tmQCD and disconnected diagrams

Alejandro Vaquero Aviles-Casco

June 3, 2013

1 Twisted mass basis and physical basis

The twisted mass basis and the physical basis are related through the transformation

tm Basis → Physical basis

ψ = χeiωγ5

τ3

2

ψ = eiωγ5

τ3

2 χ,

where ω is the twisted mass angle, which relates the standard mass to the twisted mass through

tanω =µ

m.

We will usually work at maximal twist, that is, ω = π2.

2 Basis transformation for the pseudoscalar iψγ5ψ

Applying the transformation, and taking into account that

[

γ5, eiKγ5

]

= 0 K = Constant

we find

iψγ5ψ = iχueiω

γ5

2 γ5eiω

γ5

2 χu + iχde−iω

γ5

2 γ5e−iω

γ5

2 χd = iχueiωγ5γ5χu + iχde

−iωγ5γ5χd =

= iχγ5χ cosω − χτ3χ sinω.

At maximal twist, only the second term survives

iψγ5ψ = −χτ3χ. (1)

Figure 1: Disconnected diagram associated to the pseudoscalar flavour singlet.

1

2.1 The one-end-trick

Now we have to face the problem of computing the following disconnected diagramwhich appears associated to the contractions of quark-antiquark pairs in the same point. In order to

evaluate this diagram W , we need to calculate the propagators in points 0 and x

WP = −D−1u +D−1

d , (2)

according to expression (1) in the physical basis. The one-end-trick consists of taking into account thefollowing relationship for twisted mass fermions

Du −Dd = 2iµγ5, (3)

therefore we can write (2) as

WP =Du −Dd

DuDd

= D−1u (2iµγ5)D

−1

d . (4)

This way we remove the difference of propagators, which is bound to introduce large errors and reducethe signal-to-noise ratio. Let’s see this in a more detailed fashion. What we want to compute is

Tr (WP ) = −Tr [Su(x, x)− Sd(x, x)] = Tr [Su(x, y) (2iµγ5)Sd(y, x)] =

= Tr[

Su(x, y) (2iµγ5) γ5S†u(x, y)γ5

]

= 2µiTr[

Su(x, y)S†u(x, y)γ5

]

=

= 2µi∑

r

∑

x

〈φ∗r(x)|γ5|φr(x)〉 , (5)

which should enable us to compute the disconnected diagram without the need of all-to-all propagators.

3 Basis transformation for the neutral pion iψγ5τ3ψ

Again, we use the transformation rules to find

iψγ5τ3ψ = iχueiω

γ5

2 γ5eiω

γ5

2 χu − iχde−iω

γ5

2 γ5e−iω

γ5

2 χd = iχueiωγ5γ5χu − iχde

−iωγ5γ5χd =

= iχγ5τ3χ cosω − χχ sinω,

and assuming maximal twist, we find

iψγ5τ3ψ = −χχ. (6)

Minus sign??

3.1 The one-end-trick

The situation in this case is completely different, when compared to the pseudoscalar. The problem isthe following, now we have to evaluate

Wπ0 = D−1u +D−1

d , (7)

which is no difference, so the twisted mass relations between the up and down quarks would give rise to

Du +Dd = 2DW , (8)

where DW is the Dirac operator without the twisted mass term, so in the end, the application of thisrelationship would result in

Wπ0 =Dd +Du

DuDd

= 2D−1u DWD−1

d . (9)

which, to tell the truth, I’m not sure why is not so profitable.

2

Let’s analyse this last result in detail. The evaluation in terms of propagators is

Tr (Wπ0) = −Tr [Su(x, x) + Sd(x, x)] = −2Tr [Su(x, z1)DW (z1, z2)Sd(z2, x)] =

= −2Tr[

Su(x, z1)DW (z1, z2)γ5S†u(x, z2)γ5

]

(10)

where we have used

Su(x, y) + Sd(x, y) = 2DW (x, y),

Sd(x, y) = γ5S†u(y, x)γ5. (11)

Now we apply the non-trivial relationship

Su(x, y)DW (y, z)γ5 = γ5DW (x, y)Su(y, z), (12)

for there exists a rule not very difficult to prove. First we multiply both sides by Du at both ends

DuSuDW γ5Du = DW γ5Du,

Duγ5DWSuDu = Duγ5DW .

