物 理 8017: 量 子 場 論 一
潘傑森，D00222032
作業 1
Contents
Ex. 1
Ex. 2
Ex. 3
Ex. 4
Ex. 5
Ex. 6
Ex. 7
Ex. 8
References
1
2
3
4
4
5
6
6
6
Equations numbers refer to [2], however they have been changed from, for example, (1) to
(4.1) in order to more clearly indicate the section for the ease of reference of the reader. All
problems in these solutions are from part I of [2].
Exercise 1 (I.2.1). Verify
−iHT qI =
qF e
Z
Z
Dq(t) exp i
0
T
1
dt mq˙2 + V (q)
2
,
(2.5)
where H = pˆ2 /2m + V (ˆ
q ).
Proof. This is carried out in almost precisely the same manner as the case of a free particle
Hamiltonian, the relevant steps of which we will reproduce for the edification of the reader.
T
Indeed, we begin by dividing the time interval into N segments of length δt = N
, so that
+
* !
N
X
−iHT qF e
qI = qF exp −iH
δt qI
n=1
+
* N
Y
−iHδt = qF e
qI .
n=1
By inserting a complete set of position eigenstates between each term in the above product, and
setting qF = qN and qI = q0 , we obtain
Z
Z
−iHT qF e
qI =
· · · dq1 · · · dqN −1 qN e−iHδt qN −1 qN −1 e−iHδt qN −2 · · · q1 e−iHδt q0
R
R
(∼
= 2.3)
In what follows, all integrals will be over all of R (unless otherwise indicated), so we will drop
the bounds on the integrals below. Just as in the case with the free particle, we now turn our
attention to evaluating the matrix elements begin multiplied in the above integration; however,
this is where we need to begin taking into account the nonzero potential term in our Hamiltonian,
so the divergence of the discussions will become apparent. Inserting a complete set of momentum
1
物理8017, 潘傑森
eignestates and expanding the exponential to first order (since we are interested in the limit
sup{δt} → 0 as it is) yields
Z
−iHδt dp −iH(ˆp,ˆq)δt qj e
p hp| qj−1 i
qj e
qj−1 =
2π
Z
dpj
=
hqj |1 − iH(ˆ
p, qˆ)δt| pj i hpj | qj−1 i
2π
Z
dpj pj (qj −qj−1 )
1 − iH(pj .qj−1 )δt
e
=
2π
Z
dpj pj (qj −qj−1 ) −iH(pj ,qj−1 )δt
=
e
e
2π
Z
dpj ipj (qj −qj−1 )−H(pj ,qj−1 )δt
=
e
2π
From here the remaining steps are identical to free particle case. Putting together what we have
so far yields
Z
Z
−iHT qF e
qI = · · · dq1 · · · dqN −1 qN e−iHδt qN −1 qN −1 e−iHδt qN −2 · · · q1 e−iHδt q0
!
Z
Z
N Z
Y
dpj ipj (qj −qj−1 )−H(pj ,qj−1 )δt
= · · · dq1 · · · dqN −1
e
2π
j=1
!
!
Z
Z
N Z
N
Y
X
dpj
= · · · dq1 · · · dqN −1
exp i
pj (qj − qj−1 ) − H(pj , qj−1 )δt .
2π
j=1
j=1
Taking the limit sup{δt} → 0, the sum becomes an integral
and
we are
−iHδt
left
with the desired
expression. It should be noted that in the evaluation of qj e
qj−1 one expands the
potential V (q) as a power series and uses a few Gaussian integral formulas (like thse in the
appendix for I.2 in [2]) in order to see that
pj (qj −qj−1 )
hqj |1 − iH(ˆ
p, qˆ)δt| pj i hpj | qj−1 i = e
1 − iH(pj .qj−1 )δt .
Exercise 2 (I.2.2). Verify
hxi xj · · · xk xl i =
X
(A−1 )ab · · · (A−1 )cd ,
(2.24)
Wick
where we have defined
R∞
hxi xj · · · xk xl i =
−∞
R∞
1
· · · −∞ dx1 · · · dxN e− 2 x·A·x xi xj · · · xk xl
R∞
R∞
1
· · · −∞ dx1 · · · dxN e− 2 x·A·x
−∞
(2.25)
and where the set of indices {a, b, ..., c, d} represent a permutation of {i, j, ..., k, l}. The sum in
(2.24) is over all such permutations or Wick contractions.
