Putting in some i's (A -iA, JiJ), we find the generalization of (22) + +o To...

  

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Putting in some i's (A →-iA, J→→iJ), we find the generalization of (22) +∞ •+∞o To fo dx₁dx₂. •dxx e(i/2)x.A.x+iJ.x -∞ = (2лi) N det[A] Lo (x; x ;): = ... Jk, and Xi The generalization of (18) is also easy to obtain. Differentiate (22) p times with respect to J₁, Jj J₁, and then set J = 0. For example, for p = 1 the integrand in (22) becomes e-xxx, and since the integrand is now odd in x; the integral vanishes. For p = 2 the integrand becomes e-**A** (x;x;), while on the right hand side we bring down A¹. Rearranging and eliminating det[A] (by setting J = 0 in (22)), we obtain e-(i/2)J.A-¹.J f+∞ +∞...S+x dx jd x₂. dxN e √∞ ∞...x dx₁dx₂. = Σ(A-¹) ab... (A-¹) cd Wick where we have defined (X₁X; • Xxx1) Just do it. Doing it is easier than explaining how to do it. Then do it for p = 3 and 4. You will see immediately how your result generalizes. When the set of indices i, j, ..., k, I contains an odd number of elements, (x¡xj. xx₁) vanishes trivially. When the set of indices i, j, ... , k, I contains an even number of elements, we have (X;X;• • • Xxx₁) s+∞ +∞o. •fdx₁dx₂. dxy e ... dx e-x · A.x xix j +∞ S ... st dx ₁dx ₂ ··· dx e - ² x ·A•x - 1x ·A•x, = A¹ = (23) •XkX1 (24) (25) and where the set of indices (a, b,, c, d] represent a permutation of {i, j, , k, l}. The sum in (24) is over all such permutations or Wick contractions. For example, (x¡x jxkxj) = (A¯¹)¡¡(A−¹)k1 + (A¯¹)¡¡ (A¯¹) jk + (A−¹)ik (A¯¹) ji (26) (Recall that A, and thus A-¹, is symmetric.) As in the simple case when x does not carry any index, we could connect the x's in (xxxx) in pairs (Wick contraction) and write a factor (A¹)ab if we connect xa to xp. Notice that since (x;xj) = (A-¹)¡¡ the right hand side of (24) can also be written in terms of objects like (x¡xj). Thus, (xixxx₁) = (x;x;) (xxx₁) + (x;x₁)(x;xk) + (x;xx) (xjx₁). Putting in some i's (A →-iA, J→→iJ), we find the generalization of (22) +∞ •+∞o To fo dx₁dx₂. •dxx e(i/2)x.A.x+iJ.x -∞ = (2лi) N det[A] Lo (x; x ;): = ... Jk, and Xi The generalization of (18) is also easy to obtain. Differentiate (22) p times with respect to J₁, Jj J₁, and then set J = 0. For example, for p = 1 the integrand in (22) becomes e-xxx, and since the integrand is now odd in x; the integral vanishes. For p = 2 the integrand becomes e-**A** (x;x;), while on the right hand side we bring down A¹. Rearranging and eliminating det[A] (by setting J = 0 in (22)), we obtain e-(i/2)J.A-¹.J f+∞ +∞...S+x dx jd x₂. dxN e √∞ ∞...x dx₁dx₂. = Σ(A-¹) ab... (A-¹) cd Wick where we have defined (X₁X; • Xxx1) Just do it. Doing it is easier than explaining how to do it. Then do it for p = 3 and 4. You will see immediately how your result generalizes. When the set of indices i, j, ..., k, I contains an odd number of elements, (x¡xj. xx₁) vanishes trivially. When the set of indices i, j, ... , k, I contains an even number of elements, we have (X;X;• • • Xxx₁) s+∞ +∞o. •fdx₁dx₂. dxy e ... dx e-x · A.x xix j +∞ S ... st dx ₁dx ₂ ··· dx e - ² x ·A•x - 1x ·A•x, = A¹ = (23) •XkX1 (24) (25) and where the set of indices (a, b,, c, d] represent a permutation of {i, j, , k, l}. The sum in (24) is over all such permutations or Wick contractions. For example, (x¡x jxkxj) = (A¯¹)¡¡(A−¹)k1 + (A¯¹)¡¡ (A¯¹) jk + (A−¹)ik (A¯¹) ji (26) (Recall that A, and thus A-¹, is symmetric.) As in the simple case when x does not carry any index, we could connect the x's in (xxxx) in pairs (Wick contraction) and write a factor (A¹)ab if we connect xa to xp. Notice that since (x;xj) = (A-¹)¡¡ the right hand side of (24) can also be written in terms of objects like (x¡xj). Thus, (xixxx₁) = (x;x;) (xxx₁) + (x;x₁)(x;xk) + (x;xx) (xjx₁).

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