Gaussian integrals and Feynman diagrams. February 28

Transcription

1 Gaussian integrals and Feynman diagrams February 28

2 Introduction A mathematician is one to whom the equality e x2 2 dx = 2π is as obvious as that twice two makes four is to you. Lord W.T. Kelvin to his students

3 Introduction A mathematician is one to whom the equality e x2 2 dx = 2π is as obvious as that twice two makes four is to you. A physicist must be one to whom the formula Lord W.T. Kelvin to his students is equally obvious.

4 Introduction Generalizing P (x)e x2 2 dx e x2 2 dx = 2π: P (x)e x,ax 2 dx Here, A is a positively-defined R n symmetric matrix and, is the standard inner product on R n. P (x)e x,ax 2 r 3 r/2 1 1 r! Br(x,...,x) dx, where is a R n parameter.

5 Introduction Generalizing P (x)e x2 2 dx e x2 2 dx = 2π: P (x)e x,ax 2 dx Here, A is a positively-defined R n symmetric matrix and, is the standard inner product on R n. P (x)e x,ax 2 r 3 r/2 1 1 r! Br(x,...,x) dx, where is a R n parameter.

6 Introduction Generalizing P (x)e x2 2 dx e x2 2 dx = 2π: P (x)e x,ax 2 dx Here, A is a positively-defined R n symmetric matrix and, is the standard inner product on R n. P (x)e x,ax 2 r 3 r/2 1 1 r! Br(x,...,x) dx, where is a R n parameter.

7 Motivation In classical mechanics, the motion of a point-particle of mass m in a potential field V = V (x) is determined by the Newton s equation mx = V (x). Alternatively, one can state the law of motion in the form of the stationary action principle. One starts by introducting the Lagrangian of the system L(x, x, t) = (kinetic energy) (potential energy) and the action functional S[x] = b a L(x, x, t) dt. Stationary action principle The trajectory of a particle x = x(t) (t [a, b]) has to be an extremum of the action functional S.

8 Motivation In classical mechanics, the motion of a point-particle of mass m in a potential field V = V (x) is determined by the Newton s equation mx = V (x). Alternatively, one can state the law of motion in the form of the stationary action principle. One starts by introducting the Lagrangian of the system L(x, x, t) = (kinetic energy) (potential energy) and the action functional S[x] = b a L(x, x, t) dt. Stationary action principle The trajectory of a particle x = x(t) (t [a, b]) has to be an extremum of the action functional S.

9 Motivation In classical mechanics, the motion of a point-particle of mass m in a potential field V = V (x) is determined by the Newton s equation mx = V (x). Alternatively, one can state the law of motion in the form of the stationary action principle. One starts by introducting the Lagrangian of the system L(x, x, t) = (kinetic energy) (potential energy) and the action functional S[x] = b a L(x, x, t) dt. Stationary action principle The trajectory of a particle x = x(t) (t [a, b]) has to be an extremum of the action functional S.

10 Motivation In classical mechanics, the motion of a point-particle of mass m in a potential field V = V (x) is determined by the Newton s equation mx = V (x). Alternatively, one can state the law of motion in the form of the stationary action principle. One starts by introducting the Lagrangian of the system L(x, x, t) = (kinetic energy) (potential energy) and the action functional S[x] = b a L(x, x, t) dt. Stationary action principle The trajectory of a particle x = x(t) (t [a, b]) has to be an extremum of the action functional S.

11 Motivation Stationary action principle The trajectory of a particle x = x(t) (t [a, b]) has to be an extremum of the action functional S. This means that such a trajectory has to be a solution of the variational problem δs = 0. The latter can be reduced to solving the Euler-Lagrange equation (or rather a system of EL equations). In the one-dimensional case it has the form L x ( ) L t x = 0, x(a) = x 0, x(b) = x 1. Example The Lagranian for a mass on a spring system is L = mx 2 2 kx2 2. The EL equation reads: kx t mx = 0. That is, mx = kx (with some boundary conditions).

