livro felsager geometry particles fields
TRANSCRIPT
GEOMETRY, PARTICLES AND FIELDS
Based upon lectures given by BJ0RN FELSAGER Odense University, Mathematics Department and The Niels Bohr Institute, Copenhagen
Edited with the help of CARSTEN CLAUSSEN Odense University, P"'ysi.cs Department.~,~~~
ODENSE UNIVERSITY PRESS 1981 2. edition 1983
PREFACEThe present book is an attempt to present modern field theory in an elementary way. It is written mainly for students and for this reason it presupposes little knowledge in advance except for a standard course in calculus (on the level of multiple integrals) and a standard course in classical physics (including classical mechanics, special relativity and electrodynamics). The main emphasize is laid on the presentation of the central concepts, not on mathematical rigour. Hopefully this textbook will prove useful to high-energy physicists, who want to get acquainted with the basic concepts of difderential geometry. Mathematicians may also have fun reading about the applications of central concepts from differential geometry in theoretical physics. To set the stage I have in the first part included a self-contained introduction to field theory leading up to recent important concepts like solitons and instantons. There are two main themes in part one: One the one hand I discuss the structure of a gauge theory, exemplified by ordinary electromagnetism. This includes a derivation of the Bohm-Aharonov effect and the flux quantization of magnetic vortices in a superconductor. One the other hand I discuss the structure of a non-linear field theory, exemplified by the ~~-model and the sine-Gordon model in (l+l)-dimensional space-time. This includes the construction of a topological charge, the particle interpretation of the kink-solution,and finally the relevance of the kink-solution for the tunnel effect in quantum mechanics is pointed out. Although the present text deals mainly with the classical aspects of field theory I have also touched the quantum mechanical aspects using pathintegral techniques. In part two I have included a self-contained introduction to differential geometry. The main emphasize is laid on the so-called exterior calculus of differential forms, which on the one hand permits the construction of various differential operators - the exterior derivative, the co-differential and the Laplacian - on the other hand the construction of a covariant integral. But I also investigate metrics and various related concepts, especially Christoffel fields, geodesics and conformal mappings.
Apart from the introduction to the basic concepts in differential geometry the second part contains a number of illustrative applications. The Lagrangian formalism is put on covari~nt (i.e. geometrical) form. A detailed discussion of magnetic monopoles, including the Lagrangian formalism for monopoles and the quantization of magnetic charges, continues the investigation of gauge theories initiated in part one. Further examples of non-linear models are presented: The Heisenberg ferromagnet, the exceptional ~~-model and the abelian Higgs' model (including a discussion of the Nambu strings and their relationship with the Nielsen-Olesen vortices). Finally symmetry transformations and their associated conservation laws (i.e. Noether's theorem) are investigated in great detail.
ACKNOWLEDGEMENTSA project like this would never have been completed were it not for the moral and financial support of a great number of persons and institutions. From mathematics department, Odense University, I would especially like to thank my scientific advisors Erik Kjrer Pedersen and Hans J~r gen Munkholm, who have followed the project through all its various stages. I am also grateful to Ole Hjort Rasmussen for many stimulating discussions about geometry. From the Niels Bohr Institute I would like to thank my scientific advisor Paul Olesen (who originally suggested me to take a closer look at the geometrical and topological structure of gauge theories and who encouraged me to give the lectures upon which the book, is based). I am also grateful for moral support from Torben Huus. Helge Kastrup Olesen look~d over a priliminary version of the manuscript and taught me a lot about english grammar. Carsten Claussen has been of invaluable help to me. He has been reading the whole manuscript in several versions and has suggested innumerable improvements. I would also like to thank the secretaries, Lisbeth Larsen at Odense University and Vera Rothenberg at the Niels Bohr Institute, who, with great patience and professional skill, typed major parts of the manuscript. Finally I would like to thank Odense Universitets publikationskant a for financial support to the printing of the manuscript, and L~ rup and Holck's fonde (at the Niels Bohr Institute) for a generous donation to typing assistance.
CONTENTS
PART I:Chapter 1:1.1
BASIC PROPERTIES OF PARTICLES AND FIELDSELECTR1ETRIES AND CONSERVATION LAI,/SCONSERVATION LAWS Conserved currents. The tube lemma. Conservation of electric charge: The first half of Abraham's theorem. Conservation of energy and momentum: The second half of Abraham's theorem. SYMMETRIES AND CONSERVATION LAWS IN QUANTUM MECHANICS Unitary transformations generated by isometries. The infinitesimal generator associated with a one-parameter group of unitary transformations. The characteristic vector field associated with a one-parameter group of isometries. Symmetry transformations in quantum mecha nics. Conservation laws in quantum mechanics. CONSERVATION OF ENERGY , MOMENTUM AND ANGULAR MOMENTUM IN QUANTUM MECHANICS Conservation of momentum as a consequence of translationalinvariance. Conservation of angular momentum as a consequence
11.2
592
11.3
596
of rotational invariance. The intrinsic spin of a vector particle. Angular momentum for a charged particle in a monopole field. The infinitisemal generator necessary to compensate for rotations. The relationship between the infinitisemal generator and the correction term for the orbital angular momentum. 11.4 SYMMETRIES AND CONSERVATION LAWS IN CLASSICAL FIELD THEORY Symmetry transformations in a classical field theory. The characteristic vector field associated with a one-parameter family of diffeomorphisms. Noether's theorem for space-time transformations, internal transformations and generalized symmetry transformations. Noether's theorem for scalar fields. ISOMETRIES AS SYMMETRY TRANSFORMATIONS poincare transformations as symmetry transformations. Conformal transformations as symmetry transformations. Conservation of energy and momentum as a consequence of translational invariance. Conservation of angular momentum as a consequence of Lorentz invariance. Symmetry of the canonical energy-momentum tensor in a scalar field theory. Spin contribution to the canonical energy ~momentum tensor in a vector field theory. THE TRUE ENERGY -MOMENTUM TENSOR FOR VECTOR FIELDS Non-covariance of the canonical energy-momentum tensor for a vector field. Construction of the true energy-momentum tensor. Noether's theorem for vector fields. The conservation laws corresponding to conformal invariance. 600
11 .5
606
11 .6
611
xx11.7: ENERGY-MOMENTUM CONSERVATION AS A CONSEQUENCE OF COVARIANCEThe metric energy-momentum tensor as the functional derivative of the action with respect to the metric. The metric energymomentum tensor for electromagnetic fields. The metric energymomentum tensor for scalar fields, vector fields and relati~ vistic particles. Covariant conservation of the metric energymomentum tensor.
614
11.8
SCALE INVARIANCE IN CLASSICAL FIELD THEORIESScale transformations in Minkowski space. Internal scale transformations. The scaling dimensions for scalar fields and vector fields. Scale invariance of massless field theories in Minkowski space. The conformal energy-momentum tensor for a scalar field.
618
11.9
CONFORMAL TRAl,SFORMATIONS AS SYMMETRY TRANSFORMATIONSThe scale associated with an arbitrary diffeomorphism. Internal scale transformations on an arbitrary manifold with a metric. Conformal invariance of massless scalar field theories in Minkowski space. The conservation laws associated with conformal transformations.
