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					This article is for finding out an eigenstate with energy zero of the Hamiltonian derived in the following article. Discretized Classical Mechanical Model@Theory of Quantum History Entangled in Time-like Direction@Products H = -[m/(2ε)]Σk=-∞∞(Xk+1 - Xk)2 + [1/(2mεα2)]Σk=-∞∞[Pk - mα(Xk+1 - Xk)]2 　= [1/(2mεα2)]Σk=-∞∞[(Pk)2 - mαPk(Xk+1 - Xk) - mα(Xk+1 - Xk)Pk] 　= [1/(2mεα2)]Σk=-∞∞[(Pk)2 + mα(PkXk + XkPk)] - [1/(εα)]Σk=-∞∞ PkXk+1 ・・・(1). Each term of the first sum can be interpreted as the Hamiltonian of each particle. Each term of the second sum can be interpreted as the interaction between two particles adjoining each other. I am glad to see it here because this interaction is of the same form as one proposed by John von Neumann to represent measurement. --- --- Now I first try to find out an eigenstate of (Pk)2 + mα(PkXk + XkPk) without summing over k. This problem reduces to solving the following equation. -h2(d/dx)2ψ(x) - ihmα[(d/dx)x + x(d/dx)]ψ(x) = eψ(x). This equation reduces as follows. -h2(d/dx)2ψ - ihmα(2x d/dx + 1)ψ= eψ, ∴ h2(d/dx)2ψ + 2ihmαx(d/dx)ψ + (e + ihmα)ψ = 0 ・・・(2). --- To solve equation (2), I write ψ in the Fourier expansion form. ψ(x) =∫-∞∞dp f(p) exp(ipx) ・・・(3). Substituting expression (3) for ψ in equation (2), we obtain the equation for f. [-h2p2 - 2ihmα(d/dp)p + e + ihmα]f(p) = 0 ・・・(4)', ∴ [-h2p2 - 2ihmα + e + ihmα - 2ihmαp(d/dp)]f(p) = 0 ・・・(4). I suppose that equation (4) has a solution of the following form. f(p) = pnexp(kp2) (n is not zero) ・・・(5). Substituting expression (5) for f(p) in equation (4), we obtain a condition to determine n and k. h2p2 + ihmα - e + 2ihmα(n +2kp2)= 0, ∴ (h2 + 4ihmαk)p2 + (ihmα - e + 2ihmαn) = 0, ∴ h2 + 4ihmαk = 0, ihmα - e + 2ihmαn = 0, ∴ k = ih/(4mα), n = e/(2ihmα) - 1/2 ・・・(6). These values of k and n are to be used in the expression ψ(x) = ∫-∞∞dp pnexp(kp2 + ipx) ・・・(7) which is a combination of (3) and (5). Expression (7) for the solution of equation (2) is not certain because partial integration was used to obtain expression (4)', but expression (7) is not used later. --- Now I try to