Topics in Field Theories in Lower Dimensions
by
Edwin Richard Karat
Submitted to the Department of Physics
in partial fulfillment of the requirements for the degree of
Doctor of Philosophy in Physics
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
June 1999
© Massachusetts Institute of Technology 1999. All rights reserved.
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Department of Physics
April 20, 1999
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Roman Jackiw
Professor
Thesis Supervisor
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Thomas J. reytak
Associate Department Head for E ucation
MASSACHUSETTS INSTITUTE
MARzE1999
LIBRARIES
Topics in Field Theories in Lower Dimensions
by
Edwin Richard Karat
Submitted to the Department of Physics
on April 20, 1999, in partial fulfillment of the
requirements for the degree of
Doctor of Philosophy in Physics
Abstract
In this thesis, we discuss three problems in gauge theory.
First, we discuss the Hamiltonian for a nonrelativistic electron with spin in the
presence of an abelian magnetic monopole and note that the Hamiltonian is not selfadjoint in the lowest two angular momentum modes. We then use von Neumann's
theory of self-adjoint extensions to construct a self-adjoint operator with the same
functional form. In general, this operator will have eigenstates in which the lowest
two angular momentum modes mix, thereby removing conservation of angular momentum. However, consistency with the Dirac equation limits the possibilities such
that conservation of angular momentum is in fact not lost. Because the same effect
occurs for a spinless particle with a sufficiently attractive inverse square potential,
we also study this system. We use this simpler Hamiltonian to compare the eigenfunctions corresponding to a particular self-adjoint extension with the eigenfunctions
satisfying a boundary condition consistent with probability conservation.
Second, we examine a system with charged vector mesons interacting with a constant external magnetic field in 2 + 1 dimensions. We look at the eigenvalue problem
of the operator corresponding to the equations of motion, and use these solutions to
obtain the propagator.
Third, we consider a charged scalar field propagating in a constant external electric
field in 1 + 1 dimensions. The interesting property of this simple system is that it
has a central charge in its classical Poincar6 algebra. In order to check Poincar6
invariance, we modify the energy-momentum tensor such that it satisfies the DiracSchwinger relation. Next, we quantize the system and show that the massless case is
exactly solvable. Finally, we show that the algebra after quantization holds without
anomalies.
Thesis Supervisor: Roman Jackiw
Title: Professor
2
Acknowledgments
We would like to thank E. Farhi and R. Jackiw for discussions about our work.
We would also like to thank M. Schulz for collaboration on the work on self-adjoint
extensions. This work is supported in part by funds provided by the U.S. Department
of Energy (D.O.E.) under cooperative research agreement #DF-FC02-94ER40818.
This material is based upon work supported under a National Science Foundation
Graduate Research Fellowship.
3
Contents
0.1
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Self-Adjoint Extensions of the Pauli Equation in the Presence of a
7
Magnetic Monopole
2
5
1.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.2
Pauli Equation
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.3
General - Potential
. . . . . . . . . . . . . . . . . . . . . . . . . . .
15
Charged Vector Mesons in a Constant External Magnetic Field in
24
2 + 1 Dimensions
2.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
2.2
Eigenvalues
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
2.3
The Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
2.3.1
Two Useful Green's Functions . . . . . . . . . . . . . . . . . .
38
2.3.2
Sum Over n . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
3 An Example of Poincare Symmetry with a Central Charge
48
3.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
3.2
The Classical Symmetries
. . . . . . . . . . . . . . . . . . . . . . . .
49
3.3
The Dirac-Schwinger Relation . . . . . . . . . . . . . . . . . . . . . .
53
3.4
Solution of the Equations of Motion (for Zero Mass) . . . . . . . . . .
56
3.5
Quantization of the Algebra . . . . . . . . . . . . . . . . . . . . . . .
57
3.6
Current Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
4
0.1
Introduction
This thesis covers three separate problems solved over the course of our graduate
studies. The first studies a nonrelativistic electron with spin in the presence of an
abelian magnetic monopole in 3 + 1 dimensions. The second looks at charged vector
mesons propagating in a constant magnetic field in 2 + 1 dimensions. The third
considers charged particles moving in a constant electric field in 1 + 1 dimensions.
First, we discuss the Hamiltonian for a nonrelativistic electron with spin in an external magnetic field, as would be created by the presence of a fixed abelian magnetic
monopole. We note that the Hamiltonian is not self-adjoint in the lowest two angular
momentum modes and then use von Neumann's theory of self-adjoint extensions to
construct a self-adjoint operator with the same functional form. In general, this operator will have eigenstates in which the lowest two angular momentum modes mix,
thereby removing conservation of angular momentum. However, the solutions to the
Dirac equation do not need quite as large extension. In fact, if we limit the solutions
to the Pauli equation to be consistent with the nonrelativistic limit of the solutions
to the Dirac equation, the remaining extentions are limited such that conservation of
angular momentum is restored. Because the same effect occurs for a spinless particle
with a sufficiently attractive inverse square potential, we also study this system. We
use this simpler Hamiltonian to compare the eigenfunctions corresponding to a particular self-adjoint extension with the eigenfunctions satisfying a boundary condition
consistent with probability conservation.
Second, we examine a system with charged vector mesons interacting with a constant external magnetic field in 2 + 1 dimensions. We look at the classical eigenvalue
problem of the operator corresponding to the equations of motion. This gives us a
complete, orthonormal basis of eigenvectors of this operator. By taking an appropriate sum we can construct the operator's inverse, which is the system's classical
propagator.
Third, we consider a charged scalar field propagating in a constant external electric
field in 1 + 1 dimensions. The interesting property of this simple system is that it
5
has a central charge in its classical Poincare algebra. In order to check Poincard
invariance, we modify the energy-momentum tensor such that it satisfies the DiracSchwinger relation. Next, we quantize the system and show that the massless case is
exactly solvable. Finally, we show that the algebra after quantization holds without
anomalies.
Though the three problems are different, are solved in a variety of settings (quantum mechanics, classical field theory, and quantum field theory), and each lie in progressively fewer dimensions, there are some common threads. Each problem is that
of a charged particle propagating in an external electromagnetic field. Furthermore,
each problem is exactly solvable (all in terms of Bessel functions).
6
Chapter 1
Self-Adjoint Extensions of the
Pauli Equation in the Presence of a
Magnetic Monopole
1.1
Introduction
First, we examine the Pauli Equation for an electron in the field of a magnetic
monopole. This equation has appeared in the literature before, and it is well known
that an extension is needed to make the Hamiltonian self-adjoint in the zero angular
momentum
(j
=
0) sector[1]. What seems to have gone unnoticed is that, for minimal
monopole charge, the domain to be extended includes the
j
=
1 sector as well. With
the inclusion of this sector, the structure of the extensions becomes richer, and the
number of parameters required to describe them jumps from 1 to 16. While the parameters may be chosen to be consistent with angular momentum conservation, this
is not required: a pure incoming s-wave can come out with a p-wave component, even
though the functional form of the Hamiltonian is spherically symmetric. However, if
we require our states to match the states of the Dirac equation in the nonrelativistic
limit, we shall only have 1 free parameter, and angular momentum will be conserved.
To better understand the effect of the extension parameters and their relation to
7
angular momentum conservation, we also consider a simpler Hamiltonian of the form
H
2p
2pr2'
(1.1)
where c is an arbitrary constant. (Spin is not essential to this discussion and is omitted.) This Hamiltonian has all the essential features of the monopole Hamiltonian, so
we can use it to investigate how the extension parameters arise. To accomplish this,
we look at this Hamiltonian in 3-space minus a sphere of radius ro around the origin.
We compare the results of imposing a boundary condition consistent with probability
conservation with the results of creating a self-adjoint extension. Finally, we compare
the case of a nonzero radius ro with that of a zero radius ro.
1.2
Pauli Equation
Working in units where the speed of light and Planck's constant are both equal to
one, the Hamiltonian for an electron (with spin) in an electromagnetic field is
H = - - -a -B,
2p 2p
(1.2)
where 7 is the kinematic momentum P'- eAl, A is the vector potential, and B is the
magnetic field. For a point magnetic monopole of strength g sitting at the origin[2],
we have
-osCos 0)q 5
-g(1
A
=
(1.3)
r sin 0
B
(1.4)
=
H =
egd -
22
2p
8
ego
2pr2
(1.5)
where we have chosen a particular gauge to determine A. Using L = f x T - eg?, the
Hamiltonian can be rewritten as
1
H= f
-.
2L_
7+
e2g2 -egd*?
Here L is the orbital angular momentum satisfying [Li, Vi]
(1.6)
.
-
iciikVk for any spin-
independent vector operator. Note that [L, H] 0 0. The angular momentum operator
that commutes with H is the total angular momentum J = L + -5!.
In appendix I,
(1.6) is shown to simplify to
K(K + 1)
1 1 02
r+
2
2pr2
2p r Or
H =
(1.7)
with the further definition
K
=
-1 - j x
7
(1.8)
-a
[3]. We can find simultaneous eigenstates of K, J 2 , and J, involving only angular
variables. Such "monopole harmonics" will be denoted by Qm(0, q) and satisfy
=
KQKm
J 2Qrm
=
j(j + 1)QKm
(1.10)
JzQ#m
=
mQam
(1.11)
dQ IQ m(0,
where
j
(1.9)
KQKm
0)12 = 1
takes on the values 0, 1, 2, . .. and , is related to
j(j
K=:
for eg = }.
(More generally, K =
See appendix I.) Note that for
j
(1.12)
j
by
+1)
(1.13)
(j+ })2 _ e2 g2 with j
> 0 (or
j
> eg
-
}in
=
eg
-
}eg + }.
the general case), there are
two sets of m-multiplets for each value of j, one corresponding to
,
> 0, and one
corresponding to K < 0. In terms of Bessel functions and the monopole harmonics,
9
the E > 0 solutions to the eigenvalue equation H4'
qNmE =
1
=
JKm
EVb, are
=
r-- J, (A )Qm(0, q)
(1.14)
r Y, (Ar) Q r,,,(0, )
(1.15)
A=
2pE,
=
K+
2
and the E < 0 solutions are
_=r
K (Ar) Qm(,
(1.16)
)
(There is another set of E < 0 solutions which has I, instead of K,, but they
grow exponentially at large distances and need not be considered.)
At this point
we mention that any dependence on the gauge is contained entirely in the form of
the Q,,m. The radial part of these solutions and the eigenvalues of the operators are
left invariant under a guage transformation. The set (1.14) (N for Nonsingular) of
solutions vanishes at the origin, while the sets (1.15) (S for Singular) and (1.16) (B
for bound) are singular at the origin. Although the TN are not square integrable over
all space, they can be 6-function normalized and are square integrable over any finite
region. However, the AFs and
"B
are only normalizable (6-function normalizable in
the case of the Ts) when v, < 1 4
j
K = 0, -V/2
= 1 triplet). If v,, > 1, the Ts and
containing the origin. (The cutoff UK
=
'JB
(i.e. for the j = 0 singlet and one
are not square integrable over any region
1 is equivalent to a coefficient of (K+ 1) =
for the 1 term in (1.7). This coefficient is a general cutoff for a - potential [6] [7] [8].)
