A Exam Presentation:Instantons and the U(1) problem

Christian Spethmann

May 1st, 2007

Outline

1 Instantons In Particle MechanicsEuclidean path integralsDouble WellPeriodic Potential

2 Topology Of Non-Abelian Gauge TheoriesInstantons and the Winding Numbern and theta-Vacuum States

3 The Chiral U(1) Problem and its SolutionThe Anomalous Axial CurrentThe Gauge-Variant Conserved CurrentThe U(1) ProblemChiral Ward identities in QCDQuark Zero Modes and Index TheoremChiral U(1) is spontaneously brokenThe chiral Goldstone boson in one-flavor QCDQCD with two massless quarks

Euclidean Transition Amplitudes

Consider a spinless particle of unit mass, moving in one dimension,with a Euclidean Hamiltonian

H =p2

2+ V (x).

We want to calculate the probability for the particle to move from aninitial position xi to a final position xf during an imaginary time intervalT . This probability is given by

〈xf |e−HT/~|xi〉 = N∫

[dx ] e−S/~,

using the usual path integral approach.

Recovering the Ground State Energy

Here S is the Euclidean action

S =

∫ T/2

−T/2dt

[12

(dxdt

)2

+ V

]

and [dx ] means integration over all functions that satisfy the boundaryconditions x(−T/2) = xi and x(T/2) = xf .The transition amplitude can be expanded in energy eigenstates,which gives

〈xf |e−HT/~|xi〉 =∑

n

e−EnT/~〈xf |n〉〈n|xi〉

For large T the lowest energy eigenstate will dominate the sum⇒ The real-world energy of the ground state can be determined fromthis Euclidean amplitude.

Integrating over Orthogonal FunctionsTo evaluate the path integral, we assume that the action has astationary point x̄(t),

δSδx̄

= −d2x̄dt2 + V ′(x̄) = 0.

A general function x(t) that satisfies the boundary conditions canthen be expressed by

x(t) = x̄(t) +∑

n

cnxn(t),

where the xn(t) are a set of real orthonormal functions that vanish atthe boundaries and are eigenfunctions of the second variationalderivative of S at x̄ ,

−d2xn

dt2 + V ′′(x̄) xn = λnxn.

Functional Determinant

The integral measure can then be defined by

[dx ] =∏

n

(2π~)−1/2 dcn,

and the action can be expanded in a Taylor series, so that thefunctional integral becomes

N∫

[dx ]e−S(x)/~ = Ne−S(x̄)/~∏

n

λ−1/2n [1 +O(~)]

= Ne−S(x̄)/~ [det

(−∂2

t + V ′′(x̄))]−1/2

[1 +O(~)]

including terms up to quadratic order in the cn’s. The result is just oneover the square root of the functional determinant, as expected forbosonic degrees of freedom.

Instanton in a Double Well

We now assume that the potential is a double well, with minima at ±aand V ′′(±a) = ω2, and we want to calculate

〈±a|e−HT/~| ± a〉 and 〈±a|e−HT/~| ∓ a〉

The first step is to find the stationary point of the action. Thisequation describes a particle moving in an inverted potential −V (x).From this it is clear that there is a solution with zero energy, so that

dx/dt = (2V )1/2,

or

t = t1 +

∫ x

0dx ′(2V )−1/2.

This is the so-called instanton.

Dilute Gas Approximation

The action integral over the instanton solution is

S0 =

∫dt

[12

(dx/dt)2 + V]

=

∫dt

(dxdt

)2

=

∫ a

−adx (2V )1/2.

For large |t |, the particle is exponentially close to the minimum:

(a− x) ∝ e−ωt ,

so a chain of isolated instantons and anti-instantons gives anapproximate solution to the problem.

Let us now try to find the functional determinant corresponding tosuch a solution. In this approximation, the action is just the sum of ninstanton actions, or nS0, since the action at the minima is zero.

Evaluating the Functional Determinant

To find the determinant we consider the following points:• In the interval between the instantons, the particle is very close

to the minima, so V ′′ = ω2. If the particle would always be there,the determinant would be

N[det(−∂2

t + ω2)]−1/2

=( ω

π~

)1/2e−ωT/2.

• Let the correction to this result from a single instanton oranti-instanton be K . The determinant then gets an additionalfactor of K n.