12 Motivation Stationary action principle The trajectory of a particle x = x(t) (t [a, b]) has to be an extremum of the action functional S. This means that such a trajectory has to be a solution of the variational problem δs = 0. The latter can be reduced to solving the Euler-Lagrange equation (or rather a system of EL equations). In the one-dimensional case it has the form L x ( ) L t x = 0, x(a) = x 0, x(b) = x 1. Example The Lagranian for a mass on a spring system is L = mx 2 2 kx2 2. The EL equation reads: kx t mx = 0. That is, mx = kx (with some boundary conditions).

13 Motivation Stationary action principle The trajectory of a particle x = x(t) (t [a, b]) has to be an extremum of the action functional S. This means that such a trajectory has to be a solution of the variational problem δs = 0. The latter can be reduced to solving the Euler-Lagrange equation (or rather a system of EL equations). In the one-dimensional case it has the form L x ( ) L t x = 0, x(a) = x 0, x(b) = x 1. Example The Lagranian for a mass on a spring system is L = mx 2 2 kx2 2. The EL equation reads: kx t mx = 0. That is, mx = kx (with some boundary conditions).

14 Motivation Remark The stationary action principle is often referred to as the principle of the least action, but, in general, classical trajectories do not have to minimize the action. In classical field theory one studies not the motion of point-particles, but rather motion of a continuum of particles (e.g. a string, a membrane, a jet of fluid etc.) Now, the trajectory of a system is described by a classical field φ = φ(x, t). The stationary action principle still holds: one starts with the Lagrangian L = L(φ,... ) and looks for φ s that give the extremum to the action functional S[φ] = D L(φ,... ) dx dt. Example The Lagrangian[ for a string has the form L(z, z x, z t) = 1 2 m ( ) z 2 ( t λ z ) 2 ] x. The EL equation for the Gaussian integrals 2 z and Feynman λ 2 z diagrams

15 Motivation In classical field theory one studies not the motion of point-particles, but rather motion of a continuum of particles (e.g. a string, a membrane, a jet of fluid etc.) Now, the trajectory of a system is described by a classical field φ = φ(x, t). The stationary action principle still holds: one starts with the Lagrangian L = L(φ,... ) and looks for φ s that give the extremum to the action functional S[φ] = D L(φ,... ) dx dt. Example The Lagrangian[ for a string has the form L(z, z x, z t) = 1 2 m ( ) z 2 ( t λ z ) 2 ] x. The EL equation for the variational problem δs = 0 takes the form 2 z λ 2 z t 2 m = 0. x 2

16 Motivation In classical field theory one studies not the motion of point-particles, but rather motion of a continuum of particles (e.g. a string, a membrane, a jet of fluid etc.) Now, the trajectory of a system is described by a classical field φ = φ(x, t). The stationary action principle still holds: one starts with the Lagrangian L = L(φ,... ) and looks for φ s that give the extremum to the action functional S[φ] = D L(φ,... ) dx dt. Example The Lagrangian[ for a string has the form L(z, z x, z t) = 1 2 m ( ) z 2 ( t λ z ) 2 ] x. The EL equation for the variational problem δs = 0 takes the form 2 z λ 2 z t 2 m = 0. x 2

17 Motivation In classical field theory one studies not the motion of point-particles, but rather motion of a continuum of particles (e.g. a string, a membrane, a jet of fluid etc.) Now, the trajectory of a system is described by a classical field φ = φ(x, t). The stationary action principle still holds: one starts with the Lagrangian L = L(φ,... ) and looks for φ s that give the extremum to the action functional S[φ] = D L(φ,... ) dx dt. Example The Lagrangian[ for a string has the form L(z, z x, z t) = 1 2 m ( ) z 2 ( t λ z ) 2 ] x. The EL equation for the variational problem δs = 0 takes the form 2 z λ 2 z t 2 m = 0. x 2

18 Motivation Let F be a space of fields (for us: scalar or vector-valued functions of space-time). An observable f is a function f : F R. On the quantum level, the behavior of physical systems is no longer deterministic and we cannot use the SAP directly. Instead, we just hope to find the expectation values of observables: for f : F R, f = 1 f(φ)e i S[φ] Dφ, Z F where Z = F e i S[φ] Dφ is the normalizing factor (partition function).

