Quantum Field Theory II

Transcription

1 Quantum Field Theory II GEORGE SIOPSIS Department of Physics and Astronomy The University of Tennessee Knoxville, TN U.S.A. Spring 8

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3 Contents Interactions. Scattering theory Wick s theorem Feynman diagrams Reaction rates Unitarity Path Integrals

4 iv CONTENTS

5 UNIT Interactions When you include interactions, the plot thickens. Until now, we have only considered free fields, described by a Hamiltonian H H, which we were able to diagonalize (find all eigenvalues and corresponding eigenstates). But that is no physics! It is time to do some serious physics. When you include interactions, the Hamiltonian is modified to H H + H I (..) where H I describes interactions. It is no longer possible to solve the eigenvalue problem for H, except in very few special cases (mostly in two dimensions). The only thing we can do is perturbation theory, assuming H I is small. This does not answer deep questions, such as what is a proton?, but it provides a method for very accurate calculations (e.g., the magnetic moment of the electron is known to about significant digits both theoretically and experimentally, and they agree with each other!). Experimental results are obtained primarily through scattering. To compare with them, we need to develop scattering theory. It turns out that this type of processes (scattering) holds all the information of a quantum field theory.. Scattering theory Recall in quantum mechanics, H H + V (..) where H is the kinetic energy and V the potential. If V has compact support, then at times t ±, H H, because V. Thus, we may define incoming and outgoing states, in (at t ) and out (at t + ), respectively, which are eigenstates of H. We shall attempt to do the same in quantum field theory. Consider a state in. It evolves in time as e iht in (..) As t, H H (no interactions), so asymptotically, our state approaches a state in the Hilbert space constructed from H. Call that state in,. It evolves in time as Then the statement that this is asymptotic to the state in amounts to e iht in, (..3) e iht in e iht in,, as t (..4) Therefore, in lim t eiht e iht in, (..5)

6 UNIT. INTERACTIONS The operator U(t) e iht e iht (..6) (a unitary map, U U I) maps a state in the Hilbert space of H ( in, ) to a state in the Hilbert space of H ( in ) by in lim U (t) in, (..7) t Notice that if H and H commute ([H, H ] ), then we may write U(t) e ih I t, where we used (..). But this rarely happens. Thus, in general, U(t) is a very complicated object. Similarly, in the infinite future, we may map out where out ( out, ) is in the Hilbert space of H (H ). We wish to calculate the amplitude of the process lim U (t) out, (..8) t + in out (..9) i.e., where out in lim out, U(t +)U (t ) in, out, S in, (..) t ± ± S lim U(t +)U (t ) (..) t ± ± is the S-matrix. S maps in-asymptotes to out-asymptotes (these two Hilbert spaces may, in general, be distinct, but not here). In fact, S contains all the information of the quantum field theory. It is an important object to study. Let us bring it into a more convenient form. To this end, we need to establish some properties of U(t) first. We have du(t) dt ih e iht e iht + e iht ( ih)e iht ie iht (H H )e iht ie iht H I e iht ie iht H I e iht e iht e iht ih I (t)u(t) (..) Thus U(t) is a true evolution operator with (time-dependent) Hamiltonian H I (t). The latter is obtained from H I by evolving with H. The first-order ODE (..) together with the initial condition U() I uniquely determine U(t). To solve (..), convert it into an integral equation, and then use iteration, t du(t ) dt dt i t U(t) I i t dt H I (t )U(t ) dt H I (t )U(t ) (..3) U (t) I, U n (t) I i t dt H I (t )U n (t ) (n ) (..4)

7 . Scattering theory We obtain U (t) I U (t) I i U (t) I i t t dt H I (t ) t t dt H I (t ) + ( i) dt dt H I (t )H I (t ) (..5) etc.. This may be brought into a more compact form. Look at the second-order term. In it, t t t. The two-dimensional integral is over a triangle in the (t, t ) plane, which is half of the square t, t t. The other half gives the same answer, but with t and t interchanged. It follows that the two-dimensional integral can be written in terms of a time-ordered product as t t dt dt H I (t )H I (t ) For the nth term, we similarly obtain t dt It follows that tn dt n H I (t ) H I (t n ) n! t t t t dt dt T [H I (t )H I (t )] (..6) dt t t dt n T [H I (t ) H I (t n )] (..7) ( i) n U(t) T dt dt n H I (t ) H I (t n ) n! n T [e i ] t dt H I (t ) (..8) which is Dyson s formula. Next, define the generalized evolution operator U(t, t ) U (t )U(t ) T It is easily seen from its definition that it has the following properties [ e i t t dt H I (t ) ] (..9) U(t, t )U(t, t ) U(t, t ), U (t, t ) U(t, t), U (t, t )U(t, t ) I (..) The S-matrix can be written as S lim U(t +, t ) T t ± ± Various important properties of the S-matrix follow. The S-matrix is unitary, [e i + dth I (t) ] (..) S S I (..) This may look like a trivial statement (direct consequence of the unitarity of U(t +, t )), but we need to take limits t ± ±, and that may spoil unitarity, unless the range of S is the entire Hilbert space. The latter is a consequence of the requirement that probability be conserved: everything that comes in should go out. There are cases where this is not true and particles may be trapped into bound states. The S-matrix has a way to address these issues. For the most part, we shall assume that no trapping occurs and the S-matrix is unitary. S commutes with the free Hamiltonian H, i.e., a scattering process conserves unperturbed energy. [S, H ] (..3)

8 4 UNIT. INTERACTIONS Proof. e iɛh Se iɛh lim t ± ± eiɛh e it H e it H e it+h e it+h e iɛh lim t ± ± eih(t +ɛ) e ih(t +ɛ) e ih(t++ɛ) e ih(t ɛ) lim t ± ± U(t + ɛ)u (t + + ɛ) But as t ± ±, adding an ɛ makes no difference, so e iɛh Se iɛh S (..4) Expanding in ɛ, at first order we obtain the desired result (..3). S is Lorentz-invariant. Proof. The Lagrangian density (a Lorentz-invariant quantity) can be split as L L + L I (..5) If L I contains no time derivatives, then H I d 3 xl I (..6) Therefore, using (..), S T [e i ] dth I (t) T [e i ] d 4 xl I (..7) which is manifestly Lorentz-invariant. The integral is over the entire space-time. It should be noted that time-ordering does not spoil Lorentz invariance. Indeed, time-ordering is frame-dependent only for spacelike separations, and fields commute at spacelike separations (causality). Corollary: By Lorentz invariance, it follows from (..3) that S commutes with the unperturbed four-momentum, [S, P µ ] (..8) This implies momentum conservation in scattering. Indeed, consider an incoming (outgoing) state of free particles with momenta p, p,... (p, p,... ), i.e., in, p, p,..., out, p, p,... (..9) (notation as in eqs. (..7) and (..8)). These are eigenstates of unperturbed momentum P µ in, (pµ + pµ +... ) in,, P µ out, (p µ + p µ +... ) out, (..3) It follows that ( out, [S, P µ ] in, p µ i ) p µ i out in and so p µ i p µ i (..3) i.e., momentum is conserved.

