By V.B. Berestetskii E. M. Lifshitz

Landau Lifshitz Vol. four half 1

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Then we show the computation with Grassmann integrals which, once the properties of Grassmann integration are known, fits in three lines. 2) On the other hand, according to chapter 3, the perturbation series for Z(λ) is given by ∞ Z(λ) = (−λ)n n! [0,β]n ds1 · · · dsn TH (s1 ) · · · H (sn ) 0 n=0 ∞ = (−λ)n n! β 0 β x1 σ·1 · · 0 ds1 M1d dsn M1d xn σn Tψx+1 σ1 (s1 )ψx1 σ1 (s1 ) · · · n=0 ∞ = · · · ψx+n σn (sn )ψxn σn (sn ) (−λ)n n! 4) we wrote the integral as a limit of a Riemannian sum, s = h1 j, 0 ≤ j ≤ βh − 1.

8) et2 H0 [t1 ,t2 ]n We claim that for arbitrary λ d n (t2 −t1 )Hλ e dλ = et2 Hλ [t1 ,t2 ]n where Vλ (s) = e−sHλ V esHλ . 7). 8) is proven by induction on n. 8) reduces to the lemma above. 8) is correct for n − 1. Since T[Vλ (s1 ) · · · Vλ (sn )] is a symmetric function, one has [t1 ,t2 ]n ds1 · · · dsn T[Vλ (s1 ) · · · Vλ (sn )] = n! = n! 9) t1 and the induction hypothesis reads d n−1 (t2 −t1 )Hλ e = dλ t2 s (n − 1)! t1 ds1 · · · t1n−2 dsn−1 e(t2 −s1 )Hλ V e(s1 −s2 )Hλ · · · V e(sn−1 −t1 )Hλ Thus d n (t2 −t1 )Hλ e dλ t2 t1 = (n − 1)!

In particular, x0 − x0 ∈ (−β, β). On this interval, C is an antisymmetric function. Namely, let x0 − x0 ∈ (−β, β) such that also x0 − x0 + β ∈ (−β, β). That is, x0 < x0 . 62) Since C is only defined on (−β, β), we may expand it into a 2β-periodic Fourier series. This gives us ‘frequencies’ k0 ∈ 2π 2β Z. 62), we only get the odd frequencies k0 ∈ 2π (2Z + 1). 59) is then equivalent to 2β β −β eik0 x0 C(x0 , k) dx0 = 2β β C(k0 , k) = 2C(k0 , k) where C(x0 , k) = e−x0 ek χ(x0 ≤ 0) χ(x0 > 0) 1 , C(k0 , k) = − βe −βe k k 1+e 1+e ik0 − ek This is checked by direct computation.

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