625
SOLUTIONS OF WORKED EXERCISES
632
PARr I
A~ 2(~~
-+
{-YjX
>UttttJ]
o
&It I : ~I I" I IIIIIIIII
I IIII II
~ B=O
-
BASIC PROPERrlES OF PARTICI.ES AND FIELDS
2
I
GENERAL REFERENCES TO PART I:
R.P.Feynman, R.B.Leighton and M.Sands, "The Feynman Lectures on physics", McGraw-Hill, New York (1965). P.A.M. Dirac,"The Principles of Quantum mechanics", second edition, Oxford (1935). L.I.Schiff,"Quantum Mechanics", McGraw Hill Kogakusha ltd (1968) de Gennes, "Superconductivity of Metals and Alloys", W.A. Benja-
min Inc, New York (1966) A.O. Barut, "Electrodynamics and Classical Theory of Field and particles", MacMillan, New York (1964) Coleman, "Classical Lumps and Their Quantum Descendants" in
Erice Lectures, "New Phenomena in Subnuclear Physics", Plenum Press, New York and London (1975). Coleman, "The Uses of Instantons", in Erice Lectures, " The Whys of Subnuclear Physics", Plenum Press, New York and London (1977) R.P. Feynman and A.R.Hibbs, "Quantum Mechanics and Path Integrals", McGraw-Hill, New York (1964). L.S. Schulman,"Technique and Applications of Path Integration", Wiley, New York (1981).
3
I
ehllpter I
ELECTROMAGNETISM1.1 THE ELECTROMAGNETIC FIELD-+ -+ -+
The fundamental quantities of the electromagnetic field are the fieZd
strengths:
E(t,x)
-+
and
B(t,x) This vector may depend
The electric field strength is an ordinary vector field, i.e. to each point in space we have attached a vector. on time, too. The magnetic field strength is pseudo-vector field, i.e.
to each point in space we have attached a pseudo-vector which may depend on time, too.Although we are going to discuss mathematical concepts in greater detail later on, let us clarify the situation a little. An ordinary vector (like is nothing but a directed line segment PQ connecting two points in the Euclidian space. It is defined before we have introduced a coordinate system. But a pseudo-vector can only be specified if we also specify an orientation. Consider two orderei s~ts-+of linea~ilz i~ dependent vectors (UI,U2,U3) and (VI,V 2 ,V3). We can decompose one set on the other set in Q the following way:
E)
Vj = DiAij We say that they are equivalent if det(Ai.O. z I~ this way all possible ordered sets J (U,V,W) are divided into two classes. We p arbitrarily choose one of these classes to represent positive oriented sets, and the other class to represent negative oriented sets. Once we have chosen a specific orientation, the pseudo-vector is represented by an y ordinary vector B = (Bx,By,B z ) but if we exchange the orientation, then the pseudox vector is represented by the opposite vector: Fig. 1 !'= -! = (-Bx,-B ,-B z ) The most well-known example of a pseudovector is the cross product of two ordinary vectors: D x It is sometimes denoted U A V and referred to as the wedge product, but we shall avoid this notation, which we r~serve for differential forms. Once we have chosen a specific orientation, then D x V is specified by the following requirements:
V.
1.
The length of D x V is equal to the a~ea of the parallellogram spanned by U and V
2.3.
Ux V isspanned by (D,
pe~pendicular to the parallellograrnU and V.x
V,
D
V)
generates a positive orientation.
(See fig. 2)
4
I
z+
V
U
+~Fig. 2+ +
3x
y
Back to business! ic field. a force locity(1.1)V+
The field strengths
E
and
B
can be measured it will experience+
using the interaction between charged particles and the electromagnetIf the particle carries the chargeF :+
q,
+
which depends both on the particle's position
x
and ve-
F+ +
q(E + v x
B)B+
(Observe that vector!)
v
x
B
is an ordinary vector because
is a pseudo-
This force is known as the Lorentz force.
using the prin-
ciples of special relativity, this means the particle has the following equation of motion:(1.2)
E dt
p=
)1- ::
mv
+
(c
=
velocity of light)
This relation is of extreme importance and actually it serves to define the electromagnetic field. If the charge is small, we may negUsing lect the influence of the particle on the electromagnetic field. may analyse their motion and hence determine the field strengths: E(t,i) and 13(t,i). The field strengths themselves evolve in space and time hence they, too, obey some equations of motion, the Maxwell equations:
a swarm of test particles moving through the electromagnetic field, we
5
I
(1.3)
+
V
+
B
0
(No magnetic poles)
(1.4)
at
3B + V x + E
-0LEO~"t-
(Faraday's Law)
(1.5)
V 3E - c 2v x at
EB+
(Gauss' Law)
(1.6)
EO
(Ampere's Law)
Here
p
is the charge density,
-T
J
is the current and
EO
is the
permittivity of empty space. Furthenrore we have introduced the vector operator '" 3 3 3 v =(3x'3y'a-z)' In the following table we have collected some useful formulas:
q,(t,~)
is a scalar field; is a vector field; is a pseudo-vector field;
Vq,(t,~)
is a vector field: The gradient. is a scalar field: llie divergence.
E(t,~)B(t,~)
V'E(t,~l
vxB(t,~)
is a vector field: The curl.
{:"
2 2 3 32 3 = -+ - + -2 2 3y2 3x 3z
:
Laplacian
(1.
7lA A x+ +
+ + + A x B = -B x A
(1.8) (1. 9) (1.10) (1.11) (1.12) (1.13) (1.14)
(BXC) = (AxE) C (EXC) = E(A'C)
+
-
(A'E)C
(AxE) V
x C = (A'C)E - A(E'C) (Vq,)
= M = -0 = =0
VxV
(Vq,)(vxB)
V x (vxB)
V(V'E)
- {:"i3
6
I
Gauss' theorem:
f ~'E(l (l
dV =
f E'~S
dA
(1.15)
S
fl
inside region bounded by S Closed surface Unit normal pointing outwards
Stokes' theorem:
f ('7S
->- ->-
x B)ft dA
,( ->
J Bdr r
->
(1.16)
S
~r
r 8
Surface bounded by r Closed loop Unit normal pointing outwards
Theorem of line-integrals:
p
~
f 9~ 'd~= ~(Q)-~(P)rQ
r
Smooth curve with end-points P and Q.
let us deduce sane of the irnp:lrtant consequences
of Maxwell' s equations.
If
we take the divergence of (1.6), we get:
9
(dE)
dt 1
c2
v
(9 x B)0
=
1
e:
9 9
.
-T
J
O-T
d at (9'E) 1
1
e: O e: o
J
.. ..
e: o at
.!Q.
__I_V
.
