135
5.2. The quantum field: (ii) Lagrange–Hamilton formulation
Since δφ is an arbitrary function, the requirement δS = 0 yelds the Euler–
Lagrange field equation
(
)
(
)
∂L
∂
∂L
∂ ∂L
−
−
= 0.
(5.96)
∂φ
∂x ∂(∂φ/∂x)
∂t ∂φ ˙
The generalization to three dimensions is
(
)
(
)
∂L
∂L
∂ ∂L
− ∇ ·
−
= 0.
(5.97)
∂φ
∂(∇φ)
∂t ∂φ ˙
As an example, consider
( ) 2
( ) 2
1
∂φ
1
∂φ
L ρ = ρ
− ρc
2
(5.98)
2
∂t
2
∂x
where the factor ρ (mass density) and c (a velocity) have been introduced to
get the dimension of L right. Inserting this into the Euler–Lagrangian field
equation (5.96), we obtain
∂
2 φ
1 ∂
2 φ
−
= 0
(5.99)
∂x 2
c 2 ∂t 2
which is precisely the wave equation (5.30) for the one-dimensional string,
now obtained via the Euler–Lagrange field equations. Note that the Lagrange
density L has the expected form (cf (5.48)) of ‘kinetic energy density minus
potential energy density’.
For the final step – the passage to quantum mechanics for a field system
– we shall be interested in the Hamiltonian (total energy) of the system,
just as we were for the discrete case. Though we shall not actually use the
Hamiltonian in the classical field case, we shall introduce it here, generalizing
it to the quantum theory in the following section. We recall that Hamiltonian
mechanics is formulated in terms of coordinate variables (‘q’) and momentum
variables (‘p’), rather than the q and ˙
q of Lagrangian mechanics. In the
continuum (field) case, the Hamiltonian H is written as the integral of a
density H (we remain in one dimension)
∫
H = dx H
(5.100)
while the coordinates q r (t) become the ‘coordinate field’ φ(x, t). The question
is what is the corresponding ‘momentum field’ ?
The answer to this is provided by a continuum version of the generalized
momentum derived from the Lagrangian approach (cf equation (5.44))
p = ∂L/∂q. ˙
(5.101)
5.2. The quantum field: (ii) Lagrange–Hamilton formulation
Since δφ is an arbitrary function, the requirement δS = 0 yelds the Euler–
Lagrange field equation
(
)
(
)
∂L
∂
∂L
∂ ∂L
−
−
= 0.
(5.96)
∂φ
∂x ∂(∂φ/∂x)
∂t ∂φ ˙
The generalization to three dimensions is
(
)
(
)
∂L
∂L
∂ ∂L
− ∇ ·
−
= 0.
(5.97)
∂φ
∂(∇φ)
∂t ∂φ ˙
As an example, consider
( ) 2
( ) 2
1
∂φ
1
∂φ
L ρ = ρ
− ρc
2
(5.98)
2
∂t
2
∂x
where the factor ρ (mass density) and c (a velocity) have been introduced to
get the dimension of L right. Inserting this into the Euler–Lagrangian field
equation (5.96), we obtain
∂
2 φ
1 ∂
2 φ
−
= 0
(5.99)
∂x 2
c 2 ∂t 2
which is precisely the wave equation (5.30) for the one-dimensional string,
now obtained via the Euler–Lagrange field equations. Note that the Lagrange
density L has the expected form (cf (5.48)) of ‘kinetic energy density minus
potential energy density’.
For the final step – the passage to quantum mechanics for a field system
– we shall be interested in the Hamiltonian (total energy) of the system,
just as we were for the discrete case. Though we shall not actually use the
Hamiltonian in the classical field case, we shall introduce it here, generalizing
it to the quantum theory in the following section. We recall that Hamiltonian
mechanics is formulated in terms of coordinate variables (‘q’) and momentum
variables (‘p’), rather than the q and ˙
q of Lagrangian mechanics. In the
continuum (field) case, the Hamiltonian H is written as the integral of a
density H (we remain in one dimension)
∫
H = dx H
(5.100)
while the coordinates q r (t) become the ‘coordinate field’ φ(x, t). The question
is what is the corresponding ‘momentum field’ ?
The answer to this is provided by a continuum version of the generalized
momentum derived from the Lagrangian approach (cf equation (5.44))
p = ∂L/∂q. ˙
(5.101)