanism.
EDIT: if there are corrections, let me know.