-T
J
where we have used (1.13) and (1.5) Omitting the trivial common factor we have thus shown:
(1.17)
o
7
I
This equation is known as the equation of continuity. portant consequence of charge-conservation. total charge of the system:
It has the im-
To see this, consider the
where we have integrated over all space at a specific time. sity and all currents_ vanish sufficiently far away. tegrals are well-defined. cause the charge-density Now formally Q
We assume,
of course, that the system is confined in space so that the charge-denHence, all our inBut might depend on time be-
p(t,~)
may very well depend on time!
due to the equation of continuity, we get: dQ ; dt
J~ at
3 d x
Jv . jj
d 3x
Using Gauss' theorem (1.15) this can be rearranged as
~; - J AL.(A is typically a L few hundred Angstrom.)
Observe that (2.76) implies that not only
B vanishes
inside the
superconductor, the same thing holds for the Cooper current
3-
Consider now a superconducting ring as shown on fig. 34. Inside the ring at the curve r the Cooper current vanishes, as does the magnetic field! therefore get:Us~
(2.75) we
o
=
~ m
(Vc.p-
9.Ii.
lA)
But now we can compute the magnetic flux inside the ring beccluse: (2.79)
pit.r
dr
Ii..! q!ll
r
V,n' '"
dr
You might be tempted to say that it is zero! But let us look a little closer at the phase lP. The wave function ljJ(r,t)
=
jl p(r,t) I exp[
ilP(r,t) 1 that it
is only non-trivial in the ring, and all we can demand is
is single valued. But then nothing can prevent lP from making a jump of 2nn. If that is the case, VlP will contain a 6 -like singularity
77
I
and the integral need not vanish!
(Compare the discussion in sec. 1.4)
TO compute the line integral we assume that V~ makes the jump at the pOint B. If B+ and B- are pOints extremely close to B on each side of B we get: h
P A d r
h0(xn,t)]
2
- mgr[l-Coslj>(xn,t)]
(Compare with the Lliscussi =cP+"1') TriA
~Af
Iil
If
J~(1-C052t)
dt
=
~Jst!f~Tr
1>.
!IT
i.e.(4.24)
We can also easily determine the energyof the kink. Inserting the ansatz (4.18) in the energy functional (4.6) this reduces to +co
J(~~2(~~)2+
U[fj)dt;
~ing
the equation of motion (4.20) this reduces to
+co (4.25)
138
+co
+co y J /2U[f] df -co
I
Thus the kink energy can be computed directly from the pctential. We need not know the explicit form of the kink at all! Once again we look at some specific examples: In the "-model we get (4.26)
+]1/1XH[kink] = yJ I~(~ - i')df -]1/ IX271 / A ,-,,.-_ __2
Y-3- A
2/1 1!...3
while in the sine-Gordon model we obtain (4.27) H[kink1
= yI
~r~(I-COSAf)
df
Y~
oNotice that the solitary waves constructed above interpclate between different vacua. Consequently they belong to the non-pertubative sectors Since they actually interpolate between neighbouring vacua, they carry the smallest possible non-trivial topological charge. The corresponding sectors are called the kink sea tors respectively the anti-kink seators.
4.4
GROUND STATES FOR THE NON-PERTUBATlVE SECTORS
Now that we have gained some familiarity with the solitary waves we will rederive them from a somewhat different point of view. As w~ have seen the space of all finite energy configurations, E, breaks ~p into disconnected sectors characterized by the different possibilities for the asymptotic behaviour of the finite energy configurations. In each sector we will now look for a ground state, i.e. a configuration o(x) which has the lowest po~sible static energy in that sector. Since the ground state minimizes the static energy Hstatic[] =
J~(~!)2
+ U[(x)] dx
it must satisfy the associated Euler-Lagrange equation
Notice that the ground state extends to a static solution (x,t)=o(x) of the full field equations (3.30). Actually the following holds: The ground state for a given seator aorresponds to the solution
Of the field equations whiah has the lowest possible energy.
139 Proof: To see this let~(x,t)
I
be a non-static solution in the corresponding
sector. Then there is a space time point (xo,t ) such that o
~(Xo,to) f'and
0
Consider then another solution specified by the initial data
It has the same asymptotic behaviour at infinity as to the same sector.
~.
Thus it belongs~
Furthermore it has the same static energy as
but
it misses the kinetic energy! Consequently a non-static solution cannot minimize the energy in a given sector.
0
Okay, let us look for possible ground states. In a vacuum sector the ground state is simply given by the corresponding classical vacuum. But what about the non-pertubative sectors? To look for possible ground states in these sectors we will first examine the static solutions in our model (since a ground state is necessarily a static solution although the converse need not be true). The field equation for a static solution reduces to(4.28 )
2 d " = U' :::.....z 2 dx
[~l
This can be integrated once (using the same trick as before), whereby we get
(~) 2 = 2U[~ldx Thus we get two first order equations:(4.29)~ dx
v'zU[~l
-2
-1
.-
dx -
~- -v'zU[~l
~.~,
0 monotone
~-axis~
~2:constant
They have two kind of solutions:
solution
solution
(a) On the one hand there are constant solutions, corresponding to the zeroes of U, i.e. the constant solutions reproduce the classical vacua. (b) On the other hand there are monotone solutions, which interpolate between two adjacent vacua:
(4.30)
x - Xo
=
~Jv2atl
They correspond precisely to the static kink and the static anti-kink. This suggests that the kink is the groundstate in the kink sector!
140
I
Notice that there are no static solutions which pass through a classical vacuum. If the model possesses more than two classical vacua there will consequently be sectors (consisting of configurations which interpolate between non-neighbouring vacua) which has no ground state! Observe that if~(x,t)
is any solution to the field equations, then
so is the Boosted configuration: (4.31)~(x,t)
=
~[y(x-vt) ;y(t+vx)]
due to the Lorentz-invariance. If we apply this to the static solution~(x)
we thus produce a solitary wave:~(x,t)
=
~[y(x-vt)]
We can therefore easily get back our solitary waves once we have determined the static solutions! Next we want to tackle the problem of whether the static kink reallyis a ground state configuration for the kink sector. This will be shown
using a beautiful trick going back to Bogomolny.
He showed that in many
interesting models the static energy can be decomposed as follows: (4.32)
Here
P(~;~)
is a first order differential operator acting on
~ while
the topological term only depends upon the asymptotic behaviour 01 the field at infinity, i.e. it is constant through out each sector. Provided the topological term is positive we therefore get the following bound for the static energy (4.33) Hstatic
~
{TOPOlOgiCal term}~
Furthermore this bound is only saturated provided order differential equation ; (4.34 )
solves the first
Any solution to this first order differential equation is thus a ground
state. A decomposition of the type (4.32) is known as a Bogomolny decomposition. The associated first order differential equation (4.34) is known as the ground state equation. Let us try to apply this to our (l+l)-dimensional models. Guided by the kink equations (4.29) we try the following decomposition
141
I
The rest term is given by](4.4~
Since they consist of two far separated solitons moving with velocity
(4.44)
=
2~ly
where
M
is the rest mass of a single soliton.