Proof. We will begin by writing, for the sake of clarity in what follows (at least for a mathematician...),
N X
N
X
J · A−1 · J =
(A−1 )ji Jj Ji .
作業 1
j=1 i=1
物理8017, 潘傑森
Following the suggestion given in [2], we can rewrite the numerator of the expression we are
interested in as a collection of derivatives with respect to the components of J, and then set
J = 0, i.e.
Z
Z
Z
Z
∂p
− 12 x·A·x
− 21 x·A·x+J·x · · · dx1 · · · dxN e
· · · dx1 · · · dxN e
xi xj · · · xk xl =
,
∂Ji ∂Jj · · · ∂Jk ∂Jl
J=0
where p is the number of terms in xi , xj , · · · , xk , xl . For notational simplicity, we will use
∂p
∂p
≡
.
∂Jij···kl
∂Ji ∂Jj · · · ∂Jk ∂Jl
Now it should be clear that we can apply (2.22) to the remaining integral, from which we obtain
r
Z
Z
1
(2π)N
∂p
1
− 2 x·A·x
−1
· · · dx1 · · · dxn e
·
exp
J ·A ·J
xi xj · · · xk xl =
det A ∂Jij···kl
2
J=0
!! r
N
N
(2π)N
∂p
1 X X −1
.
=
·
exp
(A )mr Jm Jr
det A ∂Jij···kl
2 m=1 r=1
J=0
Thus, using (2.22) with J = 0 to evaluate the denominator of (2.25), the constant cancels so
that we find
!! N
N
∂p
1 X X −1
.
exp
(A )mr Jm Jr
hxi xj · · · xk xl i =
∂Jij···kl
2 m=1 r=1
J=0
Carrying out the remaining derivative with this level of generality will quickly become a notational nightmare (even the simple case considered below is quite cumbersome), so we will simply
consider a specific example in the case p = 2 and N = 2. Then we have that
∂2
1
−1
2
−1
−1
−1
2
hxi xj i =
(A )11 J1 + (A )12 J1 J2 + (A )21 J2 J1 + (A )22 J2
exp
∂J21
2
J=0
1
∂ 1
−1
2(A−1 )11 J1 + [(A−1 )12 + (A−1 )21 ]J2 e 2 J·A ·J =
∂J2 2
J=0
1 −1
1
1
=
[(A−1 )12 + (A−1 )21 ] e 2 J·A ·J +
2(A−1 )11 J1 + [(A−1 )12 + (A−1 )21 ]J2
2
2
1 J·A−1 ·J 1
−1
−1
−1
2
×
[(A )12 + (A )21 ]J1 + 2(A )22 J2 e
2
J=0
= (A−1 )12 ,
where we have use the symmetric nature of A in the last step. It should be clear that for larger
values of p and n one would have a term (A−1 )ab · · · (A−1 )cd for each way of pairing the indices
{i, j, · · · , k, l}, and all other terms will have a power of some component of J which will vanish
when evaluating the derivative at J = 0; thus, we will be left with a sum of terms of the above
type over all Wick contractions, as desired.
Exercise 3 (I.3.1). Verify that D(x) decays exponentially for spacelike separation.
Proof. For spacelike separated points we can always find a reference frame so that x0 = 0. In
particular, this tells us that x2 = −~x2 and that px = −~p · ~x = −|~p| · |~x| cos θ, where θ is the angle
between p~ and ~x. This allows us to right the propagator in polar coordinates as
Z 2π
Z π
Z
p2
dp
−i|p|·|x| cos θ
p
D(x) =
dϕ
dθ e
2(2π)3 p2 + m2
0
0
Z π
Z
dp
p2
−i|p|·|x| cos θ
p
=
dθ e
.
