Method of (covariant) symbols
General formalism
Suppose we want to calculate the trace of a differential operatorf(D,M) whereDµ≡∂µ+Aµ(x)
is a (covariant) derivative operator,Aµ(x),M(x) are background fields, andf is a function that
we assume to be analytic in both its arguments. (Series expansion is one general way how to define functions of operators.) Both background fields can in principle have some additional matrix structure (internal degrees of freedom). If these are constant the task is simple, for we just need to diagonalize the derivative operator by going to momentum space. In the general case, it is unfortunately impossible to evaluate the trace in a closed form. If we nevertheless assume that we background fields vary sufficiently slowly in spacetime, we can calculate the trace in a (covariant) gradient expansion. The method of (covariant) symbols is a suitable tool to accomplish this task.
In order to calculate the trace we need to choose a basis. It is convenient to choose a direct product basis, consisting of independent bases in the space of test functions of the coordinate
x, denoted as χn(x), and in the internal space. Provided the basis χn(x) is orthonormal, the
full trace becomes
Trf(D,M) =X
n
Z
x
tr
χ∗n(x)f(D,M)χn(x)
, (1)
where “tr” refers to a trace over the internal space only. The integration measure is set by the scalar product in the space on which the differential operators act. We will keep using this general notation, but have implicitly in mind flat Euclidean spacetime, either at zero temperature where R
x ≡
R
ddx with d the spacetime dimension, or at nonzero temperature
with Rx≡R1/T 0 dx0
R
dd−1x.
We can choose any basis we want, in particular the basis of plane waves, in which the trace becomes
Trf(D,M) = Z
x,p
tre−ip·xf(D,M)eip·x. (2)
The summation over the basis elements now takes the form Rp ≡R
ddp/(2π)d at zero
temper-ature, and Rp ≡T P
p0
R
dd−1p/(2π)d−1 at nonzero temperature. The Matsubara frequency p 0 assumes the values 2nπT or (2n+ 1)πT with integern, depending on the boundary condition in the temporal direction (periodic for bosons vs. antiperiodic for fermions). As the next step, ob-serve thate−ip·xD
eip·xψ(x)
= (D+ ip)ψ(x) for any functionψ(x) so thate−ip·xDeip·x =D+ ip
holds as an operator identity. This leads us to the general formula for the operator trace within the method of symbols,
Trf(D,M) = Z
x,p
trf(D+ ip,M)11. (3)
The unit matrix 11 is what is left of the basis wave functionχp(x); we keep it to remember that
we are calculating a matrix element of a differential operator.
As long as M(x) transforms in the same manner, the trace Trf(D,M) obviously is gauge-invariant. Unfortunately, this property is no longer manifest in Eq. (3). The reason for this is the presence of the differential operator ∂µ in the covariant derivative Dµ. In particular when
commutation rules are used to bubble such a covariant derivative to the far right off(D+ip,M) where it hits the unit matrix, the derivative ∂µ annihilates it so that only the Aµ piece of the
covariant derivative is left. It will be shown on an explicit example below that the covariance of the trace is only restored after the integral over momentum is performed.