It is tempting to let the Hamiltonian operator (1.7) act on all linear combinations
of the normalizable (including 6-function normalizable) solutions. However, this is
incompatible with the Hermiticity condition (H#, 'b)
=
(0, HVb) for all 0, 4 in the
domain of H. Furthermore, we seek a Hamiltonian which is not only Hermitian, but
also self-adjoint, for only then are its eigenfunctions complete. At this point, we need
to become more precise in our usage of the term operator, from now on including
10
the domain of an operator as part of the operator's definition. Precise definitions of
Hermitian and self-adjoint will be employed, and can be found in appendix II. To
start with, let H1 be the operator given functionally by H in (1.7), and defined on
all functions V) in the Hilbert space such that H4' is in the Hilbert space. We expect
the Hamiltonian to be Hermitian; however,
(H1'V, #) = (V, H1 ) +
rim-
r0o2p I
r 2dQ
(Or
# -
*
(1.17)
Or
from integration by parts, so H1 is not Hermitian. Next, define H 2 to be the operator
identical to H, except that the domain is further restricted to
dom(Hi) lim
r-O f
r2 d#
r
-
*
Or)
=0
for all 0 E dom(H).
(1.18)
By comparing (1.17) with (1.18), H2 is seen to be Hermitian. In other words, the
domain of H 2 is the set of # in the domain of H1 such that
(V, H 2 0) = (H1 0, )
(1.19)
for all V4 in the domain of H 1 , which means that Ht = H 1 . (H1 is the adjoint
of H 2 .) Since the domain of H1 is larger than that of H2 , H 2 is not self-adjoint;
however, the domain of H2 can be extended through a method of von Neumann to
create an operator that is self-adjoint. According to the von Neumann theory of selfadjoint extensions, we need to look at the number n+ of normalizable solutions to the
equation H 10+ = +ip+ and the number n- of normalizable solutions to the equation
H10- = -ip#.
(Note that the use of 1 is arbitrary and chosen only to provide the
correct units. Any positive real constant may be used instead.) We can index these
solutions and denote them by 0', where i ranges from 1 to ni±. If n+ = nthen H2 can be made self-adjoint by introducing the n vectors
#b =
n,
#$ + U,'L into
its domain, where U is an arbitrary unitary n x n matrix (with n 2 real parameters).
Thus, a general element of the domain of the extended Hamiltonian, HU, is of the form
Z ciz
+
, where q is in the domain of H 2 . (The reader can check that the extended
11
Hamiltonian satisfies the self adjointness criterion of appendix II.) For the monopole,
the domain of H 2 consists of those functions in the domain of H1 that vanish at the
origin at least as fast as r 1 / 2 . Then, the normalized solutions to Hlq4 =
2
_ 8p
cos(vslr/2) r-1
2
are
t±
KvK ((1 - i)iur)Qnm,
(1.20)
7F
j
0, K = 0, m = 0 or
j
= 1, , = -v 2, m = 0, ±1.
There are 4 of each, so von Neumann's Theorem tells us that we need 42
=
16
parameters to describe each extension. Given 16 parameters in the form of a unitary
matrix Ujm/,,
(j, j'
m' = -j',.
= 0,1; m= -j,...,j;
..
,j'), the Hamiltonian can be
made self-adjoint by introducing the 4 vectors
0
=
m
(1.21)
+
into its domain. (On a cautionary note: The superscripts
be considered merely labels.
jm
on the Ym should
The states Ym defined above and the WIFJn defined
below are not necessarily eigenstates of angular momentum.) The qfPm are not energy
eigenstates; however, for each
/flm
and for each positive energy eigenvalue E, there
exists one energy eigenstate, Ti", that differs from
/Y m by
an element in the domain
of H 2 . When UP!", is diagonal, the IjNm are simultaneous eigenfunctions of J 2, j,
and Hu, but in general this is not the case: The V
m
will be eigenfunctions of HU
only, since the simultaneous eigenstates of H 1 , J 2 , and J, corresponding to
j
=
0,1
are not in the domain of HU. Similarly, not all of the eigenstates of J 2 and J, will be
eigenstates of Hu. In other words, a pure angular momentum eigenstate with
j
= 0, 1
will in time evolve as a superposition of states with mixed angular momenta. Thus,
angular momentum is not conserved in the j
=
0, 1 sector for general U (though it
is conserved for all higher angular momentum states). Now we can construct the
energy eigenstates. Because the T'" differ by Ofm by an element in the domain of
H2 , which vanishes at the origin at least as fast as r-, we only need to consider the
behavior of the solutions (1.14), (1.15), and (1.16) at the origin. To get a particular
12
energy eigenstate, pick a value of energy E and a particular Ofm. Then, look at its
behavior and the behavior of the above solutions for small r. Any linear combination
of solutions whose small r behavior matches the small r behavior of the particular q'"
(up to a part that vanishes at least as fast as r") is an eigenvalue of the self-adjoint
operator HU. For positive values of E, there is precisely one energy eigenstate for
each /"'.
However, most negative values of E fail to yield an eigenstate. When we
try to match the solution to Y', we derive the following relation between the energy
E and the diagonal matrix element Um
1 + i"KU
E =
where, as before, v, =i + 11and
ested in r,
l
v-p"
i
,
(1.22)
j(j + 1). (Specifically, we are only inter-
=
0 for j = 0 and r, = -V/'2 for j = 1.)
For the bound state to exist,
we require that E be real and negative. The reality condition requires Uj
2
= 1.
Since U is unitary, this requires the the row and column corresponding to j and m
consist entirely of zeros, except for the diagonal element, which must be of the form
eiO, where 6 is real. (Note that this also implies that the bound state must be an
angular momentum eigenstate.) This allows us to simplify (1.22) to
E= -[P [cos(7rv,,/2) + cos(O)
11
(1.23)
+ cos(O - 7rvr,/2)_
So we see that we also have the additional condition on the diagonal element:
cos(6) > -cos(7rvK,/2).
(1.24)
There exists one bound state for each such row; therefore, there can be anywhere from
zero to four bound states, depending on the particular self-adjoint extension chosen.
Now we compare our results with a similar treatment for the Dirac equation that has
13
already appeared in the literature[9][10]. We work with the Dirac Hamiltonian
.-7r +
H =
(1.25)
and in a basis where
,
(0=
,
(i
S0
0
(1.26)
-
and 0" are the Pauli matrices. The advantage of this basis is that the Dirac spinor
can be separated into upper and lower bispinors, where the lower bispinor is dropped
in the nonrelativistic limit. When appropriately separated, the eigenvalue equation
H@b = E4 becomes
g
(p
-
E)fKmE
i(Or +
Solving for
f
mE,
)
LQ
Z
(;f
(1.27)
-mE
+
-(Or
K)9mE
+ (y + E)gKmE
+
r
0
(1.28)
0.
(1.29)
we get the familiar solutions for E > p
1(1.30
anE
=r
v,~Ar
.(Ar)
fKmE =r-
(1.31)
and for F < p
A=
p2 - E,02
KmE
-r
=f
1
14
2K,,(Ar)
1
1
(1.32)
These are the same solutions as for the Pauli equation, except that A has the different
+ E' and identify E' with the nonrelativis-
expression above. However, if we let E =
tic energy, we get A = I2pE'+ E'2
12pE'l1/2 in the nonrelativistic E' << p
1/2
limit, and we recover the solutions to the Pauli equation in this limit. One additional
difference between the solutions to the Pauli and Dirac equations is that we now have
an additional function
gmE
_i(r
+ (1+ )r-1) fKmE/(,(
+ E), which we require to
be square integrable over a finite region containing the origin. One may easily check
for
K =
-
V'2 that
gSmE
is not square integrable over this region, even though fmE
is. Thus, we no longer have the singular solution for K =-/
as we did in the Pauli
case above. When we look for self-adjoint extensions, we find that we no longer need
an extension for the j = 1 sector; only the j = 0 sector requires an extension. Thus,
angular momentum modes may not mix; furthermore, we only need one parameter
to specify the extension. From another point of view, consistency with the nonrelativistic limit of the Dirac equation requires us to fix 15 of the 16 parameters of the
Pauli equation.
1.3
General ' Potential
The Pauli equation in the presence of a magnetic monopole is just one example of a
Hamiltonian where the choice of a self-adjoint extension can lead to non-conservation
of angular momentum. Looking at the form of (1.7), we can see that the same essential
behavior can be obtained from a spinless particle with a sufficiently attractive inverse
square potential, as in (1.1). Analysis of this new Hamiltonian is simple and analagous
to the analysis of the magnetic monopole. Similarly, the arguments in this section
can be modified to include the monopole Hamiltonian or any similar Hamiltonian
that needs an extension. With this simpler Hamiltonian, we can investigate how the
extension parameters arise. To do this, consider a space with a sphere of radius ro and
centered around the origin removed. We impose conservation of probability at the
boundary and seek a relation between the extension parameters and the boundary
conditions.
Now, consider a spinless particle governed by the Hamiltonian (1.1),
15
whose functional form is given by (using L = -i'
H =
1 1 d2
2p r dr
r +
2
x V)
-+L
2pr2I
(1.33)
r > ro.
-
This Hamiltonian has appeared in the literature before. [6] The solutions to Ho = E0
are
N
WSmE
BjmE
A =
2pjEj',
1
,(AT)ylm (Q)
r 2J7
=lm
(1.34)
(Ar) Ym(Q)
(1.35)
_ - Ka (Ar)Ym(Q)
(1.36)
=
r-
2Y
1
61 + j(j + 1) - C,
are the solutions for E > 0 and TlmE is the only solution
where T1 'mE and
IjpmE
for E < 0.
and j1gmE are still singular at the origin; nevertheless, because the
4FmE
sphere r < ro is no longer a part of our space, the singularity of solutions at r = 0
is no longer important: All of the E < 0 solutions can be normalized, and all of the
E > 0 solutions can be 6-function normalized. Now we wish to select those solutions
that are consistent with probability conservation. Probability is conserved at the
boundary r = ro if
I
dQ
L
(1.37)
0.
-ro
To ensure this, we impose the most general boundary condition at r
=
ro that is
consistent with (1.37) by introducing a function gro(Q, Q'):
0
(r,Q)
=
JdQgro (Q,Q')V)(ro,
')
(1.38)
ro
with the requirement that gro (Q, Q') is Hermitian, i.e.,
g*o (Q', Q) = gr 0 (Q, Q').