• The locations of the instantons are arbitrary, so we have tointegrate over the center positions. This gives a factor of∫ T/2

−T/2dt1

∫ t1

−T/2dt2 . . .

∫ tn−1

−T/2dtn = T n/n!.

Double Well Transition Amplitude

• If we start at −a and end at +a (or vice versa), we need an oddnumber of instantons. If we get back to the same place where westarted, we need an even number of instantons. We thereforehave

〈−a|e−HT/~| − a〉 =( ω

π~

)1/2e−ωT/2

∑even n

(Ke−S0/~T )n

n!

and the corresponding sum over odd n’s for 〈a|e−HT/~| − a〉.The result for the path integral is therefore

〈±a|e−HT/~| − a〉

=( ω

π~

)1/2e−ωT/2 1

2

[exp(Ke−S0/~T )∓ exp(−Ke−S0/~T )

].

Energy Eigenstates in the Double Well

Comparing this to the expansion of the transition amplitude, we seethat there are two states |+〉 and |−〉 with energies

E± =12

~ω ± ~Ke−S0/~.

Now we have to calculate the value of K . We start by looking at theequation for the functions xn. It has an eigenfunction of eigenvaluezero,

x1 = S1/20 dx̄/dt .

The integration over the corresponding expansion coefficient c1 canbe rewritten as an integration over the time variable t1 that defines thecenter of the instanton,

(2π~)−1/2 dc1 = (S0/2π~)1/2 dt1.

Double Well Functional Determinant

The integration over this zero eigenvalue function has already beendone above. To get the correct value of the determinant K , wetherefore have to include a factor of (S0/2π~)1/2 and omit the explicitintegration over this zero eigenvalue function. The one-instantoncontribution to the matrix element becomes

〈a|e−HT | − a〉 = NT (S0/2π~)1/2e−S0/~ (det ′[−∂2

t + V ′′(x̄)])−1/2

.

Comparing this to the one-instanton term in the transition amplitudesum, we see that

K = (S0/2π~)1/2∣∣∣∣ det(−∂2

t + ω2)

det′(−∂2t + V ′′(x̄))

∣∣∣∣1/2

.

Instantons in a Periodic Potential

In a periodic potential with minima at integer positions, the calculationis almost the same. The only difference is that now instantons andanti-instantons don’t have to alternate. The transition matrix elementfrom position j− to position j+ is then given by

〈j+|e−HT/~|j−〉 =( ω

π~

)1/2e−ωT/2

∞∑n=0

∞∑n̄=0

1n!n̄!

(Ke−S0/~T

)n+n̄δn−n̄−j++j− ,

where n and n̄ are the numbers of instantons and anti-instantons.This can be rewritten by using

δab =

∫ 2π

0

dθ2π

eiθ(a−b).

Continuous Energy Eigenstates

We then find

〈j+|e−HT/~|j−〉 =( ω

π~

)1/2e−ωT/2∫ 2π

0ei(j−−j+)θ dθ

2πexp

[2KT cos θe−S0/~

].

There is a continuum of eigenstates with energies given by

E(θ) =12

~ω + 2~K cos θ e−S0/~.

This result is very familiar from the energy bands encountered in solidstate periodic potentials.

Classical Vacua in Yang-Mills Theory

A pure gauge theory has the Euclidean action

S =1

4g2

∫d4x(Fµν ,Fµν)

with the field strength tensor

Fµν = ∂µAν − ∂νAµ + [Aµ,Aν ]

where the gauge potentials are defined as matrices by

Aµ = gAaµT a.

We now want to find field configurations with finite action. The aboveformulas imply that the field strength tensor has to be of O(1/r3) infour dimensional Euclidean space-time.

Homotopy Classes of Classical Vacua

This means that the gauge potentials have to be of the form

Aµ = g∂µg−1 +O(1/r2),

where the first term is a gauge transform of zero.

This corresponds to a continuous mapping from thethree-dimensional hypersphere S3 (at infinite distance in fourEuclidean dimensions) to the gauge group.

In general, it is not possible to eliminate this term by a gaugetransformation that is continuous over all four-space, because thosekinds of gauge transformations correspond to the mappings from S3

to g that can be reached by continuously deforming from the trivialmapping S3 → 1. The pure gauge field configurations therefore fallinto gauge-invariant homotopy classes.

