PHY 610 QFT, Spring 2015 HW10 Solutions

QFT HW10 SOLUTIONS

1. (a) [Ψ] = (d − 1)/2

(b) [gn] = d − n(d − 1) = n − (n − 1)d

2 (c) A scalar ﬁeld has dimension [ϕ] = d/2 − 1 from the kinetic term (∂µϕ) , so [gm,n] = d − n(d − 1) − m(d/2 − 1) = n + m − (n + m/2 − 1)d.

(d) The only renormalizable interaction in d = 4 is g1,1ϕΨΨ¯ (the Yukawa interaction).

2. We have computed the one loop corrections to the propagator and vertex in ϕ4 theory in the previous homework; they are, in the MS scheme,  Z = 1 + O[λ]2,  ϕ  α Z = 1 + + O[λ]2, m  3α Z = 1 + + O[λ]2, λ

where α = λ/(4π)2. Recall that the beta function is deﬁned as the dependence of the coupling α on the renormalization 4 4 1/2 scale µ. In terms of bare parameters, Zλλµ˜ ϕ = λ0ϕ0, with ϕ0 = Zϕ ϕ, so the bare coupling is −2 α0 = ZλZϕ µ˜ α. The bare coupling is µ-independent, so

d ∂ log Z ∂ log Z 1 d log α 0 = log α = λ − 2 ϕ + + . d log µ 0 ∂α ∂α α d log µ

Multiplying both sides by α, we have

∂ log Z ∂ log Z dα α λ − 2α ϕ + 1 + α = 0. ∂α ∂α d log µ

In a renormalizable theory, dα/d log µ is an afﬁne function of . The linear piece is ﬁxed by the above renormalization group equation to be −α; while the -independent piece is the β function:

∂ log Z ∂ log Z α λ − 2α ϕ + 1 (−α + β(α)) + α = 0 ∂α ∂α

Comparing both sides at order 0 gives the one loop beta function

β(α) = 3α2 + O[α]3.

(If you had worked with λ instead of α = λ/(4π)2, you would have obtained β(λ) = 3λ2/(4π)2 + O[λ]3.)

Similarly, the anomalous dimension of m is deﬁned as γm = d log m/d log µ, the dependence of the 2 −1 2 mass on scale µ. This may be computed using the µ-independence of the bare mass m0 = ZmZϕ m , so that d 1 ∂ log Z 1 ∂ log Z dα d log m 0 = log m = m − ϕ + . d log µ 0 2 ∂α 2 ∂α d log µ d log µ

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Hence 1 γ (α) = α + O[α]2. m 2

Finally, the anomalous dimension of ϕ is

d log Z1/2 1 ∂ log Z γ (α) = ϕ = ϕ (−α + β(α)) = 0 + O[α]2. ϕ d log µ 2 ∂α

3. No counterterm of the form Yχχ is needed to cancel χ tadpoles due to the Z2 symmetry χ → −χ, which ensures that χ tadpoles vanish (assuming that the vacuum does not break this symmetry). Diagrammatically, this can be seen by the fact that it is not possible to draw a diagram for a χ tadpole.

(a) • Corrections to ϕ propagator:

2 iΠϕ(k ) = + +

(igµ˜/2)2 Z ddl −i −i (ihµ˜/2)2 Z ddl −i −i = + −i((Z −1)k2+(Z −1)m2) 2 (2π)d l2 + m2 (l + k)2 + m2 2 (2π)d l2 + M 2 (l + k)2 + M 2 ϕ m

The general method for computing these integrals were done step by step in homework 9. For the sake of brevity, I will quote the result (14.30)

Z 1 2 1 2 2 2 iΠϕ(k ) = − αg Γ(−1 + /2)(k /6 + m ) + dx Dm log(µ /Dm) 2 0 Z 1 1 2 2 2 2 2 − αh Γ(−1 + /2)(k /6 + M ) + dx DM log(µ /DM ) − i((Zϕ − 1)k + (Zm − 1)m ), 2 0

2 2 where Dm = x(1 − x)k + m . In the MS scheme we therefore have, at one loop,

1 1 M 2 Z − 1 = − (α + α ),Z − 1 = − α + α . ϕ 6 g h m g m2 h

• Corrections to χ propagator:

2 iΠχ(k ) = +

Z ddl −i −i =(ihµ˜/2)2 − i((Z − 1)k2 + (Z − 1)M 2) (2π)d l2 + m2 (k + 1)2 + M 2 χ M Z 1 2 2 2 2 2 2 = − αh Γ(−1 + /2)(k /6 + m /2 + M /2) + dx Dm,M log(µ /Dm,M ) − i((Zχ − 1)k + (ZM − 1)M ), 0

2 2 2 where Dm,M = x(1 − x)k + xm + (1 − x)M . Therefore, at one loop,

α 1 m2 Z − 1 = − h ,Z − 1 = − 1 + α . χ 3 M M 2 h

• Corrections to ϕ3 vertex:

