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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.
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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.
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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,
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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
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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.
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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