Review of Generating functional and Green s functions
|
|
- Λευκοθέα Σπανός
- 6 χρόνια πριν
- Προβολές:
Transcript
1 Review of Generating functional and Green s functions Zhiguang Xiao March 26, 2017
2 Contents 1 Full Green s Function 2 Connected Green s function & Generating Functional 3 One particle irreducible Green s function 4 Amputated Green s function: G (n) Amp (x 1,..., x n ) 5 Renormalized Green s function & Bare Green s function 6 Equations For Green s Functions Schwinger-Dyson Eq. 7 Global symmetry and Ward Id. 8 Appendix: Legendre transformation and IPI(Zinn Justin)
3 Full Green s function G (n) (x 1, x 2,..., x n) = Ω T Φ(x 1)... Φ(x n) Ω [DΦ]Φ(x1)... Φ(x n) exp{ i ħ = S[Φ]} [DΦ] exp{ i S[Φ]} ħ Remarks: We have divided out the bubble diagrams.
4 Full Green s function Example. 4-point Full Green s Function: Ω T Φ(x 1)... Φ(x n) Ω
5 Generating functional for full Green s functions Z[J] = = [DΦ] exp{ i (S[Φ] + ħ d 4 xj(x)φ(x))} ħ [DΦ] exp{ i S[Φ]} ħ i n d 4 x... d 4 x n G (n) (x 1,..., x n)j(x 1)... J(x n) n! n=0 = Ω Ω J Ω Ω (1) G (n) (x 1,..., x n) = 1 i n δ n Z[J] δj(x 1)... J(x n) J=0 Expansion of Z[J] around J = 0, generating the full Green s functions. Z[0] = 1, G (0) = 1.
6 Connected Green s functions G (n) conn(x 1,..., x n) only includes connected diagrams Example: Four-point connected Green s function in ϕ 4 :
7 Generating functional for Connected Green s functions Generating functional for Connected Green s functions: using the full generating functional Z[J] W [J] = ħ ln Z[J] i = ħ i n i n! G (n) n=1 conn(x 1,..., x n) = 1 ħ d 4 x 1... d 4 x n G (n) conn(x 1,..., x n)j(x 1)... J(x n) 1 i n 1 δ n W [J] δj(x 1)... J(x n) J=0 We now take ħ = 1, G (n) conn(x 1,..., x n) is symmetric with respect to idential J(x i).
8 Symmetric factor for a connected diagram L = 1 2 µϕ µ ϕ 1 2 m2 ϕ ! λϕ3 = L 0 + L 1 { Z[J] exp i V =0 [ 1 i V! P =0 1 P! ( )} d 4 1 δ xl 1 Dϕe is 0+Jϕ i δj ( ) ] 3 V 3! λ d 4 δ x iδj(x) ( 1 d 4 x(ij(x))( i (x, y))(ij(y)) 2 Combine equal terms: Permutations of vertices, cancel V!; permutations of propagators, cancel P!; permutations of δ/δj, cancel 3!; permutations of J, cancel 1/2. Overcounting must be divided symmetric factor Feynman rules already including all the contractions (permutations) for the same field, eg. ϕ 3 vertex iλ, 1/3! cancelled. ) P
9 Symmetric factor for a connected diagram For the coefficients with J in the generating functional, the external legs with J s permutations of external J, overcounted, should be divided; for amplitude, or greens function, extra external different δ/δj counted as different, not overcounted.
10 iw generates all connected GF (Srednicki chapt 9) C I connected components, n I times in one disconnected diagrams: the term in Z[J] for this diagram 1 D ni = (C I) n I S D S D = I Exchange all vertices, propagators... for n I C I components. Overcount n I! times. The full Z[J] I n I! Z[J] D ni {ni } {ni } I n I =0 exp{ C I} I exp{iw [J]} I 1 n I! (CI)n I I 1 n I! (CI)n I exp(c I) Since we normalize Z[0] = 1, all the should be =. So iw [J] = I CI, generating all the connected diagrams.
11 Example: 2-point conn. GF in ϕ 3 theory δ δ Ω T Φ(x 1)Φ(x 2) Ω conn = ln Z[J] iδj(x 1) iδj(x 2) ( ) δ Ω Φ(x 2) Ω J = iδj(x 1) Ω Ω J J=0 ( Ω Φ(x1)Φ(x 2) Ω J = Ω Φ(x1) Ω J Ω Ω J Ω Ω J J=0 Ω Φ(x 2) Ω J Ω Ω J ) J=0
12 One particle irreducible Green s function 1PI: amputated diagrams remains connected after cutting an arbitrary internal propagator. in ϕ 4 theory, 1PI 4-point function: Γ(x 1,..., x 4) =
13 Generating functional for 1PI We expect the Generating functional to be: 1 Γ[Φ] = d 4 x 1... d 4 x nγ (n) (x 1... x n)φ(x 1)... Φ(x n) n! n=1 Γ (n) δγ[φ] (x 1... x n) = δφ(x 1)... δφ(x n) Φ=0 In fact, there are some subtleties for the generating functional we will construct: The generating funtional generates 1PI for n 3 at Φ = 0 for vev Φ(x) = 0. for n = 2, Γ (2) (x, y) = (G (2) conn) 1, inverse propagator (as operator inverse). for n = 1,???.
