Feynman Path Integral Formulation

28 1 Continuum Formulation
and the spin zero or trace part
are such that their sum gives the original field h
with the quantity P μν defined as
h T μν = 1 3 P μνP αβ h αβ , (1.144)
h = h TT + h L + h T , (1.145)
P μν = η μν − 1
∂ 2 ∂ μ∂ ν , (1.146)
or, equivalently, in k-space P μν = η μν − k μ k ν /k 2 .
One can learn a number of useful aspects of the theory by looking at the linearized
form of the equations of motion. As before, the linearized form of the action
is obtained by setting g μν = η μν +h μν and expanding in h. Besides the expressions
given in Eq. (1.45), one needs
√ gR 2 =(∂ 2 h λ λ − ∂ λ ∂ κ h λκ ) 2 + O(h 3 )
√
gRλμνκ R λμνκ = 1 4 (∂ μ∂ κ h νλ + ∂ λ ∂ ν h μκ − ∂ λ ∂ κ h μν − ∂ μ ∂ ν h κλ ) 2 + O(h 3 ) ,
(1.147)
from which one can then obtain, for example from Eq. (1.135), an expression for
√ g(Rμν ) 2 ,
√
gRαβ R αβ = 1 4 (∂ 2 h μ μ∂ 2 h α α + ∂ 2 h μα ∂ 2 h μα − 2∂ 2 h μ μ∂ α ∂ β h αβ
−2∂ α ∂ ν h ν μ∂ α ∂ β h μβ + 2∂ μ ∂ ν h μν ∂ α ∂ β h αβ )+O(h 3 ) .
(1.148)
Using the three spin projection operators defined previously, the action for linearized
gravity without a cosmological constant term, Eq. (1.7), can then be re-expressed as
∫
I lin = 1 4 k dx h μν [P (2) − 2P (0) ] μναβ ∂ 2 h αβ . (1.149)
Only the P (2) and P (0) projection operators for the spin-two and spin-zero modes,
respectively, appear in the action for the linearized gravitational field; the spin-one
gauge mode does not enter the linearized action. Note also that the spin-zero mode
enters with the wrong sign (in the linearized action it appears as a ghost contribution),
but to this order it can be removed by a suitable choice of gauge in which the
trace mode is made to vanish, as can be seen, for example, from Eq. (1.13).
It is often stated that higher derivative theories suffer from unitarity problems.
This is seen as follows. When the higher derivative terms are included, the corresponding
linearized expression for the gravitational action becomes