solve equation (2) by using another method. At first, I write ψ in a power series expansion form. ψ(x) = Σn=0∞ anxn ・・・(8). As a preparation, I write down the results of some operations for ψ. (d/dx)ψ(x) = Σn=1∞ nanxn-1 = Σn=0∞ (n+1)an+1xn. x(d/dx)ψ(x) = Σn=1∞ nanxn. (d/dx)2ψ(x) = Σn=2∞ n(n-1)anxn-2 = Σn=0∞ (n+2)(n+1)an+2xn. Taking into account these results, equation (2) reduces to the following recurrence relations for an. h2(n+2)(n+1)an+2 + (e + ihmα)an = 0 (n is zero), h2(n+2)(n+1)an+2 + 2ihmαnan + (e + ihmα)an = 0 (n is not zero). These recurrence relations are rewritten as follows. 2h2a2 + (e + ihmα)a0 = 0 ・・・(9), h2(n+2)(n+1)an+2 + [e + ihmα(2n + 1)]an = 0 (n is not zero) ・・・(10). Seeing these conditions, we know that all coefficients of xn for odd values of n are independent from all coefficients of xn for even values of n. So, every eigenspace belonging to each value of e has two dimension. The solution of recurrence relations (9) and (10) are as follows. a2(e) = -(e + ihmα)a0(e)/(2h2) ・・・(11) a2n(e) = (-1/h2)n-1 a2Πk=2n [e + ihmα(4k - 3)]/[(2k)(2k - 1)] (n≧2) 　= (-1/h2)n-1 a2 (4・6・・・2n)-1[3・5・・・(2n-1)]-1(4ihmα)n-1 Πk=2n [k - 3/4 + e/(4ihmα)] 　= (-1/h2)n-1 a2 [(1/2)(2n)!]-1(4ihmα)n-1 Γ(n + 1/4 + e/(4ihmα))/Γ(5/4 + e/(4ihmα)) (n≧1) ・・・(12) ∵ Γ(n + 1/4 + e/(4ihmα)) = Γ(2 - 3/4 + e/(4ihmα)) Πk=2n [k - 3/4 + e/(4ihmα)]. Expression (12) proves itself correct even when n = 1. a2n+1(e) = (-1/h2)n a1Πk=1n [e + ihmα(4k - 1)]/[(2k + 1)(2k)] 　= (-1/h2)n a1 [3・5・・・(2n+1)]-1(2・4・・・2n)-1(4ihmα)n Πk=1n [k - 1/4 + e/(4ihmα)] 　= (-1/h2)n a1 [(2n + 1)!]-1(4ihmα)n Γ(n + 3/4 + e/(4ihmα))/Γ(3/4 + e/(4ihmα)) ・・・(13) ∵ Γ(n + 3/4 + e/(4ihmα)) = Γ(1 - 1/4 + e/(4ihmα)) Πk=1n [k - 1/4 + e/(4ihmα)]. --- --- As a second step, I use a direct product of the above functions as a basis vector, and I expand an eigenstate of the whole Hamiltonian by it. Φ[x] = [Πs=-∞∞Σi(s)=12∫-∞∞des] A[e, i] Πk=-∞∞ue(k)i(k)(xk) ・・・(14) where ue1(x) = Σn=0∞ a2n+1(e) x2n+1 ・・・(15) ue2(x) = Σn=0∞ a2n(e) x2n ・・・(16). To express the effect of PkXk+1, here I try to write the coefficients of xuei(x) and -ih(d/dx)uei(x) in terms of the coefficients of