14 Construct Γ from W Legendre transformation: 1 Recall the classcial mechanics: From Lagrangian to Hamiltonian L(q, q) H(q, p), we define p solve q = q(p, q) and insert into H(q, p) p q L[q, q(p, q)]. We change from q, q to p, q formalism. 2 For J[x] 0, (L[q, q] W [J], q J,) we define classical field: L[q, q] q, φ cl (x) = Ω Φ(x) Ω J Ω Ω J = δw [J] δj(x). φ cl (x) is a functional of J(x). Solve J(x) = J x[φ cl ]. 3 define Γ[φ cl ] W [J[φ cl ]] Γ is the generating functional we need. d 4 xj[φ cl ]φ cl (x),
15 Generating functional Γ Γ[φ cl ] W [J[φ cl ]] d 4 xj[φ cl ]φ cl (x), φ cl (x) = Ω Φ(x) Ω J Ω Ω J = δw [J] δj(x). 1 δγ[φ cl ] δφ cl = = J(x) 2 if J = 0, φ cl = Ω Φ(x) Ω Ω Ω d 4 y δj(y) δw [J] δφ cl (x) δj(y) J(x) δw [J] = δj[x] ϕ c, the v.e.v. of Φ. J=0 δγ[φ cl ] δφ cl (x) = 0 ϕc gives the equation of motion for ϕ c, the vev for Φ. 3 Γ[φ cl ] is also called Quantum Effective Action. 4 Vacuum preserves Poincaré symmetry, and no SSB, ϕ c = 0. 5 Perturbation theory: perturbation around φ = ϕ c = 0. Γ (1) = δγ[φ cl] δφ cl ϕc = 0. d 4 y δj(y) δφ cl (x) φ cl(y)
16 Generating functional Γ 1 n = 2, from J = δγ[φ cl] δφ cl δ (4) (y x) = δj(x) δj(y) = δ2 Γ[φ cl ] δj(y)δφ cl (x) = d 4 z δφ cl(z) δ 2 Γ[φ cl ] δj(y) δφ cl (z)δφ cl (x) = d 4 δ 2 W [J] δ 2 Γ[φ cl ] z δj(y)δj(z) δφ cl (z)δφ cl (x) [ ] Γ (2) δ 2 Γ[φ cl ] 1 (x, y) = δφ cl (x)δφ cl (y) = δ2 W [J] φcl =ϕ c δj(x)δj(y) J=0 2 in momentum space: iγ (2) inverse propagator. G (2) (k 2 )Γ (2) (k 2 ) = i G (2) (k 2 i ) = k 2 m 2 M(k 2 ) Γ (2) (k 2 ) = k 2 m 2 M(k 2 ) = [ ig (2) conn] ( 1) (x,y)
17 Generating functional Γ 1 Expand around ϕ c 1 Γ[Φ] = d 4 x 1... d 4 x nγ (n) (x 1... x n)(φ(x 1) ϕ c)... (Φ(x n) ϕ c) n! n=2 2 n = 3, G (3) iδw [J] (x, y, z) = (iδj(x))(iδj(y))(iδj(z)) J=0 = ig (2) (x x )ig (2) (y y )ig (2) (z z ) x,y,z δ 3 Γ[φ cl ] δφ cl (x )δφ cl (y )δφ cl (z ) iγ (3) (x, y, z) is 1PI. 3 for general n, iγ (n) (x 1,..., x n) = i δ n Γ[φ cl ] δφ cl (x 1 )...δφ cl (x n) ϕc is 1PI, for n 3. ϕc
18 Amputated Green s function: G (n) Amp (x 1,..., x n ) G (n) conn(x 1,..., x n) = d 4 y 1... d 4 y ng (2) (x 1, y 1)... G (2) (x n, y n)g (n) Amp (y1,..., yn) In general, amputated Green s function is not 1PI. In momentum space: G (n) conn(p 1,..., p n) = G (2) (p 1)... G (2) (n) (p n) G Amp (p1,..., pn)
19 Renormalized Green s function & Bare Green s function Bare field Φ B(x) and renormalized field Φ R(x): Φ B(x) = Z 1/2 Φ R(x) Bare Green s function: G (n) B (x1,..., xn) Ω ΦB(x1)... ΦB(xn) Ω If Φ s are different, differnt Z s are used. = (Z 1/2 ) n Ω Φ R(x 1)... Φ R(x n) Ω = Z n/2 G (n) (x1,..., xn) R Generating functional for Bare. S B = d 4 xl[φ B] = d 4 xl R[Φ R] + L ct = S R + S ct (2) Z B[J B] = [DΦB] exp{i(s[φ B] + d 4 xj B(x)Φ B(x))} [DΦB] exp{is[φ B]} (3) G (n) 1 δ n Z B[J B] B (x1,..., xn) = i n δj B(x 1)... J B(x n) JB =0
20 Renormalized Green s function & Bare Green s function Renormalized Green s funct: define J R = Z 1/2 J B, J BΦ B = J RΦ R [DΦR] exp{i(s R[Φ R] + S ct + d 4 xj R(x)Φ R(x))} Z R[J R] = Z B[J B] = [DΦR] exp{is[φ R] + S ct} G (n) 1 δ n Z R[J R] R (x1,..., xn) = i n δj R(x 1)... J R(x n) JR =0 = 1 δ n Z B[J B] i n Z n/2 δj B(x 1)... J B(x n) = Z n/2 G (n) (x1,..., xn) B JB =0 Similar for Connected Green s function. For on-shell renormalization scheme: in momentum space, near physical pole G (2) i R (p2 ) p 2 m 2 ph + iϵ G (2) iz B (p2 ) p 2 m 2 ph + iϵ