Perring and Skyrme found two other interesting analytic solutions. They can be obtained from(4.41)
using various "analytic continuations". (x,t)~
First we apply the "symmetry-transformation" the singular complex valued solution:
(it,ix)
and obtain
Sinh[~-1-'Tg[AP(t,x)]= v 4
t]
COSh[~--V,I1_v 2
(f:V2
x]v
)whereby we obtain the regular
Then we perform the SUbstitution solution:(4.45)
v ~1
-(t,x) Tg[ A4> S 5 ] = I Sinh["yvt] ,,_ 4 v Cosh[~yxl
Exercise 4.5.3
Problem: (a) Show that (4.45) represents a scattering process between a soliton and an anti-soliton by investigating the asymptotic form of the solution as t+-oo and t->+oo. (b) Show that its energy is the same as the 2-soliton solution: (4.46)
Performing the 'analytic continuation"
v
-+
iv
(4.45)
can further-
more be transformed into the periodic solution:(4.47)
with
r
I
,1l+v 2
This solution is strictly localized at all times within the region ixl < (~r)-l since it falls off exponentially due to the denominator. 0, so it belongs to the vacuum sector. It reIt has winding number
presents clearly a periodic oscillating configuration, called a bpeat-
148her. The energy of the breather can also be obtained by "analytic conti-
I
nuation"2M
Slowly moving breather.
Il+v2 Observe that it is less than the total energy of a far separated solitonanti-soliton pair. The breather is interpreted as a bound state of a soliton and an anti-soliton. For this reason the particle it represents is called a bion. (Bion is an abbreviation for bi-soliton bound state). When investigation the breather it is in fact more natural to introduce the cyclic frequency wand the "amplitude" n: (4.49)W
= Ilrv
n
v
!
In terms of these variables the breather takes the form: (4.50) Sin(wt) nCos h (nwx) withn
Notice that when t=nX (n integer) the breather momentarily degenerates to a classical vacuum. At these times the energy of the breather is thus purely kinetic.Exercise 4.5.4Problem: (4.51 ) Show, by explicit calculation, that the energy of the breather at the time t~O is given byE = 2M(l - ~2)2
w2
1
where M is the soliton mass, cf. the earlier obtained result (4.48).
Now suppose that
w
Il
Let us furthermore assume that Sin[wt) is positive, say get the following asymptotic expansions:
O a e +'" 2 iAx dx Tn . 1T Je -~'4 e when A < a
r
=/ =i
jU 1m
Now, going back to the action of the Fourier path, (5.37) exp lii S 1 N-1 = exp {imT 4n IA
2 (k 2 7T an --:2 - w ) k=1 T
2 2}k < wT/1T, and "negaThe
we observe that the analogue of
is negative when
positive when k > WT/1T. Consequently there are Ent[wT/1Tl ti ve terms" (where Ent [x 1 denotes the entire part of x). multiple integral is therefore actually given by
For the free-particle propagator this ambiguity does not occur. corrected formula for the propagator thus comes out as follows:
The
(5.38)
But that is precisely the Feynman-Soriau formula!Now that we have seen how the limiting procedure based upon Fourier paths work, let us return for a moment to the problem of the integration measure. If you want to calculate a path-integral directly, i.e. you do not calculate a ratio any longer, you must necessarily incorporate an integration measure in the limiting procedure. Feynman suggested that one could use the same integration measure as for the piece-wise linear path. This assumption leads to the formula:
188
I
x(T)=O(5.39)
J x(O)=O
exp{i}[~(dX)2_ TIO 2 dtlim(V21iThi)a>
V(x)]dt} D[x(t)] J . N-1 k t ... exp{i' S[ L akSirr.f-]}dx ... dx N -1 k=1
r;;- N J
N ...
limN ...a>
But here the right hand side diverges as you can easily see, when you try to calculate it in the case of a free particle, cf. exercise 5.3.1 below:
Exercise 5.3.1 Introduction: Consider the matrix A involved in the linear transformation (5.34), i.e. 'k Sin [,T I ] j,k 1, ... ,N-l AjkN
a) Show that AXT
N-1
=2
N
I
and
Det LA.]=
[~]-Z
b) Show that the series involved in the limiting procedure (5.39) is given by limN-""
1
[N]
r (N)
TI
N-1~21TihT
and use Stirlings' formula for the r-function to verify that it diverges. In fact, by considering the case of the free particle propagator, it is not difficult to construct the correct integration measure, which (when you integrate over the Fourier components) turns out to be given by:
XjT) =0 i exp{[ S[x(t)]}x(O)=O
D[x(t)]
=
Nl!~ (vrz)
1T N-1
r(N)[V~]
rm
NJ
_jexp{~[k:1akSln.y- d
r
1
. N-1
.
k1Tt]} N-1
a
(As a consistency check you can u~ this formula to rederive the propagator of the harmonic oscillator). So now you 'see why we prefer to neglect the integration measure: Every time we introduce a new limiting procedure, i.e . a new denumerable complete set of paths, we would have to introduce a new integration measure!
5,4
ILLUSTRATIVE EXAMPLE: THE TIME-DEPENDENT OSCILLATOR
As another very important example of an exact calculation of a propagator we now look at the quadratic Lagrangian: L dx 2 = tm(at) - tmW(t)x 2
189
I
Since it is quadratic we can as usual expand around a classical solution, so that we need only bother about the calculation of the following path-integralx(tb)~O
x(t }~Oa
J
exp{211 [(CIt)
t imfb dx 2 taPerforming
Let us first rewrite the action in a more suitable form. a partial integration we get tbS [x (t)]
=- ~ J [taboundar~
Here we have neglected the tions x(t a )
terms due to the boundary condi-
= x(t b ) =
O.
Inserting this we see immediately that
the above path-integral is actually an infinite dimensional generalization of the usual Gaussian integral since it now takes the form:
x(tb)=O
J
im b dZ exp{- 2fJ. fx (t) [dt Z
t
+
W(t)]x(t)dt}D[X(t)]
x(ta)=O
ta
To'compute it we ought therefore to diagonalize the Hermitian operator-
d
22+ WIt)
dt
At the moment we shall however proceed a little different. the free particle action! differential equation(5.40)
By a
beautiful transformation of variables we can change the action to Let f be a solution the second order
d2 {-2 + W(t)} fIt) dtf
=
a
i.e.
belongs to the kernel of the above differential operator. The only thing we will assume is
The solution can be chosen almost completely arbitrarily within the two-dimensional solution space. that f does not vanish at the initial endpoint:f (t a ) " O.