This unpleasant feature of the method of symbols can be fixed by further manipulating the trace. Introduce an arbitrary vector operator Ξµ acting in thex-space and rewrite Eq. (2) as
Trf(D,M) = Z
x,p
tr
e−iΞ·∇eiΞ·∇e−ip·xf(D,M)eip·xe−iΞ·∇
| {z }
11
. (4)
Here we used the shorthand notation∇µ ≡∂/∂pµ. Note that adding the exponential on the far
right does not change anything since it acts on the unit matrix, while the two extra exponentials before f(D,M) combine to a unity. Therefore, this is an identity which in particular holds regardless of the temperature. The underbraced expression is now a similarity transformation of the original operator, defined generally as
v ≡eiΞ·∇e−ip·xv eip·xe−iΞ·∇. (5)
It can thus be applied separately to each factor in the operator, f(D,M) = f(D,M). As the last step, we expand the remaining exponential e−iΞ·∇ in powers and note that its spatial part produces zero since it gives an integral of a total derivative in momentum space. (This observation relies on the fact that Ξµ acts in the coordinate space and thus commutes with
∇µ.) The final result for the trace of the differential operator within the method of covariant
symbols therefore reads
Trf(D,M) =
∞
X
k=0 (−i)k
k! Z
x,p
tr(Ξ0∇0)kf(D,M)11. (6)
The sum over k only matters at nonzero temperature; at zero temperature all k 6= 0 terms are integrals over a total p0 derivative, ∇0, hence only thek = 0 term survives. To see why Eq. (6) constitutes a convenient starting point for calculations, we use the freedom in the choice of the operator Ξµ. Setting simply Ξµ=Dµ, we obtain
Dµ=eiD·∇(Dµ+ ipµ)e−iD·∇ = ipµ+ ∞
X
n=1
in
(n+ 1)(n−1)!Fα1···αnµ∇α1· · · ∇αn
= ipµ+
i
2[Dα,Dµ]∇α+· · · ,
M=eiD·∇Me−iD·∇ =M+
∞
X
n=1 in
n!Mα1···αn∇α1· · · ∇αn
=M+ i[Dα,M]∇α−
1
2[Dα,[Dβ,M]]∇α∇β +· · · ,
(7)
where Fα1···αnµ ≡ [Dα1,[· · ·[Dαn,Dµ]· · ·]] and Mα1···αn ≡ [Dα1,[· · ·[Dαn,M]· · ·]]. We can see that by the additional similarity transformation, the “free” derivative in Dµ+ ipµ is removed
strength tensor Fµν and its covariant derivatives. The derivatives with respect to coordinate
have been traded for ones with respect to momentum. As a result, the objectf(D, M) is gauge-covariant. At zero temperature, the integrand in Eq. (6) is then gauge-invariant before the momentum integral is performed. At nonzero temperature, there are noncovariant contributions coming from the k 6= 0 terms in the sum, as we will see below. The above series expansion of
Dµ and M allows us to calculate the trace of f(D,M) systematically order by order in the
covariant gradient expansion.
Explicit example
Let us illustrate the above rather abstract derivation on a concrete example. We will calculate the trace of a bosonic propagator on a background of gauge and scalar fields, 1/(−DµDµ+
M), with the same notation as before. For the sake of clarity, we will first assume that the temperature is zero. In order to appreciate the computational simplicity of the method of
covariant symbols, we will start with the method of symbols.
Zero temperature: method of symbols
According to Eq. (3), we are to compute the trace of 1/[−(Dµ+ ipµ)(Dµ + ipµ) +M]. We
rewrite the denominator as G−1−Σ and expand the propagator as
G−1−Σ−1
=G+GΣG+GΣGΣG+· · · , (8)
where G = 1/(p2 +M) and Σ = 2ip· D+D
µDµ. To second order in the covariant derivatives,
we thus obtain
Tr 1
−DµDµ+M
= Z
x,p
tr
G+GDµDµG −4G(p· D)G(p· D)G
11 , (9)
where we discarded the term linear inDµ since it is odd in momentum and thus vanishes upon
integration. In the second term, we use the isotropy of the momentum integral to replace
pµpν → 1dp2δµν so that the whole term becomes −4dp2GDµGDµG. As the next step, we have to
work out the action of the derivatives. Since they simultaneously have a matrix structure, it is most convenient to do so in terms of commutators. We rewrite theDµG factor on the far right
as [Dµ,G] +GDµand replace this with [Dµ,G] +GAµ since the derivative part ofDµ annihilates
