Lehmann–Symanzik–Zimmermann (LSZ) Reduction Formula

Earlier in class (see my notes) I explained the two-point correlation functions and showed how their poles are related to the physical masses of particles and the strengths of bare ﬁelds. The LSZ reduction formula focuses on the n> 2 correlation functions

def (x ,...,x ) = Ω TΦ(ˆ x ) Φ(ˆ x ) Ω (1) Fn 1 n h | 1 ··· n | i

— where the quantum ﬁelds Φ(ˆ x) are in the Heisenberg picture of quantum mechanics — and relates the poles of their Fourier transforms

(p ,...,p ) = d4x eip1x1 d4x eipnxn (x ,...,x ), (2) Fn 1 n 1 ··· n ×Fn 1 n Z Z to the S–matrix elements. The poles happen when any of the n momenta pi approaches the mass shell, p2 M 2 , and the most interesting pole is the simultaneous pole when all n i → phys momenta go on-shell. Speciﬁcally, let

p0 +E(p ) = + p + M 2, ..., p0 +E(p ) (3) 1 → 1 1 k → k p for some k < n, but

p0 E(p ), ..., p0 E(p ). (4) k+1 → − k+1 n → − n

In this limit, Lehmann–Symanzik–Zimmermann formula (published in 1955) gives us

n √ i Z ˆ n(p1,...,pn) 2 2 pk+1,..., pn S p1,...,pk . (5) F −−−−→on shell pi M + iǫ × h− − | | i Yi=1 −

The field-strength factors √Z in this formula stem from the in eq. (1) being the correlation Fn of the bare fields. If we re-define it as the correlation function of the renormalized fields — or equivalently, use the counterterm perturbation theory to calculate the correlation function, —

1 then we would get

renormalized = Z−n/2 bare (6) Fn Fn

and consequently

n renormalized i ˆ n (p1,...,pn) 2 2 pk+1,..., pn S p1,...,pk . (7) F −−−−→on shell pi M + iǫ × h− − | | i Yi=1 −

Perturbation Theory for the Correlation Functions

Before deriving the LSZ reduction formula, let me show how the poles at p2 M 2 arise i → from formal resummation of the perturbation theory. For simplicity, let’s focus on the connected correlation functions, which in perturbation theory obtain as

all connected diagrams conn(p ,...,p ) = . (8) Fn 1 n with n external vertices X

Topologically, a general diagram of this kind has an amputated core, plus any number of external leg bubbles on any of the n external legs, thus

2 1PI 1PI

1PI 1PI

amputated core

1PI 1PI

(9)

Each external leg bubble here is one-particle irreducible (1PI) — if we cut any propagator internal to the bubble, it stays connected. As to the amputated core, if we cut any propagator internal to the core, it may stay connected or break into two disconnected parts, but if it breaks then each part must remain connected to at least two external legs.

A general diagram (9) contributing to the connected n-point function may have:

Any kind of amputated core with n external legs. • Any number N =0, 1, 2,... of leg bubbles in each of the n legs. • i And any such bubble may be any kind of 1PI subgraph with 2 external legs. • ⋆ Most importantly, we may chose any amputated core we like and any leg bubbles we like completely independently from each other.

3 Consequently, when we formally sum over all connected Feynman diagrams with n external vertices, the sum factorizes into a product of a sum over the cores and a sum over the bubbles in each leg,

n connected amputated external conn(p ,...,p ) = = . (10) Fn 1 n diagrams cores × leg factors X X Yi=1 where each external leg factors includes the blue propagators and the leg bubbles, and should be summed over all numbers of bubbles of any kinds. In general, for Ni bubbles we have Ni +1 blue propagators with ﬁxed momentum pi, thus

∞ i i external i N +1 single N = 2 2 leg factor i pi mb + iǫ × bubbles NXi=0 − X ∞ Ni+1 Ni i 2 (11) = 2 2 iΣ(pi ) pi mb + iǫ × − NXi=0 − h i i 2 = = 2(p ), p2 m2 Σ(p2)+ iǫ F i − b − exactly as in the two-point correlation function. Therefore,

n amputated conn(p ,...,p ) = (p2) . (12) Fn 1 n F2 i × cores Yi=1 X

This formula explains where the poles in the correlation functions come from: When any of the momenta p goes on-shell, p2 M 2 , the corresponding (p2) has a pole, i i → phys F2 i

2 iZ 2(p ) = + ﬁnite, (13) F p2 M 2 + iǫ i − which translates into the pole of the whole product (12). When all of the n momenta go on