16
(1.39)
We allow the boundary condition to have a continuous dependence on ro through the
explicit appearance of the subscript ro in (1.38) and (1.39). If gr(Q, Q') and V(Y) are
expanded in spherical harmonics as
6,#)(.0
gro (G, ') = YM(0, #)gom''y,1m,
1M1'M'(1.41)
(1.41)
(O) = O1"'(r)Ym(O, #),
then (1.38) and (1.39) take on the matrix form
d V'm(r)
dr
=
lml'm'Vl')m'(ro)
(1.42)
lml'm'
(1.43)
ro
l'm'Im*
Now, to construct the eigenfunctions of the Hamiltonian, we simply take the linear
combinations of the above solutions for a given energy E which are consistent with the
boundary condition. The above procedure is self-contained and distinct from the von
Neumann procedure. It describes a boundary condition that restricts wavefunctions
from the entire Hilbert space to a subspace on which probability is conserved at the
origin. To make contact with the von Neumann procedure, we seek a relation between
the boundary condition (1.42) and the unitary matrix that needs to be specified to
apply the von Neumann theory. Instead of imposing the boundary condition (1.38)
on all solutions to the eigenvalue equation, we can choose a particular self-adjoint
extension HU. To constuct Hu, start with H 1 , the operator with the functional form
of H in (1.33) and domain consisting of functions 7P in the Hilbert space such that
HV' is in the Hilbert space. Then, create a new operator H 2 as we did in (1.19) and
look at the solutions to the equation H1 q4 = ±4tqp±. This time, we obtain a solution
01"v()= O"(r)Yim(0, #)
(no sum)
(1.44)
for each l,m. Since there are infinitely many 1, m, n+ and n_ are infinite, and we
can create a self-adjoint operator HU by extending the domain of H 2 to include the
17
infinite collection of vectors {jlm()}, where each vector is of the form
() = 5+"lm(-) + Uj,
(lm
(1.45)
"'
and where U is an infinite-dimensional unitary matrix. (Again, the qlm are not angular
momentum components in the sense of (1.40) and (1.41).) We now seek a relationEnforcing the hermiticity condition, ($m, HUV) =
ship between Ulml'and g,'
(Hyq5m,V)), for all V) in the domain of Hu, we have
'"*(7)HU4(7)=
r>!ro d
(
- Im dG"m (7)
1ro
3
'd(H
= :1m]U
"'m())* (Y)
Idlm * v()
(7)r
OQ
r
ro
(1.46)
ro
using the explicit functional form of the Hamiltonian, (1.33).
Expanding V) as in
(1.41), we can take advantage of the orthogonality of the angular momentum harmonics to integrate them out, leaving only the radial part of the angular momentum
components:
1
It'mI
+
+
d
Uj/,#'"/'/ ))* dl''d"'(r)
U111M(r
dr
ro
(r)n,
±
(r))* y"'m'(r)
+'mm
(no sum on 1m). (1.47)
ro
(The "Im" of (1.46) has been dropped since the phase of 0 is arbitrary.) We define
am
(r)= (#"(r)6f,"1 + U
01g''"'(r))
(no sum).
(1.48)
Note that aml'm'*(r) is the i'm' angular momentum component of qlm(Y), i.e.,
) = am"'
7xl"(
(r)Ypm, (Q).
18
(1.49)
Viewing a as a matrix in 1m and i'm' which depends on r, we can write (1.47) as
(r)=
a-(r)--a(r)
dr
'mI'(r)
,
(1.50)
dr
where the inverse and product inside the parentheses are a matrix inverse (but not a
functional inverse with respect to r) and a matrix product. Since this is of the same
form as (1.38),
(a-i
~ (r)+a(r).
dr
gi'm"'
9
ro
(1.51)
r
Furthermore, if we take Vp = 01"""in (1.50) (so that Vlm(r) = al"/ml"*(r)), we have
da
l"lm"*
"(r)
=
-1 (r
a(r)
((r)
--
a(r)
mma
=a-'(r)
a(r) ,
"'(
(1.52)
((1.53)
which shows that gM''m' is Hermitian. Equation (1.51) is the desired link between
the extension and the boundary condition. Now that we have found it, we ask what
happens in the limiting case ro = 0. To this effect, first note that, for a fixed value of
c, most of the singular Bessel function solutions cease to be normalizable in this limit.
The same holds for the #7(7): Since many of the ql7(') are no longer normalizable,
any self-adjoint extension defined in this section which adds non-normalizable
1± (1)
to the domain of H1 is no longer valid. To determine which of the solutions Om(5),
VmE(z), and XFimE(') are still at least 6-function normalizable, we require, as
in the
previous section, that the coefficient of the - term in the Hamiltonian be less than
3.
Then, our requirement becomes 1 < crit, where
3
'crit(icrit+ 1) - C = -.
4
(1.54)
Those extensions that remain valid have U diagonal for 1 > 1crit, with entries such
that the linear combinations of
qJi 2E
for 1
1Crit.
1 J1mE
and
ThnJE
within the domain of HU are purely
At this point, the reader may object that we are including an
19
infinite number of vectors in these extensions while the von Neumann indices are now
finite. This is not a problem, since the
#5Im(s)
with 1 > 1crit that we are including
already exist within the domain of H 2 . Returning to equation (1.51), it follows that
the only boundary conditions admissible in the ro -+ 0 limit are those for which g1ir'm'
is diagonal, except possibly for the block with 1,1' < 1crit as ro -+ 0. One should
note that the relationship between gro and U becomes singular as ro -+ 0 due to the
singularities of the 0s"(r) in al'm'. In general, finite entries in U lead to singular
entries in gr0ao. Thus, for ro = 0, it is more convenient to describe the domain of the
Hamiltonian through U than through a boundary condition at the origin. However, if
one asks for any physical description of the choice of extension, the formulation (1.42)
is more valuable. It tells us, for instance how the radial flux leaving the boundary
through the 1m channel is related to the probability amplitudes at the boundary for
each channel:
JIM
-=
lm* (r)d 1 M()
E
4
ro
lm*(ro)gm'm' Il'm'(ro)
(no
sum on
Im).
(1.55)
In summary, we can deal with the singularity at the origin by removing a small sphere
of radius ro from around the origin. When we do this, we must impose a boundary
condition consistent with probability conservation, and we need a periodic function
in two angular variables to describe this. With certain restrictions on the boundary
condition when ro = 0, the angular momentum components of this function are in
direct correspondence with the elements of the unitary matrix required in the von
Neumann theory. It is in this sense that the boundary condition is equivalent to a
choice of self-adjoint extension, or alternatively, that the self-adjointness condition is
equivalent to probability conservation at the boundary.
20
Appendix I
We start with a Hamiltonian of the form (1.6)
H =(
Using -eg-
=
2
=
, we can rewrite the contents of the curly braces as
- - e2 2
2 +.
U)2 --
-(j
(1.57)
x i).-
+ -
x
Following the convention of [4] and [5], we define K = -1
eigenstate of J 2 (where J
(1.56)
.
2pr2
2p-r2r
2
(L - rx
{}
{L 2 _ 22 - eg( -f)}
1..
Ir(T-)+
e2
(1.58)
2.
- (r' x
)
For an
i.
L + .5), this then becomes
{}=
J2
(1.59)
+ K + 4 e22.
The eigenvalues of J 2 will be of the form
j(j+1) for j
=
We shall
eg - 1, eg + ,.
show below that the eigenvalues of K are
±
t(j
K
+
)2 _e 2g 2.
(1.60)
Given this relation between the eigenvalues of K and J 2 , the operator represented by
the terms in curly braces has eigenvalues r,(K + 1), where , is the eigenvalue of K.
This demonstrates the equivalence of the terms in (1.6) and (1.7). Next, the vector
potential (1.3) for the magnetic monopole has no radial component, so 7 -r'
and
j'-
'
fp -r
f - , where jis just the mechanical momentum operator given by -iV.
Thus, the first term of (1.6) is just the usual -
r term. To prove (1.60), first
write
((r
x
7)
--
--
-.
-,
eg -..
r
= (L + egr) x (L + eg?) = L x L + -(L x r+
21
x
L).
(1.61)
Now, L satisfies [L', Vi]
= ijiikVk
for any vector operator that does not depend on
spin coordinates, so the relations L x L = iL and L x r'+ r' x L = 2ii? hold, and
( x ') x ( x ')
=
(1.62)
i(L + 2eg?).
If we now square the operator K, we see that
K2
1+2
=
2'
using the identity (o-A)2
((? x
-)
2
= L2 _ e 2 g ),
=
-(x
) + (, . (( x
)) 2
(1.63)
(x2
)]
x ( x
A 2 +id- (Ax A). Making the substitution rx 7
(1.64)
=
L+egy
this finally becomes
K2 =1+
-+
_2
eL2 2
j2 ±
e2 2.
(1.65)
Now, K commutes with J, since it is a total angular momentum scalar. Therefore,
we can find eigenfunctions Qm of K and J with respective eigenvalues K and m.
(Although we do not prove it here, these eigenfunctions are complete.) Then, from
(1.65),
K 2 Q0m = (J 2 + 1 - e2 2 )aKm,
(1.66)
4
which shows that QKm is also an eigenfunction of J 2 . Replacing J2 in (1.66) with its
eigenvalue j(j + 1), we then arrive at
2
= j(j+1) + - - e2
2=
(j + -)2 _e2g2.
(1.67)
2
4
This proves (1.60), as long as we can show that r takes on both positive and negative
values. The latter follows from the anticommutation of K with (a
K(9 - ) + (04 - )K = 0,
22
)
(1.68)
as the reader may verify. Then, if KQm = Kcm,
K(-
i
,QKM
=
-(-
-)Krm
= -K(0-
(the m value is unchanged through multiplication by a'
-
QKM
Q
).
(1.69)
Furthermore, it can be
shown that
(or5)0
-
(1.70)
= Qoo.
So, from (1.69) and a phase convention that is consistent with (1.70),
Q-Km
(1.71)
= (0 - )QKm.
Appendix II
The definitions of Hermitian and self-adjoint are[11:
An operator H is Hermitian
if its domain is dense, meaning every state in the Hilbert space can be arbitrarily
well-approximated by states in the domain, and if
(H, #) = (0, H#)
for every
# and V) in the
(1.72)
domain of H. The adjoint of a densely defined operator H,
denoted Ht, is defined for any
4 such
that there is an n for which
(0, HO) = (TI, #),
(1.73)
for all 0 in the domain of H. In this event HtV' = j. H is self-adjoint if H = Ht.
A crucial part of this definition is that the domains of H and Ht be equal. For a
Hermitian H, the domain of Ht is always at least as big as the domain of H, and
Ht#$= H# if # is in the domain of H.
23
Chapter 2
Charged Vector Mesons in a
Constant External Magnetic Field
in 2 +1 Dimensions
2.1
Introduction
We want to study charged vector mesons in 2 + 1 dimensions.
The mesons,
1",
interact with a constant magnetic field via the Lagrangian
L
(Gv)tG,,
(2.1)
= DIbv - DV<I
(2.2)
= O, + ieA,
(2.3)
-
2
where
GM"
DA
Ao
0
(2.4)
A1 = 0
(2.5)
= xB
(2.6)
A2
24
The classical equations of motion are
-DAGA"
= 0
(2.7)
A quantity of interest is the propagator. In order to calculate this, we shall first
solve the eigenvalue problem for the inverse propagator. Then we shall calculate the
propagator P via the relations
> = A
P
>
P =
(2.8)
(2.9)
A
For simplicity, we shall do our calculations in Euclidean space.
2.2
Eigenvalues
We want to find the solutions to the eigenvector equation
-DG""
-
AV"
(2.10)
Under some changes of variables
(t, x, y)
=
x
=
E
=
A =
2krdE eiky+iEtI
eBj - k/eB
X/
eB1,
D
=
eBx
=
Z eBP
=
iM(eB)
D2
i
25
(2.12)
(2.14)
eBE
=
(2.11)
(2.13)
eBju
Do
(E, x, k)
(2.15)
(2.16)
eBIX
(2.17)
where E(eB) denotes the sign of eB and we have defined P = -i~x such that [P, X]
-i. If we also absorb a factor of c(eB) into V)Y, such that e(eB)V@y
X2 + p
2
-eP
-EP
E2 +X
4'y,
we get
-eX
2
(2.18)
= up
-Xp
E2 + P2
-PX
-eX
-+
Defining
=
XiX - PObY
(2.19)
(2.20)
and taking the appropriate linear combinations of equations, we see that
x
2
+ p
2
I
-C
0
-
= 0
0
-i
-W(X2 +
62
i
P 2)
_
(2.21)
From this we can see that we can reduce our eigenvalue equations to a system of
b^, and
C-number equations if we expand V/),
Vi/
in terms of the simple harmonic
oscillator eigenfunctions f,(X) =< X~n > such that
(X 2 + P 2 ) In > = Nn >= (2n +I1)In >
(2.22)
for all nonnegative integers n.