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iVϕ3 = + +

iα gµ˜/2 Z iα h Z = g Γ(/2) dF (µ2/D0 )/2 + h Γ(/2) dF (µ2/D0 )/2 + i(Z − 1)gµ˜/2, 2 3 m 2 3 M g 0 2 2 2 2 where Dm = x1x3k1 + x3x2k2 + x1x2k3 + m , and dF3 = dx1 dx2 dx3 δ(x1 + x2 + x3 − 1). (These integrals are computed on p.124.) Therefore, in MS, we have ! 1 α3/2 Z − 1 = − α + h . g g 1/2 αg • Corrections to ϕχ2 vertex:

iVϕχ2 = + +

iα gµ˜/2 Z iα h Z = h Γ(/2) dF (µ2/D00)/2 + h Γ(/2) dF (µ2/D000)/2 + i(Z − 1)hµ˜/2, 2 3 2 3 h so 1 Z − 1 = − α + α1/2α1/2 . h h g h (b) Due to the scale independence of the bare couplings, we have  d log g   ∂G ∂G   dg  0 ! g g/2 1 + g g  d log µ   ∂g ∂h  d log µ 0 =  d log h  = +  ∂H ∂H   dh  , h 0 h/2 h 1 + h d log µ ∂g ∂h d log µ which we may invert to yield  dg   ∂G ∂G −1 ! 1 + g g g/2 d log µ  ∂g ∂h   dh  = −  ∂H ∂H  . h 1 + h h/2 d log µ ∂g ∂h For a renormalizable theory, dg/d log µ and dh/d log µ are afﬁne functions of . The inverse may be computed by inspection in orders of −1. At order , we see that that pieces linear in are dg dh = −g/2 + ..., = −h/2 + ..., d log µ d log µ and the constant pieces are obtained by substituting the above back in and looking at the order 0 terms, yielding  dg g g ∂G1 ∂G1  = − + + , d log µ 2 2 ∂ log g ∂ log h dh h h ∂H1 ∂H1  = − + + . d log µ 2 2 ∂ log g ∂ log h

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(c) From the results of part (a), we have  3 1 3 1 2 2 2 h G1 = − − (g + h ) − g + ,  (4π)3 2 6 g 1 1 2 2 1 1 2 2 H1 = − − (g + h ) + − − h − (h + gh) ,  (4π)3 6 2 3

so1  1 3 3 3 1 2 βg(g, h) = − g − h + gh ,  (4π)3 4 4 1 7 3 2 1 2 βh(g, h) = − h − gh + g h .  (4π)3 12 12

(d) Redeﬁne ϕ and h so that g is positive, and write the beta functions as

 3 3 2!  g 3 h 1 h βg = − − + ,  (4π)3 4 g 4 g 2 2 !  g h 7 h h 1 βh = − − + .  (4π)3 12 g g 12 √ Both βg and βh/h are negative for h/g > (−g + 43)/7 = 0.08. Therefore, in that range, βh is also negative. Physically, a negative beta function means that the coupling decreases as the energy scale µ increases. This means that perturbation theory (an expansion in the orders of the coupling con- stants) can be used to probe the short distance behavior of such a theory. This is known as asymptotic freedom.

4. (a) Recall that β(g) = lim→0 dg/d log µ, so for the given redeﬁnition of coupling constant we have dg˜ dg˜ β(˜g) = lim = β(g). →0 d log µ dg

Now, taking derivative with respect to g˜ yields d dg˜ dg˜ dg β0(˜g) = β(g) + β0(g) = β0(g), dg˜ dg dg dg˜ so when evaluating at g˜ = g = 0 we see that the linear coefﬁcients coincide (and are equal to zero). Taking a second derivative with respect to g yields dg β00(˜g) = β00(g), dg˜ and at g˜ = g = 0, dg/dg˜ = 1, so the quadratic coefﬁcients coincide. A third derivative yields

dg d dg dg dg 2 β000(˜g) = β00(g) + β000(g) = β000(g), dg˜ dg dg˜ dg˜ dg˜ so the cubic coefﬁcients coincide. (b) This generalizes readily to multiple dimensionless couplings by treating g˜ as a vector, and the re-

deﬁnitions are such that dg/dg˜ |g=˜g=0 is the identity matrix. Then the proof that the beta function has the same form proceeds identically as before.

1 −1/2 −1 /2 Note that there is a typo in earlier versions of the text; (28.44) should read h0 = Zϕ Zχ Zhhµ˜ .

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5. We pick up with the expression for the one loop diagram for the one loop correction to the Yukawa coupling,

" # g3 1 1 1 Z 1 Z N˜ iV one-loop(p0, p) = = − − dF log(D/µ2) γ + dF , Y 8π2 4 2 3 5 4 3 D

2 02 0 2 2 ˜ where D = x1(1 − x1)p + x2(1 − x2)p − 2x1x2pp + (x1 + x2)m + x3M and N = (−x1p/ + (1 − 0 0 x2)p/ + m)(−(1 − x1)p/ + x2p/ + m)γ5. This is to be cancelled by the counterterm −(Zg − 1)gγ5, subject

to the renormalization condition iVY (0, 0) = −gγ5.

