Friday, September 23, 2011

Gauge Theory and Higgs Mechanism


(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 = ½ mv2 – V(x). The corresponding Hamiltonian is, H = ½ mv2 + V(x). Quantizing this, (ℏ =1),we get the Schroedinger equation:

i∂Ψ(x)/∂t =( -½m∆2 + V(x))Ψ(x).

(4) In Relativity, the energy equation is:

E2= p2c2 + m2c4.

Quantizing this, (c =1) yields the K-G equation:

½(∂μΦ)(∂μΦ) + ½mΦ2 = 0.

From this, the Lagrangian can be deduced as:

L = ½ (∂μΦ) 2 – ½mΦ2.

(5) In QFT, the general Lagrangian is:

L = ½ (∂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 transformed as Φ → eiqXΦ
(8) the operator ∂μ → ∂μ + iqAμ
(9) the electromagnetic field Aμ → Aμ - ∂μX

(10) In QED, in equation (5), V(Φ) → - ¼ Fμν Fμν,
where Fμν = ∂μAν - ∂νAμ

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.

Higgs Mechanism:

Basically, I will only look at U(1) symmetry. Electroweak interactions need a U(1) x SU(2) symmetry, but SU(2) requires 2 by 2 matrices, and the software on this forum is inadequate to deal with matrices. But you can get the flavor just by doing U(1) symmetry and how mass is introduced in the Lagrangian of equation (5).

I will rewrite this equation as:

(11) L = ∂μΦμΦ - ¼ Fμν Fμν – V(ΦΦ).

(12) where V(ΦΦ) = (m2)/(2φ2) {ΦΦ - φ2} 2

Three important things to note:

(13) The field Φ is now a complex number, denoted by (Φ1, Φ2) or Φ = Φ1 + iΦ2 ( i being the imaginary number, square root of – 1), and Φ = Φ1 – iΦ2.

(14) the minimum field energy is obtained when ΦΦ = φ2.

(15) The number of possible vacuum states is infinite. We break this symmetry by requiring that Φ is real, we take the vacuum state to be (φ,0), and expand:

Φ = φ + (½ ½)h

Substituting 7,8,9, 12, and 15 into 11, we get

(17) L = {(∂μ - iqAμ)( φ + (½ ½)h)}{( ∂μ + iqAμ)( φ + (½ ½)h} - ¼ Fμν Fμν - (m2)/(2φ2) {2½φh + ½h2}2

After calculating the Lagrangian, we separate it into two parts:

(18) L = Lfree + Lint


(19) Lfree = ½∂μh∂μh - m2h2 - ¼ Fμν Fμν + q2φ2AμAμ

All the remaining terms are lumped into Lint, which offer no interest.

So, we can see that by breaking the symmetry, we end up with two massive particles. In equation 19, the second term refers to a scalar particle with mass equal to 2½m, associated with h (the higgs field) and the fourth term, a vector boson with mass 2½qφ, associated with Aμ( the electromagnetic field).

NOTE: in the Weinberg electroweak theory, equation 19 would have three extra terms for the vector boson instead of a single term, each one was identified with the W+, W -, and Z bosons, with 2x2 matrices that would be groups obeying the algebra under SU(2). This prediction, which was confirmed subsequently in the following years, earned Weinberg, Salam and Glashow the Nobel prize.
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