This inspires the definition of the following basis of Ox and O/
(
X
_1
-P
n
VN +
2
VN -1
-iN+1
26
( )
In -1 >
In +
>
(2.23)
X +
v)2
iNP
'IN3-N (iNX
vN+1-
-N
VN -
in -1>
iN
ivdN -1I
In+1I >
I
1 is
well defined for n > 0 and
+I
02
is well defined for n > 1, but
)
02
(2.24)
is not defined for
n = 0. Instead, we define
1=0 >
v-
(2.25)
Together, these form a complete, orthonormal basis for V)' and V1.
We can now start looking for solutions. We let V)' and Viy be some linear combination of V51 and
V)2
and let V)' be some multiple of In > for n > 1. For some choice
of constants K and C
(
400
4,;
-C~n >
(2.26)
K+N
1
N 2 -1Kv
2
X -iZKP
(2.27)
-i KX - P )i
our eigenvalue equation (2.10) implies
N- u
-EP
--CP
x2
-Ex
+2
-PX
-Ex
-U
C
X -iKP
-XP
p2 + E2 _
-iKX - I )
(N -u)C+iE+iKNE
(N + K+62 _u)X +(-EC -iK(E 2 _ u))P
(-C
=
- iK(E 2 -u))X - (N + K +E 2 _u) p
0
27
I
I
(2.28)
These equations imply
u
=
K+N+e2
(2.29)
C
=
K(K + N)
(2.30)
where K is one of the three solutions to the cubic equation
K 3 + K2(N + E2) _
E2
=0
(2.31)
These solutions are
K 1 = a(1 + 2cos(#))
(2.32)
K2 =
a(1 - cos(#) + x/_sin(#))
(2.33)
K3
a(1 - cos(#) - V3sin(o))
(2.34)
=
e=
cos(30)
=I
1(N + E2)
3
(2.35)
+
(2.36)
23
2a
For c2 > 0 note that -1 < 1 + '
<
I so
#
is well-defined up to an irrelevant
ambiguity that interchanges the three roots) and that the three roots of the cubic
equation are real. Moreover, the above solutions are well defined and independent
for n > 1 and E2 > 0. Finally, the existence of three independent solutions for each
n > 1 implies that these solutions span V) with the exception of the space spanned by
V)= 10 >, V'@, and
021.
For
= iCf0 >
=
01=
28
(2.37)
(
(2.38)
we require
1 - u
-EP
X
2
-EP
-EX
iC
+ E2 - u
-XP
x
P 2 ±62
-PX
-eX
|0 >
-P
_
iC(1 - u)+ i
-i6CP + (1+E2 -u)X
-iECX - (1 +
|0
u)P
62
i(C(1 - U) + E)J
_(EC + 1 + E2 _
>
I >/
2
I
(2.39)
from which we obtain
U
= E2 +
C2 + CC - 1
C =
-)
(2.40)
C + 1
(2.41)
0
2
2FG+1
Since this has two independent solutions, only
(2.42)
remains. For this case
2
-0
(
(2.43)
= v)2
71)
=
(2.44)
and so
X
2
-EX
+ p2 -u
X2+
62
-PX
29
-XP
_
P2
+ 62 _
0
-iE(X + iP)
1
u- C2-_x(X+i1P)
=
i(2
S
_u)+ P(X +iP))
0
10 >
-2
-If2)
-
-0
~
(2.45)
which simply requires
U
(2.46)
2
-
so we expect this to be all of the solutions.
We shall now summarize the solutions that we have just found to the eigenvector
problem (with the proper normalizations). For n > 1, we have
00
i K(K+N)
= A
X
X -iKP
(2.47)
|n >
-iKX - P
Y
u
K+N+c
2
(2.48)
A-
1
AK2(N +E2)+
3K + 2N
-
2(2.49)
This defines three solutions for each value of n > 1, with K for each solution taking
on one of the three real solutions to the cubic equation
K 3 + K2(N +E2)
30
_2
= 0
(2.50)
In addition, we need to include
00
iC
1=A
IPX
X
|0>
(2.51)
-P)
U
ec+
+
=
(2.52)
I
1
(2.53)
This defines two solutions, with C for each solution taking on one of the two real
solutions to the quadratic equation
C 2 +EC-
=
(2.54)
0
Finally, we have one last solution
\I
0
2/
U
For the case e
=
=
-1
I0>
(2.55)
(2.56)
f2
0, the last three solutions are well defined, but the solutions for
n > 1 are not. However, the solution in the limit as e -- 0 is both well defined and
still a solution of the equation with c = 0. Explicitly,
K,=
-N + 0(c2)
(2.57)
K2
=
-N-1/26 + O(C2)
(2.58)
K3
=
+N-1/26+O(62)
(2.59)
to lowest order in E. For K = -N, we get
-
31
0
(2.60)
(
,~v)2
(2.61)
n~
(2.62)
which satisfies
N
0
0
X2
-XP
0
-PX
P2
0
0
X + NP
=
|n >
0
(2.63)
iNX- P)
±N- 1/ 2C, we have
For the final two solutions K
i
7p)
(
(2.64)
vf2n
(2.65)
u = N
(2.66)
which satisfy
0
0
0 X2 -N
--XP
X
-pX
p2-N
-P)
0
2.3
0
|n>
=0
(2.67)
The Propagator
We now seek to find the propagator by calculating the sum (2.9). We have
P1-"
dE
IA
f2wF
2w
2w
Ie
2d
AI
X
ik(y-y')+iE(t-t')
A
ik(y-y')+ic
32
eB(t-t')
__
(2.68)
Note that the extra factor of
eBl was added to the sum over V) to compensate for
the normalization difference between X and x.
E V)P(X)V)vl(xl)
=
276(X - X')6P"
1
27r6(x
-
-
(2.69)
x)61"
eBl
First, we need to calculate
1
1
K 2 g+3K+ 21 - 3C2K+N
n>1 K
1
-
(
C2
2
2
p x2
Cj0 >< 11
)
ill ><
2ill >< 1
0
0
0
0
1
i1
0
p1x
III ><1
Cl1 >< 0|
+
2 0 ><
±+
PiXi
CI0 >< 11
Cl1 >< 0
-
p2x1
+1
C 2 10 >< 0l
-
P1x2
1
C 2 + 1c2 +C
+
Plxi
(2.70)
-Zi 1
K 2 (K + N) 2
-=>
(2.71)
K(K+N)
= 2e
In>
(
(K -1)
v/N--I< n- 11
(K +1)V/N
+1< n
(2.72)
p
2
2
xl
p x
2
K(K+N)
2c
i
-1
( (
(
i
-i
_1 -1)
(K -1)v/,,IN-
I n -1I>
(K +1)v/-N- +11N +1I>
(K-i1) N-1In-I>
-1)
(K+1)VN+1|n+1>)
33
)
(2.73)
-(K - 1),IN - I < n - 11 (K + 1)VrN + I < n + 1
-i
Ki
-q
(2.74)
-1)
where N = N + 62 and the sums over K and C are the sums of the roots of the cubic
and quadratic equations in K and C respectively.
We need to evaluate the sums over the roots. We begin by observing that the
symmetric sums of products of roots Si through S5 satisfy
-
(K - K1 )(K - K 2 )(K - K3 )
=
0
(2.75)
S1 = K1+K2+K3
S2
-N
= K 1 K 2 +K 1 K+ K 2 K
-
0
(2.77)
K 1 K 2 K3
S3=
=
c2 +EC - I
(2.76)
-
=
2
(2.78)
(C -C
1
)(C -C
2)
0
(2.79)
S4= C1+C
2
(2.80)
=-f
S5 =
-
C1C
2
-1
(2.81)
We can also simplify expressions cubic or higher in K and quadratic or higher in C.
(K 2 N + 3K + 2N - 3E2 )(K + N)
(C2 + 1)(EC + E2 +1)
34
=
3K
=
EC + E2 +2
2
+ K(5N + 262) + 2NN
(2.82)
(2.83)
K2(K + N) 2
2
=
K(K + N)(K + 1)
2(1
-
K(K+N)(K -1)
=-K
C2
2
K
+K+N-6
_ C2)
(2.84)
(2.85)
+ KN + E2
-KN
2)
2(1 +
2
+E2
(2.86)
1-EC
-
(2.87)
To evaluate the sums, we also need the sums of products of
Di
=
3K2 + Kz(5N + 26 2 ) + 2NN
D1 D2 D3
=
27S3 + 9(5N + 2E2 )S 2 S3
+18NN(S2
+3(5N +
-
2S 1 S2 +6S
(2.88)
3)
262)2SiS 3
+6NN(5N + 2E2 )(SIS
2
-
S 3 ) + (5N + 2E2 ) 3 S 3
+12N 2N 2 (S2 - 2S2) + 2NNR(5N + 2E2 ) 2S 2
+4N 2 N 2 (5N + 2E2 )S 1 + 8N 3N3
=
D1 D 2 + D1 D 3 + D 2 D 3
=
2 (4N 3 -
27, 2 )(N
2
-
(2.89)
1)
9(S2-2S1 S 2 + 6S 3 )
+3(5N + 2E2) (S1S2 - 3S3) + 12NN(S2 - 2S 2 )
K1 D 2 D 3 + D 1 K 2 D 3 + D1 D 2K 3
+(5N +
262)2S2
+12N 2N
2
3-
=
N(4N
=
9S2S3 +
+ 4NN(5N + 2E2)S1
(2.90)
27E2)
6(5N + 2E2 )SiS
+3(5N +
2E2)2 S3
3
+ 6NN(SS
2 -
3S 3 )
+ 4NN(5N +2E2)S2
+4N 2N 2 S
= -N
Ei
2 (4N 3
=
EC, + 62
=
E2+4
- 27E2)
+2
(2.91)
(2.92)
E1 E2
(2.93)
35
cS
4
=
E2
+4
+4
2
+2c
=
(2.94)
which gives us
K
K
1
K2 + K(5N + 262)+ 2NN
K
3
K2 + K(5N + 262)+ 2NN
2N
(2.95)
N2 -_1
3
2N2
-6-2
=
I
(2.96)
-
= 1
El
CC + E2+ 2
(2.97)
(2.95) and (2.96) imply
E
K
K2
3K
2
K(5N + 2E2) + 2N
3K
K(5N + 2E2) + 2NN
1-
+K(5N +2E 2 )+ 2NN
2+
K
2N 2
1+
=
(2.98)
_
So our original sums are
K
E2 +
2
N 2 -1
N +
=
2
K + K + N -E
3K 2 + K(5N + 26 2 )+ 2NN
N 3 - N 2 c- 2 + N 2 E-2 - N
C
=
2
(2.99)
2
-(1 + E2 )K2 - KN + E
2E
3K 2 +K(5N +2f 2 )+ 2NN
N 3 (1+6 2 )+N 3 +E 2 N
2/N--1 (_(1+62)+
N/-1z
K
S
E-2
2-
N2 - 1
=-(N + 1) sNR2
=
26
3K
2E
K
\/N +1
-
-
(2.100)
2+
K(5N +
2
(1-2+
(A-i1)
2/6+1
36
-
2 (1
-
2E2)+
2
2NN
)N 3 - N3 +6 2 N
)
(2.101)
(K -1)
N-i
R3
4
K
3K2+K(5N+2E2)+2NN
SN-(1
2N3+2N2+N
N2 - i
4
1
-(N - I + E-2N(N + 1))
4
=
N+i
44
(K +1)
4
i
2
2
N3 -2N2+N
N2 - i
(N + 1 + E-2N(N - 1))
K2
N2
=
(2.102)
3K2+K(5N+22)+2NN
N+1
4
R
2
4
(2.103)
i
3K2+K(5N+2E2)+ 2NN
K
vIN 2 - I
N3 - N
(1+ N 2 - i
4
I
(N + 1) v-iN2 _
=
(2.104)
11
C2
E EC + f2+2
3 + 62
-2+E
6C+62+2
C
=
E EC + E2+2
2
- (2
=
C
+ 1
(2.105)
-E
C
C2+62+2)
(2.106)
c
S
=
60+E2+2
(2.107)
i
After summing over K and C, we have
PE=
(I + f2)J0 >< 01
7261><
n>O
+0 ><
+
2
E2
O1
>< 01
261
0
0
0
0
1
i
0
-l
1|
-E|0 >< 11
2 1><1
i|
62l ><
i|
(2.108)
1)
37
PO
fRoIn >< ni
(2.109)
-iln > (R 1 < n - 11 - R2< n + 1 )
(2.110)
PSO
=
P2
= -
P
=
>(fi<rn- 1+
R2< n + 11)
(2.111)
(2.112)
i(RoIn - I > -Rijn + 1 >)
PSI= R 3 in- I >< n - II+ R 4 In + I >< n + II
-R 5 (in - 1 >< n +
PS 2=
1
=
3
-iR
5 (n-1
i
3
>< n + 1 - in + 1 >< n -)
=
(2.115)
In + I > < n - 1)(2.116)
R 3 In- 1 >< n - II+ R4In +1 >< n +li
n + 1 > < n - 1)(2.117)
+ R5 (In - 1 > < n + 1-
2.3.1
(2.114)
n - 1 >< n - I - iR4In + 1 >< n +1
-iR5 (In - I > < n + 1+
PS2
(2.113)
n - 1 >< n - II - iR4n + 1 >< n + lI
-iR
-(Rijn - 1 > +R 2 n +I1 >) < n
p 0
P
+I In+l >< n - 1)
Two Useful Green's Functions
We are now faced with several sums over n. All of these sums we need to do can be
broken down in terms of the following two sums
En>1 Nin>< nj
(2.118)
En>i N-1
(2.119)
in >< ni
These sums will be calculated by finding the inverse operator that is the Green's
function of the corresponding differential operator.