0 2 2 2 2 At p = p = 0 the expression greatly simpliﬁes, with D = (x1 + x2)m + x3M = (1 − x3)m + x3M ˜ 2 R R 1 R 1−x3 R 1 and N = m γ5. The integral dF3 becomes 2 0 0 dx2dx3 = 2 0 (1 − x3)dx3. The dx3 integral can be done using

Z 1 a 1 a a + b

log(a + bx) dx = x log(a + bx) − x + log(a + bx) = −1 − log a + log(a + b), 0 b 0 b b

yielding

g3 1 1 M m2 M iV one-loop(0, 0) = γ + − log − log , Y 8π2 5 2 µ M 2 − m2 m

so that g2 1 1 M m2 M Z − 1 = + − log − log + O[g]4. g 8π2 2 µ M 2 − m2 m

This may be substituted back into the vertex to yield " # g3 3 1 Z D m2 M 1 Z N˜ iV (p0, p) = −gγ − + dF log − log γ − dF , Y 5 8π2 4 2 3 M 2 M 2 − m2 m 5 4 3 D

which is independent of µ2 and .

6. A scalar ﬁeld in d spacetime dimensions has dimension [ϕ] = d/2 − 1, so a ϕn interaction is renormalizable if n(d/2 − 1) ≤ d.

For d = 2, interactions to all orders are renormalizable. A very important class of theories with interactions to all orders is the nonlinear sigma model, where the couplings may be taken to be the series expansion of the metric of some manifold. In d = 2, such theories are renormalizable.

For d = 3, ϕ3, ϕ4, ϕ5, ϕ6 interactions are renormalizable.

For d = 4, ϕ3, ϕ4 are renormalizable.

For d = 5, 6, ϕ3 is renormalizable.

For d > 6, only the mass term is renormalizable.

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7. (a) The calculations in the N flavor ϕ4 theory will be largely similar to that in the N = 1 ϕ4 theory, which was done in the previous problem set, but with the addition of some flavor factors. I will 4 use the results from the N = 1 ϕ theory, which shall be denoted with a subscript 0. 2 2 The Feynman rule for the propagator from flavor i to j is −iδij/(k + m − i), and the Feynman rule attached to the four point vertex is

j l

= −iλ(δijδkl + δikδjl + δilδjk).

i k

The correction to the propagator therefore has the extra ﬂavor factor k

2 iΠi,j(k ) = + i j i j

2 =2(δijδkl + δikδjl + δilδjk)δkliΠ0(k ) 2 =2(N + 2)iΠ0(k ).

(The factor of 2 is because there is no longer a symmetry factor of 1/2, which was present in the N = 1 case.) Therefore, λ Z − 1 = 2(N + 2) + O[λ]2,Z − 1 = O[λ]2. m (4π)2 ϕ The correction to the 4 point vertex similarly acquires ﬂavor factors. For the s channel diagram this goes as

j m l s = 2(δijδmn + δimδjn + δinδjm)(δmnδkl + δmkδnl + δnkδml)iV4,0

i n k

s = 2 [(N + 4)δijδkl + 2δikδjl + 2δilδjk] iV4,0.

(Once again, the factor of 2 comes from the lack of symmetry factor here compared to the N = 1 case.) Adding up the contributions from the t and u channels, we obtain λ Z − 1 = 2(N + 8) . λ (4π)2 Notice that these reduce to the correct limits (cf. problem 2) when N = 1. Following problem 2, we ﬁnd the beta function and anomalous mass dimension 2(N + 8)λ2 (N + 2)λ β(λ) = , γ (λ) = . (4π)2 m (4π)2

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(b) The Wilson-Fisher ﬁxed point is a nontrivial solution to dλ/d log µ = −λ+β(λ) = 0, dm2/d log µ = 0, which is at (4π)2 λ = , m2 = 0. ∗ 2(N + 8) ∗

(c) This scalar ﬁeld theory may be viewed as an effective description of some statistical model. The massive ﬁeld ϕ gives rise to a Yukawa potential −e−mr/r, which is corrected in interacting

γm −γm theories by the renormalization ﬂow d log m/d log µ = γm, ie. (m/m0) = (µ/µ0) = (r/r0) , where we have taken the mass scale to be µ = 1/r. Therefore, the corrected Yukawa potential is −(r/ξ)1/2ν −2ν 1−2ν −e , where ν = 1/2(1 − γm) is the critical exponent and ξ = m0 r0 is the correlation length. To obtain the critical exponent in d = 3 we can set = 1 (this is certainly not small, but it seems to give good estimates in certain theories). Then 1 1 N + 8 ν = = 2 = . 2(1 − γm) 2(1 − (N + 2)λ∗/(4π) ) N + 14

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