21 Renormalized & Bare Amputated Green s function Amputated Green s funct: G (n) B,conn (x 1,..., x n) = G (n) R,conn (x 1,..., x n) = using G (n) B d 4 y 1... d 4 y ng (2) B (x 1, y 1 )... G (2) (xn, yn)g(n) B B,Amp (y 1,..., y n) d 4 y 1... d 4 y ng (2) R (x 1, y 1 )... G (2) (xn, yn)g(n) R R,Amp (y 1,..., y n) (x1,..., xn) = Zn/2 G (n) (x1,..., xn), G(2) R G (n) B,Amp (y1,..., yn) = Z n/2 G (n) R,Amp (y1,..., yn) B = ZG(2) R : Bare & Renormalized 1PI: similar as before J R = Z 1/2 J B Γ B[φ B,cl ] W B[J B[φ B,cl ]] d 4 xj B[φ B,cl ]φ B,cl (x) = W R[J R[φ R,cl ]] d 4 xj R[φ R,cl ]φ R,cl (x) Γ R[φ R,cl ] φ B,cl = δwb[jb] = Z 1/2 δw R[J R] = Z 1/2 φ R,cl (x) δj B δj R Γ (n) R (x1,..., xn) = δ n Γ R[φ R,cl ] δφ R,cl (x 1)... δφ R,cl (x = n) Zn/2 Γ (n) B (x1,..., xn)
22 Equations For Green s Functions Schwinger-Dyson Eq. Consider path integral:(bare fields) Z[J] = [Dϕ]e i(s[ϕ]+ d 4 yj a (y)ϕ a (y)). Change the integration variable: ϕ a ϕ a (x) = ϕ a (x) + δϕ a (x), the integral is invariant. Assume [Dϕ] = [Dϕ ]. Just like d(x + a) = dx. 0 = δz[ϕ] = [Dϕ ]e i(s[ϕ ]+ d 4 yj a (y)ϕ a (y)) ( ) = d 4 x [Dϕ]e i(s[ϕ]+ d 4 yj a (y)ϕ a (y)) δs i δϕ a (x) + ij(x) δϕ a (x) [Dϕ]e i(s[ϕ]+ d 4 yj a (y)ϕ a (y)) Take n functional derivative w.r.t ij aj (x j), set J = 0 [ 0 = [Dϕ]e is d 4 δs n x i δϕ a ϕ ai (x i ) (x) i=1 n + ϕ ai (x 1 )... δ a,aj δ 4 (x x j )... ϕ an (x n) δϕ a(x). j=1 Since δϕ(x) is arbitrary, 0 = Ω it δs δϕ a (x) + n ϕ ai (x i ) Ω i=1 n Ω Tϕ ai (x 1 )... δ a,aj δ 4 (x x j )... ϕ an (x n) Ω j=1
23 Equations For Green s Functions Schwinger-Dyson Eq. Schwinger-Dyson Eq. 0 = Ω it δs δϕ a (x) + n ϕ ai (x i ) Ω i=1 n Ω Tϕ ai (x 1 )... δ a,aj δ 4 (x x j )... ϕ an (x n) Ω j=1 δs δϕ(x) = 0, the classical EOM. Free scalar: ( 2 x + m 2 )ϕ = 0 Two point function, scalar: i( 2 x + m 2 ) Ω T ϕ(x)ϕ(y) ω = δ 4 (x y) For x x i, Ω it δs n ϕ δϕ a ai (x i) Ω = 0. (x) For more detailed discussion, see Itzykson and Zuber,Quantum field theory, or Arxiv: , Swanson, A primer on functional methods and the Schwinger-Dyson Equation. i=1
24 Global symmetry and Ward Id. Noether theorem: Continuous symmetry conserved current. Classical global continuous symmetry: ϕ a ϕ a (x) = ϕ a (x) + iα ϕ a (x), α : an infinitesimal constant. L[ϕ (x)] = L[ϕ(x)] + iα µk µ If α α(x), depends on x, only nonzero in a finite region δl = L[ϕ (x)] L[ϕ(x)] = L L ϕ a (x) iα(x) ϕa (x) + ( µϕ a (x)) i(α(x) µ ϕa (x) + ( µα(x)) ϕ a (x))) = iα(x) µk µ L + i( µα(x)) ( µϕ a (x)) ϕa (x) = i µα(x) j µ + iα(x) µk µ For ϕ satisfying EOM 0 = δs[ϕ] = i d 4 x µα(x) j µ + α(x) µk µ (integrate by part) = i d 4 x α(x) µ( j µ + K µ ) = d 4 xα(x) µj µ Since α(x) is arbitrary, define J µ = i(j µ K µ ), we have conserved current: µj µ = 0. Current conservation is a local property: α(x) can be non-zero in any arbitrary small region. Charge conservation is a global property.