(Notice that
fIt)
is not an admissible path, since it breaks the Using f we then construct the following
boundary conditions!).
linear transformation, where x(t) is replaced by tha path y(t):
190t
r
(5.41 )
x(t)
f (t)
Jta
~ f (s)
ds
Here we assume that the transformed function y(t)also satisfies the boundary condition y(t ) = 0 a Differentiating (5.41) we obtaint
x' (t) = f' (t)
J Yf'(~sl
ds + y' (t)
"""f'(t) x(t) + y' (t)
f' (t)
ta so that the inverse transformation is given byt
(5.42)
y(t) = x(t) -
I
fTs)
f' (s) x (s) ds
ta We can now show that the above transformation has the desired effect. Using thatt
x"
(t)
f"
(t)
J Lill f(s)
d
s
+ ft (t)f (t) fIt
+ y" (t)
we obtain:
{;i$
t
+ W(t)}x(t) ={f"(t) + W(t)f(t)}
I[~'(~S)]dS
+
f'n~r(t)
+y"(t)
ta But here the first term vanishes on account of (5.40). the action reduces to: S[x(t)] = Consequentlyt
-1 J
~[F(t)f' (t)y' (t) + F(t)f(t)y"(t)] dt with F(t)
= J L...!& f(s) ds
ta Performing a partialintegrat~on
on the second term this can be
further rearranged as:
s[x(t)
1
- 1 [X(t)Y'(t) Jtx(t).
tb
a
But here the boundary terms vanish due to the boundary conditions satisfied by Thus we precisely end up with the free-particle y(t)!
action in terms of the transformed path
191
I
There is only one complication associated with the above transformation, and that is the boundary condition associated with the final end-point path y(t): tb tb' The boundary conditions satisfied by x(t) are transformed into the following conditions on the transformed
Jta handle directly. the identity, Ii (x(t )) = -,}; b cf. end-point: x(1)=O
~dS
o
The second boundary condition is non-local and therefore not easy to We shall therefore introduce another trick! Using
J exp
{-ia x(t ) }da
b
(5.5), we can formally introduce an integration over the final
J exp{iS[X(t)]}D[X(t)]x(ta)=O
2~
x( t ) arbitrary b
Jf~eXP[-iaX(~)]eXP{kS[x(t)]}da D[x(t)]a
x(ta)~O
This is because the integration over
now produces a Ii-function (Notice that if we x ). N Changing
which picks up the correct boundary condition! must now also integrate over the final end-point variables we then get
attempt to calculate the path integral by a limiting procedure we
Here the infinite dimensional generalization of the Jacobi-determinant is independent of linear. y(t) because the transformation (5.41) is "Completing the square" The remaining integral is Gaussian.~
the whole formula therefore reduces to
OX = #ed liy ]
1
J-'"
exp{ -
2iif
ill Z 2
y(tb ) arbitrary b dt im bdy 2 f (\)f -z-}da exp{crifCat) dt}D[y(t)] t f (t) t a yet )=0 a a t
J
with yet)
t
~
yet) -
~f(~)ff~)ta
192 At this point we get a pleasant surprise: the a-integration! In fact we get: We can actually carry out
I
Furthermore the remaining path-integral is with-
in our reach, since it only involves the free particle propagator.
y(1)
arbitraIb
Jexp{~ {(~~)2dt}D[y(t)]y(ta)=O a
== f-0>
K o (x;1b IO ;ta )dx2
==
lz C 1b7Tih
-t )
a
LeXP[~
0>.
(1, ~tidx ==
This should hardly come as a surprise since, by construction, the above path-integral represents the probability amplitude for finding the particle anywhere at the time tb. The total path integral thus collapses into the simple expression:
(5.43)
- W(t)x 2]dt}D[X(t)]
It remains to calculate the Jacobiant!using a very naive approach. dure!). The paths points
We shall calculate it
(The following argument is only includ-
ed for illustrative purposes, it is certainly not a rigorous proceAs in the approximation procedure for path-integrals we x(t) and y(t) are then replaced by the multidimensional discretize the linear transformation by introducing a time-slicing.
The linear transformation (5.42) can then be approximated by T
- N
k~ 1 ~ ---'2"---:":"--:""
n
f' (t k )
(x k " x k - 1 )
(This is actually the delicate pOint, since the discrete approximation of the integral is by no means unique, and the Jacobi-determinant turns out to be very sensitive to the choice of the approximation procedure). Okay, so the Jacobi matrix has now been replaced The determinant thus comes exclusively by a lower diagonal matrix. from the diagonal, i.e.
I
Taking the limit
N
-T
we then find:N
limN +a>
exp[log n (1 k= 1
N
limN +a>
exp[ E log(1 - ! ( t )k= 1
'(t k ) Tk
N) 1
tb
f'J11 exp[-! T(t)dtl
f ta
Consequently
Inserting this into
(5.43)
our formula for the path-integral fi-
nally boils down to the following remarkable simple result:
The path-integraZ corresponding to the quadratic action(5.44)t t m b dx 2 2 m b d2 S[x(t)l == 2 f[(dt) - W(t)x ldt == - 2 fx(t) [dt2 ta ta+
W(t)lx(t)dt
is given byx(~)=O
(5.45)
fwhere
exp{~[x(t)l}D[x(t)l
x(ta)=Of(t) is an almost azobitrary soZution to the differentiaZ equation
{~2
2
+
W(t)}(t)
=
0~
the onZy constraint being that
f(t a )
O.
Notice that the above formula in fact includes the free-particle propagator (with W(t)=O) and the harmonic oscillator (with W(t)~2). You can easily check that (5.45) reproduces our previous findings f(t)=Cosw(t-t ). a in these two cases by putting f(t):1, respectively
194
I
5.5
PATH-INTEGRALS AND DETERMINANTSNow that we have the formula for the propagator corresponding to
a quadratic Lagrangian at our disposal, we will look at it from a somewhat different point of view. tb Sex (t) Since the action is quadratic,
1
;:J
x(t)
{~+
d
2
W(t)} x(t)dt
dt For this purpose we consider the Hermitian
we can "diagonalize" it. operator
- -:-2 - WIt)dt which acts upon the space of paths, x(t ) = x(t ) = O. b a normalized eigenfunctions n(t): ary conditions: x(t), all satisfying the boundIt possesses a complete set of
d
2
A given path,
x (t), N
can now be approximated by a linear combination tb ann (t) with an
xN(t)
n=1
I
ta
f
n(t)x(t)dt
Notice that the corresponding action of the approximative path reduces to
The summation over approximative path's can therefore immediately be carried out since it is just a product of ordinary Gaussian integrals:
I. f exp{~sIn the limit where by
[XN (t)
1 }d3., ... da Nthe approximative paths fill out the whole
N
+
=
space of paths and the path-integral is therefore essentially given
By analogy with the finite-dimensional case we define the determinantof the Hermitian operator eigenvalues.
_a t2
- WIt)
to be the infinite product of
Of course the determinant will in general be highly
195
I
divergent but we can "regularize" it in the usual way by calculating the ratio of two determinants. From the above calculation we then learn the following important lesson:
The path integral corresponding to a quadratic Lagrangian is essentially given by the determinant of the associated differential Operator, i.eX(s,)=o ~(5.46)
f
exp{~fX(t)[-:tr2ta
W(t)]x(t)dt}D[x(t)] =
MDet[-~2- W(t)J}-~
x(t )=0 a
hlhere the right hand side should actually be interpreted as a limiting procedure (i.e. it is a short hand version of the follOhling expression:(5.47)N _12
limN ->- co
L',(N)[ IT;')n=1
=
lim L',(N)f.N ->- co
fexp[~ l: Ana~l(vj~~/dNaN n"'111
Notice that we have included a factor
(12 ~ft f in
the integration
over the generalized Fourier components, i.e. the proper integration
measure for evaluating the determinant is given by(5.48)
k=1
~
[/2:Hi da k ]L',(N)
Apart from that we still need a further integration measure
since the determinant is only proportional to the path-integral. Coleman [1977] .