2(2π)2 p2 + m2
0
作業 1
物理8017, 潘傑森
Making the change of variable ρ = cos θ we then obtain
Z 1
Z
dp
p2
−i|p|·|x|ρ
p
D(x) =
dρ e
2(2π)2 p2 + m2
−1
Z
1
p sin(|p| · |x|)
= 2
dp p
4π |x|
p2 + m2
m
= 2 K1 (m|x|),
4π |x|
where K1 is a modified Bessel function of the second kind. It is well-known that K1 can be
written in the form
K1 (x) = e−ax P (x),
where a is a positive constant and P (x) is a power series; thus the above expression for the
propagator reveals that it will decay exponentially.
Exercise 4 (I.3.2). Work out the propagator D(x) for a free field theory in (1 + 1)-dimensional
spacetime and study the large x1 behavior for x0 = 0.
Proof.
Exercise 5 (I.3.3). Show that the advanced propagator defined by
Z
d4 k
eik(x−y)
Dadv (x − y) =
(2π)4 k 2 − m2 − isgn(k0 )ε
(where the sign function is defined by sgn(k0 ) = +1 if k0 > 0 and sgn(k0 ) = −1 if k0 < 0) is
nonzero only if x0 > y0 . In other words, it only propagates into the future. [Hint: both poles
of the integrand are now in the upper half of the k0 -plane.] Incidentally, some authors prefer to
write (k0 − ie)2 − ~k 2 − m2 instead of k 2 − m2 − isgn(k0 )ε in the integrand. Similarly, show that
the retarded propagator
Z
d4 k
eik(x−y)
Dret (x − y) =
(2π)4 k 2 − m2 + isgn(k0 )ε
propagates into the past.
Proof. The hint essentially gives away the entire argument; however we will attempt to provide
some level of detail for the arguments from complex analysis that the physics student reader
may not be familiar with (although we will not be entirely rigorous).
The primary portion of the integral defining the advanced propagator of interest to us is the
k0 integral, so we will focus our attention on
Z 0
Z R
Z
dk
eik0 (x0 −y0 )
dk eik0 (x0 −y0 )
dk eik0 (x0 −y0 )
= lim
+
.
2
2
(2π) k02 − m2 − isgn(k0 )ε R→∞ −R (2π) k02 − m2 + iε
0 (2π) k0 − m − iε
We will utilize the standard machinery from complex analysis to rewrite the above integrals in
a form more suitable for these considerations, i.e. we will consider k0 as a complex variable
and will examine an integral over a contour which will include the real axis portion that we are
interested in. An important observation to make at this point about our choice of contour is that
in the event that x0 > y0 , the above integral is highly divergent if the contour we integrate over
extends infinitely into the upper half plane (Re(k0 ) > 0), as the growth of the exponential term
will dominate the shrinking contribution of the quadratic in the denominator; hence we should
consider contours which avoid the upper half plane as much as we can. Similarly, if x0 < y0 , then
we want to avoid the lower half plane (Re(k0 ) < 0). Thus, we consider the following contours in
these scenarios, respectively:
作業 1
物理8017, 潘傑森
Using the residue theorem (see for example, [1], Theorem V.2.2) we obtain that
Z
√
√
X
eik0 (x0 −y0 )
dk
i[Res(+ ) + Res(− )],
=
i
Res(Pole
#i)
=
2
2
0, C = Cp ,
C (2π) k0 − m − isgn(Re(k0 ))ε
C = Cf
i
where the sum is over all the poles enclosed by C, which would be both for Cf and neither for
Cp . In the above formulas we did not explicitly calculate the residues, nor write out where the
poles are located, as all we care about for this problem is the fact that they lie in the upper
half plane. In order to use this information to understand our original integral, we consider now
the integral over the semi-circular portion of the contour C (C can be either Cf or Cp : the only
difference will be a sign change that will prove inconsequential). We will denote this portion by
CR . We have
Z
dk
eik0 (x0 −y0 )
2
2
(2π)
k
−
m
−
isgn(Re(k
))ε
0
CR
0
Z π/2 iθ
Z π iθ
iReiθ (x0 −y0 )
iReiθ (x0 −y0 )
Re dθ e
Re dθ e
≤
+
2
2iθ
2
2
2iθ
2
(2π) R e − m − iε
(2π) R e − m + iε 0
π/2
Z π/2
Z π
1
Rdθ
1
Rdθ
≤
+
2
2
2
2π 0
R + m + ε 2π π/2 R + m2 + ε
1
R
= · 2
.