Definition of the Winding Number

Let us assume that the gauge group is SU(2). A general groupelement can be parameterized by

g = a + ib · σ

where a2 + |b|2 = 1. SU(2) is therefore homomorphic to to S3, andwe have to consider mappings from S3 to S3.

All possible mappings are homotopic to a member of the family givenby

g(ν)(x) = [(x4 + ix · σ)/r]ν ,

where ν is called the winding number. The trivial mapping e.g.corresponds to ν = 0, the identity mapping to ν = 1.

Winding Number as TopologicalCharge

For the two-dimensional representation of SU(2), the winding numbercan be calculated from the gauge configuration by

ν = − 124π2

∫dθ1 dθ2 dθ3 Tr εijk g∂ig−1 g∂jg−1 g∂k g−1,

where the θi are angles that parameterize the 3-sphere. UsingGauss’s theorem, it can be shown that the winding number can alsobe calculated by the volume integral∫

d4x (F , F̃ ) = 32π2ν,

whereF̃µν =

12εµνλσFλσ

is the dual of the field strength tensor.

The BPST Solution

The BPST solution is the explicit form of the ν = 1 instanton, whichcan be written as

Aµ =r2

r2 + λ2 g(1)−1(x)∂µg(1)(x).

where

r =4∑

i=1

x2i .

As it should be, this solution is a continuous function that is zero atthe origin.

Definition of the |n〉-Vacua

There exists an infinite family of classical field configurations |n〉 thatare analogous to the degenerate minima in the 1-D periodic potentialproblem. Those states are pure gauge configurations in 3-space, withthe additional constraint that g(r) has to approach the same constantvalue as r →∞. 3-space then becomes topologically equivalent toS3 and the same arguments as above can be used.

If a spatial field configuration is continuously deformed to a differenthomotopy class, the space of pure gauge transformations has to beleft. Therefore an energy barrier exists and the situation is analogousto the periodic potential.

Energy Eigenstates

We now consider a large 4-dimensional box of Euclidean spacetime.Let the integral over all configurations inside the box with windingnumber n be

F (V ,T ,n) = N∫

[dA] e−S δνn,

where we have used some gauge-fixing condition and set ~ = 1.For large times T1 and T2,

F (V ,T1 + T2,n) =∑

n1+n2=n

F (V ,T1,n1)F (V ,T2,n2),

because the winding number can be written as the integral over alocal density. This equation tells us that the |n〉-vacua are not theenergy eigenstates, because the transition element is not a simpleexponential with a multiplicative composition law.

Definition of the |θ〉-Vacua

Using a Fourier transform, it is possible to bring this expression intothe desired form:

F (V ,T , θ) =∑

n

einθ F (V ,T ,n) = N∫

[dA] e−S eiνθ.

This can now be interpreted as the expectation value of e−HT in anenergy eigenstate, called the |θ〉-vacuum:

F (V ,T , θ) ∝ 〈θ|e−HT |θ〉 = N ′∫

[dA] e−Seiνθ.

This new vacuum state can be written as a sum over the classicalfield configurations

|θ〉 ∝∑

n

eiθn|n〉.

〈θ|e−HT |θ〉 Transition Amplitude

Using the same methods as for the 1D particle case, let us now try tocalculate the energy density of this new vacuum state. In the dilutegas approximation, the integral over all gauge configurations with thesame winding number ν is replaced by a sum over n instantons and n̄anti-instantons, so that ν = n − n̄. This gives

〈θ|e−HT |θ〉 ∝∑n,n̄

(Ke−S0)n+n̄(VT )n+n̄ ei(n−n̄)θ/(n!n̄!)

= exp(

2KVTe−S0 cos θ).

The energy density of a |θ〉-vacuum is then given by

E(θ)/V = −2K cos θ e−S0 .

Action of a Yang-Mills Instanton

The action S0 of a single instanton can be found using the Schwartzinequality:

4g2S0 =

∫(F 2) =

[∫(F ,F )

∫(F̃ , F̃ )

]1/2

≥∣∣∣∣∫ (F , F̃ )

∣∣∣∣ = 32π2|ν|,

so that

S0 ≥8π2

g2 |ν|.