4 shell at the same time, all n of the (p2) factors develop poles, thus F2 i

n conn iZ amputated n (p1,...,pn) 2 2 ) . (14) F −→ pi M + iǫ × cores Yi=1 − X Note the combined residue of this n-fold pole,

conn n amputated Residue [ (p1,...,pn)] 2 2 = (iZ) . (15) Fn all pi →M × cores X

Thus far, we have explained the poles of the Lehmann–Symanzik–Zimmermann formula (5). In a moment, we should compare the residues. But ﬁrst let’s cluster-expand the LHS of the LSZ formula into connected correlation functions while the S-matrix element on the RHS likewise expands into connected and disconnected pieces. For example, for n = 4 we have

(p ,p ,p ,p ) = conn(p ,p ,p ,p ) F4 1 2 3 4 F4 1 2 3 4 + (p ,p ) (p ,p ) + (p ,p ) (p ,p ) (16) F2 1 2 ×F2 3 4 F2 1 3 ×F2 2 4 + (p ,p ) (p ,p ), F2 1 4 ×F2 2 3 and at the same time

p , p Sˆ p ,p = (2π)4δ(4)(p + p + p + p ) p , p i p ,p h− 3 − 4| | 1 2i 1 2 3 4 × h− 3 − 4| M| 3 4i + p Sˆ p p Sˆ p + p Sˆ p p Sˆ p (17) h− 2| | 1i × h− 4| | 3i h− 3| |c1i×h− 4| | 2i + p Sˆ p p Sˆ p . h− 4| | 1i × h− 3| | 2i In terms of the LSZ formula for n = 4, we get 4 terms — 1 connected and 3 disconnected — on each side of the equation. The corresponding disconnected terms on the left-hand and the right hand sides match each other by the LSZ formula for n = 2, so the connected terms should also match each other,

4 conn i√Z 4 (4) 4 (p1,p2,p3,p4) 2 2 (2π) δ (pnet) p3, p4 i p1,p2 , F −−−−→on shell pi M + iǫ! × × h− − | M| i Yi=1 − c (18)

5 Likewise, for n> 4

n conn i√Z 4 (4) n (p1,...,p4) 2 2 (2π) δ (pnet) F −−−−→on shell pi M + iǫ! × (19) Yi=1 − p ,..., p i p ,...,p × h− k+1 − n| M| 1 ki c 0 0 0 0 for p1,...,pk > 0 while pk+1,...,pn < 0.

Now let’s compare the residue of the combined pole here,

conn √ n 4 (4) Residue [ n (p1,...,pn)]on shell = (i Z) (2π) δ (pnet) F × (20) p ,..., p i p ,...,p , × h− k+1 − n| M| 1 ki c to the residue (15) of the same connected correlation function which obtains from the Feynman rules. Matching the two expressions, we see that the LSZ formula implies

amputated (2π)4δ(4)(p ) p ,..., p i p ,...,p = Zn/2 . (21) net × h− k+1 − n| M| 1 ki × cores X c And this is why we calculate the scattering amplitudes using only the amputated Feynman diagrams!

The factor Zn/2 in eq. (21) follows from using the correlation functions for bare ﬁelds and hence the bare perturbation theory for the amputated diagrams. In the counterterm perturbation theory, this factor goes away and we are left with

amputated cores (2π)4δ(4)(p ) i (on shell p ,...,p ) = . (22) net × M 1 n with n external lines X

6 Deriving the LSZ Reduction Formula

Now that we know what is the LSZ reduction formula good for, let’s prove it. Let’s start by focusing on a single momentum, say p1, and look for the quantum origin of the poles when that momentum goes on shell, p0 E(p ). To simplify our notations, we keep the (x ,...,x ) 1 → ± 1 2 n coordinates of the n-point correlation function in the coordinate basis, only the x1 gets Fourier transformed to the momentum basis, thus

(p ; x ,...,x ) = d4x e−ip1x1 (x , x ,...,x ). (23) Fn 1 2 n 1 ×Fn 1 2 n Z Let’s split the time integral here over t = x0 into 3 integration ranges: Range (i) from to 1 1 −∞ some very early but ﬁnite time T1; range (II) from T1 to some very late but ﬁnite time T2; and range (III) from T to + . Thus, 2 ∞ 3 0 0 (p ; x ,...,x ) = dt e−ix1p1 d3x e−ix1p1 (x , x ,...,x ). (24) Fn 1 2 n 1 × 1 ×Fn 1 2 n Xi=1range#Z i Z

Note that in the coordinate space, the (x , x ,...,x ) is an analytic function of the xµ. Fn 1 2 n 1 Consequently, integrating the phase over a finite range#2 of time cannot possibly produce Fn × a pole — all such integrals are analytic and finite. Instead, the poles at p0 = E(p ) must 1 ± 1 come from integrating over the semi-infinite time ranges #1 and #3. So let’s take a closer look at these time ranges.