25 Global symmetry and Ward Id. Consider correlation function: n i=1 ϕ(xi) = 1 Z [Dϕ] n i=1 ϕ(xi)eis[ϕ]. A global symmetric trans: ϕ a ϕ a (x) = ϕ a (x) + iα ϕ a (x), S[ϕ] = S[ϕ ]. α α(x), S[ϕ ] = S[ϕ] d 4 xα(x) µj µ. n n 0 = [Dϕ ] ϕ (x i )e is[ϕ ] [Dϕ] ϕ(x i )e is[ϕ] i=1 ( n = [Dϕ] ϕ (x i )e is[ϕ ] i=1 i=1 ) n ϕ(x i )e is[ϕ] i=1 i=1 ( n n ) = [Dϕ] ϕ(x 1 )... iα(x i ) ϕ(x i )... ϕ(x n) i d 4 xα(x) µj µ ϕ(x i ) e is[ϕ] i µ TJ µ (x) n ϕ(x i) = i=1 In momentum space: ϕ(x) = k µ TJ µ (k) i=1 n Tϕ(x 1)... δ 4 (x x i)(i ϕ(x i))... ϕ(x n) i=1 n ϕ(k i) = i=1 d 4 k (2π) 4 e ik x φ(k). (outgoing) n Tϕ(k 1)... i ϕ(k i + k)... ϕ(k n) i=1 For S-Matrix: from LSZ, for each external leg, lim p 2 m 2 p2 m 2 +iϵ Zi, there is no on-shell pole for leg ϕ i(k + k i), the right hand side 0, k µm µ = 0
26 Global symmetry and Ward Id. QED: consider U(1) symmetry ψ e ieα ψ, J µ = e ψγ µ ψ, correlation funtion J µ (k)ψ( (p + k)) ψ(p), Ward Id: k µ J µ (k)ψ( (p + k)) ψ(p) ( =ie ψ( p) ψ(p) ψ( (p + k)) ψ(p ) + k) where S(p + k)[ ik µγ µ (p + k, p)]s(p) = S(p) S(p + k), S(p) = ψ( p) ψ(p) = i p/ m Σ(p). ik µγ µ (p + k, p) = S 1 (p + k) S 1 (p). Γ µ (p + k, p) Z 1 1 γµ as k µ 0, S(p) i Z 2 p/ m, at around p2 = m 2, k 0 iz 1 1 k/ = iz 1 2 k/ Z 1 = Z 2 Note: ( ψγ µ ψ)ψ( (p + k)) ψ(p) γ µ αβ ψ( (p + k)) ψ αψ β ψ(p), no minus sign
27 Local gauge symmetry and Ward Id. in generating functional formulation L = 1 4 (F i 0,µν) 2 + i( ψ 0 /ψ 0 + ie 0 q 1 ψ0 A/ 0,µ ψ 0 ) m ψ,0 ψ0 ψ 0 1 2ξ 0 ( µa µ 0 )2 = Z (F i µν )2 + iz ψ 2 ( ψ /ψ + ieq 1 Z ψ 1 Z ψ 2 e 0 = e Z 1 = e Z 2 Z 1/2 3 Z 1/2 3. Only depends on Z 3. ψa/ψ) mz ψ 0 ψψ 1 2ξ ( µaµ ) 2 Bare gauge trans: ψ 0 e iαq 1 ψ 0, A 0,µ A 0,µ 1 e 0 µα. Renormalized gauge trans: ψ e iαq 1 ψ, A µ A µ 1 µα. ē = Zψ 1 e = e ē Z ψ 2 Consistent with A 0,µ = Z 1/2 3 A µ. the gauge fixing term is not invariant. We will see that there is no counterterm for gauge fixing ξ term.
28 iqeᾱ δz δz iqe iδᾱ iδα α µ j µz 1 2 µ δz ξ 0 iδj µ = 0 Local gauge symmetry and Ward Id. in generating functional formulation Adding source term for bare fields: (omit the subscript 0.) (α, ᾱ : grassman variables.) L[A, j, α, ᾱ] = L + ᾱψ + ψα + j µa µ Under gauge transformations A A = A µ + 1 µϵ, ψ e ψ = e iϵq 1 ψ: source terms and gauge fixing terms not invariant, all others including the path integral measure are inv. Z[A, ψ; j, α, ᾱ] =N [dad ψdψ] exp{is[a, ψ, ψ] + i (ᾱψ + ψα + j µa µ )} (rename ) =N (Gauge inv) =N 0 = δz =N [dad ψdψ] =N [dad ψdψ] [da d ψ { dψ ] exp [dad ψdψ] { exp i S[A, ψ, ψ] + i 1 µa µ νa ν} 2ξ 0 is[a, ψ, ψ ] + i (ᾱψ + ψ α + j µa µ ) (ᾱψ + ψ } α + j µa µ ) ( i ᾱδψ + δ ψα + j µδa µ 1 µa µ νδa ν) exp{is +... } ξ 0 ( dx ϵ(x) iqᾱψ + iq ψα 1 e µ j µ 1 2 µa µ) exp{... } eξ 0 iqeᾱψ + iqe ψα µ j µ 1 ξ 0 2 µa µ j,α,ᾱ = 0
29 Local gauge symmetry and Ward Id. in generating functional formulation iqeᾱ δz δz iqe iδᾱ iδα α µ j µz 1 2 µ δz ξ 0 iδj = 0 µ In connected Green Functional iqeᾱ δw δᾱ In 1PI generating functional: + iqe δw δα α + µ j µ + 1 ξ 0 2 µ δw δj µ = 0 iqe δγ ψ c + iqe δψ ψ δγ c c δ ψ + µ c where Γ = W j µa µ ᾱψ ψα, A µ c = δw δj µ, j µ = δγ δa cµ, δγ δa µ c δw ψc = δᾱ, α = δγ δ ψ c, + 1 ξ 0 2 µ A cµ = 0 ψc = δw δα ; ᾱ = δγ δψ c.
30 Local gauge symmetry and Ward Id. in generating functional formulation δγ iqe δψ ψc(z) + iqe ψ δγ c(z) c(z) δ ψ + δγ c(z) µ δa µ c (z) µ A cµ(z) = 0 ξ 0 δ ψ c(x) ψ c(y), Aµ c, ψ c, ψ c 0,: δγ δγ δγ iqe δψ c(x)δ ψ δ(x z) + iqe c(y) δψ c(x)δ ψ δ(y z) + µ c(y) δa µ c (z)δψ c(x)δ ψ = 0 c(y) iqe( is 1 (x, y)δ(x z) + is 1 (x, y)δ(y z)) + iqe µ z Γ µ(z, x, y) = 0 ik µγ µ (p + k, p) = S 1 (p + k) S 1 (p)
31 Local gauge symmetry and Ward Id. in generating functional formulation eg: δ δj ν (z) iqeᾱ δw δᾱ α,ᾱ,j µ 0 + iqe δw δα α + µ j µ + 1 ξ 0 2 µ δw δj µ = 0 : x µ (g µνδ(x y)) µ δw x = 0 ξ 0 δj µ (x)δj ν (y) j µ =0 δw =i TAµ(x)Aν(y) = δj µ (x)δj ν ig(2) (y) We then have: =i d 4 k µν (x y) = i e ik (x y) G (2) (2π) 4 [ d 4 k ( ) e ik (x y) g (2π) 4 µν kµkν A(k 2 ) + B(k 2 ) kµkν k 2 k 2 ik ν + ib(k 2 ) ikνk2 G (2) µν (k) = ξ 0 = 0 B(k 2 ) = iξ0 k 2 ( ) g µν kµkν A(k 2 ) + iξ0 k 2 k 2 k µk ν k 2 ( ) Notice: Free photon propagator: G (2) µν = i g k 2 µν kµkν iξ 0k µk ν. k 2 k 4 µν (k) ] The ξ 0 term is not renormalized. Renormalized propagator G R(2) µν = Z 1 3 G(2) µν, ξ = ξ 0Z 1 3.
