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Master's Thesis in Theoretical Physics - Universiteit Utrecht

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wherey ≡ y ++ (x, x ′ ) = −(|η − η′ | − iε) 2 + |⃗x −⃗x ′ | 2ηη ′ = ∆x2 (x, x ′ )ηη ′ . (5.94)Note that ε is used for the pole description, whereas ɛ is def<strong>in</strong>ed as −Ḣ/H 2 . The pole prescriptionis such that we calculate the Feynman propagator. Let us stress that the propagatoris strictly speak<strong>in</strong>g <strong>in</strong>f<strong>in</strong>ite (hence the <strong>in</strong>dex ∞), because <strong>in</strong> the <strong>in</strong>frared (for smallvalues of k <strong>in</strong> Eq. (5.92)) the propagator diverges. These <strong>in</strong>frared divergencies can be removedby different procedures and this has been done <strong>in</strong> [33] and [34]. For this thesis weleave the discussion of <strong>in</strong>frared divergencies aside and cont<strong>in</strong>ue to br<strong>in</strong>g the propagator toa simpler form. We use the series expansion of the hypergeometric function2F 1 (a, b, c, d) =Γ(c) ∞∑ Γ(a + n)Γ(b + n) z nΓ(a)Γ(b) Γ(c + n) n!n=0(5.95)and the transformation formula2F 1( D − 12=++ ν D , D − 12− ν D ; D 2 ;1 − y 4Γ( D 2 )Γ( D 2 − 1) ( D − 1Γ( 1 2 − ν D)Γ( 1 2 + ν F 1 + ν D , D − 1 − ν D ; DD) 2 2 2 ; y 421 Γ( D 2 )Γ( D 2 − 1)( y) D2 −1 Γ( D−142 − ν D)Γ( D−11)2+ ν D ) 2F 1( 12 + ν D, 1 2 − ν D;2 − D 2 ; y 4Also we use properties of the gamma function such as xΓ(x) = Γ(1 + x) to show that)). (5.96)We then get for the propagatorΓ(1 − D 2 )Γ( D 2 ) = −Γ( D 2 − 1)Γ(2 − D ). (5.97)2i∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D 2 − 1)Γ(2 − D 2 )Γ( 1 2 + ν D)Γ( 1 2 − ν D)(4π) D 2[ ∞∑ Γ( 1 2×+ ν D + n)Γ( 1 2 − ν D + n) ( yn=0 Γ(2 − D 2 + n)Γ(n + 1) 4−∞∑n=0Γ( D−12 + ν D + n)Γ( D−12 − ν D + n)) n−D2 +1Γ( D 2 + n)Γ(n + 1) ( y4) n ]. (5.98)Now, <strong>in</strong> the first sum, we change our summation variable n → n +1, such that the sum nowruns from n = −1 to ∞, and we split off the n = −1 term. The reason for this will becomeclear <strong>in</strong> a moment. The propagator can then be written asi∆(x, x ′ ) ∞ = [(1 − ɛ)2 HH ′ ] D 2 −1Γ( D {( y) 1−D(4π) D 2 2 − 1) 2+4∞∑ [ Γ( 3 2×+ ν D + n)Γ( 3 2 − ν D + n) ( yΓ(3 − D 2 + n)Γ(n + 2) 4n=0Γ(2 − D 2 )Γ( 1 2 + ν D)Γ( 1 2 − ν D)) n−D2 +2D−1Γ(2− + ν D + n)Γ( D−12 − ν D + n) ( y) n ]}Γ( D 2 + n)Γ(n + 1) . (5.99)4

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