19 Motivation Let F be a space of fields (for us: scalar or vector-valued functions of space-time). An observable f is a function f : F R. On the quantum level, the behavior of physical systems is no longer deterministic and we cannot use the SAP directly. Instead, we just hope to find the expectation values of observables: for f : F R, f = 1 f(φ)e i S[φ] Dφ, Z F where Z = F e i S[φ] Dφ is the normalizing factor (partition function).

20 Motivation Let F be a space of fields (for us: scalar or vector-valued functions of space-time). An observable f is a function f : F R. On the quantum level, the behavior of physical systems is no longer deterministic and we cannot use the SAP directly. Instead, we just hope to find the expectation values of observables: for f : F R, f = 1 f(φ)e i S[φ] Dφ, Z F where Z = F e i S[φ] Dφ is the normalizing factor (partition function).

21 Motivation f = 1 f(φ)e i S[φ] Dφ, Z F Intuitively, we counting contributions from all possible φ s, but the ones which are closer to the extrema of the action functional S[φ] yield greater contributions (if φ deviates a lot from a classical trajectory, then oscillations of e i S[ ] near φ will cancel each other out). Stationary phase formula Let f = f(x) be a smooth function with a unique critical point c (a, b), f (c) 0 and g = g(x) be a smooth function with vanishing derivatives at x = a, b. Then b a g(x)e i f(x) dx = 1/2 e if(c)/ I( ), where I is a smooth function g(c) on [0, ) such that I(0) = 2πe ±πi/4. f (c)

22 Motivation f = 1 f(φ)e i S[φ] Dφ, Z F Intuitively, we counting contributions from all possible φ s, but the ones which are closer to the extrema of the action functional S[φ] yield greater contributions (if φ deviates a lot from a classical trajectory, then oscillations of e i S[ ] near φ will cancel each other out). Stationary phase formula Let f = f(x) be a smooth function with a unique critical point c (a, b), f (c) 0 and g = g(x) be a smooth function with vanishing derivatives at x = a, b. Then b a g(x)e i f(x) dx = 1/2 e if(c)/ I( ), where I is a smooth function g(c) on [0, ) such that I(0) = 2πe ±πi/4. f (c)

23 Motivation The (very) bad thing about the formula f = 1 Z F f(φ)e i S[φ] Dφ: since the space of fields F is, in general, huge (infinitely-dimensional), Dφ is ill-defined. Yet, it makes sense to approximate such integrals by their finite-dimensional analogs. This leads to the problem of studying oscillating integrals of the form P (x)e i S[x] dx R n and their real counterparts R n P (x)e 1 S[x] dx

24 Motivation The (very) bad thing about the formula f = 1 Z F f(φ)e i S[φ] Dφ: since the space of fields F is, in general, huge (infinitely-dimensional), Dφ is ill-defined. Yet, it makes sense to approximate such integrals by their finite-dimensional analogs. This leads to the problem of studying oscillating integrals of the form P (x)e i S[x] dx R n and their real counterparts R n P (x)e 1 S[x] dx

25 Motivation The (very) bad thing about the formula f = 1 Z F f(φ)e i S[φ] Dφ: since the space of fields F is, in general, huge (infinitely-dimensional), Dφ is ill-defined. Yet, it makes sense to approximate such integrals by their finite-dimensional analogs. This leads to the problem of studying oscillating integrals of the form P (x)e i S[x] dx R n and their real counterparts R n P (x)e 1 S[x] dx

26 Steepest descent and stationary phase Stationary phase formula Let f = f(x) be a smooth function with a unique critical point c (a, b), f (c) 0 and g = g(x) be a smooth function with vanishing derivatives at x = a, b. Then b a g(x)e i f(x) dx = 1/2 e if(c)/ I( ), where I is a smooth function g(c) on [0, ) such that I(0) = 2πe ±πi/4. f (c) Steepest descent formula Let f = f(x) be a smooth function with a unique minimum point c (a, b), f (c) > 0 and g = g(x) be a smooth function. Then b a g(x)e 1 f(x) dx = 1/2 e f(c)/ I( ), where I is a smooth g(c) function on [0, ) such that I(0) = 2π. f (c)