For the first sum (2.118) we want to find the solution to
(-a
+ X 2 + 1)F(X, X')
=
6(X - X')- < X|0 >< X'|0 >
38
(2.120)
where < Xn > is the nth normalized harmonic oscillator eigenfunction
< X10 >=<
IX >
< X~n >=-< n|X >
=
7gr-1/4e-X2/2
(2.121)
2
CnHn(X)e -X /2
(2.122)
with H, the nth Hermite polynomial and Cn is a normalization constant. Our solution
is subject to the conditions
F(X, X')
=
I dX'F(X, X') < X'0 >
F(X', X)
(2.123)
= 0
I dX'F(X, X') < X'|1 >
(2.124)
4 < X|1 >
(2.125)
We start the solution by writing
F(X, X') =
e+(x2+x, 2 )/2f(X, X') -
1 e-(x2+x'
2, fi
2
)/2
(2.126)
which yields the equation
(2Xix +Ok)f (X, A
=
-e-X
2
(2.127)
6(X - X')
Writing
axf (X, X')
=
k(X,
X')eX
2
(2.128)
we have
Oxk(X, X')
=
-6(X - X')
(2.129)
1
k(X, X') = ko(X') - IE(X - X')
2
Oxf (X, X')
=
(ko (X') - IE(X - X'))eX
= fo(X') +ko(X')I(X)
f(X, X')
39
(2.130)
2
(2.131)
1
2
-
X') (I- (X) -- I_ (X'))
(2.132)
I_ (X)
=
e-S2ds
(2.133)
Since F(X, X') = F(X', X) we have
2
-ex,
6(X - X')
= (2X'Ox, + ak,)f(X, X')
= (2X'Ox, + O2,)fo(X') + (2X'Ox, + O,)ko(X')
-ex,26(X - X')
(2.134)
(2X'Ox, + O,)fo(X')
(2X'Ox, + ak,)ko(X')
(2.135)
=0
(2.136)
fA(X')
=A
+ BI (X')
(2.137)
ko(X')
=C
+ DI (X')
(2.138)
One final check of F(X, X') = F(X', X) requires us to have B = C. Checking (2.124),
0
=
=
dX'F(X, X') < X'|0
>
Coex2/2 f dXI(A + BL_(X) - I(I-(X)
2
-A
CoeX 2/2(2A(A + BI_(X)) 2
e
f
dse-2
-
L-(X/))c(X
-
X'))
-
EeX
2
/2
2
2
dX' - 1
dse-, I
2 x,
A
dX'
2
_x /2
CoAeX 2 /2(2A + 2BI_ (X) + from which we conclude that A = --
(2.139)
dse-, 2 )
and B = 0. Checking our final condition
(2.125),
4< X|I>
=
dX'F(X, X') < X'|1 >
=
CieX 2 /2
JdX'(Dl(X)I_(X')
=
CeX2/2
(DI_(X)
(- I
dse-
40
1
- (X)
2
2
dX'X'
ISA
-
-I-(X')) C(X
-
')X
±A
Jf dse2 -A
dse2
x
=
CieX2 /2
A2
-
from which we conclude D = - 1-
>
n>1
1
< Xn >< nIX'>
N+ I
=
dX'X'))
2
f
AdX'X'-
1) I- (X)
-- D +
dse~s 2 IA dX'X')
4
2)
Xe~x2)
(2.140)
So we end up with our sum
e+(x2+x'
2
)/2
(
I
V-Fi
(4
-
E(X
- X')(I_ (X) - I_ (X'))
(
(2.141)
/
(2X/
-
2
I_(X)I_(X')
The other sum is similar but slightly more complicated since 10 >< 01 is in the
kernel of the differential operator. We want to solve
(-ak + X 2
1)G(X, X')
-
=
6(X - X')- < X|0 >< X'|0 >
(2.142)
subject to the conditions
G(X, X') = G(X', X)
I dX'G(X, X') < X'|0 >
=0
I dX'G(X, X') < X'|1 >
=
(2.143)
(2.144)
2
< X|1 >
(2.145)
We start the solution by writing
G(X, X')
=
e -(X2+X,2)/2g(X, X')
(2.146)
which yields the equation
(2XOx - Oi)g(X, X')
=
41
eX26(X - X') -
I
(2.147)
Writing
Ox g(X, X')
= h(X, X')ex 2
(2.148)
we have
9xh(X, X')
=-6(X - X') + I e-
h(X, X')
I+(X)
I-+(X)
Since G(X, X')
e x,26(X
-
=
2
=ho(X') - 1E(X - X') +
E(X - X') +
(ho(X') -
axg(X, X')
g(X, X')
X
=
go(X') + ho(X')I+(X) 1
+
I-+(X)
I-(X)
(2.150)
I_(X))e-X2
(2.151)
E(X
-
X')(I+(X) - I+(X'))
(2.152)
Xes2ds
=
=
(2.149)
J0
(2.153)
es2I (s)ds
(2.154)
G(X', X) we have
X')
1-
=
(2X'Ox,
-
O,)g(X, X')
(2X'ax, - ak,)go(X') + (2X'Ox, -
+e x, 26(X
(2X'Ox,
--
(2.155)
- X')
1
aX,)go(X')
(2.156)
(2X'ax, - ak,)ho(X') =0
g 0 (X')
=
h0 (X')
=C
One final check of G(X, X')
0 =
=
Oi,)ho(X')
(2.157)
A+BI+(X')+
(2.158)
I+
+ DI (X')
G(X', X) requires us to have B
JdX'G( X, X') < X'|0 >
42
(2.159)
=
C. Checking (2.144),
=
CoeX 2 /2
(A + BI+(X) - -(I+(X)
dX'e
-
I+(X'))C(X - X')
+ I (I-+(X) - I-+(X')))
=
CoeX
2
(-
+
>
ds I
1
dX'eX
j
SAds
dte-,2)
2
dX/e8
_x/2)
Coe7X 2/2 ((A + BI+(X))V+I-+(X)
2
Coe-X
2
(
\i
dt
dX'
f0
dt f dX' lo ds)
2
13)
(2.160)
2_,2
dseS2
eS2_t2_X,2
x dse, (I?
+ I
+I(s))
((A + BI +(X)) /-
/2
/A
0
2
dse
A
ds+
dt
(fAdX'
j
=
(2
dseS2 _X,2
X'
A
J-
-
=
1vI-+(X)
/2 \ F (A
\ + BI+(X) +
(2.161)
0
from which we conclude that A = -- 13 and B = 0. To determine 13
A
3
fo
o
dO j
JA r2dr
f=
7r/4
(
p7 4 dO
rsinO
-
a)
dses2-r 2
drr2e-r2cos2a)
c
p3/2
_an
16F 4
(1:r4 do)
V/jcosr /
a tancz-
='K71lr(2)0
so A= -
_X,2
dyacosae-r2Cos2
j
dacosa (IA
da
2
20
A1 rd
-
dse's
d
dX'
r/4
V ln(s)
(2.162)
ln(2).