Then we notice thatDu(x, y)γ5DW (y, z) = DW (x, y)γ5Du(y, z) (13)

which is very straightforward

Du(x, y)γ5DW (y, z) = DW (x, y)γ5DW (y, z) + 2iµDW (x, z),

DW (x, y)γ5Du(y, z) = DW (x, y)γ5DW (y, z) + 2iµDW (x, z).

Thence we can go on with the calculation

Tr (Wπ0) = −Tr[

Su(x, z1)DW (z1, z2)γ5S†u(x, z2)γ5

]

= −Tr[

γ5Du(x, z1)Su(z1, z2)S†u(x, z2)γ5

]

=

= −∑

r

∑

x

〈φ∗r(x)| (DWφr) (x)〉 , (14)

where we have called

|φr〉 = D−1u |ηr〉 , (15)

and the |ηr〉 represent the vector sources.

4 Basis transformation for a general bilinear iψΓψ

Now we are considering a general bilinear with an operator Γ, which can have dirac and flavour indices.We will explicitly write the flavour index in our calculations, referring to Γf1f2 to the specific componentif1Γf1f2f2 ∈ iψΓψ. In our case, fi = u, d. In order to compute the transformation to the physical basis,what we need to know is the commutator [γ5,Γ] and the anticommutator {γ5,Γ}, which will depend onthe particular choice for Γ. With this information, we can proceed

iψΓψ = iχueiω

γ5

2 Γuueiω

γ5

2 χu + iχueiω

γ5

2 Γude−iω

γ5

2 χd + iχde−iω

γ5

2 Γdueiω

γ5

2 χu + iχde−iω

γ5

2 Γdde−iω

γ5

2 χd.

At this point the calculation becomes incredibly tedious. We shall face it by parts. Let’s start with theuu term

iχueiω

γ5

2 Γuueiω

γ5

2 χu = iχuΓuuχu cos2 ω

2− iχuγ5Γuuγ5χu sin

2 ω

2− χu

{γ5,Γuu}

2χu sinω.

Imposing maximal twist,

3

iχueiω

γ5

2 Γuueiω

γ5

2 χu = iχuγ5[γ5,Γuu]

2χu − χu

{γ5,Γuu}

2χu

The combination dd yields almost the same result, except for a small sign change in the term involvingthe anticommutator

iχde−iω

γ5

2 Γdde−iω

γ5

2 χd = iχdγ5[γ5,Γdd]

2χd + χd

{γ5,Γdd}

2χd.

And the cross terms ud and du switch the place of the commutator and the anticommutator

iχueiω

γ5

2 Γude−iω

γ5

2 χd = iχuγ5{γ5,Γud}

2χd − χu

[γ5,Γud]

2χd,

iχde−iω

γ5

2 Γdueiω

γ5

2 χu = iχdγ5{γ5,Γdu}

2χu + χd

[γ5,Γdu]

2χu.

sectionTwist-two operator conventions The twist-two operator, also called 〈x〉 is defined as

Kµν = iψγµDνψ, (16)

where the Dν stands for the naive fermionic operator. This Kµν are usually symmetrized, and madetraceless, so the final operator becomes

Oµν =Kµν +Kνµ

2−δµν

4

∑

λ

Kλλ. (17)

The quantity 〈x〉 is one of this Oµν , and take non-zero values at vanishing momentum.There are two different versions of 〈x〉: 〈x〉u+d, which is the one I just exposed, and 〈x〉u−d, where a

flavour matrix τ3 is inserted.