uei(x). xue1(x) = xΣn=0∞ a2n+1(e)x2n+1 = Σn=0∞ a2n+1(e)x2n+2 = Σn=1∞ a2n-1(e)x2n. This must be equal to ∫de' ξ1(e, e')ue'2(x). a2n-1(e) = ∫de' ξ1(e, e')a2n(e') (n is not zero) ・・・(17), 0 = ∫de' ξ1(e, e')a0(e') ・・・(18). Condition (18) determines the form of a0 as a function. xue2(x) = xΣn=0∞ a2n(e)x2n = Σn=0∞ a2n(e)x2n+1. This must be equal to ∫de' ξ2(e, e')ue'1(x). a2n(e) = ∫de' ξ2(e, e')a2n+1(e') ・・・(19). Conditions (17) and (19) determine ξ1 and ξ2. -ih(d/dx)ue1(x) = -ih(d/dx)Σn=0∞ a2n+1(e)x2n+1 = -ihΣn=0∞ (2n+1)a2n+1(e)x2n. This must be equal to ∫de' η1(e, e')ue'2(x). ∴ -ih(2n+1)a2n+1(e) = ∫de' η1(e, e')a2n(e') ・・・(20). -ih(d/dx)ue2(x) = -ih(d/dx)Σn=0∞ a2n(e)x2n = -ihΣn=1∞ 2n a2n(e)x2n-1 = -ihΣn=0∞ (2n+2) a2n+2(e)x2n+1. This should be equal to ∫de' η2(e, e')ue'1(x). ∴ -ih(2n+2)a2n+2(e) = ∫de' η2(e, e')a2n+1(e'). ∴ -ih(2n)a2n(e) = ∫de' η2(e, e')a2n-1(e') ・・・(21). Conditions (20) and (21) determine η1 and η2. ξ1, ξ2, η1, η2 are to be used in the following expressions. Xue1 = ∫de' ξ1(e, e')ue'2 ・・・(22), Xue2 = ∫de' ξ2(e, e')ue'1 ・・・(23), Pue1 = ∫de' η1(e, e')ue'2 ・・・(24), Pue2 = ∫de' η2(e, e')ue'1 ・・・(25). The concrete expressions of ξ1, ξ2, η1, η2 are as follows. Expression (13) can be rewritten as follows when n ≧ 1. a2n-1 = (-1/h2)n-1 a1 [(2n - 1)!]-1(4ihmα)n-1 Γ(n - 1/4 + e/(4ihmα))/Γ(3/4 + e/(4ihmα)) 　= (-1/h2)n-1 a1 2{[n - 1/4 + e/(4ihmα)] + [1/4 - e/(4ihmα)]}[(2n)!]-1(4ihmα)n-1 Γ(n - 1/4 + e/(4ihmα))/Γ(3/4 + e/(4ihmα)) 　= (-1/h2)n-1 a1 2[(2n)!]-1(4ihmα)n-1{Γ(n + 3/4 + e/(4ihmα)) + [1/4 - e/(4ihmα)]Γ(n - 1/4 + e/(4ihmα))}/Γ(3/4 + e/(4ihmα)) ・・・(26). Substituting expressions (26) and (12) in condition (17), we know that ξ1(e, e') = a1(e)Γ(3/4 + e/(4ihmα))-1{δ(e + 2ihmα - e') + [1/4 - e/(4ihmα)]δ(e - 2ihmα - e')}a2(e')-1Γ(5/4 + e'/(4ihmα)) ・・・(27). We can directly verify that expression (27) satisfies condition (17) when n = 1. ∫de' ξ1(e, e') a2(e') = a1(e)Γ(3/4 + e/(4ihmα))-1{Γ(7/4 + e/(4ihmα)) + [1/4 - e/(4ihmα)]Γ(3/4 + e/(4ihmα))} 　= a1(e){[3/4 + e/(4ihmα)] + [1/4 - e/(4ihmα)]} 　= a1(e). Expression (12) is deformed as follows. a2n = (-1/h2)n-1 a2 [(1/2)(2n)!]