27 Steepest descent and stationary phase Let B be a box region in R n and f = f(x) be a smooth function with a unique critical point c B such that... Stationary phase formula... Hf(c) 0. Then if g = g(x) is a smooth function with vanishing derivatives at the boundary of B, we have g(x)e i f(x) dx = n/2 e if(c)/ I( ), where I is a smooth function B on [0, ) such that I(0) = (2π) n/2 e ±πiσ/4 g(c) Hf(c) Steepest descent formula... Hf(c) > 0. Then if g = g(x) is a smooth function, we have g(x)e 1 f(x) dx = n/2 e f(c)/ I( ), where I is a smooth B function on [0, ) such that I(0) = (2π) n/2 g(c) Hf(c)

28 Asymptotic expansion B g(x)e 1 f(x) dx = n/2 e f(c)/ I( ) }{{}? What do we want: find a power series expansion for I( ) = A 0 + A 1 + A (Not so) bad thing: Although, I = I( ) is smooth, it is not analytic at in general: the Taylor series for I at = 0 may have the zero radius of convergence. How to fix it: We still can try to find a formal power series expansion for I. Then under some pretty mild conditions on A i s there is canonical way to sum the series up (key term: Borel summation). Thus, we can focus on finding the power series coefficients A i.

29 Asymptotic expansion B g(x)e 1 f(x) dx = n/2 e f(c)/ I( ) }{{}? What do we want: find a power series expansion for I( ) = A 0 + A 1 + A (Not so) bad thing: Although, I = I( ) is smooth, it is not analytic at in general: the Taylor series for I at = 0 may have the zero radius of convergence. How to fix it: We still can try to find a formal power series expansion for I. Then under some pretty mild conditions on A i s there is canonical way to sum the series up (key term: Borel summation). Thus, we can focus on finding the power series coefficients A i.

30 Asymptotic expansion B g(x)e 1 f(x) dx = n/2 e f(c)/ I( ) }{{}? What do we want: find a power series expansion for I( ) = A 0 + A 1 + A (Not so) bad thing: Although, I = I( ) is smooth, it is not analytic at in general: the Taylor series for I at = 0 may have the zero radius of convergence. How to fix it: We still can try to find a formal power series expansion for I. Then under some pretty mild conditions on A i s there is canonical way to sum the series up (key term: Borel summation). Thus, we can focus on finding the power series coefficients A i.

31 Asymptotic expansion B g(x)e 1 f(x) dx = n/2 e f(c)/ I( ) }{{}? What do we want: find a power series expansion for I( ) = A 0 + A 1 + A (Not so) bad thing: Although, I = I( ) is smooth, it is not analytic at in general: the Taylor series for I at = 0 may have the zero radius of convergence. How to fix it: We still can try to find a formal power series expansion for I. Then under some pretty mild conditions on A i s there is canonical way to sum the series up (key term: Borel summation). Thus, we can focus on finding the power series coefficients A i.

32 Wick s theorem Theorem [Gian-Carlo Wick] Let A be a symmetric, positively-defined d-by-d matrix and l i s be linear forms on R d (i = 1, 2,..., m). Then R d l 1 (x)... l m (x)e x,ax /2 dx = (2π)d/2 l i, A 1 l σ(i), det A where σ s run through all pairings on {1, 2,..., m} (that is, σ is an involution on {1, 2,..., m}). σ i

33 Wick s theorem Theorem [Gian-Carlo Wick] Let A be a symmetric, positively-defined d-by-d matrix and l i s be linear forms on R d (i = 1, 2,..., m). Then R d l 1 (x)... l m (x)e x,ax /2 dx = (2π)d/2 l i, A 1 l σ(i), det A where σ s run through all pairings on {1, 2,..., m} (that is, σ is an involution on {1, 2,..., m}). Example m = 0, d = 1, A = 1: R e x2 /2 dx = 2π σ i