5 Several basis transformations

An abreviated way to perform a basis transformation over the bilinear ψXψ is to calculate Y =1

2(1 + iγ5τ3)X (1 + iγ5τ3), so the resulting bilinear will be ψY ψ. Let’s apply this rule to the bilinears

we are interested in.

5.1 Local Axial quantities

These are of the formψγ5γkψ with k = 1, 2, 3; (18)

and should be combined with the projectors 1

4(1 + γ0) iγ5γk for the two-point functions. The result of

the basis transformation is

1

2(1 + iγ5τ3) γ5γk (1 + iγ5τ3) = γ5γk, (19)

so it is invariant, and requires the generalized one-end trick. The standard one-end trick can still be usedto compute the insertion γ5γkτ3.

5.2 Local Vector quantities

These areψγµψ with µ = 0, 1, 2, 3; (20)

and should be combined with the projectors 1

4(1 + γ0) for any µ, and

1

4(1 + γ0) iγ5γk for µ = k = 1, 2, 3.

The basis transformation gives

1

2(1 + iγ5τ3) γµ (1 + iγ5τ3) = γµ, (21)

so it is also invariant, demanding the application of the generalized one-end trick. The standard one-endtrick applies to the insertion γµτ3

4

5.3 One derivative Vector

The derivative insertion makes its appearance. When computing the basis transformation, it behaves asthe identity, because it has trivial γ–structure

ψγµDνψ with µ, ν = 0, 1, 2, 3; (22)

and should be combined with the projectors 1

4(1 + γ0) for any µ, ν combination, and 1

4(1 + γ0) iγ5γk

whenever one of the indices is a spatial one. After the transformation

1

2(1 + iγ5τ3) γµDν (1 + iγ5τ3) = γµDν , (23)

so it remains invariant and we should apply the generalized one-end trick. The standard version appliesto γµDντ3.

All the remarks explained in 4 apply here.

5.4 One derivative Axial

This case is quite similar to the last one,

ψγ5γµDνψ with µ, ν = 0, 1, 2, 3; (24)

and should be combined with the projectors 1

4(1 + γ0) iγ5γk whenever one of the indices is a spatial one.

This is a requirement, we never compute the temporal-temporal component. The transformation resultsin

1

2(1 + iγ5τ3) γ5γµDν (1 + iγ5τ3) = iγ5γµDν , (25)

thus, like the former case, it remains invariant, and the generalized one-end trick will do. The standardversion applies to γ5γµDντ3.

The remarks found in 4 are also valid here.

5.5 Summary

The next table summarizes my results for the different bilinears. The standard one-end trick will beappliable whenever the transformed bilinear has a τ3 matrix within.

Bilinear Transforms to Standard Generalized(Physical basis) (Twisted basis) One-end Trick One-end Trick

ψψ iψγ5τ3ψ X ✗

ψτ3ψ iψγ5ψ ✗ X

iψγ5ψ −ψτ3ψ X ✗

iψγ5τ3ψ −ψψ ✗ X

ψγµψ ψγµψ ✗ X

ψγµτ3ψ ψγµτ3ψ X ✗

ψγ5γµψ ψγ5γµψ ✗ X

ψγ5γµτ3ψ ψγ5γµτ3ψ X ✗

iψγµDνψ iψγµDνψ ✗ X

iψγµDντ3ψ iψγµDντ3ψ X ✗

iψγ5γµDνψ iψγ5γµDνψ ✗ X

iψγ5γµDντ3ψ iψγ5γµDντ3ψ X ✗

Table 1: Application of the standard and the generalized one-end trick to different bilinears.

5

5.6 Cosas

Proton:

ǫαβγ[

dαT (x, t)Cγ5uβ(x, t)

]

uγ(x, t)

Correlador

⟨

P (x, t) |γ5γµ|P (0, 0)⟩

=⟨

ǫαβγ uγ(x, t)γ0

[

uβ(x, t)γ0γ5Cdαγ0

]

γ0γ5γµǫλµν[

dλT (0, 0)Cγ5uµ(0, 0)

]

uν(0, 0)⟩

6