-1(4ihmα)n-1 Γ(n + 1/4 + e/(4ihmα))/Γ(5/4 + e/(4ihmα)) 　= (-1/h2)n-1 a2 2(2n + 1)[(2n + 1)!]-1(4ihmα)n-1 Γ(n + 1/4 + e/(4ihmα))/Γ(5/4 + e/(4ihmα)) 　= (-1/h2)n-1 a2 4{[n + 1/4 + e/(4ihmα)] + [1/4 - e/(4ihmα)]}[(2n + 1)!]-1(4ihmα)n-1 Γ(n + 1/4 + e/(4ihmα))/Γ(5/4 + e/(4ihmα)) 　= (-1/h2)n-1 a2 4[(2n + 1)!]-1(4ihmα)n-1{Γ(n + 5/4 + e/(4ihmα)) + [1/4 - e/(4ihmα)]Γ(n + 1/4 + e/(4ihmα))}/Γ(5/4 + e/(4ihmα)) (n≧1)・・・(28). Substituting expressions (28) and (13) in condition (19), we know that ξ2(e, e') = [-h/(imα)] a2(e)Γ(5/4 + e/(4ihmα))-1{δ(e + 2ihmα - e') + [1/4 - e/(4ihmα)]δ(e - 2ihmα - e')}a1(e')-1Γ(3/4 + e'/(4ihmα)) ・・・(29). We can directly verify that expression (29) satisfies condition (19) when n = 0. ∫de' ξ2(e, e') a1(e') = [-h/(imα)] a2(e)Γ(5/4 + e/(4ihmα))-1{Γ(5/4 + e/(4ihmα)) + [1/4 - e/(4ihmα)]Γ(1/4 + e/(4ihmα))} 　= [-h/(imα)] a2(e){1+[1/4 - e/(4ihmα)]/[1/4 + e/(4ihmα)]} 　= [-h/(imα)] a2(e)(1/2)/[1/4 + e/(4ihmα)] 　= [-h/(imα)](2ihmα)a2(e)/(e + ihmα) 　= a0(e) ∵(11) Substituting expressions (12) and (13) in condition (20), we know that η1(e, e') = -2mαa1(e)Γ(3/4 + e/(4ihmα))-1δ(e + 2ihmα - e')a2(e')-1Γ(5/4 + e'/(4ihmα)) ・・・(30) We can directly verify that expression (30) satisfies condition (20) when n = 0. ∫de' η1(e, e') a0(e') = -2mαa1(e)Γ(3/4 + e/(4ihmα))-1∫de' δ(e + 2ihmα - e')a0(e')a2(e')-1Γ(5/4 + e'/(4ihmα)) 　= -2mαa1(e)Γ(3/4 + e/(4ihmα))-1∫de' δ(e + 2ihmα - e')[-2h2/(e' + ihmα)]Γ(5/4 + e'/(4ihmα)) ∵(11) 　= -2mαa1(e)(3/4 + e/(4ihmα))[-2h2/(e + 3ihmα)] 　= -iha1(e) Substituting expression (12) and the first row of expression (26) in condition (21), we know that η2(e, e') = -2iha2(e)Γ(5/4 + e/(4ihmα))-1δ(e + 2ihmα - e')a1(e')-1Γ(3/4 + e'/(4ihmα)) ・・・(31) --- δ(e ± 2ihmα - e') appearing in the above expressions are not ordinary because its variable is not real. However, the expression δ(x) = (2π)-1∫-∞∞dp exp(ipx) straightly suggests a natural extension to the delta function of a complex variable. δ(z) = (2π)-1∫-∞∞dp exp(ipz). Perhaps it is the analytic continuation of the existing delta function. Let f represent a function of a complex variable, then ∫de' f(e') δ(e + 2ihmα - e') = (2π)-1∫-∞∞dp∫de' f(e') exp[ip(e + 2ihmα - e')] ・・・(32). Now I extract