34 Wick s theorem Theorem [Gian-Carlo Wick] Let A be a symmetric, positively-defined d-by-d matrix and l i s be linear forms on R d (i = 1, 2,..., m). Then R d l 1 (x)... l m (x)e x,ax /2 dx = (2π)d/2 l i, A 1 l σ(i), det A where σ s run through all pairings on {1, 2,..., m} (that is, σ is an involution on {1, 2,..., m}). Example [ ] 2 1 m = 2, d = 2, A =, l = x, l 2 = y: R xy e 2 (x2 +xy+y 2) dxdy = 2π σ i

35 Wick s theorem Theorem [Gian-Carlo Wick] Let A be a symmetric, positively-defined d-by-d matrix and l i s be linear forms on R d (i = 1, 2,..., m). Then R d l 1 (x)... l m (x)e x,ax /2 dx = (2π)d/2 l i, A 1 l σ(i), det A where σ s run through all pairings on {1, 2,..., m} (that is, σ is an involution on {1, 2,..., m}). Example m = 4: R d (... ) l 1, A 1 l 2 l 3, A 1 l 4 + l 1, A 1 l 3 l 2, A 1 l 4 σ i + l 1, A 1 l 4 l 2, A 1 l 4

36 Wick s theorem It is convenient to represent pairings using graphs. Namely, for a pairing σ on a set {1, 2,..., m}, consider a graph with vertices indexed by 1, 2,..., m and connect vertices i,σ(i) with an edge. Example The graph represents the pairing 1, 3, 2, 5, 4, 6.

37 Wick s theorem Example ( (2π) d/2 det A ) 1 l 1 (x)... l 4 (x)e x,ax /2 dx = R d

38 Wick s theorem: sketch of the proof The main idea: reduce everything to the one-dimensional case, where all computations can be performed in terms of the Gamma-function. Useful facts: Any real symmetric matrix A can be diagonalized. Polarization identity. Let P : V V R be a symmetric polylinear form. Then P (x 1, x 2,..., x n ) = 1 n! I {1,...,n}( 1) n I P ( x i,..., x i ). i I i I Γ ( ) 2k+1 k = (2k)! 4 k k! π

39 Wick s theorem: sketch of the proof The main idea: reduce everything to the one-dimensional case, where all computations can be performed in terms of the Gamma-function. Useful facts: Any real symmetric matrix A can be diagonalized. Polarization identity. Let P : V V R be a symmetric polylinear form. Then P (x 1, x 2,..., x n ) = 1 n! I {1,...,n}( 1) n I P ( x i,..., x i ). i I i I Γ ( ) 2k+1 k = (2k)! 4 k k! π

40 Wick s theorem: sketch of the proof The main idea: reduce everything to the one-dimensional case, where all computations can be performed in terms of the Gamma-function. Useful facts: Any real symmetric matrix A can be diagonalized. Polarization identity. Let P : V V R be a symmetric polylinear form. Then P (x 1, x 2,..., x n ) = 1 n! I {1,...,n}( 1) n I P ( x i,..., x i ). i I i I Γ ( ) 2k+1 k = (2k)! 4 k k! π

41 Wick s theorem: sketch of the proof R d l 1 (x)... l m (x)e x,ax /2 dx = (2π)d/2 l i, A 1 l σ(i), det A σ i We can assume that m is even. Step 1. Both sides of the desired identity are symmetric and polylinear forms with respect to l i s. Thus, it suffices to prove it for l 1 = = l m. Step 2. The desired identity is stable under a linear change of variables. Thus, we can choose a basis in such a way that A becomes diagonal. Moreover, by rescaling, we can make A = E. Then x, Ax = x x2 d Step 3. We find x 2k e x2 /2 dx = 2π (2k)! 2 k k! by substituting y = x 2 /2 and reducing it to the Gamma-function.

42 Wick s theorem: sketch of the proof R d l 1 (x)... l m (x)e x,ax /2 dx = (2π)d/2 l i, A 1 l σ(i), det A σ i We can assume that m is even. Step 1. Both sides of the desired identity are symmetric and polylinear forms with respect to l i s. Thus, it suffices to prove it for l 1 = = l m. Step 2. The desired identity is stable under a linear change of variables. Thus, we can choose a basis in such a way that A becomes diagonal. Moreover, by rescaling, we can make A = E. Then x, Ax = x x2 d Step 3. We find x 2k e x2 /2 dx = 2π (2k)! 2 k k! by substituting y = x 2 /2 and reducing it to the Gamma-function.

43 Wick s theorem: sketch of the proof R d l 1 (x)... l m (x)e x,ax /2 dx = (2π)d/2 l i, A 1 l σ(i), det A σ i We can assume that m is even. Step 1. Both sides of the desired identity are symmetric and polylinear forms with respect to l i s. Thus, it suffices to prove it for l 1 = = l m. Step 2. The desired identity is stable under a linear change of variables. Thus, we can choose a basis in such a way that A becomes diagonal. Moreover, by rescaling, we can make A = E. Then x, Ax = x x2 d Step 3. We find x 2k e x2 /2 dx = 2π (2k)! 2 k k! by substituting y = x 2 /2 and reducing it to the Gamma-function.

44 Wick s theorem: sketch of the proof R d l 1 (x)... l m (x)e x,ax /2 dx = (2π)d/2 l i, A 1 l σ(i), det A σ i We can assume that m = 2k. Step 4. What is (2k)!? It s the number of pairings on a set 2 k k! {1, 2..., 2k}.

45 Feynman s theorem Back to our main problem: find the coefficients of the asymptotic power series expansion of the integral P (x)e S(x)/ dx, B where P is a polynomial and S is a smooth function having a unique minimum critical point c B.

46 Feynman s theorem Back to our main problem: find the coefficients of the asymptotic power series expansion of the integral l 1 (x)... l m (x)e S(x)/ dx, B where S is a smooth function having a unique minimum critical point c B.

47 Feynman s theorem Back to our main problem: find the coefficients of the asymptotic power series expansion of the integral l 1 (x)... l m (x)e S(x)/ dx, B where S is a smooth function having a unique minimum critical point c B. WLOG, we can assume c = 0, S(c) = 0. Then S(x) has a (formal, at least) power series expansion S(x) = x, Ax 2 + r 3 1 r! B r(x,..., x) where B r s are symmetric polylinear forms.

48 Feynman s theorem Make a change of variables x x : l 1 (x)... l m (x)e S(x)/ dx = B m/2 l 1 (x)... l m (x)e x,ax B m/2 l 1 (x)... l m (x)e x,ax R d 2 r 3 r/2 1 Br(x,...,x) 2 r 3 r/2 1 Br(x,...,x) r! dx = r! dx + o( )

49 Feynman diagrams Definition A Feynman diagram with m external vertices is a graph (possibly, with loops and multiple edges) with m vertices of degree 1 labeled by 1, 2,..., m and finitely many unlabeled vertices of degrees 3. G 3 (m) := {isomorphism classes of Feynman diagrams with m external vertices}. Here, an isomorphism of a labeled graph is supposed to preserve the labeling.

50 Feynman diagrams Definition A Feynman diagram with m external vertices is a graph (possibly, with loops and multiple edges) with m vertices of degree 1 labeled by 1, 2,..., m and finitely many unlabeled vertices of degrees 3. G 3 (m) := {isomorphism classes of Feynman diagrams with m external vertices}. Here, an isomorphism of a labeled graph is supposed to preserve the labeling.

51 Feynman diagrams Definition A Feynman diagram with m external vertices is a graph (possibly, with loops and multiple edges) with m vertices of degree 1 labeled by 1, 2,..., m and finitely many unlabeled vertices of degrees 3. G 3 (m) := {isomorphism classes of Feynman diagrams with m external vertices}. Here, an isomorphism of a labeled graph is supposed to preserve the labeling. Example

52 Feynman s theorem Theorem l 1... l m := m/2 l 1 (x)... l m (x)e x,ax R d where = (2π)d/2 det A Γ G 3 (m) 2 r 3 r/2 1 Br(x,...,x) r! dx b(γ) Aut(Γ) F Γ(l 1,..., l m ), b(γ) = Edges of Γ internal vertices of Γ ; Aut(Γ) is the number of automorphisms of a graph Γ which leave the external vertices fixed; F Γ (l 1,..., l m ) is the Feynman amplitude of the graph Γ computed as follows.