For the ﬁrst time range x0 < T , the x point is earlier than all the other n 1 points 1 1 1 − x2,...,xn, hence

T Φ(ˆ x ) Φ(x ) = T Φ(ˆ x ) Φ(x ) Φ(ˆ x ) (25) 1 ··· n 2 ··· n × 1 and therefore

(x , x ,...,x ) = Ω T Φ(ˆ x ) Φ(x ) Φ(ˆ x ) Ω Fn 1 2 n h | 2 ··· n × 1 | i = Ω T Φ(ˆ x ) Φ(x ) Ψ Ψ Φ(ˆ x ) Ω , (26) h | 2 ··· n | i×h | 1 | i |Ψi X where the sum is over all quantum states Ψ. Similar to what we did in an earlier class for the two-point functions (see my notes, pages 9–11), we restrict the sum to the quantum states

7 which can be created by the ﬁeld Φˆ from the vacuum Ω , and then we label such states as ψ, q | i | i where qµ is the net momentum of the state while ψ denotes the rest of its quantum numbers, discrete or continuous. Consequently,

d3q 1 = (27) (2π)3 2E(q; ψ) X|Ψi Xψ Z for

q0 = +E(q, ψ) = + q2 + M 2(ψ). (28) q Also, the x and the q dependence of the matrix element ψ, q Φ(ˆ x ) Ω obtain as simply 1 h | 1 | i

ψ, q Φ(ˆ x ) Ω = e+iqx1 ψ Φˆ Ω . (29) h | 1 | i ×h | | i

Plugging these formulae into eq. (26) and hence into the range#1 contribution to the (p ; x ,...,x ), we arrive at Fn 1 2 n

T1 range#1 d3q 1 = dt d3x e−ix1p1 Ω T Φ(ˆ x ) Φ(x ) ψ, q contribution 1 1 (2π)3 2E(q; ψ) h | 2 ··· n | i Z Z ψ Z −∞ X ψ Φˆ Ω e+iqx1 . ×h | | i × (30)

Now let’s integrate over the x1 before integrating over q and summing over ψ. The only x -dependent factors here are the e+ip1x1 e−iqx1 , so the space integral 1 ×

d3x e+ip1·x1 e−iq·x1 = (2π)3δ(3)(q p ) (31) 1 × − 1 Z

0 sets q = p1, and hence q = +E(q, ψ)=+E(p, ψ). Consequently, the time integral over the range#1 becomes

T1 T1 0 0 0 0 T1(−ip1+iE(p)+ǫ) −iT1(p1−E(p)) −it1p1 +it1E(p) t1(−ip1+iE(p)+ǫ) e ie dt1 e e = dt1 e = 0 = 0 . × ip1 + E(p)+ ǫ p1 E(p)+ iǫ −∞Z −∞Z − − (32)

8 Consequently, the big integral (30) reduces to

0 −iT1(p1−E(p)) range#1 1 ie ˆ ˆ = 0 Ω T Φ(x2) Φ(xn) ψ, p ψ Φ Ω contribution 2E(p1, ψ) × p1 E(p)+ iǫ ×h | ··· | i×h | | i Xψ − (33) Note that for discrete state ψ there is a pole at p0 =+E(p, ψ)=+ p2 + M(ψ)2. In particular, the one-particle state of physical mass M contributes the pole p

0 ie−iT1(p1−E(p)) Ω T Φ(ˆ x ) Φ(x ) 1 : p √Z (34) 2E(p)(p0 E(p)+ iǫ ×h | 2 ··· n | i × − where the √Z factor comes from 1 Φˆ Ω = √Z. Moreover, near the pole | i | i

0 ie−iT1(p1−E(p)) i i = + ﬁnite = + ﬁnite. (35) 2E(p)(p0 E(p)+ iǫ (p0)2 E2(p)+ iǫ p2 M 2 + iǫ − 1 − 1 − phys so we may rewrite the pole in usual the relativistic form.

