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Units, Constants and Useful Formulas
Lagrangians in Quantum Field Theory
-----------------------------------
Quantum field theory frequently makes use of the Lagrangian formalism from
classical field theory. This formalism is analogous to the Lagrangian formalism
used in classical mechanics to solve for the motion of a particle under the
influence of a field. In classical field theory, one writes down a Lagrangian
density, involving a field, φ, and its first derivatives ∂φ/∂t (kinetic energy)
and ∂φ/∂x (potential energy), and then applies a field form of the
Eulerâ€“Lagrange equation.
In the canonical approach, φ are the field operators composed of creation
and annihilation operators.
Lagrangian of a System of Classical Fields
------------------------------------------
L = L(φ,∂_{μ}φ) μ = t, x, y, z
Euler-Lagrange for a field:
∂/∂x^{μ}(∂L/∂φ_{μ}) - ∂L/∂φ = 0
In long form this is:
∂/∂t(∂L/(∂φ/∂t) - ∂/∂x(∂L/(∂φ/∂x) - ∂/∂y(∂L/(∂φ/∂y) - ∂/∂z(∂L/(∂φ/∂z) - ∂L/∂φ = 0
One simple scalar field in 1D:
L = (1/2)(∂φ/∂t)^{2} - (1/2)(∂φ/∂x)^{2} - V(φ) (L = T - V)
Apply E-L:
∂{∂((1/2)(∂φ/∂t)^{2}}/∂t = ∂^{2}φ/∂t^{2}
∂{∂((1/2)(∂φ/∂x)^{2}}/∂x = ∂^{2}φ/∂x^{2}
∂L/∂φ = -∂V/∂φ
..
φ - ∂^{2}φ/∂x^{2} = -∂V/∂φ
Make V(φ) = (m^{2}/2)φ^{2} - the Field Energy
..
φ - ∂^{2}φ/∂x^{2} = -m^{2}φ
Simple quantum field:
Assume φ = exp(i(kx - ωt))
-ω^{2}φ + k^{2}φ = -m^{2}φ
ω^{2} = k^{2} + m^{2}
=> E^{2} = p^{2} + m^{2} assuming h = c = 1
2 simple scalar fields in 1D:
L = = (1/2)(∂_{μ}φ)^{2} + (1/2)(∂_{μ}ρ)^{2} - (m^{2}/2)φ^{2} - (M^{2}/2)ρ^{2}
..
φ - ∂^{2}φ/∂x^{2} = -m^{2}φ
..
ρ - ∂^{2}ρ/∂x^{2} = -M^{2}φ
2 simple interacting scalar fields in 1D:
Add an interaction term ρφ^{2} to illustrate what is going on
L = = (1/2)(∂_{μ}φ)^{2} + (1/2)(∂_{μ}ρ)^{2} - (m^{2}/2)φ^{2} - (M^{2}/2)ρ^{2} - ρφ^{2}
..
φ - ∂^{2}φ/∂x^{2} = m^{2}φ - 2ρφ
..
ρ - ∂^{2}ρ/∂x^{2} = M^{2}φ - φ^{2}
The wave equations are now coupled and there is an interaction between
the two. Now, if φ^{3} was replaced by φ^{3} then the equations would be
non-linear and φ and ρ would scatter each other.
Consider:
L = iψ^{†}∂ψ/∂t + ψ^{†}α∂ψ/∂x + ψ^{†}βψm
Applying the E-L equation would generate the Dirac equation:
iψ = iα∂ψ/∂x + mβψ
Add scattering term:
L = iψ^{†}∂_{t}ψ + ψ^{†}α∂_{x}ψ + ψ^{†}βψm + ψ^{†}βψφ
Now φ scatters ψ and vice versa.
Lagrangian of a System of Quantum Fields
----------------------------------------
φ is composed of creation and annihilation operators. Therefore,
something like φ^{3} gives all sorts of terms. Let's look at some of
them:
(a^{†})^{3} - creation of 3 particles at point x,
(a^{})^{3} - annihilation of 3 particles at point x
(a^{†})(a^{-})^{2} - annihilation of 2 particles and creation of 1 particle at point x
(a^{})(a^{+})^{2} - creation of 2 particles and annihilation of 1 particle at point x
What do the quadratic terms in the Lagrangian mean?
Return to our simple field:
L = (1/2)(∂φ/∂t)^{2} - (1/2)(∂φ/∂x)^{2} - (m^{2}/2)φ^{2}
Decompose the double partial differential w.r.t. x as follows:
(1/2)(∂φ/∂x)^{2} = (1/2){(φ(x) - φ(x'))/Δx}^{2}/}
= φ^{2}(x)/(Δx)^{2} + φ^{2}(x')/(Δx)^{2} - φ(x)φ(x')/(Δx)^{2} - φ(x')φ(x)/(Δx)^{2}
\ / \
stationary particle moving particle
Note: ∂φ/∂t and ∂φ/∂x are applied to the exp(±i(kx - ωt) terms in the field
operators and do not involve a^{+} and a^{-}
The φ^{2} terms simply create/annihilate a particle the same point. Likewise,
we can show that the term (m^{2}/2)φ^{2} has a similar effect. It is the
cross-terms that govern the motion of the particle by annihilating it at
one position and creating it at another (or the reverse).