the unit matrix. Similarly, we replace the far left factor ofGDµwith−[Dµ,G]+AµG. This time,
the derivative standing to the left of all the other terms corresponds to a total derivative and thus gives a vanishing contribution upon momentum integration. Working out the products of matrices and using the cyclicity of the trace,1 we arrive at the intermediate result,
Tr 1
−DµDµ+M
= Z
x,p
tr n
G −[Dµ,G]2−[∂µ,G]GAµ+ [∂µ,G]AµG − A2µG
2
+ 2AµGAµG−
−4
dp
2 −[D
µ,G]G[Dµ,G]−[∂µ,G]G2Aµ+ [∂µ,G]AµG2− A2µG3+ 2AµGAµG2
o
. (10)
1It is important to notice that the cyclicity property of the trace can be used onlyafter all “free” derivatives
We are now in the position to observe that all noncovariant terms cancel pairwise upon mo-mentum integration. To see this, we use the following identity which is easy to prove by means of integration by parts,
Z
ddp ψ(p2) =−2
d
Z
ddp p2ψ0(p2), (11)
where the prime denotes a derivative with respect to p2. Using furthermore the fact that
∂G/∂p2 = −G(∂G−1/∂p2)G = −G2, we can immediately infer that the integral of the trace of −A2
µG2 +
4
dp
2A2
µG3 as well as one of 2AµGAµG − d8p2AµGAµG2 vanishes. For the terms
containing [∂µ,G], we first rewrite [∂µ,G] =−G[∂µ,G−1]G =−G[∂µ,M]G and subsequently use
the identity (11) to manipulate the terms on the first line of Eq. (10),
Z
x,p
tr −[∂µ,G]GAµ+ [∂µ,G]AµG
=
Z
x,p
tr G[∂µ,M]G2Aµ− G2[∂µ,M]GAµ
=−2
d
Z
x,p
p2tr −2G[∂µ,M]G3Aµ+2G3[∂µ,M]GAµ
= 4
d
Z
x,p
p2tr −[∂µ,G]G2Aµ+[∂µ,G]AµG2
,
(12)
which exactly cancels similar terms on the second line of Eq. (10). Finally, we have to manip-ulate in the same way the remaining, covariant terms,
Z
x,p
tr−[Dµ,G]2+
4
dp
2[D
µ,G]G[Dµ,G]
=
Z
x,p
tr−G[Dµ,M]G2[Dµ,M]G+
4
dp
2[D
µ,G]G[Dµ,G]
=−4
d
Z
x,p
p2tr [Dµ,M]G2[Dµ,M]G3
. (13)
Putting all the pieces together, we arrive at the final result for the trace, to second order in the covariant gradient expansion,
Tr 1
−DµDµ+M
= Z
x,p
tr
1
p2+M − 4
dp
2
[Dµ,M] 1
(p2+M)2[Dµ,M] 1 (p2+M)3
. (14)
The momentum integral can actually be worked out analytically for arbitrary matrix-valued fields, but we will not do it here since our main aim is to show how the derivative expansion of the trace is performed.
Zero temperature: method of covariant symbols
To see how to derive the same result within the method of covariant symbols, recall from Eq. (7) that to second order in covariant derivatives we have
Dµ= ipµ+
i
2Fαµ∇α, M=M+ i[Dα,M]∇α− 1
2[Dα,[Dβ,M]]∇α∇β. (15) Otherwise, we follow the same steps as in the preceding calculation. This time, we find
Σ = −1
2Fαµ(pµ∇α+∇αpµ)−i[Dα,M]∇α+ 1
2[Dα,[Dβ,M]]∇α∇β. (16) The first term proportional toFαµcan be immediately discarded since the momentum derivative
in it acting on other propagators in the trace will produce a factor pα and the whole term will
vanish thanks to the antisymmetry of the field strength tensor Fαµ. We thus get
Tr 1
−DµDµ+M
= Z
x,p
tr
G+ 1
2G[Dα,[Dβ,M]]∇α∇βG − G[Dα,M]∇αG[Dβ,M]∇βG
As the next step, we apply integration by parts with respect to momentum to ∇α, using that
[∇α,G] = −2pαG2, and perform angular averaging in the momentum space, using cyclicity of
the trace,
Tr 1
−DµDµ+M
= Z
x,p
trnG+ 2p 2
d −[Dµ,[Dµ,M]]G
4
+ 2[Dµ,M]G[Dµ,M]G4
o
. (18)
Finally, we integrate the term with a double commutator by parts with respect to the coordinate, using the obvious identity
Z
x
tr X[Dµ, Y]
=−
Z
x
tr [Dµ, X]Y
, (19)
and, voil`a, this already gives us the final result (14)! Let us reiterate that in order to obtain the correct, fully covariant result, we did not have to use any identity for the momentum integrals to get rid of seemingly noncovariant terms, unlike in the case of method of symbols. Also, the amount of labor necessary to arrive at the result was now significantly smaller. In fact, the method of covariant symbols is clearly superior even if no gauge field is involved. The covariant derivatives then become ordinary ones and Dµ = ipµ. One then deals with an expansion in
derivatives ofM(x) that are all encoded inM.