43
Checking our final condition (2.145),
1 <X|1>
dX'G (X, X') < X'|1I >
=
J dX'X'e-x,
Cie-X 2 /2
I+(X'))6(X
- 2I(I+(X =
C 1 e x 2 /2
2 (DI+(X)I+(X')
(2DI+(X) j
- X')
X)
dse
dseS2 js
f- A
-
=
Cie-X 2 /2
A
dse, 2
I
dX'X'e7x, 2
I
dX'X'e-x, 2
A
dX'X'e-xt2)
(DAI+(X) + 1X
(2.163)
from which we conclude D = 0. So we end up with our sum
>< nix >
n>I1
=
e-(x2+x
2
)/2
(
+
2.3.2
2 (I+(X)
ln(2)
1
(I-+(X
-
I+(X'))E(X - X')
) - I-+(X')))
(2.164)
Sum Over n
Now that we have our two Green's functions F and G, we can calculate the sums
over n, but first we need a few identities
=
E
n><nl
n>1
=
1 -I0 >< 01
(2.165)
=
(X+iP)ln>
(2.166)
+1> =
(X-iP)ln>
(2.167)
v/N - 11n - 1 >
v/N+1-n+I
(X - iP)F(X +iP) =
Z(X-iP)N
n>1N+I
44
1
ln><n|(X+iP)
E In+ 1 >< n+11
n>1
(2.168)
1 - 10 >< 01 - 11 >< 1|
1
Z(X +iP)N - 1|n >< n|(X - iP)
(X + iP)G(X - iP)
n>1N-
n
-
1 >< n1-1
n>1
1
(2.169)
(X + iP)I
X + iP
(2.170)
I(X - iP)
X -iP
(2.171)
(X + iP)(X - iP)
N +1
(2.172)
(X - iP)(X + iP)
N- i
(2.173)
N(X - iP)
(X - iP)(N + 2)
(2.174)
N(X + iP)
(X + iP)(N - 2)
(2.175)
We also define the following
Q3=
n>1
R3n-
1 >< n - 11
E(X + ip)1(1 + E-2 N(N++1)
(X + iP)(I + E-2((N + 2)1 + 2G))(X - iF)
4
I (N + 1 + -2(N
4
Q4
)In >< nI(X - Z'P)
2
(2.176)
+ 5N + 6))
ZR4n + 1 >< n+ 1i
n>1
(X(iPIG+
n>1
1(X
4
-
4N+1I
E-2N+
1 )In >< nI(X + iP)
iP)(I + E-2((N - 2)1 + 2F))(X + iP)
1(N - 1+
4
-2(N2
- 5N + 6))
- 200><01j-
2 1>< 11
1E
1
45
(2.177)
=
Z R5Ir + 1 >< n
-2 (N 2 - 5N + 6))
1 +
-
= (X - i)4 (NE 2 + 1)I(X
1
-
(N
4
(2.178)
(2.179)
iP)
-
n>1
Q+
n>1
R5 (In - 1 >< n + 11+ In + 1>< n - 11)
I(X(N6-2 + 1)X - P(Ne- 2 + 1)P)
2
Q
R5 (n - 1 >< n + 11 - in + 1 ><n
=
(2.180)
- 11)
n>1
(X(N6-2 + 1)P + X(NE-
2
2
+ 1)P)
(2.181)
Applying all of this to our previous expression for the propagator (2.108),we get
P0
=
E Roln >< n| + (1 +
E2 )\0
>< 01
n>1
P'E,
=
(N + 62 )I + (1 + E2 )10 >< 0I
=
N+ c 2
=
1 E (Ri ln - I > -R2ln +l1>) < nl+
(2.182)
E ll>< 0l
n;>1N/2
p 20
=
-i-1(X + iPN)I +
=
-i-
=
-
1
Ell >< 01
(2.183)
(X + iPN)
> +R21n+1 >) < n +
Ell >< 01
n>1
SE- 1 (XN
-
Pe 1
- 1 (XRN+iP)
(2.184)
=Q3 + Q5 - Q+
SP
p2
+ iP)I + 1 Ell >< 0|
2
+ E-2(2
(2.185)
+ PNP)
=
iQ3 - Q5 -iQ-
=
XP +
-2 (iN
+ XNP)
(2.186)
=
p2 2 P2=Q3
+ Q5+Q+
=
X2
(2.187)
+E-2(2+XNX)
(2.188)
01 =Pf
46
pO2
P
2
p Ot
(2.189)
p2lt
(2.190)
Finally, given P, we can simply perform the inverse Fourier transform by making
the same identifications between our operators and the covariant derivatives (2.152.17).
P0
P
1
-eB1-2(D±2
=
+ D2 + D2)
-eB-2 (DODi + (D + D2 D - ieBD 2 )D- 1 )
(2.192)
(D + D D2 + ieBDI)D- 1)
(2.193)
+ (D + D1 D2 + ieBD2 )D- 1 )
(2.194)
+ (2e 2B 2 + D4 + D 1 D2D1)D - 2 )
(2.195)
p0 2
= -JeB1-2(DoD
P
=
P1
(2.191)
-eB1-2(DoD
-eB1-2(D2
2+
1
P 12
=
-JeB-2 (DID 2 + (ieB(D + D2) + D3D 2 + DjD3)D - 2 )
(2.196)
P 20
=
-eB1-2(D
2
eBD1)D-')
(2.197)
P2 1
-
-eB-(D
2 D,
+ (-ieB(Di + D2) + D2D, + D 2 D )DO2 )
(2.198)
P 22
+ (D3 + D 2 D
-
= -JeB1-2(D2 + (2e 2B 2 + D + D 2D2D 2 )D - 2 )
(2.199)
where
Do16(t - t')
Do2
-
=
-(t
2
- t')
It= t'f
2
t')
47
(2.200)
(2.201)
Chapter 3
An Example of Poincar6 Symmetry
with a Central Charge
3.1
Introduction
Central charges in symmetry algebras arise in two ways. They can be anomalies of
a quantized theory, like (in two space-time dimensions) the Virasoro anomaly (triple
derivative Schwinger term) in the diffeomorphism algebra of a diffeomorphism invariant theory [12][13] or in the (infinite) conformal algebra of a conformally invariant
theory [14][15]. However, they can also arise classically, as for example in the conformally invariant Liouville model, where the (infinite) conformal algebra possesses a
center obtained already by canonical (non-quantal) Poisson brackets[16] or as in the
asymptotic symmetry group of anti-de Sitter space in 2 + 1 dimensions[17]. Another
instance arises in the non-relativistic field theoretic realization of the Galileo group.
With the appearance of a number of systems with central charges in their symmetry algebras at the classical level, it is useful to study a simple model with this
behavior. We examine a charged, scalar field in 1+1 dimensions, interacting with
a constant external electric field. The Poincare algebra of this system has a central
charge appearing already at the classical level. As a check on Poincare invariance,
we verify the Dirac-Schwinger relation. This leads to a modified energy-momentum
tensor. Next, we take the massless case and show that we can quantize the system
48
and solve it exactly. Finally, we look at the quantized algebra of the massless case and
show that it holds without anomalies, with the electric charge operator functioning
as the central charge.
3.2
The Classical Symmetries
We begin with the Lagrangian in 1+1 dimensions for a complex scalar field of charge
e interacting with a constant external electromagnetic field, described by a positiondependent vector potential.
L = (D,#)*(DI#) - m 2 o*o
(3.1)
with
DtL# = Ou# + ieApo
1
A,=-2 "Fx",
2"
(3.2)
(3.3)
where we use a flat metric of signature (+-) and define e"
= -601 = +1. Because of
the position dependence in the externally presented vector potential, the theory does
not have manifest translation symmetry. However, since physical motion is manifestly
translation invariant, we recover invariance of the action by adding a connection term
to the transformation law of the field
#=6 Tt (a"
6 L
# under
+
translation.
1
2 Wie""xF)#
=
(D' + iee""xvF)#
=
a
[(Dv)*Dv# -
(3.4)
(3.5)
m20*#5
The associated Noether current TC"" is the energy-momentum tensor for this theory
TC
= 6t" + 2A"J",
0"
=
(DP#)*(D)
(3.6)
+ (D"#)*(Dtq#) 49
ri'" ((DO#)*(D.0)
- m20*0),
(3.7)
JM=
ie [#*(DI"q) -(D"#$)*#$].
(3.8)
(38
Using the field equations of motion,
(DPDP+ m 2 )0 = 0,
(3.9)
we see that our currents obey
OP JA = 0
il,/O"
=
OVI"
"" F J,
=
0.
T"=
(3.10)
(3.11)
(3.12)
Note that Tt" is not symmetric: it is conserved in the first index p, while the second
index v denotes the direction of the translation. Thus we arrive at our first unconventional result: Even though the fields are spinless, their canonical energy-momentum
tensor is not symmetric.
In addition, the Lagrangian has the usual Lorentz symmetry
4=
6 LL
EX,0(3.13)
=
ao
(EaxaL)
(3.14)
with the associated conserved current
M(
a-MA
=
EaWVIf
3
+ A3 JP)
= 0
(3.15)
(3.16)
We can define operators on linear functionals G of q
PI(G [0])
= G [6T'O]
(3.17)
M(G [$])
= G [L0
(3.18)
50
In this way, P' generates translations, and M generates Lorentz transformations.
Defining a bracket to be the commutator of these operators on linear functionals of
01,
[M, P"] (G
[)
G [6T6L#
=
O
-
=G ct (o + 2ieEVxF)#]
= e"Pv(G
I
[#1)
[P1, P"] (G [#])
=
G [6(6T#
=
G [ieE/IvF#]
-
(3.19)
6T 6]
-c'vFQ(G [0]),
(3.20)
where Q is the operator that multiplies by -ie, so it simply commutes with all other
operators that act on linear functionals of
#.
(3.21)
Q(G [0]) = G [-ie#] = -ieG [0]1
Thus, we have a Poincar6 algebra in 1 + 1 dimensions with a central charge [18].
[M, P"]
-
",
(3.22)
[P", P 1]
= -c"vFQ
(3.23)
This algebra can be realized using Poisson brackets with the charges
Pt
=
(3.24)
fdxTi (tx)
M = fdxM(t, x)
(3.25)
Q = JdxJO(t,x),
(3.26)
Since the charges are the spatial integrals of the time components of conserved currents, the charges are time-independent, assuming that the field
# dies
off sufficiently
rapidly. (We shall later show that the quantized versions of these operators are explic51
itly time-independent in the massless case.) We calculate r, the momentum conjugate
to
#,
to be 7 = (D'O)*. Similarly, the momentum conjugate to 0* is
7r*
= (D#).
Writing J, Tfr, and M' in terms of these quantities,
J0 (t, x)
J1 (t, x)
(3.27)
=ie(#*7r* - q7r)
ie((D1q5)*#
=
TCO(t, x)
-
Te1 (t, X)
-
irw
(3.28)
))
-#*D
+ (Diq#)*(Diq#) + m2q*q5 + xFJ0
-wr(D 1 #) -- (D1#)*7* + tFJ0
M 0 (t, x)
x (r*7r
+ (Diq)*(Diq)
(3.29)
(3.30)
+ m2*q)
+t (7r(D 1q) + (D1#)*7w*) +
2
(X2 - t2 )FJ0
(3.31)
we can now calculate the equal-time Poisson brackets we need.
-0
MJD(X),
JOMy)
JO(x),TJO
(y)
[Mo(x), JO(y)
(3.32)
-C"/ J(y)6'(x - y)
(3.33)
=
(xJ(x) - t JO (x)) 6'(x - y)
(3.34)
=
(T3o(x) + T1o(y)) 6'(x - y)
(3.35)
=
(T1(x)+ TO(y) + F(x - y) JO(y)) 6'(x - y)
(3.36)
=
(-x (T",(x) + T" (y)) + x"Tc"v(y)
1
+ (X\xx)c"vFJV(y)- E6vF(xv - yv)x'J(x)
-
+yA(XPE"VA
-
xv 6 I)FJv(y)) 6'(x - y)
(3.37)
(where the common time argument has been suppressed). Thus, the charges satisfy
[Q, M] = 0
(3.38)
[Q, P"] = 0
(3.39)
[MP/1
= fI-Pv
(3.40)
[PL,Pv
=
(3.41)
52
-iElvFQ
This is the same algebra that we obtained before in (3.22-3.23).
3.3
The Dirac-Schwinger Relation
The Dirac-Schwinger relation is a method of proving Lorentz invariance. A system is
Lorentz invariant if the energy-momentum tensor obeys the following condition (for
Poisson brackets):
[TOO (x), Too(y)] = (To1(x) + To'(y)) 6'(x - y)
(3.42)
The energy-momentum tensor we obtained in (3.6) obeys (3.42) with the indices
reversed:
[To (x),T o(y)]
1 :
Unfortunately, TC
(T0 (x)+ TCO(y)) 6'(x - y).
=
(3.43)
TCj0 is not symmetric in its indices, so the Dirac-Schwinger
condition fails.