the following integral from expression (32). ∫-∞∞de' f(e')exp[ip(e + 2ihmα - e')] = ∫-∞-2ihmα∞-2ihmα de'' f(e'' + 2ihmα)exp[ip(e - e'')] ・・・(33) where e'' = e' - 2ihmα. Integration over e'' is done along the line {x + iy|x ∈ R and y = -2hmα}. Integral (32) reduces to f(e + 2ihmα), if the following equality holds. ∫-∞-2ihmα∞-2ihmα de'' f(e'' + 2ihmα)exp[ip(e - e'')] = ∫-∞∞ de'' f(e'' + 2ihmα)exp[ip(e - e'')] ・・・(34). Equality (34) means that the shift of integration line does not change the result of integration. Dividing both sides of equation (34) by the common factor exp(ipe), we know that equation (34) is equivalent to the following equation. ∫-∞-2ihmα∞-2ihmα de'' f(e'' + 2ihmα)exp(-ipe'') = ∫-∞∞ de'' f(e'' + 2ihmα)exp(-ipe'') ・・・(35). To investigate the validity of equality (35), let's integrate the function f(z)exp[-ip(z + ib)] around the closed contour: z = -a+0i → +a+0i → +a-ib → -a-ib → -a+0i. There is no singularity inside of this contour, so the result of this intergration will be zero. 0 = ∫-a+a dx f(x)exp[-ip(x + ib)] + ∫0-b idy f(a + iy)exp[-ip(a + iy + ib)] + ∫+a-a dx f(x - ib)exp[-ip(x - ib + ib)] + ∫-b0 idy f(-a + iy)exp[-ip(-a + iy + ib)]. ∴ ∫0-b idy f(a + iy)exp[-ip(a + iy + ib)] + ∫-b0 idy f(-a + iy)exp[-ip(-a + iy + ib)] = 0 　⇒ ∫-a+a dx f(x)exp[-ip(x + ib)] + ∫+a-a dx f(x - ib)exp[-ip(x - ib + ib)] = 0 ∴ ∫-b0 dy f(-a + iy)exp[-ip(-a + iy + ib)] = ∫-b0 dy f(a + iy)exp[-ip(a + iy + ib)] ・・・(35) 　⇒ ∫-a+a dx f(x)exp[-ip(x + ib)] = ∫-a+a dx f(x - ib)exp(-ipx) ・・・(36). Equation (35) holds for infinite value of a if we chose the value of a as a = 2πn/p and lima→+∞ f(a + i(y - b)) - f(-a + i(y - b)) = 0 because ∫-b0 dy f(-a + iy)exp[-ip(iy + ib)] = ∫-b0 dy f(a + iy)exp[-ip(iy + ib)] ∵∫-b0 dy [f(a + iy) - f(-a + iy)]exp[-ip(iy + ib)] 　= ∫0b dy' [f(a + i(y' - b)) - f(-a + i(y' - b))]exp[-ip(iy')] (y' = y + b) 　= ∫0b dy' [f(a + i(y' - b)) - f(-a + i(y' - b))]exp(py'). Therefore equation (36) holds for infinite value of a if f behaves well. Substituting b = -2hmα for b in equation (36), equation (36) for infinite value of a proves equation (35). Equation (35) proves equation (34), and so integral (32) reduces to f(e + 2ihmα), and so the use of δ(e + 2ihmα - e') in the derivation of ξ1, ξ2, η1, η2 is justified. The use of δ(e - 2ihmα - e') is similarly justified. However, I am not very confident of such justification because I used the condition a = 2πn/p. --- The Hamiltonian without interaction operates as follows. [1/(2mεα2)]Σk=-∞∞[(Pk)2 + mα(PkXk + XkPk)]Φ[x] 　= [1/(2mεα2)]Σk=-∞∞[(Pk)2 + mα(PkXk + XkPk)] [Πs=-∞∞Σi(s)=12∫-∞∞des] A[e, i] Πn=-∞∞ue(n)i(n)(xn) 　= [1/(2mεα2)][Πs=-∞∞Σi(s)=12∫-∞∞des] A[e, i] Σk=-∞∞ek Πn=-∞∞ue(n)i(n)(xn) 　= [1/(2mεα2)][Πs=-∞∞Σi'(s)=12∫-∞∞de's] A[e', i'] Σk=-∞∞e'k Πn=-∞∞ue'(n)i'(n)(xn) ・・・(37). The interaction part of the Hamiltonian operates as follows. [-1/(εα)]Σk=-∞∞ PkXk+1Φ[x] 　= [-1/(εα)]Σk=-∞∞ PkXk+1 [Πs=-∞∞Σi(s)=12∫-∞∞des] A[e, i] Πn=-∞∞ue(n)i(n)(xn) 　= [-1/(εα)][Πs=-∞∞Σi(s)=12∫-∞∞des] A[e, i] Σk=-∞∞[Pkue(k)i(k)(xk)][Xk+1ue(k+1)i(k+1)(xk+1)] Πk≠n≠k+1 ue(n)i(n)(xn) 　= [-1/(εα)][Πs=-∞∞Σi(s)=12∫-∞∞des] A[e, i] Σk=-∞∞[∫de'ηi(k)(ek, e')ue', 3-i(k)(xk)][∫de''ξi(k+1)(ek+1, e'')ue'', 3-i(k+1)(xk+1)] Πk≠n≠k+1 ue(n)i(n)(xn) 　= [-1/(εα)][Πs=-∞∞Σi(s)=12∫-∞∞des] A[e, i] Σk=-∞∞[Πq=-∞∞Σi'(q)=12∫-∞∞de'q] [・・・δ(ek-1 - e'k-1)δi(k-1), i'(k-1)・ηi(k)(ek, e'k)δ3-i(k), i'(k)・ξi(k+1)(ek+1, e'k+1)δ3-i(k+1), i'(k+1)・δ(ek+2 - e'k+2)δi(k+2), i'(k+2)・・・] Πn=-∞∞ue'(n)i'(n)(xn) ・・・(38). The sum of expressions (37) and (38) must be zero to let HΦ[x] be zero, so [1/(2mεα2)]A[e', i'] Σk=-∞∞e'k - [1/(εα)][Πs=-∞∞Σi(s)=12∫-∞∞des] A[e, i] Σk=-∞∞[・・・δ(ek-1 - e'k-1)δi(k-1), i'(k-1)・ηi(k)(ek, e'k)δ3-i(k), i'(k)・ξi(k+1)(ek+1, e'k+1)δ3-i(k+1), i'(k+1)・δ(ek+2 - e'k+2)δi(k+2), i'(k+2)・・・] = 0 ∴ [1/(2mα)][Σk=-∞∞e'k]A[e', i'] = Σk=-∞∞∫dek∫dek+1 η3-i'(k)(ek, e'k)ξ3-i'(k+1)(ek+1, e'k+1)A[・・・; e'k-1, i'k-1; ek, 3-i'k; ek+1, 3-i'k+1; e'k+2, i'k+2; ・・・] ∴ [1/(2mα)][Σk=-∞∞ek]A[e, i] = Σk=-∞∞∫de'k∫de'k+1 η3-i(k)(e'k, ek)ξ3-i(k+1)(e'k+1, ek+1)A[・・・; ek-1, ik-1; e'k, 3-ik; e'k+1, 3-ik+1; ek+2, ik+2; ・・・] ・・・(39). This condition determines A. To write ξ and η more shortly, I introduce the following abbreviations. b11(e) ≡ a1(e)a2(e + 2ihmα)-1Γ(3/4 + e/(4ihmα))-1Γ(7/4 + e/(4ihmα)) ・・・(40), b12(e) ≡ [1/4 - e/(4ihmα)]a1(e)a2(e - 2ihmα)-1 ・・・(41), b21(e) ≡ [-h/(imα)] a2(e)a1(e + 2ihmα)-1 ・・・(42), b22(e) ≡ [-h/(imα)][1/4 - e/(4ihmα)] a2(e)a1(e - 2ihmα)-1Γ(5/4 + e/(4ihmα))-1Γ(1/4 + e/(4ihmα)) ・・・(43), c1 ≡ -2mαa1(e)a2(e + 2ihmα)-1Γ(3/4 + e/(4ihmα))-1Γ(7/4 + e/(4ihmα)) ・・・(44), c2 ≡ -2iha2(e)a1(e + 2ihmα)-1 ・・・(45). Then ξ and η are written as follows. ξ1(e, e') = b11(e)δ(e + 2ihmα - e') + b12(e)δ(e - 2ihmα - e') ・・・(46), ξ2(e, e') = b21(e)δ(e + 2ihmα - e') + b22(e)δ(e - 2ihmα - e') ・・・(47), η1(e, e') = c1(e)δ(e + 2ihmα - e') ・・・(48), η2(e, e') = c2(e)δ(e + 2ihmα - e') ・・・(49). By using these notations, condition (39) is rewritten as follows. [1/(2mα)][Σk=-∞∞ek]A[e, i] 　 = Σk=-∞∞∫de'k∫de'k+1 c3-i(k)(e'k)δ(e'k + 2ihmα - ek)[b3-i(k+1),1(e'k+1)δ(e'k+1 + 2ihmα - ek+1) + b3-i(k+1),2(e'k+1)δ(e'k+1 - 2ihmα - ek+1)]A[・・・; ek-1, ik-1; e'k, 3-ik; e'k+1, 3-ik+1; ek+2, ik+2; ・・・] 　 = Σk=-∞∞∫de'k+1 c3-i(k)(ek - 2ihmα)[b3-i(k+1),1(e'k+1)δ(e'k+1 + 2ihmα - ek+1) + b3-i(k+1),2(e'k+1)δ(e'k+1 - 2ihmα - ek+1)]A[・・・; ek-1, ik-1; ek - 2ihmα, 3-ik; e'k+1, 3-ik+1; ek+2, ik+2; ・・・] ∴ [1/(2mα)][Σk=-∞∞ek]A[e, i] 　 = Σk=-∞∞c3-i(k)(ek - 2ihmα){b3-i(k+1),1(ek+1 - 2ihmα)A[・・・; ek-1, ik-1; ek - 2ihmα, 3-ik; ek+1 - 2ihmα, 3-ik+1; ek+2, ik+2; ・・・] + b3-i(k+1),2(ek+1 + 2ihmα)A[・・・; ek-1, ik-1; ek - 2ihmα, 3-ik; ek+1 + 2ihmα, 3-ik+1; ek+2, ik+2; ・・・]} ・・・(50). Sufficient condition is that [1/(2mα)][sek + (1 - s)ek+1]A[e, i] = c3-i(k)(ek - 2ihmα){b3-i(k+1),1(ek+1 - 2ihmα))A[・・・; ek-1, ik-1; ek - 2ihmα, 3-ik; ek+1 - 2ihmα, 3-ik+1; ek+2, ik+2; ・・・] + b3-i(k+1),2(ek+1 + 2ihmα)A[・・・; ek-1, ik-1; ek - 2ihmα, 3-ik; ek+1 + 2ihmα, 3-ik+1; ek+2, ik+2; ・・・]} ・・・(51). Summing each side of equation (51) over k, equation (50) is reproduced. I thought that perhaps Fourier expansion of A can be used to solve this equation, but I do not think so now. For now, I give up solving equation (51). --- --- To avoid facing the entire difficulty, here I try using perturbation method. The set of conditions to be started with is as follows. Φ[x] = Σr=0∞βr[Πs=-∞∞Σi(s)=12∫-∞∞des]Ar[e, i] Πk=-∞∞ue(k)i(k)(xk) ・・・(52), H = [1/(2mεα2)]Σk=-∞∞[(Pk)2 + mα(PkXk + XkPk)] - [β/(εα)]Σk=-∞∞ PkXk+1 ・・・(53), HΦ = 0 ・・・(54). Taking the part corresponding to the zeroth power of β, we get the following condition. A0[e, i]Σkek = 0 ・・・(55). Taking the part corresponding to the first power of β, we get the following condition. [1/(2mεα2)][Πs=-∞∞Σi(s)=12∫-∞∞des]A1[e, i](Σk=-∞∞ek)Πn=-∞∞ue(n)i(n)(xn) = [1/(εα)]Σk=-∞∞ PkXk+1[Πs=-∞∞Σi(s)=12∫-∞∞des]A0[e, i] Πn=-∞∞ue(n)i(n)(xn) ・・・(56). The righthand side of equation (56) reduces as follows. [1/(εα)]Σk=-∞∞[Πs=-∞∞Σi(s)=12∫-∞∞des]A0[e, i]∫de'k∫de'k+1 ηi(k)(ek, e'k)ξi(k+1)(ek+1, e'k+1)[・・・ue(k-1)i(k-1)(xk-1)ue'(k),3-i(k)(xk)ue'(k+1),3-i(k+1)(xk+1)ue(k+2)i(k+2)(xk+2)・・・] = [1/(εα)][Πs=-∞∞Σi'(s)=12∫-∞∞des]Σk=-∞∞[Πq=-∞∞∫-∞∞de'q]A0[・・・; ek-1, i'k-1; ek, 3-i'k; ek+1, 3-i'k+1; ek+2, i'k+2; ・・・][・・・δ(ek-1 - e'k-1)η3-i'(k)(ek, e'k)ξ3-i'(k+1)(ek+1, e'k+1)δ(ek+2 - e'k+2)・・・]Πn=-∞∞ue'(n)i'(n)(xn) = [1/(εα)][Πs=-∞∞Σi(s)=12∫-∞∞de's]Σk=-∞∞[Πq=-∞∞∫-∞∞deq]A0[・・・; e'k-1, ik-1; e'k, 3-ik; e'k+1, 3-ik+1; e'k+2, ik+2; ・・・][・・・δ(e'k-1 - ek-1)η3-i(k)(e'k, ek)ξ3-i(k+1)(e'k+1, ek+1)δ(e'k+2 - ek+2)・・・]Πn=-∞∞ue(n)i(n)(xn) = [1/(εα)][Πs=-∞∞Σi(s)=12]Σk=-∞∞∫-∞∞de'k∫-∞∞de'k+1[Πq=-∞∞∫-∞∞deq]A0[・・・; ek-1, ik-1; e'k, 3-ik; e'k+1, 3-ik+1; ek+2, ik+2; ・・・]η3-i(k)(e'k, ek)ξ3-i(k+1)(e'k+1, ek+1)Πn=-∞∞ue(n)i(n)(xn) So, condition (56) is rewritten as follows. [1/(2mεα2)]A1[e, i]Σk=-∞∞ek = [1/(εα)]Σk=-∞∞∫-∞∞de'k∫-∞∞de'k+1A0[・・・; ek-1, ik-1; e'k, 3-ik; e'k+1, 3-ik+1; ek+2, ik+2; ・・・]η3-i(k)(e'k, ek)ξ3-i(k+1)(e'k+1, ek+1) ∴ A1[e, i] = 2mα(Σs=-∞∞es)-1Σk=-∞∞∫-∞∞de'k∫-∞∞de'k+1A0[・・・; ek-1, ik-1; e'k, 3-ik; e'k+1, 3-ik+1; ek+2, ik+2; ・・・]η3-i(k)(e'k, ek)ξ3-i(k+1)(e'k+1, ek+1) ・・・(57) To see how entanglement yields, I start with A0 such that A0[e, i] = Πk=-∞∞ δ(ek - e0k) where Σk=-∞∞ e0k = 0 ・・・(58). A1[e, i] = 2mα(Σs=-∞∞es)-1Σk=-∞∞[・・・δ(ek-1 - e0k-1)η3-i(k)(e0k, ek)ξ3-i(k+1)(e0k+1, ek+1)δ(ek+2 - e0k+2)・・・] = 2mαΣk=-∞∞(ek - e0k + ek+1 - e0k+1)-1[・・・δ(ek-1 - e0k-1)η3-i(k)(e0k, ek)ξ3-i(k+1)(e0k+1, ek+1)δ(ek+2 - e0k+2)・・・] ・・・(59) ∵ ・・・ + e0k-1 + ek + ek+1 + e0k+2 + ・・・ = Σse0s + (ek -e0k) + (ek+1 - e0k+1) = (ek -e0k) + (ek+1 - e0k+1).
				Author Yuichi Uda, Write start at 2014/10/06/18:05JST, Last edit at 2014/12/28/22:41JST