Feynman Path Integral Formulation

230 7 Analytical Lattice Expansion Methods
L sym = − 1 2 ∂ λ h λμ ∂ μ h νν + 1 2 ∂ λ h λμ ∂ ν h νμ
− 1 4 ∂ λ h μν ∂ λ h μν + 1 4 ∂ λ h μμ ∂ λ h νν . (7.18)
The latter can be conveniently split into two parts, as was done already in Eq. (1.67),
as follows
L sym = − 1 2 ∂ λ h αβ V αβμν ∂ λ h μν + 1 2 C2 (7.19)
with
V αβμν = 1 2 η αμη βν − 1 4 η αβη μν , (7.20)
or as a matrix,
⎛ 1
4
− 1 4
− 1 4
− 1 ⎞
4
0 0 0 0 0 0
− 1 1
4 4
− 1 4
− 1 4
0 0 0 0 0 0
− 1 4
− 1 1
4 4
− 1 4
0 0 0 0 0 0
− 1 4
− 1 4
− 1 1
4
4
0 0 0 0 0 0
V =
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
, (7.21)
0 0 0 0 0 0 1 0 0 0
⎜ 0 0 0 0 0 0 0 1 0 0
⎟
⎝
0 0 0 0 0 0 0 0 1 0
⎠
0 0 0 0 0 0 0 0 0 1
with metric components 11,22,33,44,12,13,14,23,24,34 more conveniently labeled
sequentially by integers 1...10, and the gauge fixing term C μ givenbythe
term in Eq. (1.68)
C μ = ∂ ν h μν − 1 2 ∂ μh νν . (7.22)
The above expression is still not quite the same as the lattice weak field action, but
a simple transformation to trace reversed variables ¯h μν ≡ h μν − 1 2 δ μνh λλ leads to
L sym = 1 2 k λ ¯h i V ij k λ ¯h j − 1 2 ¯h i (C † C) ij¯h j , (7.23)
with the matrix V given by
V ij =
( 12
)
β 0
0 I 6
, (7.24)
with k = i∂. Nowβ is the same as the matrix in Eq. (7.13), and C is nothing but
the small k limit of the matrix by the same name in Eq. (7.14), for which one needs
to set ω i − 1 ≃ ik i . The resulting continuum expression is then recognized to be
identical to the lattice weak field results of Eq. (7.12).
This concludes the outline of the proof of equivalence of the lattice weak field
expansion of the Regge action to the corresponding continuum expression. To summarize,
there are several ingredients to this proof, the first of which is a relatively
straightforward weak field expansion of both actions, and the second of which is the
correct identification of the lattice degrees of freedom ε i (n) with their continuum