Equality holds for F = ±F̃ .The symmetry group of the solution has eight parameters: the size ofthe instanton, the 4D location of its center, and three global gaugetransformation parameters. From this we get a factor of (1/

√~)8, or

equivalently 1/g8.

|θ〉-Vacuum Energy-Density

Using this, the formula for the vacuum energy becomes

E(θ)/V = − cos θ e−8π2/g2g−8

∫ ∞

0

d%%5 f (%M),

where M is a mass-dimension parameter that defines therenormalization scale. From RGE analysis it follows that observablequantities can only depend on

1g2 − β1 log M +O(g2),

where β1 can be calculated from one-loop perturbation theory, and ishere given by 11/12π2. Taking this into account, the energy densitybecomes

E(θ)/V = −A cos θ e−8π2/g2g−8

∫ ∞

0

d%%5 (%M)8π2β1

[1 +O(g2)

].

Axial U(1) symmetryWe now include massless fermions in our gauge theory. TheLagrangian in Minkowski space becomes

L = − 14g2 (Fµν ,Fµν) +

∑f

ψ̄f iDµγµψf

where f is a flavor index and Dµψ = ∂µψ + Aµψ the usual covariantderivative. This Lagrangian is symmetric under chiral rotations of thefermion fields

ψf → e−iαγ5ψf ⇒ j5µ =∑

f

ψ̄fγµγ5ψf .

It is well known that this current is not conserved in the quantumtheory because of the Adler-Bell-Jackiw anomaly:

∂µj5µ =Nf

32π2 εµνλσ (Fµν ,Fλ,σ).

Gauge-Variant Conserved Current

The anomalous term is proportional to the divergence of the current

Gµ = 2εµνλσ (Aν ,Fλσ − 23

AλAσ),

with∂µGµ = (Fµν , F̃µν) =

12εµνλσ (Fµν ,Fλσ).

It is therefore possible to define an anomaly free (but gaugedependent) axial current

J5µ = j5µ −

Nf

16π2 Gµ ⇒ ∂µJ5µ = 0.

The conserved charge corresponding to this current is given by

Q5 =

∫d3x J5

0 .

Rotation of the |θ〉-Vacuum

It can be shown that the effect of a gauge transformation in the n-thhomotopy class is to change the value of this charge by 2n.We therefore get

Q5|n〉 = 2n|n〉,

because the charge of the trivial vacuum Aµ = 0 is obviously zero.The axial charge is the generator of a U(1) symmetry group, and theeffect of a rotation with angle θ′ on the |θ〉-vacuum is therefore

exp(

12

iθ′Q5)|θ〉 = exp

(12

iθ′Q5) ∑

n

eiθn|n〉

=∑

n

ei(θ′+θ)n|n〉

= |θ + θ′〉.

Symmetries of Massless QCD

Let us consider QCD with two quarks:• In the massless limit, the Lagrangian of this theory is symmetric

under U(2)× U(2), because the left- and righthanded quarkfields can be independently rotated with arbitrary unitarymatrices.

• The observed spectrum on the other hand has only(approximate) U(1)B × SU(2)V symmetry, where U(1)Bcorresponds to baryon number and SU(2)V to the vectorsubgroup (with identical rotation of left- and righthanded fields).

• Since the axial SU(2)A and preudoscalar U(1)P symmetries arenot realized, this symmetry has to be spontaneously broken, andone would expect to see four Goldstone bosons corresponding tothose generators.

Broken Chiral Symmetries

However, only three low-mass pseudoscalar bosons (the pions) areobserved, corresponding to the generators of SU(2)A. As Weinbergshowed, the mass of the missing boson would have to be less than√

3mπ, so that it can not be any other of the observed mesons. Thismissing Goldstone boson is the U(1) problem.

Kogut and Susskind found the solution of this problem by looking atthe 1+1 dimensional Schwinger model. It has two massless fields φ+

and φ−, one with the usual propagator, one with minus the usualpropagator. All gauge invariant operators couple to the sum of thosefields, so that the propagators cancel and no Goldstone pole isobserved. The gauge-variant, conserved current couples to thegradient of the difference ∂µ(φ+ − φ−), so that a Green’s function forone gauge-variant current and a string of gauge-invariant fields hasthe expected pole. Following Coleman’s analysis, I will now show thatthis also happens in QCD.

Green’s Function for Fermion FieldsThe Euclidean action for massive fermions in a gauge field is given by

S = −i∫

d4x ψ̄(γµDµ −M)ψ.

