Feynman Path Integral Formulation

230 7 Analytical Lattice Expansion MethodsL 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 followsL sym = − 1 2 ∂ λ h αβ V αβμν ∂ λ h μν + 1 2 C2 (7.19)withV αβμν = 1 2 η αμη βν − 1 4 η αβη μν , (7.20)or as a matrix,⎛ 14− 1 4− 1 4− 1 ⎞40 0 0 0 0 0− 1 14 4− 1 4− 1 40 0 0 0 0 0− 1 4− 1 14 4− 1 40 0 0 0 0 0− 1 4− 1 4− 1 1440 0 0 0 0 0V =0 0 0 0 1 0 0 0 0 00 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 1with metric components 11,22,33,44,12,13,14,23,24,34 more conveniently labeledsequentially by integers 1...10, and the gauge fixing term C μ givenbytheterm 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, buta simple transformation to trace reversed variables ¯h μν ≡ h μν − 1 2 δ μνh λλ leads toL 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 byV ij =( 12)β 00 I 6, (7.24)with k = i∂. Nowβ is the same as the matrix in Eq. (7.13), and C is nothing butthe small k limit of the matrix by the same name in Eq. (7.14), for which one needsto set ω i − 1 ≃ ik i . The resulting continuum expression is then recognized to beidentical to the lattice weak field results of Eq. (7.12).This concludes the outline of the proof of equivalence of the lattice weak fieldexpansion of the Regge action to the corresponding continuum expression. To summarize,there are several ingredients to this proof, the first of which is a relativelystraightforward weak field expansion of both actions, and the second of which is thecorrect identification of the lattice degrees of freedom ε i (n) with their continuum