53 Feynman s theorem Theorem l 1... l m := m/2 l 1 (x)... l m (x)e x,ax R d where = (2π)d/2 det A Γ G 3 (m) 2 r 3 r/2 1 Br(x,...,x) r! dx b(γ) Aut(Γ) F Γ(l 1,..., l m ), b(γ) = Edges of Γ internal vertices of Γ ; Aut(Γ) is the number of automorphisms of a graph Γ which leave the external vertices fixed; F Γ (l 1,..., l m ) is the Feynman amplitude of the graph Γ computed as follows.

54 Feynman s theorem Theorem l 1... l m := m/2 l 1 (x)... l m (x)e x,ax R d where = (2π)d/2 det A Γ G 3 (m) 2 r 3 r/2 1 Br(x,...,x) r! dx b(γ) Aut(Γ) F Γ(l 1,..., l m ), b(γ) = Edges of Γ internal vertices of Γ ; Aut(Γ) is the number of automorphisms of a graph Γ which leave the external vertices fixed; F Γ (l 1,..., l m ) is the Feynman amplitude of the graph Γ computed as follows.

55 Feynman s theorem Theorem l 1... l m := m/2 l 1 (x)... l m (x)e x,ax R d where = (2π)d/2 det A Γ G 3 (m) 2 r 3 r/2 1 Br(x,...,x) r! dx b(γ) Aut(Γ) F Γ(l 1,..., l m ), b(γ) = Edges of Γ internal vertices of Γ ; Aut(Γ) is the number of automorphisms of a graph Γ which leave the external vertices fixed; F Γ (l 1,..., l m ) is the Feynman amplitude of the graph Γ computed as follows.

56 Feynman amplitude of a graph For a connected graph Γ, Step 1. put the linear form l i at the i-th external vertex (i = 1,..., m); Step 2. put the polylinear form B r at each internal vertex of degree r (r = 3, 4,... ); Step 3. take contractions of these forms along the edges of Γ using the pairing, A 1. If Γ has several connected components Γ i, F Γ is defined to be the product of F Γi s.

57 Feynman amplitude of a graph For a connected graph Γ, Step 1. put the linear form l i at the i-th external vertex (i = 1,..., m); Step 2. put the polylinear form B r at each internal vertex of degree r (r = 3, 4,... ); Step 3. take contractions of these forms along the edges of Γ using the pairing, A 1. If Γ has several connected components Γ i, F Γ is defined to be the product of F Γi s.

58 Feynman amplitude of a graph For a connected graph Γ, Step 1. put the linear form l i at the i-th external vertex (i = 1,..., m); Step 2. put the polylinear form B r at each internal vertex of degree r (r = 3, 4,... ); Step 3. take contractions of these forms along the edges of Γ using the pairing, A 1. If Γ has several connected components Γ i, F Γ is defined to be the product of F Γi s.

59 Feynman amplitude of a graph For a connected graph Γ, Step 1. put the linear form l i at the i-th external vertex (i = 1,..., m); Step 2. put the polylinear form B r at each internal vertex of degree r (r = 3, 4,... ); Step 3. take contractions of these forms along the edges of Γ using the pairing, A 1. If Γ has several connected components Γ i, F Γ is defined to be the product of F Γi s.

60 Feynman theorem Example Let S(x) = x, Ax (free theory). Then l 1... l N = (2π)d/2 det A = (2π)d/2 det A Γ G 3 (N) pairings b(γ) Aut(Γ) F Γ(l 1,..., l N ) N/2 1 l i, A 1 l σ(i) That s basically the statement of Wick s theorem. i

61 Feynman amplitude of a graph Example Let d = 1, S(x) = x x3. Then x x x x = 2π Γ G 3 (4) 2 1 F Γ(x, x, x, x)

62 Feynman digrams in physics