However, if we modify the energy-momentum tensor to make it symmetric, the
condition then holds. By adding a superpotential &II"aOVV, we obtain a new energymomentum tensor
TIV"
=
TO"
+ 6pV"
=0"v + 6avx, FJ + E11130V".
(3.44)
Requiring T" to be symmetric, we obtain
a v"
=
-Fxl
J.
(3.45)
Since JP is conserved, we can define a new variable h by
J
-
EIvah
53
(3.46)
With this new expression for JP, we obtain
0j,V" = a,,(Fxuc""h)
(3.47)
For solutions to (3.47) and (3.46), we can take
V" = FxEl"h
h(t, x)
1
[00
=
(3.48)
0 dy E(x - y)J(t, y)
2
]- 0
(3.49)
where E(x - y) denotes the sign function. For these solutions, T"' simplifies to
T
=
-v 091"
-
rIhF
(3.50)
Checking our commutation relations, we see that
[TOO (x), T'O(y)] = (To1 (x) + To1 (y))6'(x - y)
(3.51)
so the Dirac-Schwinger condition indeed holds for Tv.
Alternately, we can derive the energy-momentum tensor (3.50) by varying the
metric in a generally covariant version of the Lagrangian (3.1). We start by writing
the generally covariant Lagrangian
L= y
(gv(Dq$)*(Dv#) - m2q#*o)
(3.52)
with AP no longer an external quantity, but a functional of the metric satisfying the
relation[19]
OpAv - OvA, = EIFvr--g
(3.53)
Varying the metric, we get
6L
=
1
/14,gOAjg1." 2
54
V1-_gJ1h6A1L
(3.54)
Owing to the covariant conservation of J, we can write
-J"
= 0'"M,
(3.55)
as a defining relation for h. Substituting (3.55) in (3.54) in the variation of the action
and integrating by parts, we get
6S =
d
-gO,,6g"v - 0"O,6Avh).
(3.56)
Using our relation (3.53) and simplifying, we end up with our result
2
OL
Tg = ,/~ g- agll" = ol
(3.57)
- gthF
This lends credence that this is the correct energy-momentum tensor to use for the
Dirac-Schwinger relation.
Unfortunately, the momentum associated with T"v differs from the momentum
associated with Tb".
dx (To' - T')
f dxa1V1
=
V 1 1x
= --tQ
(3.58)
Furthermore, by taking the time derivative of this difference, we see that both momenta cannot be conserved simultaneously, except possibly in the uncharged sector.
(We shall later calculate the charges associated with our original energy-momentum
tensor TC"! in the zero-mass case and see explicitly that they are time-independent.
Therefore, the momentum associated with Th" is not conserved, except in the uncharged sector.)
55
3.4
Solution of the Equations of Motion (for Zero
Mass)
For the remainder of this paper, we shall take our field
#
to be massless by setting
m = 0. We shall also absorb e into F, replacing eF by F. Then, we shall use the
equations of motion and cannonical commutation relations to quantize the system.
Putting
# in the
form
c
T
=
Jk7
(f(T)ak + f(T
4
-bk)
FI
,
F (Ft - k)
(3.59)
(3.60)
and setting the mass to zero, our equations of motion (3.9) reduce to
d )2 + T 2 )f(T)=0
(3.61)
dT
which has a closed form solution in terms of Bessel functions
f(T) = A(k)f,(T) + B(k)f 2 (T),
fi(T) =
(3.62)
ITVI/2J_ 1 / 4 ( T2),
(3.63)
f 2 (T) = e(T)IT1/2JI/ 4 ( T2).
(3.64)
A(k) and B(k) are arbitrary functions that parameterize the creation and annihilation
operators, ak and bt, which satisfy the usual commutation relations.
We now impose canonical quatization relations. Our form for
cally satisfies [7, *]
=
[r, ir*] = 0. The condition [0, -r] = [0*, 7r*]
f (T)f'(T)* - f'(T)f (T)* = i,
56
#
=
(3.59) automatii6(x - y) require
(3.65)
which with the help of the Bessel function identity
J-34 (x)J-14 (x) + J 3 14 (x)Ji/4 (x) =
(3.66)
7X
implies
A(k)B*(k) - B(k)A*(k) =
-7i.
4
(3.67)
Apart from this constraint (which can be seen as a normalization condition), the
choice of these functions remains arbitary below. If one insists on choosing a parameterization, two useful parameterizations are the constant parameterization and
the parameterization which reduces to the standard free-field expression in the limit
where F goes to zero
A(k)
=
B(k)
=
|F
1/
k| (J
3 / 4 (k2/2|F|
- ic(Fk)J1/4 (k2/2|FD))
-i 'FI -1/2 kl (J_1/ 4 (k2/21FI) - ic(Fk)J3/ 4 (k2/21FI)) .
(3.68)
(3.69)
This second parameterization satisfies the requirement (3.67) by the Bessel function
identitiy (3.66).
Quantization of the Algebra
3.5
In this section, we shall use the original energy-momentum tensor Tj" (3.6) and verify
that its associated charges are time-independent. Inserting the solution for < (3.59,
3.62) into the classical expressions for the charges (3.24-3.26), and normal ordering,
we get expressions for the quantized charges in terms of creation and annihilation
operators.
Q
=
M
M
J
-e atak- bt b]
Ik
k -A
(3.70)
-O 27r=
=
00
-k2 12F -
27 [(rI
-/2iF(aAaB* - aBaA*)
57
aakn-
bt b~g
iF (A*OB - BOA* + A&B* - B*DA)
(4 (Oaj-
81F (at (02a)k
-
(Ob)
kb~k
(Oat)kak + bt k(Ob)_k)
-
(O2 bt)_kb-k - 2(Oat)k(Da)k
+ 2(Obt)_k(Ob)_k +
+ ---
iF (OAB - AOB) ((Oa)kb-k + ak(Ob)_k)
(DAB - AOB)* (-at (Obt )_k - (Oa)kbt
+-iF
J
P0=
(02 at)kak - bt (2 b)_k)
-oo 27[
(3.71)
v2 F (A*OB - BOA* + AOB* - B*OA) atak + bkbk)
k
7Fr
+ IiF (at(Oa)k + (Obt)_kb-k - (Oat)kak - bt~(Ob
+
P1
=L
F (OAB - AOB) akb-k +
F00 27r
F (OAB - AOB)* a bt
k[ak ak - btkbk]
k
-
(3.72)
(3.73)
We note that these charges are manifestly time-independent, as claimed earlier. These
charges satisfy the algebra
[Q, P]
=
[Q, MI = 0
(3.74)
[M, P"]
=
-ic""PV
(3.75)
ic"vFQ
(3.76)
[P",P"] =
with no anomalies.
As a note of caution, we formed the charges first and then calculated the commutators, rather than calculating the commutators of the local currents and then
integrating over space. In addition to the Virasoro anomaly, the currents have other
anomalies. These other anomalies depend on time as well as the particular parameterization chosen for A(k) and B(k) in (3.62). Furthermore, when these anomalies
are integrated over space, the results are ill-defined and depend on how the spatial in-
58
tegration is evaluated. As an illustrative example, let us sketch the calculation of the
commutator between the quantized charges
Q and
M by integrating the commutator
between : JO : and : MO :.
S dxdy [: J0 (t,x) :, : M'(t, y) :]
[Q, M]
600-t(t, x) : +I (X2
2
x : 00 (t, x)
Idkl
J0 (t, x)
ir
=
2
i(kk')xi ('k-
Tk)
27r 27w
00 1(t, x)
d e i(k-k')x"(F) |F1/2
=
-S(k, k', Tk
')
--
t 2 )F : J 0 (t, x)
-
(3.78)
dk'
= kk ei(k-k')x |F|1/2 (aTkTk,
0 0 0(t, x)
S(k, k', Tk
1
00
M 0 (t, X):
(3.77)
(3.79)
S(k, k', TTk)
+ TkTk') S(k, k', Tk, Tk{o.80)
(iOTk,T
-
i&TkTk')
(3.81)
Tk')
fkf*,at,ak + f*fkbtkb
+ fkfkakb-k' + fkf*,bt
a,
(3.82)
where Tk is the expression (3.60) for T, and the added subscript allows us to denote
the same expression with the momentum appearing in the subscript substituted for k.
Similarly,
fk
denotes f(k, Tk). It must also be noted that the independent variables
with respect to the integrals are the momenta (k, k', etc.), x, and t; however, the
independent variables with respect to the derivatives are the momenta, the T variables
with respect to each of the momenta (T, Tki, etc.), and x.
The operator S satisfies
[S(k, k', Tk, Tk), S(p, p',17,, T)]
=
2,r6(k - p')
(fkf*, -
f*fp,) S(p, k', T, Tk')
- 2 7 6 (p - k') (ff*, -
f*fk')
S(k, p',TTp')
+276o(p - k')276(k - p')
.(fkfk f*f*, -
f *fkf
p
(3.83)
where the delta functions are with respect to the same independent variables as the
integration (momenta, t, and x). As a consequence of the different collections of
independent variables, derivatives of the S commutator (3.83) may not vanish, even
though (3.83) vanishes identically.
59
We can now calculate the other commutators. The source of the inconsistency
will be most evident if we do not evaluate the integrals over the delta functions in
the anomalous term yet.
[J(t, ), J O(t, y)]
= 0
Jo (t, X), 00o(t, y)
=
(3.84)
-iJ
1
(t, y)6'(x
y)
-
dkfdk' dP dp' ei(k-k')x+i(p-p')ye
J-
2 272rw
F|11 2 (&T
,
(DTk
Tk)
-
+ TPTPt)
2r6(p - k')27r6(k - p')
- (fkfk'f*f,*, -
JO (t, X), 0 1 (t, y)]
=
-iJ 0 (t, y)6'(x
Jo(t, x), M O(t
=
-i (yJ1(ty)
y)]
(3.85)
f,)
(3.86)
y)
-
-
f*f*,f~
tJO(ty)) 6'(x
fe 2wwY~
k d
-
y)
pi(k-k')x~i(p-p')yie (&Tk
F1/2 (OTPOT,
-
Tk,)
+ TT,,)
276(p - k')27r6(k - p')
(fkfklf*fp*, -
[Q,dM
MM
=
dxdy A dk' dp dp yei(k-k')x+i(-P')y
27w 27r 27r 2 7
F|1/ 2 (T
7',
(3.87)
f*f*,fPfP')
ie
(aTk
-OTk,
+
2r6(p - k')27w6(k - p')
(fkfk'f*f*
- f
(3.88)
'fff')
dp'
Ak dk' dp
fS27r
27 2-F 27r
(-i)27r6(k - k')27r'(p - p')276(p - k')27r6( k - p')
ie (aTk -
aTk,)
IF| 112 (
(fkff*f*,
f
aTP,, + TPTP')
(3.89)
This final form (3.89) illustrates the problem with evaluation - there are four delta
functions of three independent quantities (four independent momenta, but three in-
60
dependent differences of momenta).
Evaluating the charges first is equivalent to
evaluating the first two delta functions first. In this order of evaluation, the term
vanishes. Evaluating the current commutator first amounts to evaluating the last
two delta functions first. In this case, the value of the term depends on the order in
which we evaluate the remaining two delta functions. This is equivalent to the value
of the term depending on how the x and y integrations are evaluated. Since we are
interested in the commutators of the charges, it seems more reasonable to evaluate
them completely, including the spatial integration, before introducing any commutators. Fortunately, the term whose commutator gives an anomaly has a vanishing
coefficient after the spatial integration. In this way we can see that what seems to be
an ill-defined anomalous term actually vanishes.
