无限维矩阵的行列式

Determinant of Infinite Matrix Hua Zhang, Department of Physics, BeiJing Normal University 2004 08 12 TheMobius subalgebra can be expressed by infinite creation operators: L0 = rc + r cr + 1 16 L+1 = 1 4 c1/2c1/2 + p r(r + 1)c+r cr+1 L−1 = 1 4 c+1/2c + 1/2 + p r(r + 1)c+r+1cr where the index run all through the positive half integer and c+ are creation operator: [cr,c + s ] = δr,s Their commutation relation is [L+1, L−1] = 2L0, [L0, L+1] = −L−1, [L0, L−1] = L+1 where the vacuum state is: L+1 | 0 >= 0, L0 | 0 >= 1 16 | 0 > We have the following identity: exp[µL+1] exp[λL−1] = exp[ λ 1− λµL−1] exp[−2 ln(1−λµ)L0] exp[ µ 1− λµL+1] and exp[λ( 1 4 c+1/2c + 1/2+ p r(r + 1)c+r+1cr)] exp[−λ p r(r + 1)c+r+1cr] = exp[ 1 2 c+P (λ)c+] where the recurrence relation is: ∂Pr,s ∂λ = p r(r − 1)Pr−1,s + p s(s− 1)Pr,s−1 The solution is Pr,s(λ) = √ rs r + s (r − 1)r−1/2 (r − 1/2)r−1/2 (r − 1)r−1/2 (r − 1/2)r−1/2λ r+s, r, s = 1 2 , 3 2 .. where (α)k = α(α− 1)...(α− k + 1), (α)0 = 1 and the first entry is P1/2,1/2 = 1 2 λ, P3/2,1/2 = √ 3 8 λ2, P3/2,3/2 = 1 8 λ3 1 when acting on the vacuum state, we have exp[λL−1] | 0 >= exp[1 2 c+P (λ)c+] | 0 > We also have the following operator identity: exp[ 1 2 cQc] exp[ 1 2 c+Pc+] = exp[−1 2 Tr ln[1− PQ]] exp[ 1 2 c+ P 1− PQc +] exp[−c+ ln[1− PQ]c] exp[1 2 c Q 1− PQc] where the infinite matrix P and Q are symmetric and commutative: P = PT ,Q = QT , PQ = QP We define a function T with respect P and Q as: T (P,Q) =< 0 | exp[1 2 cQc] exp[ 1 2 c+Pc+] | 0 >= det[1− PQ]−1/2 By the above operator identity, we also have: T (P,Q) =< 0 | exp[λL+1] exp[λL−1] | 0 >= (1− µλ)−1/8 Comparing the above equation, we get: det[1− P (µ)P (λ)] = (1− µλ)1/4 1. Appendix 1. We have the following algebras: [L+1, L−1] = 2aL0, [L0, L+1] = −bL+1, [L0, L−1] = bL−1 We assume that: R(µ,λ) = exp[µL+1] exp[λL−1] = exp[fL−1] exp[hL0] exp[gL+1] Differentiae on both sides of this equation, we have ∂R ∂λ = RL−1 = ∂f ∂λ L−1R+RL+1 ∂g ∂λ + ∂h ∂λ exp[fL−1] exp[hL0]L0 exp[gL+1] ∂R ∂µ = L+1R = ∂f ∂µ L−1R+RL+1 ∂g ∂µ + ∂h ∂µ exp[fL−1] exp[hL0]L0 exp[gL+1] By the operator methods, we have exp[gL+1]L−1 = L−1 exp[gL+1]+2agL0 exp[gL+1]+abg2L+1 exp[gL+1] L+1 exp[fL−1] = exp[fL−1]L+1+2af exp[fL−1]L0+abf2L−1 exp[fL−1] 2 exp[hL0]L−1 = exp[bh]L−1 exp[hL0] L+1 exp[hL0] = exp[bh] exp[hL0]L+1 So we have the following equation: ∂f ∂λ = exp[bh], ∂g ∂λ = abg2, ∂h ∂λ = 2ag ∂f ∂µ = abf2, ∂g ∂µ = exp[bh], ∂h ∂µ = 2af The solution is h = −2 b ln(1− abµλ), f = λ 1− abµλ , g = µ 1− abµλ 4.We assume that: exp[ 1 2 cAc] exp[ 1 2 c+Bc+] = exp[K] exp[ 1 2 c+Fc+] exp[c+Hc][ 1 2 cGc] where A and B are symmetric matrix. Comparing the variation on both side of this equation, we get: δF = FδAF + δB exp[2H] δG = δA exp[2H] +GδBG δH = FδA+GδB δK = Tr[FδA+GδB] The solution is F = B 1−AB ,G = A 1−AB ,H = − ln[1−AB],K = −Tr ln[1−AB] 3