The
above characterization of the path-integral is the one used by e.g. Since we already know how to compute the path integral we can now extract a relation for the determinant. is given by To avoid divergence problems According to (5.45) it we calculate the ratio of two determinants.
Det[-a~- W(t)]
(5.49)
Det[-a~- V(t)]
In this formula it is presupposed that at tao singular limit, where fW and fV
fW
and
fV
does not vanish tal Let us de-
It can however be simplified considerably by going to the
does vanish at
196 note by fO
I
{- a~ -
W
the unique solution to the differential equation,
W(t)}f(t)
=
0,
which satisfies the boundary conditions
2.. fO (t ) dt w aSimilarly we denote by conditions:
= 1
f~
the solution which satisfies the boundary
oIn the above identity (5.48), we can then put fO It follows that and Finally the limit of the integral tb
W+ E W
f1
f = fO + d 1
V
V
V
fbourhood" of verges like
dt ta [f (t)]2
w
diverges, due to the vanishing of
fO at tao But since almost all W the contribution to the integral comes from an "infinitesimal neighta (in the limit where s+O), it follows that i t di-
Consequentlylim{
a Thus the identity
s
+
0
tr, f -wt
dt
f2 (t) }
I{
ftr, et
dt(t)}
(5.49)
,V a collapses into the extremely simple deter-
minantal relation:Det[ -a~-W(t) ]
(5.50)
DetH~-V(t) ]
If, eg., we put V(t)=Q
(and consequently
f~(t)=t-ta) it follows that
Det[-a~-w(t)l
DetH~l
197
I
This can be used to calculate a propagator (corresponding to a quadratic Lagrangian) relative to the free-particle prop~gator Ko: . (5.51)
K(~;tblxa;ta)
==
K(O;tbIO;t a ) exp{~[xcll}
(
Det{-d~-W(t)}]-2 KO(O;tbIO;t ) Det{-d~} am
1
-I
2TIiflf~ ( tb )
exp{~s[xCl]}
E.g. in the case of the harmonic oscillator we put W(t)=w 2 and consequently 1 fW(t); ;;Finw( t-t ). In this way we recover the by now well-known result (5.11). a/
Rema:r>k: Incidenti.lly the
investigation of determinants corresponding E.g. the
to linear operators has a long tradition in mathematics. twenties*).
basic determinantal relation (5.50) has been known at least since the This is very fortituous, since in our derivation some The passage from quotients of is actually independent dirt has been swept under the rug. gration measure n(N)
path-integrals to quotients of determinants only works if the inteintroduced in (5.47) WIt). of the potential function Since the result we have deduced,
i.e. the determinantal relation (5.50) is known to be correct, we have thus now justified this assumption. In fact the determinantal relation (5.50) beautiful reasoning which emphasizes its basic who are acquainted with more advanced analysis argument: Consider the expressions 2 Det(- a g(A)
can be proven by a very general and position. For the benefit of those we include the main line of the
WIt) - A) t 2 Det(- a - V(t) - A) t
andh (A)
It
is important for the following that A is treated as a complex variable. can be proven quite generally that a differential operator of the form,
It
- a2 t
- WIt)
has a discrete spectrum of real eigenvalues 1.. 1 ,1.. 2 ' , An'
*) J.H. Van Vleck,
Proc.Nat.Akad.Sci.
li,
178 (1928).
198
I
which are bounded below and which tend to infinity as n~. Each of these eigenvalues has multiplicity one, i.e. the associate eigenspace is one-dimensional. When A coincides with one of these eigenvalues, say A = An' it follows that the "shifted" operator 2
{-at - wit) - An}
has the eigenvalue zero, so that its determinant vanishes. As a consequence the function g(A) becomes a meromorphic function with simple zerOs at the eigenvalues A~ and simple poles at the eigenvalues A~. Consider next the function to the differential equation
f~+A(t).
By definition it is the unique solution-
{- at2
-
w(t)
A}f (t) = 0
which satisfies the boundary conditions
flta)It fo llows that
=0WIt)} if and only if
is an eigenvalue of the operatorf IW+A) It b ) = 0
o
and when that happens f(W+A)(t) is in fact the associated eigenfunction (except for a normalization factor). As a consequence the function heAl V becomes a meromorphic function with simple zeros at AW and simple poles at A k The quotient function n' g (A) /h IA) is thus an analytic function without zeroes and poles! (Thus it has a behaviour similar to e.g. exp[A]). Furthermore, from general properties of determinants (respectively solutions to differential equations) it can be shown that g(A) (respectively heAl) tends to 1 as A tends to infinity except along the real axis. The same then holds for the quotient, which as a consequence must be a bounded analytic function. But then a famous criterium of Liouville guarantees that it is constant, i.e.
Specializing to
A=O
we precisely recover the determinantal relation (5.50).
**=
1
5,6
THE BOHR-SOMMERFELD QUANTIZATION RULE
we shall now encounter the important problem of computing quantum corrections to classical quantities. Especially we shall consider quantum corrections to the classical energy. In the case of a onedimensional particle we know that it can be found by studying the trace of the propagator. Inserting the path-integral expression for the propagator, this trace can be reexpressed as
199
I
(5.52)
G(T)
j J eX~t~S[X(t)]}-00
x(T)=x x(O)=x
D[X(t)]dx o
o
== J exp{~S[X(t)]} D[X(t)]x(O)=x(T)where we consequently sum over all path's which return to the same pOint. This expression is known as the path-aum-traae integral. The There are now in principle two different approximation procedures available for the evaluation of this path-cum-trace integral. first method is the hleak-aoupling approximation. the potential has the general shape indicated on fig. 58. the potential well, we can approximate the potential byV(x)RJ
Let us assume that To calcu-
late the low-lying energy states we notice that near the bottom of2
W"
(0)x
I
iV"(O)x 2
:
E
E3
E"
E2El
Eo
X-axis Fig. 58
Thus the problem is essentially reduced to the calculation of the path-cum-trace integral for the harmonic oscillator! already solved, cf. consequently given by(5.53)
This we have
(5.19-20) and the low-lying energy states are2
with
v"(O)
mw
The second method is the so-called WKB-approximation which can be used to find the high-lying energy states. non-perturbative method. It is thus essentially a It leads to the so-called Bohr-Sommerfeld
quantization rule, which in its original form states that
200
I
(5.54) where q
f pdq
= n
h p the conjugate momentum and
is the position coordinate,
we integrate along a periodic orbit.
Notice that the left hand side
is the area bounded by the closed orbit, cf. fig. 59, so that the quantization rule states that the area enclosed in phase-space hasto be an integral multiple of
h.
The original WKB method which was But before we jump out
based upon the construction of approximative solutions to the schrodinger equation, will not be discussed here. in the path integral version of the WKB method it will be useful to take a closer look on classical mechanics and the original derivation of the quantization rule (5.54).
p-axis Phase-spacediagram.