3.6
Current Algebra
As mentioned earlier, the current algebra is not well-behaved in general; however,
for some parameterizations the algebra is well-behaved.
Since the portion of the
commutator bilinear in the fields is determined by the Poisson brackets, we shall
only calculate the anomalous, field-independent portion left after normal ordering.
Since all operators bilinear in the field will be normal ordered, we shall drop the
explicit normal-ordering notation. To calculate the current algebra, we shall write
the current operators in terms of the operator S(k, k') defined in (3.82) and make use
of the anomalous portion of the commutator of S with itself (3.83).
Even though we are ultimately concerned with the commutators of the energymomentum tensor TO"(t, x) with itself, it is more convenient to calculate the commutators of 000 (t, x), 001 (t, x), and p(t, x) = J0 (t, x) with each other. Then, it is possible
to express the commutators of the TO"(t, x) with itself in terms of these other commutators, which have been expressed in terms of the operator S in (3.79-3.81).
Denoting the anomalous portion of the commutator by []C, we obtain
[p(t, x, p(t, y)])i(kkp)(xy)
(_
61
2
) (FFpF'*
F* -
F'FF*F*
+F'FkF*F*- FFpF'*F*)
(3.90)
0
[p(t, x), 0 0 0 (t, y)]
J
,
eik-p)(x-y)(ie eF1/2) (FIFF'*Fl*- F,'FFp*F*
F'FF*F'*+ F F,',F'*F
+TPTk(-iFP|2 +
dkd
F1 12 + IF,',12
1/2)(-
i(k-p)(x-y)(ee
+TPTk|FP12
p(t, x), 00 (t, y)] ,
i|F|2
-
TPTk|Fk 2) (3.91)
dkd()ei(k-p)(x-y) (iee(eF) eF 1/2)
-(-iTp(F'FF*F*
- F FF'F*)
iT (FF'*F*
Jd
-
i(k-p)(x-y)
(000(t X),000(t, y)],
I
=
*F))
F1/2)
(ice-F
f ( 2 7) 2 e
=0
FPF
(Tp
-
TO)(FP'Fp' -
F*)
k~
Fk
(3.92)
dkdpeitk)(
|
(27) 2 e
T*F
eFITkTp(Fk FPFkFp
-
Fkl*F7IFkFp
+FkFpF'*F'* - F*F* F')
=
1000(t x),0 01(t, y)]
0
(3.93)
Idkdpik-)xy
(2 w)2 e
=
I
=
TI12_12
(-) - iT|F,'2 (_)
eF(iT,
+iTT,2 |F |2 (i)
dkdpi(k-p)(x-y)
( 2 7r) 2
2 +T2|F
k
T~k
eF(Tp(lFkl
(t, x), 00 (t, y)]C
(21(2k
2
ei(k-p)(xy)
iTTk2 |Fk|2
T
-Tk(|F',
10
-
2
12)
+ T 2 |F12 ))
(3.94)
|eF|(-TkTp)(F'FkF*F*- FFpFF'*F*
F FF'*F*- F'FF*F*)
= 0
(3.95)
To evaluate the nonzero expressions, we can evaluate the integrals over momenta
by integrating over the diffence of momenta, A, and the average momentum, s. In
terms of
Pk = |F|2+ T12 |F| 2
62
(3.96)
we obtain
[000(t, x), 00 1 (t, y)]
=
I 1 (A)
[p(t, x), 0 0 (t, y)
I2(
jdAeiA(x-)eFIi(A)
=
J
=
/
c
)=
(T-A/ 2 Ps+A/
27r
12
2
edA(x-y) eleFj1/
JA)
2,F'-A
ds121
12
-
(3.98)
- Ts+A/2Ps-A/2)
2
(3.99)
(A)
+ Ts+A/ 2 Ts-A/ 2 |Fs-/
F'+A/2
-
We can evaluate I, and
(3.97)
2 sF
Ts+A/2Ts-A/
2
2
2
Fs+A/2 2)
by cutting off the integration at A and -A,
(3.100)
shifting
variables, and calculating the contributions that remain when A goes to oc. (If we
shift variables before cutting the integral off, we could lose an important contribution
to the anomaly.) We get
11 (A)
A(A, A)
=
=
A(A, A) + B(A, A)
/-A-A/2
+
A+A/2
I
A-A/2
-A+A/2
=
AE(eF)jeFK-1/2(
I 2 (A)
=
C(A, A) + D(A, A)
C(A, A)
=
A-A/2
(3.102)
27
dk
A-A/2
+
2
P
+
+
We can calculate the integrals near A and -A
(3.103)
(3.104)
dk
)
P
A+A/2 27
A A-A/2
A+A/2 dk
-E(eF)jeF-1/ 2 -(
+
)-&OTP
2 -A-+/2
-A+/2 27r
-A-A/2
D(A, A)
) -TkPk
/A+/2
B(A, A)
/-A+A/2
(3.101)
dk
(3.105)
(3.106)
by expanding the integrand for large
A and integrating the expansion. However, this expansion (as well as the integrals
from -A to A) will depend on the parameterization chosen.
The above results hold for any choice of parameterization, but to evaluate these
expressions further, we need a choice of parameterization. For general parameterizations, the evaluations cannot be easily expressed; however, these expressions are particularly simple in the case of the constant parameterization with JA(k)j = IB(k)l
63
=
constant. Using (3.67), we can pick
A(k)
= 2-5/471/2 z
(3.107)
B(k)
= 2-5/471/2
(3.108)
A(A, A) can be evaluated by expanding the integrand. We can make use of the
Bessel function expansions
J1/ 4 ( IT2)
4
T2
4
7T 2
J_1/ 4 ( T2)
J- 3 /4 ( T2)
J- 1/4(
T2)
-
sin( T2 +
Cos(
T2
4
Cos(
T2 +
T2
4
T2
wT 2
4
T2
cos( T2
-
+
3)+
T - 2 sn ( T2
8
16
2
)
T-2cos( T2
-
8
wT2
( T2
4
WT 2
cos( T2
(3.110)
5
5 T-2 sin( 1 T 2
8
16
37)
8
8)
2 1 T-2sin(
IT2
16
2
T-2cos(
IT2
16
5-
(3.111)
2
2
cos( T 2
(3.109)
+ 3T -2sin( IT2
81) 16
2
sin( 1T2
4
J 5 / 4 ( IT2)
T2
4
WT 2
4
J 3/ 4 ( IT 2)
cos(
8)
2
T -2sin( IT 2
-
16
2
7)± 5 T-2COS( 1 T2
8
-
16
7w
8)
2
T2)
42
cos( T2 -
4
-cos( T 2
WT
2
16
wT 2
J-7 / 4 (
T2)
4
wT 2
cos(
T2
-
8
T
8
+
4
wT 2
-sin(
T2 + 8)
64
(3.113)
T-2 sin( 1T2 _
-
2
+
T-2 (-cos(T2 ++8)) + 16 T- sin(-T2
2
J3/ 4 (
(3.112)
(3.114)
T ~-2sin( T2
16
2
) + 45
-T
16
-2
1 2
sin(-T
2
-
8 16 T-2sin( 2 T2
-5
16
T-2cos( T 2
2
-))
(3.115)
+
(3.116)
(up to terms that are O(T- 5 )) in the constant parameterization expression for P
(3.96) to obtain
OTP
=
-iTK~cos(T2) + O(T-3 )
2
E(T)(v + sin(T 2 )) + O(T- 2 )
T2P
=
2|T~cos(T 2 ) - 2-1/2T- 1 + I T-'sin(T2) + O(T- 3 )
P =
v2Tj-
(3.117)
(3.119)
(3.118)
This allows us to evaluate
-A-A/2
(
A(A, A)
J-A+A/2
v/2 I
6T
=
A+A/2
+
3 (A+A/2
)-T(v21TI - -IT-cos(T 2 ) + O(T- 3 ))
27r
2
I-A±,A/2
-A-A/2)
A-A/2
12
A
l27weF
-
JA-A/2
dk
_reF
7reF
A2 A
A(eFt)2
_rV
7eF
(3.120)
To evaluate B(A, A), we note that the integrand is a total derivative for the
constant parameterization. Using
xJI(x)
=
& (
J
2(J3(x)
2
-
1
(x)J+1 (x))
(3.121)
on
P
2 -/
-
2
_IT13 (J-3/ 4 2 (T 2 /2)
+ J-1/ 4 2 (T 2 /2)
+J 1/ 4 2 (T 2 /2) + J 3 / 4 2 (T 2 /2))
(3.122)
and using the large A expansion, we obtain
B
=
-A7r2-
=
2
e(T)T 2
3 2
/
A+A/2+
v/2
+
3dA+/
A2deF
rA
only a
P
A2
+ reFA(eFt)
ra
Pk
AA/2)
2
(3.123)
Adding this to A(A, A) above, only a term cubic in A remains. We can now
65
perform the final integration to get
I
=
[00(t, X), O (t, Y)J
=
A3
67reF
(3.124)
(3.125)
-iA"'(x - y)
S6wr
which is merely the Virasoro anomaly.
To get C(A, A) we can use either method (expanding for large A before or after
integrating), as both will give us the same answer
C(A, A)
=
=
-
V(eFT) eF
1 / 2T 2 (
+
A2
A-
- E(eF)jeFj1 / 2 tA
(3.126)
7
Finally, to evaluate D(A, A) we can use the fact that the derivative with respect
to T is proportional to the derivative with respect to k. This yields
D(A, A)
=
=
v2leFj1/2A JeFt - kj
A+/2
+
AA/2)
E(eF)eF / 2 tA
(3.127)
Together, these two expressions for C(A, A) and D(A, A) cancel each other exactly,
so that there is no anomaly in [p(t, x), 90 0 (t, y)].
As a note, we shall use this last commutator, which vanishes in the constant case,
to demonstrate the ill behavior in the general case.
If we require that A(k) and
B(k) approach the constant case values as k approaches oc or -oc,
then the large
A expansions will still hold. Thus, C(A, A) maintains the same value it does in the
constant value case. The calculation of D(A, A) is similar to its calculation in the
constant case, except that OT is no longer strictly proportional to ik, since varying
A(k) and B(k) gives us a way to vary Pk independent of varying Tk. Thus, D(A, A)
equals its old value plus an integral whose integrand is proportional to &A(k) and
OB(k).
The anomaly will be equal to this integral, which has no reason to vanish.
Thus, we can see anomalies arising as we depart from the constant parameterization.
66
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[2] For a review, see S. Coleman in Magnetic Monopole - 50 years later, from International School of Subnuclear Physics, 19th, Erice, Italy (1981) 21 (HUTP82/A032).
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equation for a spherically symmetric system. For example, see [4] or [5]
[4] W. Greiner (1994) Relativistic Quantum Mechanics: Wave Equations (SpringerVerlag, New York), 171
[5] V. Brestetskii, E. Lifshitz, and L. Pitaevskii (1989) Quantum Electodynamics
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2287
A. S. Goldhaber, Phys. Rev. D16 (1977), 1815
[10] C. J. Callias, Phys. Rev. D16 (1977), 3068
[11] These definitions are taken directly from appendix A of ref. [8].
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67
[17] J.D. Brown, M. Henneaux, Commun. Math. Phys. 104 (1986), 207
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68