The Green’s function for a set φ(r) of local functions of those fields isgiven by

〈φ(1)(x1) . . . φ(m)(xm)〉A =

∫[dψ][dψ̄]e−Sφ(1)(x1) . . . φ

(m)(xm)∫[dψ][dψ̄]e−S

,

We now perform an axial rotation

ψ → e−iγ5αψ, ψ̄ → ψ̄ e−iγ5α.

or (for infinitesimal α)

δψ = −iγ5ψδα, δψ̄ = −iψ̄γ5δα.

Deriving the Ward Identity

The Lagrangian is invariant under this rotation, and we can derive thechiral Ward identity

∂µ〈j5µ(y)φ(1)(x1) . . . φ(m)(xm)〉A

+ 2〈ψ̄Mγ5ψ(y)φ(1)(x1) . . . φ(m)(xm)〉A

+ δ4(y − x1)〈∂φ(1)(x1)/∂α . . . φ(m)(xm)〉A

+ . . .

+ δ4(y − xm)〈φ(1)(x1) . . . ∂φ(m)(xm)/∂α〉A

= − iC8π2 (F (y), F̃ (y)) 〈φ(1)(x1) . . . φ

(m)(xm)〉A,

where j5µ = ψ̄γµγ5ψ and the right-hand side would be zero without theanomaly.

Integral Ward Identity Formula

Integrating this equation over y gives

2⟨∫

d4y ψ̄Mγ5ψ(y)φ(1)(x1) . . . φ(m)(xm)

⟩A

+∂

∂α〈φ(1)(x1) . . . φ

(m)(xm)〉A

= −4iCν〈φ(1)(x1) . . . φ(m)(xm)〉A

where ν is the winding number of the gauge field. The constant Cdepends on the representation of the fermions and is defined by

Tr T aT b = −Cδab.

Fermion Functional Determinant

In a gauge theory with one massless quark, the energy density of the|θ〉-vacua is still given by

E(θ)/V = −2K cos θe−S0 ,

but the determinant K now contains a term of the form

det[

iD/i∂/

]= det

[i(∂µ + Aµ)γµ

i∂µγµ

].

This factor is zero, because the Dirac field has zero modes withvanishing eigenvalues of iD/ . Let us assume that the spectrum isdiscrete,

iD/ ψr = λrψr .

Odd and Even Parity Eigenfunctions

The λ’s are real and occur in pairs of opposite sign, because γ5anticommutes with γµ. Eigenfunctions of vanishing eigenvalue can bechoosen to be also eigenfunctions of γ5

γ5ψr = ±ψr , (λr = 0).

Let n± be the number of eigenfunctions with positive/negative parity.We then have the sum rule

n− − n+ = ν,

so there are zero eigenvalues in any gauge field with nonvanishingwinding number.

Proof of the Index Theorem

To prove this, we take the integrated Ward identity for a massivequark and no φ’s,

−2iν = 2⟨∫

d4y ψ̄mγ5ψ

⟩A

=2

∫[dψ][dψ̄]e−S

∫d4y ψ̄mγ5ψ∫

[dψ][dψ̄]e−S.

The eigenfunctions of i(D/ −m) are the same as before,

i(D/ −m)ψr = (λr − im)ψr ,

so that the functional integral becomes

−2iν =2m

∑r

∫d4yψ†r γ5ψr

∏s 6=r (λs − im)∏

r (λr − im)

= 2m∑ ∫

d4ψ†r γ5ψr (λr − im)−1.

|θ〉-Vacua are Degenerate

Using the fact that eigenfunctions of a Hermitian operator withdifferent eigenvalues are orthogonal,∫

d4yψ†r γ5ψr = 0 (λr 6= 0),

∫d4yψ†r γ5ψr = ±1 (λr = 0)

we see that only the zero modes contribute, and the integral becomes

−2iν = 2i(n+ − n−) .

This is the above sum rule, so we now know that every vacuum withnonvanishing winding number has at least one zero eigenvalue. The|θ〉-vacua have therefore the same energy.

Chiral Rotations of |θ〉Let us define denominator-free Green’s functions by

〈〈φ(1)(x1) . . .〉〉A =

∫[dψ][dψ̄]e−Sφ(1)(x1) . . . ;

they obey the Ward identity[∂

∂α+ 2iν

]〈〈φ(1)(x1) . . .〉〉A = 0

The Green’s functions of one-flavor QCD are given by

〈θ|φ(1)(x1) . . . |θ〉 =

∫[dA]e−Sg eiνθ〈〈φ(1)(x1) . . .〉〉A∫

[dA]e−Sg eiνθ〈〈1〉〉A,

so that [∂

∂α+ 2

∂

∂θ

]〈θ|φ(1)(x1) . . . |θ〉 = 0.