9 . Wick s theorem. Wick s theorem Notice that in the expression for the S-matrix (..7), time evolution is accomplished with the unperturbed Hamiltonian H, and not with H. Thus the fields in L I are asymptotic (free) fields, and can be treated in the same way we have already discussed. Let us concentrate on the scalar field φ. The field φ in S does not satisfy the full equation of motion, which is usually non-linear. Instead, it obeys the Klein-Gordon equation. To distinguish it from the true φ, we shall call it φ I (in interaction picture). Since we shall be always talking about φ I, we might as well drop the subscript I. Hopefully the reader will not get too confused. Therefore, we may perform the familiar expansion φ(x) d 3 k [ (π) 3 a( ω k)e ik x + a ( k)e ik x] (..) k and define the one-particle state p Ca ( k) (..) where is the asymptotic vacuum (not the true vacuum of the whole system). When calculating vacuum expectation values φ(x)φ(y), it is convenient to normal-order. On the other hand, S contains time-ordered products. We need to relate the two types of orderings. To this end, define the contraction of two fields as the difference between the two types of ordering, φ(x)φ(y) T (φ(x)φ(y)) : φ(x)φ(y) : (..3) The contraction is a number (not an operator), because each time we commute a and a, we get a number. To calculate it, we shall take the vacuum expectation value of both sides of (..3). We obtain φ(x)φ(y) φ(x)φ(y) T (φ(x)φ(y)) : φ(x)φ(y) : The time-ordered product is the Feynman propagator, D F (x y) T (φ(x)φ(y)) d 4 k (π) 4 i k m + iɛ e ik (x y) (..4) The normal-ordered product contains terms of the form a a, a a and aa, all of which have vanishing vacuum expectation values. Therefore, the contraction is the Feynman propagator, φ(x)φ(y) D F (x y) T (φ(x)φ(y)) d 4 k i e ik (x y) (π) 4 k m (..5) + iɛ This result can be generalized to a product of n fields. To simplify the notation, denote Wick s theorem states that T n T (φ φ n ) : φ φ n : For n, Wick s theorem is the trivial statement For n, it was proved above. φ i φ(x i ), i,..., n (..6) + φ φ : φ 3... φ n : +all other terms with contraction + φ φ φ 3 φ 4 : φ 5... φ n : +all other terms with contractions +... (..7) T T (φ ) φ : φ :

10 6 UNIT. INTERACTIONS For n 3, we need to show T 3 T (φ φ φ 3 ) : φ φ φ 3 : +φ φ φ 3 + φ φ 3 φ + φ φ 3 φ 3 (..8) To show this, first arrange the indices,, 3 so that times are order as x x x 3. Then T 3 T (φ φ φ 3 ) φ φ φ 3 First, let us try to normal-order φ 3. To this end, split φ 3 φ 3+ + φ 3 (..9) where φ 3+ (φ 3 ) contains all the creation (annihilation) operators. Then T 3 φ φ φ 3 φ φ (φ 3+ + φ 3 ) φ φ φ 3 + φ 3+ φ φ + [φ, φ 3+ ] φ + φ [φ, φ 3+ ] The commutators are numbers. We have [φ, φ 3+ ] [φ, φ 3+ ] φ φ 3+ (since φ 3+ ) φ φ 3 (since φ 3 ) T (φ φ 3 ) φ φ and similarly for [φ, φ 3+ ]. We deduce Next, we normal order φ φ, T 3 φ φ φ 3 + φ 3+ φ φ + φ φ 3 φ + φ φ 3 φ φ φ : φ φ : +φ φ and use : φ φ φ 3 : : φ φ : φ 3 + φ 3+ : φ φ : together with (..9) to arrive at the desired result (..8). For n 4, with x x x 3 x 4, we have T 4 φ φ 4 φ φ φ 3 φ 4 + φ 4+ φ φ φ 3 + [φ, φ 4+ ] φ φ 3 + φ [φ, φ 4+ ] φ 3 + φ φ [φ 3, φ 4+ ] Now use our result for T 3, T 3 φ 4 + φ 4+ T 3 + φ φ 4 φ φ 3 + φ φ 4 φ φ 3 + φ 3 φ 4 φ φ T 4 : φ φ φ 3 : φ 4 + φ 4+ : φ φ φ 3 : + φ φ (φ 3 φ 4 + φ 4+ φ 3 ) φ φ 4 φ φ : φ φ φ 3 φ 4 : +φ φ : φ 3 φ 4 : φ φ 4 φ φ where the dots represent terms that are obtained by permutations of the indices (3).

11 . Wick s theorem We still need to normal order the fields in the last line. Using our result for T, we arrive at the desired result for T 4, T 4 : φ φ φ 3 φ 4 : +φ φ : φ 3 φ 4 : φ φ 4 : φ φ 3 : φ φ 4 φ φ The above argument can be generalized to arbitrary n by induction (assume Wick s theorem holds for all n < N; show it holds for N). EXAMPLE : External source The simplest interaction we can have is of the form H I d 3 xφ(x)j(x) (..) where J(x) is a given function (source) representing some fixed distribution of matter The Klein-Gordon gets modified to µ µ φ + m φ J (..) which may be solved exactly with the aid of the Feynman propagator. We obtain φ(x) d 4 yd F (x y)j(y) (..) The S-matrix is where S (n) ( i)n n! It can also be written as S T [ e i d 4 xφ(x)j(x) ] S (n) ( i)n T [S S n ], S k n! S (n) (..3) n d 4 x d 4 x n J(x ) J(x n )T (φ(x ) φ(x n )) d 4 x k J(x k )φ(x k ) (k,..., n) Of course, all S k are the same operator, but before we integrate over time (the x ks), we need to remember to time-order. Since S k is linear in φ, Wick s theorem holds for the S k s. This allows us to turn time-ordering to normal-ordering, T (S S n ) : S S n : + S S : S 3... S n : +permutations + S S S 3 S 4 : S 5... S n : +permutations +... (..4) Since all S k s are the same object, all permutations give the same result; we just need to count them. With one contraction, the number of permutations is ( ) n n! N!(n )! E.g., if you think of φ as the electromagnetic potential, J is the current that creates it. Of course the electromagnetic potential (photon) has two polarizations, so there are complications, but our discussion is still valid, as it captures the essence of the field.