32 the Legendre transformation and IPI Consider the action: S ϵ(ϕ) = 1 2 dx dyϕ(x) [ K(x, y) + ϵ ] ϕ(y) + V (ϕ) ϵ: a small parameter, expanded to the first order Propagator ϵ(x, y) : ϵ(x, z) [ K(z, y) + ϵ ] dz = δ(x y) ϵ(x, y) = (x, y) ϵη(x)η(y) + O(ϵ 2 ), η(x) = (x, z)dz Expand the Feynman diagrams w.r.t ϵ, (using ϵ(x, y)): the ϵ 1 order terms, replace (x, y) η(x)η(y), cut one internal propagator. Higher order term: irreducible when cut 2,3,...,lines. 1PI ϵ 1 term is connected
33 the Legendre transformation and IPI The generating functional Z ϵ[j] = [dϕ] exp{i(s ϵ + Jϕ)} = (1 + i 1 2 ϵ δ δ ) dx dy [dϕ] exp{i(s + Jϕ)} + O(ϵ 2 ) iδj(x) iδj(y) = (1 + i 1 2 ϵ δ δ ) dx dy exp{iw [J]} + O(ϵ 2 ) iδj(x) iδj(y) ( = 1 + i 1 ( 2 ϵ δ δ δ dx dx dy ( =Z[J] 1 + i 1 ( 2 ϵ dx δj(x) W [J] ) 2 + i 2 ϵ δ ) 2 δj(x) W [J] i + 2 ϵ iδj(x) iδj(y) δ δ dx dy iδj(x) iδj(y) ) iw [J] exp{iw [J]} ) iw [J] + O(ϵ 2 ) W ϵ[j] =W [J] + i 1 ( 2 ϵ dx δ ) 2 δj(x) W [J] i + 2 ϵ δ δ dx dy iw [J] iδj(x) iδj(y) The second term just means the disconnectness of the W [J] after cutting a propagator.
34 the Legendre transformation and IPI Legendre transformation: Γ ϵ[φ cl ] W ϵ[j[φ cl ]] d 4 xj[φ cl ]φ cl (x), φ cl (x) = Ω Φ(x) Ω J Ω Ω J = δwϵ[j] δj(x). Γ ϵ ϵ = dx J(x) φcl ϵ φ cl (x) + Wϵ φcl ϵ + J = Wϵ ϵ J dx δwϵ δj(x) J(x) ϵ φcl Thus, Γ ϵ[φ cl ] =Γ[φ cl ] + iϵ 2 [ ( ) 2 dxφ cl (x) + ( δ 2 iγ[j] ) 1 dx dy + O(ϵ )] 2 δφ cl (x)δφ cl (y) The first term in the ϵ term is just from the term ϵϕ 2 we added in the action and the second term contains just connected diagrams. Thus, Γ[φ cl ] is 1PI.
6.1. Dirac Equation. Hamiltonian. Dirac Eq.
6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2
8.324 Relativistic Quantum Field Theory II
8.324 Relativistic Quantum Field Theory II MIT OpenCourseWare Lecture Notes Hong Liu, Fall 200 Lecture 2 3: GENERAL ASPECTS OF QUANTUM ELECTRODYNAMICS 3.: RENORMALIZED LAGRANGIAN Consider the Lagrangian
Space-Time Symmetries
Chapter Space-Time Symmetries In classical fiel theory any continuous symmetry of the action generates a conserve current by Noether's proceure. If the Lagrangian is not invariant but only shifts by a
Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)
Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts
Srednicki Chapter 55
Srednicki Chapter 55 QFT Problems & Solutions A. George August 3, 03 Srednicki 55.. Use equations 55.3-55.0 and A i, A j ] = Π i, Π j ] = 0 (at equal times) to verify equations 55.-55.3. This is our third
Example Sheet 3 Solutions
Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note
Higher Derivative Gravity Theories
Higher Derivative Gravity Theories Black Holes in AdS space-times James Mashiyane Supervisor: Prof Kevin Goldstein University of the Witwatersrand Second Mandelstam, 20 January 2018 James Mashiyane WITS)
MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)
1. MATH43 String Theory Solutions 4 x = 0 τ = fs). 1) = = f s) ) x = x [f s)] + f s) 3) equation of motion is x = 0 if an only if f s) = 0 i.e. fs) = As + B with A, B constants. i.e. allowe reparametrisations
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 1 State vector space and the dual space Space of wavefunctions The space of wavefunctions is the set of all
2 Composition. Invertible Mappings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,
forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with
Week 03: C lassification of S econd- Order L inear Equations In last week s lectures we have illustrated how to obtain the general solutions of first order PDEs using the method of characteristics. We
Partial Differential Equations in Biology The boundary element method. March 26, 2013
The boundary element method March 26, 203 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet
EE512: Error Control Coding
EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3
3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β
3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle
Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.
Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action
SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions
SCHOOL OF MATHEMATICAL SCIENCES GLMA Linear Mathematics 00- Examination Solutions. (a) i. ( + 5i)( i) = (6 + 5) + (5 )i = + i. Real part is, imaginary part is. (b) ii. + 5i i ( + 5i)( + i) = ( i)( + i)
Areas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
The Simply Typed Lambda Calculus
Type Inference Instead of writing type annotations, can we use an algorithm to infer what the type annotations should be? That depends on the type system. For simple type systems the answer is yes, and
D Alembert s Solution to the Wave Equation
D Alembert s Solution to the Wave Equation MATH 467 Partial Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Objectives In this lesson we will learn: a change of variable technique
1 Classical Mechanics
From Classical to Quantum Field Theory 1 D. E. Soper 2 University of Oregon Physics 665, Quantum Field Theory 13 October 2010 1 Classical Mechanics Let φ J (t), J = 1, 2, 3, be the position of a particle
derivation of the Laplacian from rectangular to spherical coordinates
derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used
Areas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
ECE Spring Prof. David R. Jackson ECE Dept. Notes 2
ECE 634 Spring 6 Prof. David R. Jackson ECE Dept. Notes Fields in a Source-Free Region Example: Radiation from an aperture y PEC E t x Aperture Assume the following choice of vector potentials: A F = =
Second Order Partial Differential Equations
Chapter 7 Second Order Partial Differential Equations 7.1 Introduction A second order linear PDE in two independent variables (x, y Ω can be written as A(x, y u x + B(x, y u xy + C(x, y u u u + D(x, y
Variational Wavefunction for the Helium Atom
Technische Universität Graz Institut für Festkörperphysik Student project Variational Wavefunction for the Helium Atom Molecular and Solid State Physics 53. submitted on: 3. November 9 by: Markus Krammer
b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!
MTH U341 urface Integrals, tokes theorem, the divergence theorem To be turned in Wed., Dec. 1. 1. Let be the sphere of radius a, x 2 + y 2 + z 2 a 2. a. Use spherical coordinates (with ρ a) to parametrize.
HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:
HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying
CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS
CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =
On the Galois Group of Linear Difference-Differential Equations
On the Galois Group of Linear Difference-Differential Equations Ruyong Feng KLMM, Chinese Academy of Sciences, China Ruyong Feng (KLMM, CAS) Galois Group 1 / 19 Contents 1 Basic Notations and Concepts
Geodesic Equations for the Wormhole Metric
Geodesic Equations for the Wormhole Metric Dr R Herman Physics & Physical Oceanography, UNCW February 14, 2018 The Wormhole Metric Morris and Thorne wormhole metric: [M S Morris, K S Thorne, Wormholes
Section 8.3 Trigonometric Equations
99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.
Solutions to Exercise Sheet 5
Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X
= {{D α, D α }, D α }. = [D α, 4iσ µ α α D α µ ] = 4iσ µ α α [Dα, D α ] µ.
PHY 396 T: SUSY Solutions for problem set #1. Problem 2(a): First of all, [D α, D 2 D α D α ] = {D α, D α }D α D α {D α, D α } = {D α, D α }D α + D α {D α, D α } (S.1) = {{D α, D α }, D α }. Second, {D
Statistical Inference I Locally most powerful tests
Statistical Inference I Locally most powerful tests Shirsendu Mukherjee Department of Statistics, Asutosh College, Kolkata, India. shirsendu st@yahoo.co.in So far we have treated the testing of one-sided
4.6 Autoregressive Moving Average Model ARMA(1,1)
84 CHAPTER 4. STATIONARY TS MODELS 4.6 Autoregressive Moving Average Model ARMA(,) This section is an introduction to a wide class of models ARMA(p,q) which we will consider in more detail later in this
Homework 8 Model Solution Section
MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx
Finite Field Problems: Solutions
Finite Field Problems: Solutions 1. Let f = x 2 +1 Z 11 [x] and let F = Z 11 [x]/(f), a field. Let Solution: F =11 2 = 121, so F = 121 1 = 120. The possible orders are the divisors of 120. Solution: The
ST5224: Advanced Statistical Theory II
ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known
ω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω
0 1 2 3 4 5 6 ω ω + 1 ω + 2 ω + 3 ω + 4 ω2 ω2 + 1 ω2 + 2 ω2 + 3 ω3 ω3 + 1 ω3 + 2 ω4 ω4 + 1 ω5 ω 2 ω 2 + 1 ω 2 + 2 ω 2 + ω ω 2 + ω + 1 ω 2 + ω2 ω 2 2 ω 2 2 + 1 ω 2 2 + ω ω 2 3 ω 3 ω 3 + 1 ω 3 + ω ω 3 +
( y) Partial Differential Equations
Partial Dierential Equations Linear P.D.Es. contains no owers roducts o the deendent variables / an o its derivatives can occasionall be solved. Consider eamle ( ) a (sometimes written as a ) we can integrate
SPECIAL FUNCTIONS and POLYNOMIALS
SPECIAL FUNCTIONS and POLYNOMIALS Gerard t Hooft Stefan Nobbenhuis Institute for Theoretical Physics Utrecht University, Leuvenlaan 4 3584 CC Utrecht, the Netherlands and Spinoza Institute Postbox 8.195
DiracDelta. Notations. Primary definition. Specific values. General characteristics. Traditional name. Traditional notation
DiracDelta Notations Traditional name Dirac delta function Traditional notation x Mathematica StandardForm notation DiracDeltax Primary definition 4.03.02.000.0 x Π lim ε ; x ε0 x 2 2 ε Specific values
Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1
Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test
Homework 3 Solutions
Homework 3 s Free graviton Hamiltonian Show that the free graviton action we discussed in class (before making it gauge- and Lorentzinvariant), S 0 = α d 4 x µ h ij µ h ij, () yields the correct free Hamiltonian
Uniform Convergence of Fourier Series Michael Taylor
Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula
Math221: HW# 1 solutions
Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin
Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)
Phys624 Quantization of Scalar Fields II Homework 3. Homework 3 Solutions. 3.1: U(1) symmetry for complex scalar
Homework 3 Solutions 3.1: U(1) symmetry for complex scalar 1 3.: Two complex scalars The Lagrangian for two complex scalar fields is given by, L µ φ 1 µ φ 1 m φ 1φ 1 + µ φ µ φ m φ φ (1) This can be written
The kinetic and potential energies as T = 1 2. (m i η2 i k(η i+1 η i ) 2 ). (3) The Hooke s law F = Y ξ, (6) with a discrete analog
Lecture 12: Introduction to Analytical Mechanics of Continuous Systems Lagrangian Density for Continuous Systems The kinetic and potential energies as T = 1 2 i η2 i (1 and V = 1 2 i+1 η i 2, i (2 where
Tridiagonal matrices. Gérard MEURANT. October, 2008
Tridiagonal matrices Gérard MEURANT October, 2008 1 Similarity 2 Cholesy factorizations 3 Eigenvalues 4 Inverse Similarity Let α 1 ω 1 β 1 α 2 ω 2 T =......... β 2 α 1 ω 1 β 1 α and β i ω i, i = 1,...,
DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C
DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C By Tom Irvine Email: tomirvine@aol.com August 6, 8 Introduction The obective is to derive a Miles equation which gives the overall response
Chapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?
Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 3 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC DIFFERENTIAL CALCULUS) COLLECTION # Question Type A.Single Correct Type Q. (A) Sol least
Physics 582, Problem Set 2 Solutions
Physics 582, Problem Set 2 Solutions TAs: Hart Goldman and Ramanjit Sohal Fall 2018 Symmetries and Conservation Laws In this problem set we return to a study of scalar electrodynamics which has the Lagrangian
C.S. 430 Assignment 6, Sample Solutions
C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order
1 String with massive end-points
1 String with massive end-points Πρόβλημα 5.11:Θεωρείστε μια χορδή μήκους, τάσης T, με δύο σημειακά σωματίδια στα άκρα της, το ένα μάζας m, και το άλλο μάζας m. α) Μελετώντας την κίνηση των άκρων βρείτε
Differential equations
Differential equations Differential equations: An equation inoling one dependent ariable and its deriaties w. r. t one or more independent ariables is called a differential equation. Order of differential
DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.
DESIGN OF MACHINERY SOLUTION MANUAL -7-1! PROBLEM -7 Statement: Design a double-dwell cam to move a follower from to 25 6, dwell for 12, fall 25 and dwell for the remader The total cycle must take 4 sec
Integrals in cylindrical, spherical coordinates (Sect. 15.7)
Integrals in clindrical, spherical coordinates (Sect. 5.7 Integration in spherical coordinates. Review: Clindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
Every set of first-order formulas is equivalent to an independent set
Every set of first-order formulas is equivalent to an independent set May 6, 2008 Abstract A set of first-order formulas, whatever the cardinality of the set of symbols, is equivalent to an independent
Non-Abelian Gauge Fields
Chapter 5 Non-Abelian Gauge Fields The simplest example starts with two Fermions Dirac particles) ψ 1, ψ 2, degenerate in mass, and hence satisfying in the absence of interactions γ 1 i + m)ψ 1 = 0, γ
Section 9.2 Polar Equations and Graphs
180 Section 9. Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify
Inverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin
Symmetric Stress-Energy Tensor
Chapter 3 Symmetric Stress-Energy ensor We noticed that Noether s conserved currents are arbitrary up to the addition of a divergence-less field. Exploiting this freedom the canonical stress-energy tensor
Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.
Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given
Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013
Notes on Average Scattering imes and Hall Factors Jesse Maassen and Mar Lundstrom Purdue University November 5, 13 I. Introduction 1 II. Solution of the BE 1 III. Exercises: Woring out average scattering
Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequality for metrics: Let (X, d) be a metric space and let x, y, z X.
Math 446 Homework 3 Solutions. (1). (i): Reverse triangle inequalit for metrics: Let (X, d) be a metric space and let x,, z X. Prove that d(x, z) d(, z) d(x, ). (ii): Reverse triangle inequalit for norms:
Matrices and Determinants
Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z
Other Test Constructions: Likelihood Ratio & Bayes Tests
Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :
Lecture 10 - Representation Theory III: Theory of Weights
Lecture 10 - Representation Theory III: Theory of Weights February 18, 2012 1 Terminology One assumes a base = {α i } i has been chosen. Then a weight Λ with non-negative integral Dynkin coefficients Λ
Section 7.6 Double and Half Angle Formulas
09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)
Homework 3 Solutions
Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For
Concrete Mathematics Exercises from 30 September 2016
Concrete Mathematics Exercises from 30 September 2016 Silvio Capobianco Exercise 1.7 Let H(n) = J(n + 1) J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2) J(2n+1) = (2J(n+1) 1) (2J(n)+1)
Reminders: linear functions
Reminders: linear functions Let U and V be vector spaces over the same field F. Definition A function f : U V is linear if for every u 1, u 2 U, f (u 1 + u 2 ) = f (u 1 ) + f (u 2 ), and for every u U
Lecture 13 - Root Space Decomposition II
Lecture 13 - Root Space Decomposition II October 18, 2012 1 Review First let us recall the situation. Let g be a simple algebra, with maximal toral subalgebra h (which we are calling a CSA, or Cartan Subalgebra).