--+--"'------""'t'---f-"'--..... q-axis
Fig. 59
ILLUSTRATIVE EXAMPLE:
THE BOHR-SOMMERFELD QUANTIZATION RULE
Let us first collect a few useful results concerning the action. Suppose (xa;t a ) and (xb;t b ) are given space-time points. Then
S(Xb;tb[xa;t a ) will denote the action along the classical path connecting the two space-time points. Theres,ult~ng
function of the initial and final (If
space-time point is known as Hamilton's principal funation.
there is more than one classical path connecting the space-time points (xa;t a ) and (xb;t b ) it will be a multivalued function). All the partial derivatives of Hamilton's principal function have direct physical significance: (5.55) Here Pa
~
as
Pa
asataxb and E
E
at b
as
E
is the conjugate momentum at the initial point
the conjugate momentum at
xa ' P b is the energy (which is con-
served along the classical path).
201Proof: A change, 6x b , in the Epatlal position of the final space-time point will cause a change, 6x(t), in the classical path. The corresponding change in the action is given by tb 6S =
I
J [~~
6x(t) + aL Qx(t) J dt ax
ta tbt
Ja
CaL
x
-~~Jdt " aX
~6x(t)dt + [aL " aX 6x(t) ]
But here the integrand vanishes due to the equations of motion and we end up with: 6SIt
= a~(tb)QX(tb)ax follows that
-
aL(t )6x(t ) a a ax
aL ) 6X b = -;-(t b ax
Pb 6X b
6S 6X b
Pb Similarly we may consider a change,
6t , in the temporal position of the b final space-time pOint. Notice first that
The corresponding change in the action is also slightly more complicated,
[~ 6x(t)ax ta
+
~ Qx(t)Jdtax
where the first term is caused by the change in the upper limit of the integration domain.
As before we can now make a partial inte-
gration leaving us with the formula 6S
=
aL L(t b )6t b + -;-(t )6x(t b ) b dX b ax b
=
L(t b )6t b -
~(tb)x(tb)6tb
aLdX
aL' ) J 6tb = [ L(t ) - -;(tb)x(t But the energy is precisely given by
202 E =p~
I
- L
aLa~
X- L
so that we end up with the identity
6S =
-
Eot b
Le.
as3t
-
E
b
~
Notice that the above considerations were in fact anticipated in the discussion of the Einstein-de Broglie rules, cf. the discussion in section 2.5. After these general considerations we now return to the discussion of a one-dimensional particle in a potential well like the one sketched on fig. mental period T: x = xT(t) Each periodic orbit will furthermore be characterized by its energy E, which thus becomes a function of the period From (5.55) we learn that (5.56 ) where S(T) dS = _ E xT(t). We can now S(T). T, i.e. E59.
In the general case there will exist a one
parameter family of periodic orbits whi9h we can label by the funda-
= E(T).
OT
is the action of the periodic orbit
introduce the Legendre transformation of the action function It is defined by the relation: dS W(E) = S(T) - dTT=S(T) + E'T where T should be considered a function of E.
(Notice the simiIn
larity with the passage from the Lagrangian to the Hamiltonian). analogy with (5.56) we then get: (5.57) dW _ dS dT dT dT dT E dE + T + E CiE = T
dE - OT OE + T + E dE
= -
Finally we can get back the actfon function by a Legendre trans formation of(5.58)
W(E), S(T)
= W(E)
- T'E
= W(E)
dW - dE E T. Notice that
where
E
should now be considered a function of
the Legendre transformation is in fact given by the phase integral in (5.54) W(E) S(T) + E'T
f [m(E)o
T
2
T- V(x)]dt +
f [m(~~}o
2+ V(x}]dt
i.e.
203
I
(5.59) since p
WIE) dx = m at W(E)
I o
T
2m(dx) dt dt
for this simple type of theory. which we are going to quantizel
It is thus the
quantity
From (5.57) we now get, dE where w
1 = -T
dw = ~ dw 2~
with
w
2~ =T
is the cyclic frequence of the periodic orbit.
Following
Bohr we then assume that only a discrete subset of the periodic orbits are actually allowed and that the transition from one such periodic orbit to the next results in the emission of a quantum with energy ple.2~
nw, where
w
is the frequency of the classical orbit.
The
last part of the assumption is Bohr's famous correspondence princiIt follows that W can only take a discrete set of values and W can be stated as +2~ftc
that the difference between two neighbouring values is given by
= n.c
Thus the quantization rule for W(E ) = n2~(n+c)ft
=
2~n
where
is a constant which we cannot determine from the correYou can look upon this formula as the first W(En) where we expand in Thus the correct
spondence principle. powers of 1/n.
two terms in a perturbation expansion for
Bohr and Sommerfeld simply assumed it was zero, but
using the WKB-method it was actually found to be ,. quantization rule for the above case is given by (5.60)
As an example of its application we shall as usual consider the harmonic oscillator. features: This example turns out to have two remarkable T SIT) breaks down completely. As a 1) The assumption of a one-parameter family of periodic cannot be defined for a har-
solutions labelled by the period consequence the action function monic oscillator. W(E)
2) The quantization rule (5.60) is exact. A general periodic orbit
Since the action function cannot be defined, we shall define through the phase integral (5.59). x(t)= ACoswt + The associated energy is consequently is given by BSinwt
E
=
mW 2 (A2 + B2)
while the phase integral turns out to be
I
ThusW(E) = 2rr Ew
and the Bohr-Sommerfeld quantization rule reduces to the well-known result (5.20)
o
Now that we understand the Bohr-Sommerfeld quantization rule we return to the path-cum-trace integral (5.52). This time we are going to expand around a classical solution x(t) = xcI (t) + nIt) The potential energy is expanded to second order: V[Xcl(t)] + V'[xcl(t)]n(t) + Pl"[x c1 (t)]n (t) Inserting this, and using the classical equation of motion, the action then decomposes as followsRl
V[x(t)]
2
T
S[X(t)]
Rl
S[Xcl(t)] +
b{I(~)2-!V"[XCl(t)]n2(t)}dtT
+ 1bn(t){-~- ffiV"[Xcl(t)]}n(t)dt
d2
1
Consequently the path-cum-trace integral reduces to (5.61)G(T)
Rl
7-00
l:
x
exp{~[xcl(t)]}cl
n(T)=O
J expHKn{-~- ~Vff[xc1(t)]}ndt}D[n(t)]dxo
n(O)=O
where xcl(O) = xcl(T) =xo' But the remaining path integral we know precisely how to handle. According to (5.45) it is given by
j __ m_-J""-Tdt2rrifJ.f(O)(T)
with
2 (t) j
oIn fact we can relate f to the classical path: solves the Newtonian equations of motion:d XcI m~
The classical path
2
- V'[x Cl (t)]
Differenting once mOre, we thus findm
~dt
3 d x
= -
V"[x lIt)]C
205 Consequently we can put
I
Thereby the path-cum-trace integral reduces to (5.62 )dx0
IlxdP o+00
T dt
where we sum over all classical paths satisfying the constraint xcl(O) = xcl(T) = xo. Despite its complicated structure it is essentially a one-dimensional integral of the type
I
exp{if(x)} g(x)dx
Such an integral can be calculated approximatively by using thestationary phase approximation.