12 8 UNIT. INTERACTIONS With two contractions, the number of permutations is the number of ways of choosing two pairs from n objects, N ( )( ) n n n! 3 (n 4)! where we divided by because of over-counting (identical pairs). Generalizing to p contractions, the number is N p ( )( ) ( ) n n n (p ) n! p! p! p (n p)! The sum of terms with p contractions in (..4) is Its contribution to S (n) is N p S S S 3 S 4 S p S p : S p+ S n : N p (S S ) p : S n p 3 : ( i) n n! N p (S S ) p : S n p ( i) n 3 : p! p (n p)! (S S ) p : S n p 3 : Now let us collect all the terms with p contractions in S. Such terms exist in all S (n) with n p. We obtain the following contribution to S: A p n np ( i) n p! p (n p)! (S S ) p : S n p 3 : ( i) n +p p! p (n )! (S S ) p : S n ( i)p p! p (S S ) p : e is3 : 3 : Summing over p, we obtain a simple expression for the S-matrix, S A p e (SS) : e is3 : e α/ : e i d 4 xj(x)φ(x) : (..5) where the coefficient e α/ is a c-number, and α d 4 xd 4 yj(x)d F (x y)j(y) (..6) We can now study the physical properties of the system. Evidently, the source creates particles. To see this, calculate the probability amplitude that the true vacuum in will evolve to an n-particle state p... p n out. We have where the states on the right-hand side are free particle states. n. We have out p... p n in p... p n S (..7) out in S e α/ so the probability that no particle will be created is P ( ) e β, β Rα This is similar to the creation of photons (bremsstrahlung radiation) by an accelerating charged particle.

13 . Wick s theorem To show that this is an honest probability, i.e., P ( ), we need to show that β. Using (..4), we obtain d 4 k α (π) J(k) i 4 k m + iɛ J( k) in terms of the Fourier transform of J, J(x) d 4 k ik x J(k)e (π) 4 Since J(x) is real, we have J( k) J (k), therefore β d 4 [ k (π) 4 J(k) i k m + iɛ ] i k m iɛ Using [ lim ɛ + x + iɛ ] πiδ(x) x iɛ we deduce β d 4 k (π) 4 J(k) πδ(k m ) Integrating over k, we obtain two identical contributions at k ±ω k, and so d 3 k β (π) 3 ω J(k) (..8) k n. In this case only the first-order term in the expansion of the normal-ordered exponential in (..5) contributes. We obtain out p in p S e α/ ( i) d 4 xj(x) p φ(x) Using p φ(x) e ip x, which we showed a while ago, we deduce out p in ie α/ J(p ) The probability that a single particle will be created from the vacuum is d 3 p P ( ) (π) 3 out p in e β d 3 p ω p (π) 3 ω J(p ) βe β (..9) p For a general (given) n, the nth-order term in the expansion of the normal-ordered exponential in (..5) contributes. We obtain out p p n in p p n S α/ ( i)n e p p n : S S n : n! Only the term with n creation operators in : S S n : contributes. By commuting all creation operators through to the left, it is easy to see that p p n : S S n : n! J(p ) J(p n ) The probability that n particles will be created from the vacuum is P ( n) d 3 p d 3 p n n! e β (π) 3 ω p (π) 3 p p n S ω pn n! e β d 3 p d 3 p n (π) 3 ω p (π) 3 ω J(p ) J(p n ) pn βn n! e β (..)

14 UNIT. INTERACTIONS Notice that we divided by n! in the expression for the probability, because these are identical particles. Thus, we have a Poisson distribution. Notice that P ( n) n Probability is conserved and the S-matrix is unitary. The expected number of emitted quanta (photons) is n np ( n) β (..) n If the source is localized (J(x) δ 4 (x)), then the Fourier transform is J(k), so β. The divergence comes from large k (ultraviolet (UV) divergence). Thus a source can never be perfectly localized, and no experiment can measure φ(x) precisely, in accordance with Heisenberg s uncertainty principle. EXAMPLE : Static external source As a special case, consider a static source. Well, it will not be completely static, because it has to be switched on at a certain time and then switched off again, after a long time. So let J(x) f(t)ρ( x) (..) where ρ( x) represents the static ( charge ) distribution, and f(t) is a smooth function of time of compact support, so that f(t) for t > T, say, and f(t) for t < T, with a smooth transition from to near t ±T. We shall assume that interactions are turned on and off adiabatically. If the cutoff time T is very large, then the Fourier transform of f(t), f(ω) dte iωt f(t) will have narrow support (Riemann-Lebesgue Lemma). Without the smooth transition at t ±T, we have T f(ω) dte iωt sin ωt ω T It does not have compact support, but its width is ω /T (i.e., ω t ). As T, f(ω) πδ(ω), which has compact support (the point ω ), and the frequency can be measured with infinite accuracy. If we smoothen out the above f(t), then it will have compact support of width ω. Let us arrange things (choose a large enough cutoff time T ) so that f(ω) has support within the interval ω < m, i.e., f(ω), ω m (..3) To find the probability of particle production (radiation), notice that J(k) d 4 xj(x)e ik x f(ω) ρ( k) In the expression for β (..8), ω ω k k + m m, therefore, ω is outside the support of f, by our assumption (..3), and f(ω k ). It follows that β, and no particles can be created. This makes physical sense: no radiation is emitted by static charges. Thus the S-matrix is pretty boring, but we may still talk about the vacuum-to-vacuum transition amplitude S e α/

15 . Wick s theorem where α is purely imaginary (we just showed that its real part β ). The probability is e α, as expected, since this is the only possible process, but what is the physical meaning of the amplitude? To answer this question, use (..6) and (..) to write the amplitude as e α/ where we used H. For the true vacuum of our system, we have lim U(t +)U (t ) lim t ± ± t e ih(t+ t ) ± ± H phys E phys where E is the true ground state energy. From the adiabatic theorem in quantum mechanics, we know that if interactions are turned on very slowly, the state (which is the ground state in the absence of interactions) evolves into the vacuum state phys without getting excited. It follows that e α/ phys e ih(t+ t ) phys t±±t e iet This is true for large cutoff time T, more precisely, as T. We deduce E i 4 lim α T T (..4) Thus the S-matrix contains information about the ground state energy of the system. For an explicit expression, use (..6) together with (..4). The integrals over times x and y are T T dx T T dy e ik(x y ) Evidently, in the limit T, this expression has support consisting of the single point k. It follows that for large T, it is approximately T T dx T T dy e ik(x y ) πcδ(k ) The constant C is found by integrating both sides over k. We obtain C dk π T T dx T T dy e ik(x y ) dk π ( ) sin k T T k It follows that for large T, α it d 3 xd 3 yρ( x)v ( x y)ρ( y) and E d 3 xd 3 yρ( x)v ( x y)ρ( y) (..5) which is the energy of two static ( charge ) distributions interacting via the potential V ( x) d 3 k e i k x e mr (π) 3 k + m 4πr (..6) This is a Yukawa potential. As m, it turns into the Coulomb potential.

16 UNIT. INTERACTIONS EXAMPLE 3: A non-linear quantum field theory Consider three distinct Klein Gordon fields φ (i,, 3) with interaction Lagrangian density where λ is a coupling constant. The S-matrix can be expanded in powers of λ, The nth term is proportional to λ n, and S (), S () iλ L I λφ (x)φ (x)φ 3 (x) (..7) S T [e i ] d 4 xλφ φ φ 3 S (n) (..8) n d 4 xφ φ φ 3 (no need to normal order),... (..9) We shall represent S () by a diagram thusly: 3 Each line is a field. Lines that meet at a vertex have common argument which is integrated over and multiplied by iλ. S () contributes to the process (p ) (p ) + 3(p 3 ) (..3) where p µ i is the momentum of the particle of type i (i,, 3). The probability amplitude for this process is out, p ; 3, p 3, p in, p ; 3, p 3 S, p, p ; 3, p 3 S (), p +... (..3) where the dots represent contributions of higher order in λ. Using φ(x) p e ip x thrice, we obtain, p ; 3, p 3 S (), p iλ d 4 xe ip x e ip x e ip3 x iλ(π) 4 δ 4 (p p p 3 ) Therefore momentum is conserved, as expected (see discussion following eq. (..8)). For the general process i f, we define the scattering amplitude A fi by f S i f i (π) 4 δ 4 (p f p i )ia fi (..3) where p µ i (pµ f ) is the total momentum of the initial (final) state i ( f ). Notice that we subtracted the amplitude representing no scattering, f i. Earlier, for i, p, f, p ; 3, p 3, we found ia fi iλ + O(λ ) Also, momentum conservation implies that the decay (..3) is only allowed (to all orders in λ!) if m m + m 3 (..33)

17 . Wick s theorem Proof. By momentum conservation and using p i m i (i,, 3), m m + m 3 + p p 3 Also and Therefore and m p p 3 E E 3 p p 3 E E 3 p p 3 E E 3 (m + p )(m 3 + p 3 ) m m 3 + p p 3 + m p 3 + m 3 p m m 3 + p p 3 + m m 3 p p 3 m m 3 + p p 3 p p 3 E E 3 p p 3 m m 3 m + m 3 + p p 3 m + m 3 + m m 3 m + m 3 At O(λ ), we have S () ( iλ) d 4 xd 4 yt [φ (x)φ (x)φ 3 (x)φ (y)φ (y)φ 3 (y)] (..34) We need to re-express this in terms of normal-ordered products. This introduces contractions in the above expression. Here are the various possibilities. contractions 3 3 Same rules as before, but be sure to normal order! Topologically, this is a disconnected diagram. If we calculate a matrix element of S, we will get δ-functions conserving momentum (of 6 particles). contraction. Denote the contraction (propagator) φ(x)φ(y) by x y There are 3 possible contractions, each joining the two vertices, An intermediate line labeled i denotes the contraction φ i (x)φ i (y) (i,, 3). These are not Feynman diagrams (yet). contractions. There are 3 possibilities.

18 4 UNIT. INTERACTIONS The first diagram is proportional to φ (x)φ (y) φ 3 (x)φ 3 (y), etc. 3 contractions. Only possibility. 3 SYMMETRY FACTORS In general, if a diagram has a symmetry (a permutation of its vertices that does not give rise to a new contraction), we need to divide by the corresponding symmetry factor. All of the diagrams above have a symmetry: just interchange the two vertices. This explains the factor of in the expression (..34) for S(). At higher orders, this symmetry need not be present. For example, consider the contribution to S (3), 3 The vertices are distinguishable, so there is no symmetry. The symmetry factor we ought to divide by is. To see this in a little more detail, recall that there is a factor of 3! in the definition of S(3). On the other hand, there are 3! ways of doing the two contractions, here is one: Thus, we obtain a factor of 3! 3!. 3 (φ φ φ 3 )(φ φ φ 3 )(φ φ φ 3 ) EXAMPLE 3: A non-linear quantum field theory of a single field Suppose the 3 fields are identical, φ φ φ 3 φ, and let the interaction Lagrangian density be L I 3! λφ3 (..35) where we re-defined the coupling constant λ in order to expose a factor of 3! for convenience. This time, the symmetry factor is the number of permutations of vertices times the number of permutations of lines. has symmetry factor 3! (number of permutations of (3)). has symmetry factor (vertices) (two pairs of external lines) 8. has symmetry factor 3! (vertices) 3 (three pairs of external lines) 48. has symmetry factor (vertices) (internal lines) 4.

19 .3 Feynman diagrams A general graph is disconnected and consists of lower-order connected graphs. Let S (graph) be the contribution of an individual graph to the S-matrix. It can be written as S (graph) ( n!n! S (c) ) n ( S (c) ) n (..36) where S (c) i (i,,... ) is the contribution to the S-matrix of the corresponding connected subgraph. Notice that we had to divide by symmetry factors n i! (i,,... ), because we have n i identical connected sub-graphs. Be reminded that sub-graphs represent operators which do not necessarily commute with each other. However the product is normal-ordered, so the ordering of the subgraphs is immaterial. Also note that the above expression includes connected graphs, if we set n, n i (i ). The entire S-matrix is the sum over all possibilities, n S S (graph) n,n,... ( n!n! S (c) ) n ( S (c) ) n (..37) This looks hideous, but can, in fact, be neatly organized, because the sums over n, n,..., are independent of each other. Thus S can be written as an infinite product, ( ) ( ( ) S S (c) n ) ( ) S (c) n e S(c) e S (c) e S(c) (..38) n! n! n where S (c) is the connected part of the S-matrix, S (c) S (c) + S (c) +... (..39) consisting of connected graphs only. EXAMPLE: Let us re-visit the system with the external source (..) whose S-matrix we found exactly (eqs. (..5) and (..6)), albeit rather painfully. There are only two connected graphs: S (c) with rule: i d 4 xj(x)φ(x), and S (c) one contraction. The connected S-matrix is S (c) S (c) + S(c) where I included the requisite symmetry factor. The S-matrix is S e S(c) e S(c) e S (c) The two factors can be separated because they commute with each other. The first factor is a number, in fact S (c) ( i) α α. The second factor is a (normal-ordered!) operator identical to the expression (..5) derived earlier..3 Feynman diagrams So far, we have been developing tools to calculate the S-matrix, which is an operator. Now, we shall apply these rules to actual scattering processes and find the corresponding rules due to Feynman.

20 6 UNIT. INTERACTIONS Continuing to work with the interaction Lagrangian density (..35), which is physically useless, but good enough for the moment, let us consider the interaction of particles with initial momenta k µ, kµ and final momenta kµ 3, kµ 4. The amplitude is k 3, k 4 S k, k k 3, k 4 S () k, k +... (.3.) where the dots represent terms of higher order in λ. Thus, the lowest-order contribution is O(λ ). Recall that there are various contributions, with,,, or 3 contractions. Since the process of interest involves 4 particles, only the part of S () with 4 external legs will contribute (the one with contraction, of symmetry factor 3 8). We need to match the momenta with these 4 external lines. There are 3 possibilities: k k 3 k k k k k k 4 k 3 k 4 k 4 k 3 There are 8 ways to assign momenta in each of the 3 cases, so the overall symmetry factor of a diagram including momenta (i.e., a Feynman diagram) is 8 8. This symmetry factor may also be deduced by looking at the Feynman diagram itself: it is the number of permutations of internal lines and vertices (leaving external lines unchanged). EXAMPLE: The diagram contributing to k S k at O(λ ) k k has symmetry factor (for the internal lines). Recall that the corresponding S-matrix diagram has symmetry factor 4, since there are ways of assigning momenta k and k, we obtain 4, which is the correct symmetry factor of the Feynman diagram. Let us now compute the first diagram. It is ( iλ) d 4 xd 4 yφ(x)φ(y) k 3, k 4 : φ (x)φ (y) : k, k (.3.) I omitted a symmetry factor anticipating that it will eventually be. Using φ (x) p e ip x, the above expression becomes ( iλ) d 4 xd 4 yd F (x y)e ik3 x e ik4 x e ik y e ik y (.3.3) Notice that we have a factor e +ip x (e ip x ) for each outgoing (incoming) momentum. Introducing the Fourier transform of the Feynman propagator, we obtain ( iλ) d 4 xd 4 y d4 k (π) 4 e k (x y) i eik3 x k m e ik4 x e ik y e ik y (.3.4) + iɛ which yields δ-functions, δ 4 (k 3 + k 4 k) and δ 4 (k k k ), allowing us to draw the diagram k k k 3 k k 4 where the intermediate line may be thought of as representing a particle of four-momentum k µ. Momentum is conserved at each vertex, and consequently, overall, k + k k 3 + k 4 (.3.5)

21 .3 Feynman diagrams as expected on general grounds. The intermediate particle is not real. Indeed consider its decay to the two outgoing particles. For this to happen, we need m m + m 3, i.e., m m, which is not satisfied, unless m. The latter is also impossible for kinematical reasons. We say that the intermediate particle is virtual and its existence is allowed by the Heisenberg Uncertainty Principle (k m is OK in quantum mechanics). After doing all the integrals, the diagram becomes ( iλ) (π) 4 δ 4 i (k + k k 3 k 4 ) (k + k ) m + iɛ The other diagrams are similarly obtained. We deduce the scattering amplitude [ ] ia ( iλ) i (k + k ) m + iɛ + i i (k k 3 ) m + iɛ (k k 4 ) m + iɛ (.3.6) (.3.7) Notice that it is a Lorentz-invariant expression, as it ought to be. Generalizing to more complex diagrams is straightforward. One obtains the following general rules. FEYNMAN RULES (in momentum space) To each external line of a diagram assign an incoming or outgoing momentum. To each internal line assign the most general momentum consistent with momentum conservation at each vertex. Include a factor ( iλ) for each vertex. Include a factor Integrate with i k m +iɛ for each internal line of momentum kµ. d 4 k (π) 4 for each undetermined momentum. 3 Divide by the appropriate symmetry factor. These rules yield the scattering amplitude ia after removing the overall factor (π) 4 δ 4 (k f k i ). EXAMPLE: Consider the propagation of a single particle of momentum k µ. At O(λ ), the diagram that contributes is k k k k k There is arbitrary momentum, and the symmetry factor is, so ia ( iλ) d 4 k (π) 4 i k m + iɛ i (k k ) m + iɛ (.3.8) We managed to derive an expression for this amplitude without much effort, but alas, it is an infinite integral. We will come back to this later. GRAPH TOPOLOGY: The topology of a graph is related to the number of undetermined momenta, so it is important when one tries to determine the complexity of the integral(s), as well as the order in the quantum mechanical expansion of matrix elements of the S-matrix. EXAMPLES has no undetermined momenta. 3 Notice that internal ( virtual ) particles are not on the mass shell (k m ), so we integrate over all four-momenta.

22 8 UNIT. INTERACTIONS has one undetermined momentum. In general, consider a connected graph with V vertices and I internal lines. Each vertex gives a δ-function conserving momentum. One of them is the overall factor (π) 4 δ 4 (k f k i ) (for connected graphs), so we have V δ-functions constraining momenta. Each internal line provides an arbitrary momentum. Therefore, we have L integrals to do, where L I (V ) I V + (.3.9) The following theorem, relating the integrals to the topology of the graph, holds. Theorem. L is the number of loops in the graph. Proof. The proof is by induction on L. When L, we have a tree graph and I V (because each time you add a vertex, you add an internal line, so V I const. in a tree; if I, then obviously V, therefore V I ). To add a loop, join two external lines. This increases I I + and V is unchanged. So if we start with I V L, we end up with (I + ) V L (L + ), which completes the inductive step. PHYSICAL CONSIDERATIONS. Let us go back to the scattering process. In the center of mass frame, k k, k3 k 4, E + E E 3 + E 4 therefore, E E E 3 E 4 E, k k k 3 k 4 k showing that we have an elastic collision. E is the beam energy in an accelerator. The total centerof-mass energy (a Lorentz-invariant quantity) is E cm E, because E cm (k + k ) (E + E ) 3E It is also convenient to define the momentum transfers q k k 3, q k k 4 Their norms are Lorentz-invariant quantities, and found to be q (k k 3 ) k ( + cos θ), q (k k 4 ) k ( cos θ) where θ is the angle between the vectors k and k 3. The scattering amplitude (.3.7) reads λ A Ecm m + λ q + m + λ q + m (.3.) The second term can be easily understood. It is the Fourier transform of a Yukawa potential (eq. (..6)). This matches the result from non-relativistic quantum mechanics (Born approximation) for scattering off of a potential, k V ( x) k 3 d 3 xe i q x V ( x) The third term has a similar interpretation. Its presence is possible quantum mechanically because we have identical particles. The first term has poles at E cm ±m. It represents a relativistic effect. Note that the poles are at E cm ±mc, so in the non-relativistic limit, c, and the poles go away to infinity.

23 .4 Reaction rates.4 Reaction rates To get in touch with reality and calculate quantities that an experimentalist can measure, we will need the relativistic analog of Fermi s Golden Rule. To this end, it is advantageous to put the world in a box, both in space and time (recall that we need to turn off interactions for t T, where T is large; a time box is just as good and more convenient). In one dimension, we have a basis k with dk k k and x k π e ikx so x y dk x k k y dk π e ik(x y) δ(x y). We now demand x L so we get discrete eigenstates ϕ n (x) L e i πn L x x n with n n. As L, πn L k, and the difference in successive n s, π L k dk and n L π dk. In three dimensions, we repeat these substitutions 3 times with k π L (n, n, n 3 ) to get ϕ k (x) V e ik x and, as L, k V (π) d 3 k. 3 To make everything concrete, we use a K-G field ϕ(x) V k k ( e ik x a(k) e ik x a (k) ). {}}{ ϕ(x)ϕ(y) V k k e ik (x y) d 3 k (π) 3 k e ik (x y) ; [ a(k), a (k ) ] δ k,k and ϕ(x) k V k e ik x. We introduce a new Feynman rule: assign to each external leg a factor of transition amplitude is EV. The f S i ia(π) 4 δ 4 (p f p i ) j Ej V so the transition probability is f S i A ((π) 4 δ 4 (p f p i )) j A V T (π) 4 δ 4 (p f p i ) j E j V E j V where the square of the delta function is ( (π) 4 δ (k)) 4 (π) 4 δ 4 (k) (π) 4 δ 4 (k) d 4 ye ik y d 4 y (π) 4 δ 4 (k)v T because δ 4 (k) forces k. Thus the probability per unit time (the decay rate), as V, becomes A V (π) 4 δ 4 (p out p in ) E j out j j V because k V j out d 3 p jout (π) 3 E jout A V (π) 4 δ 4 (p out p in ) j in d 3 k (π) 3 as V T. All the details of the theory are contained in A. Example. One incoming particle p in and two outgoing particles p and p. The decay rate is Γ m d 3 p d 3 p (π) 3 E (π) 3 A V (π) 4 δ 4 (p in p p ) E E in V d 3 p (π) 3 A πδ(e in E E ) E E E jin V

24 UNIT. INTERACTIONS In the rest frame of the incoming particle p in (m, ), p (E, p ), and p (E, p ). Using spherical coordinates, d 3 p p dp dω and writing f(p ) m E E, we have f (p ) de dp de dp p E p E E p +E m E E p E E. Then Γ p dp dω m (π) 3 π A E E δ (p p root ) 4E E mp root p 3π m dω A In an arbitrary frame using E p 8πm A. m v Γ arb m E in Γ γ Γ γm. If the incoming particle has spin, A is independent of Ω so Γ Example. One incoming particle and three outgoing particles. d 3 p 3 (π) 3 E 3 Much the same reasoning applies. We have to include three more integrals for more delta functions δ 4 (p in p p p 3 ). In the rest frame of the incoming particle Γ d 3 p d 3 p m (π) 6 πδ(m E E E 3 ) A E E E 3 and four Use spherical coordinates with z-axis parallel to p and let θ be the angle between p and p ; d 3 p p dp dω ; dr d(cos θ )dϕ. We will eventually use spherical coordinates to represent p so we can think of the momenta as positive magnitudes. Now specialize to the case where all outgoing particles have mass. Note p i Ei. The argu- ment of the delta function m E E E 3 m p p p + p m p p p + p p p cos θ f(cos θ ); f p (cos θ ) p EE E 3. p +p pp cos θ Γ d 3 p p dp dϕ E 3 m (π) 5 A 8E E E 3 E E d 3 p p dp dϕ m (π) 5 8E A E p dp dωdp 6(π) 5 m E A 6(π) 5 de dωde dϕ A m where we have used p i Ei and converted p to spherical coordinates: d 3 p p dp dω. The delta function imposes a constraint: m E + E + E 3 E + E + E + E E E cos θ so m E + E. Furthermore (m E E ) E + E E E cos θ so m + E E me me E E cos θ which gives m(m E E ) E E ( + cos θ ) which implies a negative left hand side, i.e. E + E m. The limits on cos θ also mean m(m E E ) 4E E which rearranges to ( m E ( m ) E ). Both terms negative contradicts m E + E so E and E are less than or equal to m. Thus integration is over the upper triangle in the Dalitz plot below. In the case where all three particles have mass, the area of integration becomes a first quadrant shape (its Dalitz plot). Example 3. Two in, two out.

25 .4 Reaction rates We have the formula for the transition probability. We now calculate the cross-section. Let k (rest frame of particle ); The volume is area times length V AL; v L #particles t so flux area time. The cross-section is At v AL v V σ Γ flux V v Γ V d 3 k 3 d 3 k 4 v (π) 3 E 3 (π) 3 A V (π) 4 δ 4 (k + k k 3 k 4 ) E 4 E V E V d 3 k 3 d 3 k 4 ve E (π) 3 E 3 (π) 3 A (π) 4 δ 4 (k + k k 3 k 4 ) E 4 The integral is clearly Lorentz invariant as is the factor in front of it since the following is Lorentz invariant (k k ) m m (E m ) m m m E m m E v E E v where the left hand side has been evaluated in the rest frame of particle. The third equality comes from E γ m so E ( v ) m which gives E m E v or v k E. We also have, in the center of mass frame, k k k and k 4 k 3 k, (k k ) m m (E E + k ) m m E E + k4 + E E k m m (k + m ) (k + m ) + k4 + E E k m m k 4 + (m + m )k + E E k (k + m + m + E E )k (E + E + E E )k (E + E ) k E CM k The relative velocity of the particles is v (E+E) E E k k E + k E v v so d 3 k 3 d 3 k 4 σ 4vE E (π) 3 E 3 (π) 3 A (π) 4 δ 4 (k + k k 3 k 4 ) E 4 d 3 k 3 d 3 k 4 4E CM k (π) 3 E 3 (π) 3 A (π) 4 δ 4 (k + k k 3 k 4 ) E 4 d 3 k 3 4E CM k (π) 3 A πδ(e + E E 3 E 4 ) E 3 E 4 dω 6π k 3 E CM k E 3 E 4 f (k 3 ) A

26 UNIT. INTERACTIONS where f(k 3 ) E + E E 3 E 4 E CM m 3 + k 3 m 4 + k 3 so f (k 3 ) k3 E 3 E 3+E 4 E 3E 4 k 3 E CM E 3E 4 k 3. Therefore σ k 3 k 64π ECM dω A + k3 E 4 where A is theory dependent. In ϕ 3 theory, we ( have Feynman diagrams ) so ia ( iλ) i i i (k +k ) m + (k k 3) m + (k k 4) m +... Define the Mandelstam variables s (k + k ), t (k k 3 ), u (k k 4 ). We refer to these diagrams by the depen- dence of the propagators: the first diagram is the s-channel, the second is the t-channel, and the third is the u-channel. Cyclic permutations of the k i take s to t to u to s. Also we have crossing symmetry ia(s, t, u) ia(u, s, t). which implies s + t + u k + k + k k + k + k 3 k k 3 + k + k 4 k k 4 6m + k (k k 3 k 4 ) 6m k 4m ( ia ( iλ) i s m + i t m + ) i u m Poles develop as s, t, u m so we expect large cross sections at those values. Now let all particles be incoming so k 3 k 3 and k 4 k 4 so their energies will be negative going back in time (antiparticles). We must change the field to a complex field: two particles (b, b ) and two antiparticles (c, c ) , , (different crossing symmetries). In the center of mass frame, particles - 4 have 4-momentums (E, k), (E, k), (E, k ), (E, k ) respectively : s (E) ECM 4m ; t (k k ) ; u (k + k ). Permute to go to another channel : u 4m, s, t ; permute again to yet another channel : t 4m, s, u. If we reverse everything, there is no effect on A. In fact A i f A f i is a consequence of the CPT theorem. The CPT transformation U CP T is not unitary but anti-unitary because time reversal is anti-unitary. [U CP T, H] [U CP T, H ]. Recall S lim t± ± U(t + )U (t ) where U(t) e iht e iht. Now U CP T U(t)U CP T U( t) so a change of variables shows U CP T SU CP T lim t± ± U( t + )U ( t ) S. Under the CPT transformation (π) 4 δ (p f p i ) ia i f f S i which relates to A f i..5 Unitarity f S i U CP T f U CP T (S ) i U CP T f UCP T (S )U f S i i S f CP T UCP T i S is unitary. Define it S so f T i i f S i (π) 4 δ 4 (p f p i ) A fi. Then ( + it ) ( + it ) it + it + T T so T T it T. f T i f T i i f T T i i f T n n T i n

27 .5 Unitarity which translates to amplitudes ( Afi A ) if (π) 4 δ 4 (p f p i ) i n (π) 4 δ 4 (p f p n ) A nf (π) 4 δ 4 (p n p i ) A ni The second delta function means p n p i so A fi A if i n (π) 4 δ 4 (p i p n ) A nf A ni When f i, we then obtain a form of the Optical Theorem Im A ii (π) 4 δ 4 (p i p n ) A ni n Fock space gives a complete set of states k k...k n a (k )a (k )...a (k n ), n,... so n d 3 k (π) 3 E d 3 k d 3 k (π) 3 E... (π) 3 E k k...k n k k...k n is the identity. If two identical particles scatter to two particles that are identical to the incoming particles, the first term of Im A ii is d 3 k (π) 3 E (π) 4 δ 4 (p i k ) (>- -<); the second term is d 3 k d 3 k (π) 3 E (π) 3 E (π) 4 δ(p i k k ) (> <) and so on. This generalizes to the Optical Theorem Im A ii n n! d 3 k d 3 k n (π) 3... E (π) 3 A ni δ 4 (p i k... k n ) E n Example 4. A single incoming Klein-Gordon particle of momentum p transitioning to itself. The only Lorentz invariant parameter is p s which should not be identified with the mass of the particle because intermediate particles may not be physical. Now A is a function of s: Im A(s) d 3 k (π) 3 E (π) 4 δ 4 (p k ) A +... A which is Lorentz Invariant because the π δ(e E) E Lorentz invariant quantity δ ( k p ) δ ( E k p + p ) δ ( E p) E δ(e p ). Hence Im A(s) π δ(e E ) A πδ ( k p ) A E to first order ( A ). The higher order corrections come from diagrams ( A A ) which i p m +iε and ( A A ) which Example 5. One particle decays into two particles of the same species contributes a factor d 3 k (π) 3 E d 3 k (π) 3 E (π) 4 δ(p k k ) A. In the rest The decay rate calculated earlier is Γ m frame of the incoming particle p ( s, ), k (E, k), and k (E, k). Γ k 3π s which is Lorentz invariant if k s is. E m + k k in terms of s and divide by s to get which is Lorentz invariant. Then mγ 8π dω A ( s k s 4m s 4s 4m s ) by conservation of energy. Solve for dω A If s < 4m, decay is impossible (the initial energy is less than the sum of the masses of the decay products); Γ so Im A and A(s) A (s ).When two analytic function agree on an interval, the agree everywhere except singularities. so Im A except at singularities. Graphically there is a pole at m and a cut at parallel and slightly below the real axis for s > 4m. For n 3, the pole is the same and there is still a single cut but for s > 9m. The physical content of the theory must approach from above.

28 4 UNIT. INTERACTIONS.6 Path Integrals In one-dimensional non-relativistic quantum mechanics, we have position and momentum operators x and p with [ x, p] i. In the Heisenberg picture, x(t) e iht x()e iht and p(t) e iht p()e iht. The eigenstates of x(t) ( p(t)) are x, t e iĥt x, ( p, t e iĥt p, ) and both collections form a complete set of states with x x δ(x x ), p p πδ(p p ), and x p e ixp. To compare momentum and position at different times p, t x, t p, t e ih(t t) x, t p, t + i Ĥ (t t ) x, t for small time intervals. We can regard H as a function of p and x and write p, t + ih (t t ) x, t ( + ih(p, x) (t t )) e ixp Look at the amplitude of transition from x, t to x, t e ih(p,x)(t t) e ixp A x, t x, t ψ(x, t ) which is a solution to the Schrödinger equation; at t t, ψ(x, t ) x x δ(x x ). At a time t between t and t, insert the identity dp π p, t p, t dp dx x, t x, t π p, t p, t to get dp A π dx dp dx dp π π dp π dp π x, t p, t p, t x, t x, t p, t p, t x, t dx dp π eix p eix p, t x, t e ixp p, t x, t p e ixp p, t x, t p, t x, t but the remaining brackets are hard to evaluate since t is not necessarily close to either t or t. We can iterate by choosing t between t and t and inserting the identity dp dx x, t x, t π p, t p, t into the last bracket to get dp dx dp dx dp A π π π dp dx dp π π dx dp π eix eix p e ixp p, t x, t p, t x, t x, t p, p p, t x, t p e ixp e ixp p, t x, t p, t x, t p, t x, t Now divide the interval [t, t ] into equally spaced intervals [t i, t i+ ] of length ε with t < t <... < t n < t, then multiple iteration yields dp dx dp A π π...dx ndp n e ix p e ixp..e ixnpn p, t x, t... p n, t n x n, t n p, t x n, t n π dp dx dp π π...dx ndp n e ix p e ixp..e ixnpn e ih(p,x)(t t) e ixp...e ih(p,x n)(t t n) e ix np π dp dx dp π π...dx ndp n e i(x x n)p e i(x x)p...e i(xn xn )pn e iε(h(p,x)+...+h(pn,xn )+H(p,x n)) π where we have chosen n large enough to make ε small enough to apply the approximation of the first paragraph of this section. As ε, the choices of x i x(t i ) ultimately give a path x(t); similarly

29 .6 Path Integrals we get a path p(t). We can write H(p i, x i ) H(t i ) H(t). x i x i x(t i ) x(t i ) εẋ(t i ). Then dp dx dp A π π dp dx dp π π e i dt(pẋ H) e i dtl e is...dx ndp n e iεẋ(t )p(t ) e iεẋ(t)p(t)...e iεẋ(tn)p(tn) e iε(h(t)+...+h(tn)+h(t )) π...dx ndp n e iε((pẋ H)(t)+...(pẋ H)(tn)) π where S t t dtl is the action. This a solution of the Schrödinger equation. Note that, in the limit ε, we are actually integrating over all possible paths. If H p m + V (x) and ε has a small imaginary part If we define [dx] lim ε dp i π eiε(pẋ H) dp i π eiε dp i π eiε e iε( m ẋ V (x)) e iεl(x,ẋ) m πiε dxi, we have πiε/m ( p iẋ i p i m V (x) ) ( ) (p i mẋ) m + m ẋ V (x) dp i (p i mẋ) π e iε m x, t x, t [dx] e i dtl(x,ẋ) To look at the classical limit, divide i by and let so i dtl. If we look at S and S + δs, the difference will be very large (complete destructive interference) unless δs which implies the classical equations of motion. Green function To look at the Green function x, t x(t )x(t ) x, t with t < t < t < t, insert the identity dx x, t x, t dx x, t x, t to get x, t x(t )x(t ) x, t dx dx x, t x(t ) x, t x, t x, t x, t x(t ) x, t dx dx x x x, t x, t x, t x, t x, t x, t dx dx x x [dx] e is [dx] e is [dx] e is x,t x,t x,t x,t x,t x,t dx dx x(t )x(t )e is x,t x,t Note that x (t ) and x (t ) are operators on the left hand side but are just numbers in the result; numbers commute but operators may not. This is not a contradiction since we started with a time ordered product. We generalize to x, t T (x(t )...x(t )) x, t [dx] x(t )...x(t n )e is

30 6 UNIT. INTERACTIONS (operators on the left, numbers on the right). This is all non-relativistic quantum mechanics, but it is easily brought into QFT. (x, t) is a 4-vector, ϕ is an operator and ϕ(x) denotes the eigenstate of ϕ. ϕ(x ) T (ϕ(x )...ϕ(x n )) ϕ(x ) [dϕ] ϕ(x )...ϕ(x n )e is is a non-perturbative way of calculating any Green function. Example 6. To compare to scattering results we have already obtained, let t and t. In any theory Ω Ω + de E E where Ω is the vacuum. Let ϕ be a Klein-Gordon field. H and H k k + m k. The spectrum of H has one discrete energy (the vacuum) and all others are continuous starting at m. ϕ(x ) Ω ϕ(x ) Ω + de E ϕ(x ) E e iet ϕ(x, ) + deg(e)e iet E ϕ(x, ) E because ϕ(x, t ) e iht ϕ(x, ) and Ω and E are eigenvalues of H. The factor g(e) is the degeneracy at E. We have expressed ϕ(x ) as a function f(t ); it may be impossible to calculate, but we only need its value as t. f(t ) is the Fourier transform of the function g(e) E ϕ(x, ) E ; the Riemann-Lesbegue lemma shows that the Fourier transform of a function approaches as t ±. If E is large, the phase differences are large and, as t ±, there is full destructive interference. ϕ(x, t ) e iet ϕ(x, ) ϕ(x, t ) e iet ϕ(x, ) Ω Ω [dϕ] ϕ(x )...ϕ(x n )e is e ie(t t ) ϕ(x, ) Ω Ω ϕ(x, ) ϕ(x, ) Ω T (ϕ(x )...ϕ(x n )) Ω The last bracket is definitely a physical object, the propagator for n or the scattering amplitude. Thus [dϕ] ϕ(x )...ϕ(x n )e is Ω T (ϕ(x )...ϕ(x n )) Ω [dϕ] e is In terms of diagrams, the components of [dϕ] e is have no external legs (the diagram is a bubble) because it is a vacuum to vacuum transition. Looking at [dϕ] ϕ(x )ϕ(x )e is, the simplest graph is just a line segment, the K-G propagator. We have the connected graphs The disconnected graphs contribute which can be written as (+bubbles)(all graphs with no bubbles). When divided by (+bubbles), we have the propagator. Bubbles represent vacuum to vacuum transitions, but physical processes involve excitations of the vacuum. Therefore bubbles do not contribute to physical reality although they presumably make up the cosmological constant. Let G (n) (x,..., x n ) Ω T (ϕ(x )...ϕ(x n )) Ω [dϕ]ϕ(x)...ϕ(x n)e is [dϕ]e is (graphs without bubbles). Add an external source J so S S + Jϕ. Jϕ is shorthand for d 4 xj(x)ϕ(x). Define Z [J] [dϕ] e i(s+ Jϕ) [dϕ] e is Note e i Jϕ + i ( Jϕ + i! ) ( Jϕ + i3 3! Jϕ)

31 .6 Path Integrals so Z [J] + i [dϕ] Jϕe is [dϕ] e is + i! [dϕ] e is Jϕ Jϕ +... [dϕ] e is Note that Similarly etc. δz δj(x) J i [dϕ] ϕe is G () (x) [dϕ] e is δ Z δj(x )δj(x ) J i [dϕ] ϕ(x )ϕ(x )e is [dϕ] e is G () (x, x )