2 R + m2 + ε
In the limit R −→ ∞ in which we are interested, this upper bound forces the integral to vanish.
Thus, combining this with our previous results, we have that
Z
√
√
dk
eik0 (x0 −y0 )
i[Res(+ ) + Res(− )], C = Cf
=
0, C = Cp ,
(2π) k02 − m2 − isgn(k0 )ε
i.e. it is nonzero only if one is considering Cf , which corresponds to the condition x0 > y0 . The
argument for the retarded propagator is precisely the same, except for the poles now lie in the
lower half plane, so that the condition x0 > y0 will correspond to the vanishing integral and the
condition x0 < y0 will correspond to the integral that reduces to the poles; hence it is nonzero
only if x0 < y0 .
Exercise 6 (I.4.1). Calculate the analog of the inverse square law in a (2 + 1)-dimensional
universe, and more generally in a (D + 1)-dimensional universe.
Proof. Just as in the derivation given in [2] for the case of (3 + 1)-dimensions, we are interested
in the massless limit, in which the equation the potential must satisfy is simply given by the
(2-dimensional) Poisson equation:
∇2 V (x) = δ (2) (x).
From this, however, one can apply Gauss’ law (with the 2-dimensional Gaussian surface G, i.e.
the Gaussian circle) in order to determine that the radial gradient of V then obeys
|∇V | =
1
1
=
.
Vol(G)r
2πr
Integrating this, one obtains the potential
1
log(r),
2π
and therefore the analogue of the inverse square law will be given by
V (r) =
作業 1
1
Force ∝ .
r
物理8017, 潘傑森
Precisely the same arguments go through in (D+1)-dimensions, with the Gaussian circle replaced
by the D-dimensional Gaussian surface. Ultimately this yields
1
Force ∝ D−1 .
r
P (a)
(a)
Exercise 7 (I.5.1). Write down the most general form for
a εµν (k)ελσ (k) using symmetry
repeatedly. For example, it must be invariant under the exchange {µν ↔ λσ}. You might end
up with something like
AGµν Gλσ + B(Gµλ Gνσ ) + C(Gµν kλ kσ + kµ kν Gλσ )
+ D(kµ kλ Gνσ + kµ kσ Gνλ + kν kσ Gµλ + kν kσ Gµλ ) + Ekµ kν kλ kσ
(22)
P (a)
(a)
with various unknown A, · · · , E. Apply k µ a εµν (k)ελσ (k) = 0 and find out what that implies
for the constants. Proceeding in this way, derive (5.13).
Proof.
Exercise 8 (I.6.1). Putting in the numbers, show that the case n = 1 is already ruled out.
Proof. I’m really not sure how to approach this problem, as it seems like some sort of assumptions
must be made before carrying out any calculations and I do not know how to justify them, but
I will try nonetheless. One possible proposal, at least from what I could find, is that in ADD
theory one postulates that the true scale of gravity is on par with the electroweak scale, 10 TeV
(108 MeV). With this value for MTG , the given value of 1019 GeV (1023 MeV), and the conversion
given in class of 1 MeV−1 ) = 197 × 10−12 mm, we can use (6.3) with n = 1 in order to determine
that we should expect to see deviations from the inverse square law for r R with
M2
R = 3Pl ≈ 1022 (MeV)−1 ≈ 1011 mm.
MTG
This would certainly conflict with our everyday experience; thus ruling out the case n = 1 for
such a model.
One may object by saying that we chose MTG rather arbitrarily, and that a higher value for
the scale would result in lowering the value of R; thus one could find a value of MTG large enough
so that we would not expect deviations in even our most sensitive experiments testing gravity
on a scale of roughly 1mm and thereby save the case n = 1. Is there some upper bound on MTG
which would prevent people from making such an argument?
References
[1] J.B. Conway. Functions of One Complex Variable. Springer-Verlag, 2nd edition, 1978.
[2] A. Zee. Quantum Field Theory in a Nutshell. Princeton University Press, 2nd edition, 2010.
作業 1
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