Vacuum Expectation Value of σ±

This proves that chiral U(1) rotations, parameterized by α, rotate the|θ〉-vacuum. The energy density of the vacuum is an invariant quantityunder U(1).

To show that chiral symmetry is spontaneously broken, it suffices tofind a Green’s function that has a non-zero derivative with respect toα. We choose to calculate

〈θ|σ±(x)|θ〉 =

∫[dA][dψ][dψ̄]e−Seiνθσ±(x)∫

[dA][dψ][dψ̄]e−Seiνθ

where σ± = 12 ψ̄(1± γ5)ψ are chiral eigenfields, so that

∂σ±/∂α = ∓2iσ±.

Evaluating 〈θ|σ±(x)|θ〉

The calculation is similar to the one for the pure gauge theory. Oneimportant difference, however, is that there is now a fermiondeterminant that has n vanishing eigenvalues, if there are ninstantons and anti-instantons. The integral can therefore only give anon-zero value if there are at least 2n Dirac fields in the Green’sfunction.

The path integral in the denominator has zero Dirac fields, so the onlycontribution comes from the classical vacuum Aµ = 0, and the resultis a product of a Bose and a Fermi determinant. The numerator forσ− needs to have one instanton and no anti-instantons. The Fermiintegral gives

12ψ†0(1− γ5)ψ0

∏λr 6=0

λr = ψ†0ψ0 det ′(iD/ )

since the zero eigenvalue function has odd parity in this case.

Proof for Spontaneous SymmetryBreaking

Including all factors from the pure gauge case, and doing thecorresponding calculation for σ+, we get

〈θ|σ±(x)|θ〉 = e−8π2/g2e∓iθg−82

∫ ∞

0

d%%5 f (%M)

det ′(iD/ )

det(i∂/ ).

Since one eigenvalue has been removed from the numerator, theratio of determinants has to be of the form

det ′(iD/ )

det(i∂/ )= % h(%M),

by dimensional analysis, where % is the size of the instanton and h issome unknown function. So we now know that spontaneoussymmetry breakdown occurs.

One Flavor QCD Model

If there is a chiral Goldstone boson, the corresponding pole shouldappear in

〈θ|σ+(x)σ−(0)|θ〉.

The only contributions to this are from the zero instanton field andfrom the one instanton, one anti-instanton field configuration. Theformer is the one-loop perturbative expression, the latter gives theproduct

〈θ|σ+|θ〉〈θ|σ−|θ〉

which has no Goldstone pole either. As in the Schwinger model, thepole only appears in gauge-variant Green’s functions like

〈θ|J5µ(x)σ−(0)|θ〉 = 〈θ|j5µ(x)σ−(0)|θ〉+

i16π2 〈θ|Gµ(x)σ−(0)|θ〉.

Goldstone Pole in One Flavor QCD

The second term has a Goldstone pole if and only if∫d4x ∂µ〈θ|Gµ(x)σ−(0)|θ〉 6= 0.

But we know that ∫d4x∂µGµ = 32π2ν

and that only ν = 1 field configurations contribute, so that∫d4∂µ〈θ|Gµ(x)σ−(0)|θ〉 = 32π2 〈θ|σ−|θ〉 6= 0.

So a Goldstone pole appears in the Green’s function of onegauge-variant current and one gauge-invariant operator. However, ifwe calculate the propagator of the gauge-variant current 〈θ|JµJν |θ〉,we again find no pole, because only the classical vacuum with ν = 0contributes.

Goldstone Pole in Two Flavor QCD

For the real world case of two massless quarks, the sum rule nowbecomes

n− − n+ = 2ν.

This means that iD/ in a one instanton field now has two vanishingeigenvalues, and all quark bilinears will have zero expectation values.To prove that chiral U(1) is spontaneously broken, we can insteadcalculate quadrilinears like

ψ̄1(1− γ5)ψ1ψ̄2(1− γ5)ψ2.

Another difference is that the |θ〉-vacuum rotates twice as much underaxial rotations, as necessary to preserve SU(2)× SU(2) symmetry:[

∂

∂α+ 4

∂

∂θ

]〈θ|φ1(x1) . . . |θ〉 = 0,