12. Radon-Nikodym Theorem
Tutorial 12: Radon-Nikodym Theorem 1 12. Radon-Nikodym Theorem In the following, (Ω, F) is an arbitrary measurable space. Definition 96 Let μ and ν be two (possibly complex) measures on (Ω, F). We say
Fractional Colorings and Zykov Products of graphs
Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is
The challenges of non-stable predicates
The challenges of non-stable predicates Consider a non-stable predicate Φ encoding, say, a safety property. We want to determine whether Φ holds for our program. The challenges of non-stable predicates
Notes on the Open Economy
Notes on the Open Econom Ben J. Heijdra Universit of Groningen April 24 Introduction In this note we stud the two-countr model of Table.4 in more detail. restated here for convenience. The model is Table.4.
ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Αν κάπου κάνετε κάποιες υποθέσεις να αναφερθούν στη σχετική ερώτηση. Όλα τα αρχεία που αναφέρονται στα προβλήματα βρίσκονται στον ίδιο φάκελο με το εκτελέσιμο
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in : tail in X, head in A nowhere-zero Γ-flow is a Γ-circulation such that
Approximation of distance between locations on earth given by latitude and longitude
Approximation of distance between locations on earth given by latitude and longitude Jan Behrens 2012-12-31 In this paper we shall provide a method to approximate distances between two points on earth
The Probabilistic Method - Probabilistic Techniques. Lecture 7: The Janson Inequality
The Probabilistic Method - Probabilistic Techniques Lecture 7: The Janson Inequality Sotiris Nikoletseas Associate Professor Computer Engineering and Informatics Department 2014-2015 Sotiris Nikoletseas,
Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.
Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 +
SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM
SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM Solutions to Question 1 a) The cumulative distribution function of T conditional on N n is Pr T t N n) Pr max X 1,..., X N ) t N n) Pr max
k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +
Chapter 3. Fuzzy Arithmetic 3- Fuzzy arithmetic: ~Addition(+) and subtraction (-): Let A = [a and B = [b, b in R If x [a and y [b, b than x+y [a +b +b Symbolically,we write A(+)B = [a (+)[b, b = [a +b
Laplace Expansion. Peter McCullagh. WHOA-PSI, St Louis August, Department of Statistics University of Chicago
Laplace Expansion Peter McCullagh Department of Statistics University of Chicago WHOA-PSI, St Louis August, 2017 Outline Laplace approximation in 1D Laplace expansion in 1D Laplace expansion in R p Formal
6.3 Forecasting ARMA processes
122 CHAPTER 6. ARMA MODELS 6.3 Forecasting ARMA processes The purpose of forecasting is to predict future values of a TS based on the data collected to the present. In this section we will discuss a linear
2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.
EAMCET-. THEORY OF EQUATIONS PREVIOUS EAMCET Bits. Each of the roots of the equation x 6x + 6x 5= are increased by k so that the new transformed equation does not contain term. Then k =... - 4. - Sol.
Solutions to the Schrodinger equation atomic orbitals. Ψ 1 s Ψ 2 s Ψ 2 px Ψ 2 py Ψ 2 pz
Solutions to the Schrodinger equation atomic orbitals Ψ 1 s Ψ 2 s Ψ 2 px Ψ 2 py Ψ 2 pz ybridization Valence Bond Approach to bonding sp 3 (Ψ 2 s + Ψ 2 px + Ψ 2 py + Ψ 2 pz) sp 2 (Ψ 2 s + Ψ 2 px + Ψ 2 py)
CRASH COURSE IN PRECALCULUS
CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 01, Brooks/Cole Chapter
Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee
Appendi to On the stability of a compressible aisymmetric rotating flow in a pipe By Z. Rusak & J. H. Lee Journal of Fluid Mechanics, vol. 5 4, pp. 5 4 This material has not been copy-edited or typeset
SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM
SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM Solutions to Question 1 a) The cumulative distribution function of T conditional on N n is Pr (T t N n) Pr (max (X 1,..., X N ) t N n) Pr (max
Parametrized Surfaces
Parametrized Surfaces Recall from our unit on vector-valued functions at the beginning of the semester that an R 3 -valued function c(t) in one parameter is a mapping of the form c : I R 3 where I is some
Bessel functions. ν + 1 ; 1 = 0 for k = 0, 1, 2,..., n 1. Γ( n + k + 1) = ( 1) n J n (z). Γ(n + k + 1) k!
Bessel functions The Bessel function J ν (z of the first kind of order ν is defined by J ν (z ( (z/ν ν Γ(ν + F ν + ; z 4 ( k k ( Γ(ν + k + k! For ν this is a solution of the Bessel differential equation
Congruence Classes of Invertible Matrices of Order 3 over F 2
International Journal of Algebra, Vol. 8, 24, no. 5, 239-246 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/.2988/ija.24.422 Congruence Classes of Invertible Matrices of Order 3 over F 2 Ligong An and
Symmetries and Feynman Rules for Ramond Sector in WZW-type Superstring Field Theories
Symmetries and Feynman Rules for Ramond Sector in WZW-type Superstring Field Theories Hiroshi Kunitomo YITP 215/3/5 @Nara Women s U. Nara arxiv:1412.5281, 153.xxxx Y TP YUKAWA INSTITUTE FOR THEORETICAL
Lecture 34 Bootstrap confidence intervals
Lecture 34 Bootstrap confidence intervals Confidence Intervals θ: an unknown parameter of interest We want to find limits θ and θ such that Gt = P nˆθ θ t If G 1 1 α is known, then P θ θ = P θ θ = 1 α
Instruction Execution Times
1 C Execution Times InThisAppendix... Introduction DL330 Execution Times DL330P Execution Times DL340 Execution Times C-2 Execution Times Introduction Data Registers This appendix contains several tables
Math 6 SL Probability Distributions Practice Test Mark Scheme
Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry