I-6: the S-Matrix and Time-Ordered Products

2012 Matthew Schwartz I-6: The S-matrix and time-ordered products

1 Introduction As discussed in the previous lecture, scattering experiments have been a fruitful and efficient way to determine the particles which exist in nature and how they interact. In a typical collider experiment, two particles, generally in approximate momentum eigenstates at t = are col- lided with each other and we measure the probability of finding particular outgoing−∞ momentum eigenstates at t = + . All of the interesting interacting physics is encoded in how often given initial states produce∞ given final states, that is, in the S-matrix. The working assumption in scattering calculations is that all of the interactions happen in some finite time T

4 − ip1 x 1 2 4 ipn x n 2 f S i = i d x1 e ( + m ) i d xn e ( + m ) (1) h | | i Z Z

Ω T φ( x ) φ( x ) φ( x ) φ( xn) Ω (2) ×h | { 1 2 3 }| i with the i in the exponent applying for initial states and the +i for final states. In this formula, T − refers to a time-ordered product, to be defined below, and Ω is the ground state or vacuum{} of the interacting theory, which in general may be different from| i the vacuum in a free theory. The time-ordered correlation function in this formula can be very complicated and encodes a tremendous amount of information besides S-matrix elements. The factors of + m2 project onto the S-matrix: + m2 becomes p2 + m2 in Fourier space which vanishes for the asymptotic states. These factors will therefore− remove all terms in the time-ordered product except 1 those with poles of the form p2 − m 2 , corresponding to propagators of on-shell particles. Only the terms with poles for each factor of p2 m2 will survive, and the S-matrix is given by the residue of these poles. Thus the physical content− of the LSZ formula is that the S-matrix projects out single-particle asymptotic states from time-ordered product of fields.

2 The LSZ reduction formula [The following proof follows Srednicki. You can find different versions in various textbooks.] In the last lecture, we derived a formula for the differential cross section for 2 n scattering of asymptotic states → 1 2

dσ = d ΠLIPS (3) Q (2 E )(2 E ) Qv v |M| 1 2 | 1 − 2 | where d ΠLIPS is the Lorentz-invariant phase space, and , which is shorthand for f i , is the S-matrix element with an overall momentum-conservingM δ-function factored out: h |M| i f S 1 i = i(2 π) 4 δ4(Σ p) (4) h | − | i M

1 2 Section 2

The state i is the initial state, at t = and f is the ﬁnal state, at t = + . More precisely, | i † −∞ h | ∞ using the operators a p( t) which create particles with momentum p at time t, these states are i =√ 2 ω √2 ω a † ( ) a † ( ) Ω (5) | i 1 2 p1 −∞ p2 −∞ | i where Ω is the ground state, with no particles, and | i

† † f =√ 2 ω √2 ωn a ( ) a ( ) Ω (6) | i 3 p3 ∞ pn ∞ | i We are generally interested in the case where some scattering actually happens, so let us assume 1 f i in which case the does not contribute. Then the S matrix is | i | i

n/2 † † f S i = 2 √ω ω ω ωn Ω a p3 ( ) a p ( ) a ) a ( ) Ω (7) h | | i 1 2 3 | ∞ n ∞ p1 − ∞ p2 −∞ | This expression is not terribly useful as is. We would like to relate it to something we can compute with our Lorentz invariant quantum fields φ( x) . Recall that we had defined the fields as a sum over creation and annihilation operators 3 d p 1 − ipx † ipx φ( x) = φ( xQ , t) = a p( t) e + a ( t) e (8) Z (2 π) 3 2 ω p p p

2 2 Q where ωp = Qp + m . We also start to use the notation φ( x) = φ( x , t) as well, for simplicity. These are Heisenberg-picturep operators which create states at some particular time. However, the creation and annihilation operators at time t are in general different from those at some other time t ′. An interacting Hamiltonian will rotate the basis of creation and annihilation operators, which encodes all the interesting dynamics. For example, if H is time-independent, iH ( t − t 0 ) − iH ( t − t 0 ) iH ( t − t0 ) − iH ( t − t0 ) a p( t) = e a p( t0) e just like φ( x) = e φ( Qx , t0) e where t0 is some arbitrary reference time where we’ve matched the interacting fields onto the free fields. We

won’t need to use anything at all in this section about a p( t) and φ( Qx , t) except that these opera- Q iQ x p tors have some ability to annihilate ﬁelds at asymptotic times: Ω φ( Qx , t = ) p = Ce for some constant C, as was shown for free ﬁelds in Lecture I-2. h | ±∞ | i The key to proving LSZ is the algebraic relation

4 ipx 2 i d xe ( + m ) φ( x) = 2 ωp[ a p( ) a p( )] (9) Z ∞ − − ∞ p µ Where p = ( ωp,Q p ) . To derive this, we only need to assume that all the interesting dynamics happens in some ﬁnite time interval, T

However, we can safely assume that the ﬁelds die oﬀ at Qx = allowing us to integrate by ±∞ parts in Qx . Then,

4 ipx 2 4 ipx 2 Q 2 i d xe ( + m ) φ( x) =i d xe ∂t ∂x + m φ( x) Z Z − 4 ipx 2 2 2 =i d xe ( ∂t + Qp + m ) φ( x) Z 4 ipx 2 2 =i d xe ( ∂t + ωp) φ( x) Z Note this is true for any kind of φ( x) , whether classical ﬁeld, or operator. Also,

ipx ipx ipx 2 ∂t[ e ( i∂t + ωp) φ( x)] =[ iωp e ( i∂t + ωp) + e ( i∂t + ωp∂t)] φ( x) ipx 2 2 =ie ( ∂t + ωp) φ( x) which holds independent of boundary conditions. So,

4 ipx 2 4 ipx i d xe ( + m ) φ( x) = d x∂t[ e ( i∂t + ωp) φ( x)] (10)

Z Z Q iω p t 3 − iQ p x = dt∂t e d xe ( i∂t + ωp) φ( x) (11) Z Z The LSZ reduction formula 3

Again, this is true for whatever kind of crazy interaction field φ( x) might be. This integrand is a total derivative in time, so it only depends on the fields at the boundary † t = . By construction, our a p( t) and a ( t) operators are time-independent at late and early ±∞ p times. For the particular case of φ( x) being a quantum field, Eq. (8), we can do the xQ integral.

d3 k

Q Q Q 3 − iQ p x 3 − i p x 1 − ikx d x e ( i∂t + ωp) φ( x) = d x e ( i∂t + ωp) ak ( t) e + Z Z Z (2 π) 3 √2 ω k † ikx ak ( t) e (12)

d3 k ω ω ω ω

Q Q Q 3 k + p − ikx − iQ p x k + p † ikx − i p x = 3 d x ak ( t) e e + − ak ( t) e e (13) Z (2 π) Z √2 ωk √2 ωk

Here we used ∂tak ( t) = 0, which is not true in general, but true at t = where the ﬁelds are

±∞ 3 Q Q free, which is the only region relevant to Eq. (11). The Qx integral gives a δ ( p k ) in the ﬁrst −

3 Q

term and a δ ( Qp + k ) in the second term. Either way, it forces ωk = ωp and so we get Q 3 − iQ p x − iω p t d xe ( i∂t + ωp) φ( x) = 2 ωp a p( t) e (14) Z Thus, p

4 ipx 2 iω p t − iω p t i d xe ( + m ) φ( x) = dt∂t[( e ) 2 ωp a p( t) e ] Z Z p = 2 ωp[ a p( ) a p( ) ∞ − − ∞ which is what we wanted. Similarly (by takingp the Hermitian conjugate)

† † 4 − ipx 2 2 ωp a ( ) a ) = i d xe ( + m ) φ( x) (15) p ∞ − p − ∞ − Z p Now we’re almost done. We wanted to compute

n † † f S i =√ 2 ω ωn Ω a p3 ( ) a p ( ) a ) a ( ) Ω (16) h | | i 1 | ∞ n ∞ p1 − ∞ p2 −∞ | † and we have an expression for a p( ) a p( ) . Note that all the initial states have a opera- ∞ − −∞ p tors and and the ﬁnal states have a p operators and + , so the operators are already in time order,−∞ ∞

n † † f S i =√ 2 ω ωn Ω T a p3 ( ) a p ( ) a ( a ( ) Ω (17) h | | i 1 | { ∞ n ∞ p1 − ∞ p2 −∞ }| where the time-ordering operation T indicates that all the operators should be ordered so that those at later times are always to the{} left of those at earlier times. That is, T just man- handles the operators within the brackets, placing them in order regardless of whether{} they commute or not. Time-ordering lets us write the S-matrix element as n f S i =√ 2 ω ωn (18) h | | i 1 × † † † †

Ω T [ a p3 ( ) a p3 ( )] [ a p ( ) a p ( )][ a ( ) a ( )][ a ( ) a ( )] Ω h | { ∞ − − ∞ n ∞ − n − ∞ p1 − ∞ − p1 ∞ p2 − ∞ − p2 ∞

The time-ordering migrates all the unwanted a † ( ) operators associated with the initial states to the left, where they annihilate on f and all∞ the unwanted a( ) operators to the right, h | −∞ where they annihilate i . Then there is no ambiguity1 in commuting the a † ( ) past the | i pi ∞ a p ( ) and all the stuﬀ we don’t want drops out of this expression. j ∞ The result is then

4 − ip1 x 1 2 4 ipn x n 2 p3 pn S p1 p2 = i d x1 e ( 1 + m ) i d xn e ( n + m ) (19) h | | i Z Z

Ω T φ( x ) φ( x ) φ( x ) φ( xn) Ω (20) ×h | { 1 2 3 }| i 1. The only possible subtlety is if some momenta are identical, which would correspond to forward scattering. This ambiguity can be resolved by a careful consideration of the T → ∞ limit; the result is the same as the ana- lytic continuation of the case when all momenta are diﬀerent. Forward scattering is discussed more in lecture III- 10 on unitarity. 4 Section 2

∂ 2 where i = µ which agrees with Eq. (2). This is the LSZ reduction formula. ∂x i

2.1 Discussion The LSZ reduction says that to calculate an S-matrix element, multiply the time-ordered pro- 2 duct of fields by some + m factors and some phases. Remember that p0 = ωp for the phase space factors, indicating that the external states in i and f are on-shell. If the fields φ( x) | i | i were free fields, which they are by assumption at t = , they would satisfy ( + m2 ) φ( x, t) = 0 2 ±∞ and so the ( i + m ) terms would give zero. However, as we will see, when calculating ampli- 1 tudes, there will be factors of the propagators 2 for the one particle states. These blow up + m as ( + m2 ) 0. The LSZ formula guarantees that the zeros and infinities in these terms cancel, → leaving a nonzero result. Moreover, the + m2 terms will kill anything which does not have a divergence, which will be anything but the exact initial and final state we want. 2 That’s the whole point of the LSZ formula: it isolates the asymptotic states by adding a carefully constructed zero to cancel everything that doesn’t correspond to the state we want. It’s easy to think that LSZ is totally trivial, but it’s not. The projections are the only thing that tells us what the initial states are (the things created from the vacuum at t = ) and −∞ what the final states are (the things that annihilate into the vacuum at t = + ). Initial and ∞ final states are distinguished by the i in the phase factors. The time ordering is totally physical: all the creation of the initial states± happens before the annihilation of the final states. In fact, because this is true not just for free fields, all the crazy stuff which happens at intermediate times in an interacting theory must be time-ordered too. But the great thing is that we don’t need to know which are the initial states and which are the final states anymore when we do the hard part of the computation. We just have to calculate time ordered products, and the LSZ formula sorts out what is being scattered for us.

2.2 LSZ for operators For perturbation theory in the standard model, which is mostly what we will study in these lectures, the LSZ formula in the above form is all that is needed. However, the LSZ formula is more powerful than it seems and applies even if we don’t know what the particles are. If you go back through the derivation, you will see that we never needed an explicit form for † the full ﬁeld φ( x) and its strange creation operators a p( t) which did not necessarily evolve like creation operators in the free theory. In fact, all we used was that the ﬁeld φ( x) creates free particle states at asymptotic times. So the LSZ reduction actually implies

4 − ip1 x 1 2 4 ipn x n 2 p3 pn S p1 p2 = i d x1 e ( 1 + m ) i d xn e ( n + m ) (21) h | | i Z Z

Ω T ( x ) ( x ) ( x ) n( xn) Ω (22) ×h | {O 1 1 O 2 2 O3 3 O }| i where the i( x) are any operators which can create single particle states. By this we mean, that O p ( x) Ω = Zeipx (23) h |O | i for some number Z. LSZ does not distinguish elementary particles, which we define to mean particles which have corresponding fields appearing in the Lagrangian, from any other type of particle. Anything which overlaps with 1-particle states will produce an appropriate pole to be canceled by the + m2 factors giving a non-zero S-matrix element. Therefore particles in the Hilbert space can be produced whether or not we have elementary fields for them.

2. It should not be obvious at this point that there cannot be higher order poles, such as 1 , coming ( p2 − m 2 ) 2 out of time-ordered products. Such terms would signal the appearance of unphysical states known as ghosts, which violate unitarity. The fastest a correlation function can decay at large p2 in a unitary theory is as p− 2 , a result we will prove in Lecture III-10. Feynman propagator 5

It is probably worth saying a little more about what these operators n( x) are. The operators can be deﬁned as they would be in quantum mechanics, by their matrixO elements in a basis of states ψn of the theory Cnm = ψn ψm . Any such operator can be written as a sum over creation and annihilation operatorsh |O| i

† †

= dq1 dqn dp1 dpm a q a p a p1 Cnm( q1 , ,pm) (24) O Z q1 qn m n,mX

It is not hard to prove [cf Weinberg 4.2] that the Cnm are in one-to-one correspondence with the matrix elements of in n and m particle states. One can turn the operator into a functional of the fields, using Eq.O (14) and its conjugate. The most important operators in relativistic quantum field theory are Lorentz-covariant composite operators constructed out of elementary fields in our theory, e.g. ( x) = φ( x) ∂µφ( x) ∂µφ( x) . However, some operators, such as the Hamiltonian, are not LorentzO invariant. Other operators, such as Wilson lines (see Lecture IV-1) are non-local. Also, non-Lorentz invariant operators are essential for many condensed matter applications. As an example of this generalized form of LSZ, suppose there were a bound state in our theory, like positronium. We will derive the Lagrangian for quantum electrodynamics in Lecture II-6. We will find, as you might imagine, that it is a functional of only the electron, positron and photon fields. Positronium is a composite state, composed of an electron, a positron and lots of photons binding them together. It has the same quantum numbers as the operator P( x) = ¯ ¯ O ψe ( x) ψe ( x) where ψ ( x) and ψ( x) are the fields for the positron and electron. Thus P( x) should have some nonzero overlap with positronium, and we can insert it into the time orderedO product to calculate the S-matrix for positronium scattering or production. This is an important conceptual fact: there do not have to be fundamental fields associated with every asymptotic state in the theory to calculate the S-matrix . Conversely, even if we don’t know what the elementary particles actually are in the theory, we can introduce fields corresponding to them in the Lagrangian to calculate S-matrix elements in perturbation theory. For example, in studying the proton or other nucleons, we can treat them as elementary particles. As long as we are interested in questions which do not probe the substructure of the proton, such as non-relativistic scattering, nothing will go wrong. This is a general and very useful technique, known generally as effective field theory, which will play an important role in these lectures. Thus, quantum field theory is very flexible: it works if you use fields which don’t correspond to elementary particles (effective field theories) or if you scatter particles which do not have corresponding fields (such as bound states). It even can provide a predictive description of unstable composite particles, such as the neutron, which have neither elementary fields nor are proper asymptotic states.

3 Feynman propagator

To recap, our immediate goal, as motived in the previous lecture, is to calculate cross sections, which are determined by S-matrix elements. We now have an expression, the LSZ reduction formula, for S-matrix elements in terms of time-ordered products of fields. Now we need to figure out how to compute those time-ordered products. As an example, we will now calculate a time- ordered product in the free theory. In the next lecture, we derive a method for computing time- ordered products in interacting theories using perturbation theory. We start with the free field operator

3 d k 1 − ikx † ikx φ0( x, t) = ak e + a e (25) Z (2 π) 3 √2 ω k k 2 †

2 Q where k0 = ωk = m + k and ak and ak are time-independent (all time-dependence is in the phase). Then, usingp 0 instead of Ω to denote the vacuum in the free theory, | i | i d3 k d3 k 1 1 φ x φ x 1 2 a a † ei ( k 2 x 2 − k 1 x 1 ) 0 0( 1 ) 0( 2 ) 0 = 3 3 0 k 1 k 2 0 h | | i Z (2 π) Z (2 π) √2 ωk 1 √2 ωk 2 | |

6 Section 3

† Q 3 3 Q The 0 ak 1 a 0 gives a (2 π) δ ( k k ) so that | k 2 | 1 − 2 3 d k 1 ik ( x 2 − x 1 ) 0 φ0( x1 ) φ0( x2 ) 0 = 3 e (26) h | | i Z (2 π) 2 ωk

Now, we are interested in 0 T φ( x1 ) φ( x2 ) 0 . Recalling that time-ordering puts the later ﬁeld on the left, we get h | { }| i

0 T φ0( x1 ) φ0 ( x2 ) 0 = 0 φ0 ( x1 ) φ0( x2 ) 0 θ( t1 t2 ) + 0 φ0( x2 ) φ0( x1 ) 0 θ( t2 t1 ) h | { }| i h | 3 | i − h | | i − d k 1 ik ( x 2 − x 1 ) ik ( x 1 − x 2 ) = 3 e θ( t1 t2 ) + e θ( t2 t1 ) Z (2 π) 2 ωk − −

Q Q

Q Q Q d k 1 i k ( Qx 1 − x 2 ) − iωk τ − i k ( x 1 − x 2 ) iωk τ = 3 e e θ( τ) + e e θ( τ) Z (2 π) 2 ωkh − i (27) where τ = t t . Taking k k in the ﬁrst term leaves the volume integral d3 k invariant 1 − 2 → − and gives R

Q Q d k 1 − i k ( xQ 1 − x 2 ) iωk τ − iωk τ 0 T φ0( x1 ) φ0( x2 ) 0 = 3 e [ e θ( τ) + e θ( τ)] (28) h | { }| i Z (2 π) 2 ωk − The two terms in this sum are the advanced and retarded propagators which we saw were relevant in relativistic calculations using old-fashioned perturbation theory.

τ > 0 contour ω

−ωk + iǫ

ωk − iǫ

τ < 0 contour

Figure 1. Contour integral for the Feynman propagator. Poles are at ω = ±ωk ∓ iε. For τ > 0 we close the contour upward picking up the left pole, for τ < 0 we must close the contour downward, picking up the right pole.

The next step is to simplify the right hand side using the mathematical identity ∞ 2 ωk dω e − iωk τ θ( τ) + eiωk τ θ( τ) = lim − eiωτ (29) − ε → 0 2 π i Z ω2 ω2 + iε −∞ − k To derive this identity, ﬁrst separate out the poles with partial fractions

1 1 1 1 1 2 = = (30) ω2 ω + iε [ ω ( ωk iε)][ ω ( ωk + iε)] 2 ωk ω ( ωk iε) − ω ( ωk + iε) − k − − − − − − − − 2 Here, we have dropped terms of order ε and wrote 2 ε ωk = ε, which is fine since we will take ε 0 in the end. The location of the two poles in the complex plane is shown in Figure 1. → Integrating the first fraction from < ω < along the real line multiplied by the phase eiωτ gives 0 if τ > 0 since we close the−∞ contour up,∞ and picks up the pole if τ < 0 when we must close the contour down. That is, ∞ dω eiωτ = 2 π ieiωk τ θ( τ) + O( ε) (31) Z ω ( ωk iε) − − −∞ − − Feynman propagator 7 with the extra minus sign coming from doing the contour integration clockwise. For the second fraction, ∞ dω eiωτ = 2 π ie − iωk τ θ( τ) + O( ε) (32) Z−∞ ω ( ωk + iε) Thus, − − ∞ dω 2 π i lim eiωτ = [ eiωk τ θ( τ) + e − iωk τ θ( τ)] (33) 2 2 ε → 0 Z−∞ ω ωk + iε − 2 ωk − as desired. − Putting it together, we find 3

d k Q ω Q 1 − i k ( Qx 1 − x 2 ) 2 k 1 iωτ 0 T φ0( x1 ) φ0( x2 ) 0 = lim 3 e dω − 2 e (34) h | { }| i ε → 0 Z (2 π) 2 ωk Z 2 π i ω2 ω + iε − k Letting the limit be implicit, this is:

4 d k i ik ( x 1 − x 2 ) DF ( x1 , x2 ) = 0 T φ0( x1 ) φ0( x2 ) 0 = e (35) h | { }| i Z (2 π) 4 k2 m2 + iε − This beautiful Lorentz invariant object is called the Feynman propagator. It has a pole at k2 = m2 , exactly to be canceled by prefactors in the LSZ reduction formula in the projection onto one-particle states. Keep in mind:

Q 2 2 k k + m anymore. It is a separate integration variable. The propagating field can • 0 q be off-shell! The i comes from a contour integral. We will always get factors of i in two-point func- • tions of real fields. The ε is just a trick for representing the time ordering in a simple way. We will always • take ε 0 at the end, and often leave it implicit. You always need a +iε for time ordered products,→ but is really just short hand for a pole-prescription in the contour integral, which is exactly equivalent to adding various θ( t) factors. For ε = 0 the Feynman propagator looks just like a classical Green’s function for the • 2 4 Klein-Gordon equation ( + m ) DF ( x, y) = iδ ( x) with certain boundary conditions. That’s because it is. We are just computing classical− propagation in a really complicated way. As we saw in Lecture I-4 using old-fashioned perturbation theory, when using physical • intermediate states there are contributions from advanced and retarded propagators, both of which are also Green’s functions for the Klein-Gordon equation. The Lorentz-invariant Feynman propagator encodes both of these contributions, with its boundary condition represented by the iε in the denominator. The advanced and retarded propagators have more complicated integral representations, as you can explore on Problem 1.