Nonzero temperature
Let us now see how the calculation needs to be modified in the case of nonzero temperature. Having compared the two methods sufficiently, we will resort to the method of covariant sym-bols. To second order in derivatives, we need to evaluate the k = 0,1,2 terms in Eq. (6). Let us start with the k = 0 term. Since the integration over p0 is replaced by a mere sum, two steps we used in the preceding subsection are no longer allowed: full (d-dimensional) angular averaging and integration by parts in momentum space. We thus have to return to Eq. (17) and work out the action of derivatives first,
“k= 0 term” = Z
x,p
trG − G[Dα,[Dβ,M]](δαβG2−4pαpβG3)−
−4pαpβG[Dα,M]G2[Dβ,M]G2+ 2G[Dα,M]G[Dβ,M](δαβG2−4pαpβG3) . (20)
Next we use the identity (19) to make the replacement (valid under the coordinate integral2)
tr Gm[D
α,[Dβ,M]]
→ −tr [Dα,Gm][Dβ,M]
=−
m−1 X
n=0
tr Gn[D
α,G]Gm−n−1[Dβ,M]
=
m
X
n=1
tr [Dα,M]Gn[Dβ,M]Gm+1−n
. (21)
This leads after a short manipulation to
“k = 0 term” = Z
x,p
tr G −δαβ[Dα,M]G2[Dβ,M]G2 + 4pαpβ[Dα,M]G2[Dβ,M]G3
. (22)
2The integration by parts incoordinate space is not harmed by nonzero temperature since the bosonic fields
Since p0 is now discrete, we can only average over angles in the remaining (d−1)-dimensional Euclidean space. In other words, we can replace
pαpβ →p20δα0δβ0+ p2
d−1δαiδβjδij. (23)
Likewise, we apply the identity (11) to the δαβ term in the modification applicable to the
(d−1)-dimensional integration over p. This yields the result for the k = 0 term,
Z
x,p
tr
G − 4p
2
d−1[Dµ,M]G 2[D
µ,M]G3+ 4
p20− p
2
d−1
[D0,M]G2[D0,M]G3
. (24)
Let us proceed to the k = 1 term. He we can immediately trade the left-most covariant derivative D0 for mere A0, for its ∂0 part gives a total time derivative that vanishes upon integration. Also, we need just the term in Σ of first order in derivatives. Using repeatedly the fact that a Matsubara sum (integral) of an expression odd in p0 (p) vanishes, we obtain
“k= 1 term” =−i Z
x,p
trA0∇0(G+GΣG)
=−
Z
x,p
tr A0∇0G[Dµ,M]∇µG
=−
Z
x,p
tr 4p20A0G2[D0,M]G2−2A0G[D0,M]G2+ 8p20A0G[D0,M]G3
. (25)
Finally, thek = 2 term is easiest to evaluate. Throwing away again the left-most time derivative, one finds
“k = 2 term” =−1
2 Z
x,p
tr
(D0∇0)2G
= Z
x,p
tr (A2
0+A0∂0)(G2−4p20G 3)
, (26)
up to second order in derivatives and/or gauge fields. All that remains to be done is working out the action of ∂0. Also, in order to make the noncovariant nature of the k = 1,2 terms manifest, we rewrite [D0,M] in Eq. (25) explicitly as [∂0,M] +A0M − MA0. Finally, we once again apply the identity (11) to those terms in Eqs. (25) and (26) that do not carry a factor of
p2
0. The calculation is completely straightforward and putting all the pieces together, we arrive at the final expression for the trace of the propagator at nonzero temperature, to second order in the covariant gradient expansion,
Tr 1
−DµDµ+M
= Z
x,p
tr
G − 4p
2
d−1[Dµ,M]G 2[D
µ,M]G3+
+ 4p20− p
2
d−1
[D0,M]G2[D0,M]G3+ (27)
+[∂0,M]GA0G3−[∂0,M]G3A0G −2A0GA0G2+A20G 3
.
Note the structure of the result. The terms on the first line give the zero-temperature limit (14) since the fulld-dimensional angular average ofp2/(d−1) as well as that ofp2