So here is the bottom line: for p1 going to the positive-energy mass shell, the correlation function (p ; x ,...,x ) has a pole i/(p2 M 2 + iǫ) with residue Fn 1 2 n 1 −

residue = √Z Ω T Φ(ˆ x ) Φ(x ) 1 : p . (36) ×h | 2 ··· n | 1i This pole comes from the ﬁrst range of the time integration, < x0 < T . −∞ 1 1 The third range of time integration, T < x0 < + , can be handled ina similar manner. 2 1 ∞ To save time, let me skip the gory details of the calculation and simply give you the bottom line. This time, the pole is for p going to the negative-energy mass shell, p0 E(p, M), 1 1 → − and its residue is

residue = √Z 1:( p ) T Φ(ˆ x ) Φ(x ) Ω . (37) ×h − 1 | 2 ··· n | i

9 ⋆ ⋆ ⋆

Thus far, we have focused on the pole for a single momentum p1 going on shell. Now consider the

(p ,p ; x ,...x ) = d4x e−ip1x1 d4x e−ip2x2 (x , x , x ,...,x ) (38) Fn 1 2 3 n 1 2 ×Fn 1 2 3 n Z Z and take both momenta p and p on-shell at the same time, say p0 +E(p , M) and p0 1 2 1 → 1 2 → +E(p2, M). Similar to what we had for a single momentum, this time we get a combined pole i i (39) p2 M 2 + iǫ × p2 M 2 + iǫ 1 − 2 − which emerges from the time integrals dt and dt over the asymptotic past range, t , t 1 2 1 2 → . For two fields Φ(ˆ x ) and Φ(ˆ x ) at the asymptotic past points, we have −∞ 1 2 R R Ω TΦ(ˆ x ) Φ(ˆ x ) Ω = Ω TΦ(ˆ x ) Φ(ˆ x ) Ψ Ψ TΦ(ˆ x )Φ(ˆ x ) Ω , (40) h | 1 ··· n | i h | 3 ··· n | i×h | 1 2 | i XΨ and the pole (39) comes from Ψ = in(p ,p ) — the asymptotic incoming state of two | i | 1 2 i particles. Strictly speaking, the two particles here do not have exactly definite momenta p1 and p2 but rather wave-packet states of small δp, so in the coordinate picture these two wave packets have finite sizes. Consequently, in the asymptotic past when the two particles were very far from each other, their respective wave packets do not overlap, and the particles do not interact until they approach each other at later times. In this regime

in(q , q ) TΦ(ˆ x )Φ(ˆ x ) Ω = in(q ) Φ(ˆ x ) Ω in(q ) Φ(ˆ x ) Ω + (q q ) h 1 2 | 1 2 | i h 1 | 1 | i×h 2 | 2 | i 1 ↔ 2 = Z wavepacket(x1) wavepacket(x1) +(p1 p2) (41) × q1 × q2 ↔ Z eiq1x1 eiq2x2 + (p p ) ≈ × × 1 ↔ 2 and consequently

i√Z i√Z n(p1,p2; x3 ...,xn) 2 2 2 2 F −−−−−−−−−→p1,p2→mass shell p M + iǫ × p M + iǫ (42) 1 − 2 − Ω TΦ(ˆ x ) Φ(ˆ x ) in(p ,p ) . ×h | 3 ··· n | 1 2 i

Finally, let’s Fourier transform the remaining coordinates x3,...,xn to momenta p3,...,pn), and the take all these momenta to the negative-energy mass shell, each p E(p , M). This i → − i

10 tome, integrating each time variable t3,...,tn over the asymptotic future range, hence

Ω TΦ(ˆ x ) Φ(ˆ x ) in(p ,p ) = Ω TΦ(ˆ x ) Φ(ˆ x ) Ψ Ψ in(p ,p ) (43) h | 3 ··· n | 1 2 i h | 3 ··· n | i×h | 1 2 i XΨ where the pole i i (44) p2 M 2 + iǫ ×···× p2 M 2 + iǫ 3 − n − comes from Ψ = out( p ,..., p ) — the asymptotic outgoing state of n 2 particles. h | h − 3 − n | − Treating this state just as we treated the incoming 2-particle state, we go through a bit of algebra and eventually arrive at

n i√Z n(p1,...,pn) = 2 2 out( p3,..., pn) in(p1,p2) . (45) F −−−−−−−−−→all pi→mass shell pi M + iǫ! ×h − − | i Yi=1 −

Throughout these notes we were working in the Heisenberg picture of the quantum mechanics, so the asymptotic incoming and outgoing states in eq. (45) are the Heisenberg-picture states. Translating them into the interaction picture turns the Dirac bracket of the in and | i out states into the S-matrix element, h |

out( p ,..., p ) in(p ,p ) = out( p ,..., p ) Sˆ in(p ,p ) . (46) h − 3 − n | 1 2 iH h − 3 − n | | 1 2 iI

Consequently, eq. (45) becomes the Lehmann–Symanzik–Zimmermann formula

n √ i Z ˆ n(p1,...,pn) = 2 2 out( p3,..., pn) S in(p1,p2) . (5) F −−−−−−−−−→all pi→mass shell pi M + iǫ!×h − − | | i Yi=1 − Quod erat demonstrandum.