Thus we look for a point f' (x ) = O. o
Xo
where
the phase is stationary, i.e. around this point:
We can then expand2
fIx) "'" f (x o ) + ,f" (x o ) (x-x o ) g(x) Rl g(x ) + g' (x o ) (x-x o ) o
In this approximation the integral then reduces to a Gaussian integral:+co
(5.63)
f
exp{if(x)}g(x)dx
Rl
g(xo)eXp{if(Xo)}vI,,(:~)
Using this on the path-cum-trace integral (5.62) we see that we need only include contributions from the classical paths which produce a stationary phase, i.e.
o = x a ao
But according to (5.55) this gives
oi.e. the momenta at odic solutions! od T. t=O and
P b - Pat=T are also identical. Thus the
stationary phase approximation selects for us only the purely periThey can be parametrized by their fundamental periT we should therefore only inCorresponding to a given
clude the contributions from the periodic orbits:
206XT / 2 (t)
I
x T/ 3 (t) r .. rXT/n (t) I
Before we actually apply the stationary phase approximation we can therefore restrict ourselves to a summation over these periodic orbits:
N::>tice that the parametrization of the periodic orbit is only determined up t, i.e. we can choose the starting point quite arbitrarily on the closed path. p-axis Each of the x-values between the turning points x 1 and x 2 occurs furthermore twice on the path, cf. fig. 60. We are now ready to perform the x -integration. The x x 0 ~~____________-+____+-__-+~2~.. action of the periodic orbits does X-axis not depend upon the choice of the starting point, and neither does the expression to a translation inT
Fig. 60
J
dt(~) 2
o
Thus the only contribution to the integral comes from x2 +00 dx o 2 dt = ~ I~T(O) I x 1 Cit
J
J
f
i
n
Having taking care of the xo-integration we must then compute the integral over the reciprocal square of the velocity. Here we getT
~2(x T / n )
J o
dt
2
2n
J
(2[E(T/n) - V(xB-~) dxm
3
The same integral can however be obtained by a different reasoning! Consider the phase integral W(E), which is given by x WeE)
= 2n
J /2[Ex1
2
- vex) 1 dx
for the periodic orbit in question. respect to E we obtain:
Differentiating twice with
207
I
dT dE consequently
-3/2 (2[E - V(x)]) dx
Joreduces to:(5.64)
T
dt (x T / n )
2
_ m3/2 dT dE
Putting all this together, the path-cum-trace integral thus finally
G(T)
Rl
1.
_1_
I
m I2nifl n=1
exp{i nS - } ft
-[T] T~dEI n n dT
TT~
n
This was the hard part of the calculation! energy levels.
Now we can extract the The transformed
As usual this is done by gOing to the transformed(5.9)
path-cum-trace integral and looking for the poles. path-cum-trace integral is given by, cf. G(E)i
G(T) exp{fiET} dE
(Notice that latter as obtain
E
now both represent an integration variable and the To avoid confusion we shall write the G(T) we Inserting the above expression for
energy of the periodic orbit. Ecl I).
G(E)
Rl
iiiKv'21Til'i
i
1
n~,Jo~
""
Here it will be preferable making an exchange of variables, i.e.~
= Tin,
is the fundamental period of the periodic orbit in question:
G(E)
Rl
. 1 ~v'21Ti~
n=1
""""f ~ dT E exp{1f(ET+S[T])}T/I~Iv'noasT
The T-integration is then performed by use of the stationary phase approximation. A stationary phase requires+ S[T])
a o = 1T(EI
=E
- ~
=E
- Ecl
Thus for a given value of
E
we get the main contribution from the E!give~:
periodic orbit which precisely has the energy(5.63) the stationary phase approximation now
According to
208
I
G(E)
Rl
2- _,_
~ jZTTift
mI'lv'ZTTifl n='
n
fd2S cr:rrW(E)
_,_
Using that dE
dT
and
S (T)
+ ET (E)
the formula finally reduces to
(5.65)
G(E)
Rl
~
T(E)
n='
~
exp{%nW(E)}
i mt
T
(E)
This clearly has poles when (5.66) and thus we have precisely recovered the old Bohr-Sommerfeld quantization condition (without the half-integral correction term). simply: The reason we missed the half-integral correction term is, however, very It is due to the fact that we have not taken into account precisely as for the The exThe additional the phase ambiguity of the path integral.
harmonic oscillator we should pick up phase corrections. pression (5.62) is completely analogous to (5.38). phases then come fran the singularities in the integrand
T
-,
U f2 ~:) ] oThese singularities correspond to the zero's of to the turning pOints turning~ts
f(t)
=
x(t), i.e.
In the present calculation the x and 1 are thus analogous to the caustics. Each time we pass e
a turning pOint we therefore phase-factor
iTT/2
e~pect
, cf. xT/n
(5.29).
.
that we pick up an additional 2n turning
Since there are
points in the orbit
we consequently gain an additional phase
e- iTTn = (_1)nwhich should be included in the formula for the propagator. The transformed path-cum-trace integral (5.65) is thereby changed into
(5.67)
G(E)
Rl
~
T(E)
n=1
~
(-1)nexp {%nW(E)}
209
I
This causes a shift in the poles, whichare now given by(5.68)
and that is precisely the correct quantization rule according to the original WKB-calculation. Notice too that near a pole we get the expansion
Using the approximation 1 + exp4i W(E)}Rl
1 + exp{k W(E
)
ll
}exp{~ T(E) (E-En )} = 1 - {1
+
~ T(E) (E-En )}
it follows that
G(E)
behaves like G(E)Rl
1 E--=-En
when
E
Rl
En'
This should be compared with (5.10) and it clearly
shows that the approximative path-cum-trace integral has the correct asymptotic behaviour close to a pole. we have been working hard to derive a result which, as we have seen, can in fact almost be deduced directly from the correspondance principle! When we are going to use the path-integral technique in quantizing field theories we will actually have to work still harder Before we enter into these dreadful technicalities I want to give a few examples of almost trivial applications of the preceding machinery. The first example is concerned with the free particle.L
To find
the energy levels we enclose the free particle in a box of length and use periodic boundary conditions. Thus we have effectively replaced the line by a closed curve of length effect that the energy spectrum is discretized. uous free-particle spectrum back by letting and around the closed circumference: given by xT(t) =L
L.
This has the We get the contin-
go to infinity. x is
The free particle can now execute periodic motions by going around The periodic orbit T
Lt T
It has the following energy and action: m dx 2 _ mL 2 E(T) = Z(dE) SIT)
-;;Z
ET
210 Consequently W(E) ; SIT) + ET ; LI2mE
I
Since there are no turning pOints in this problem the quantization , rule is given by (5.66), i.e. Ll2mEn i.e.(5.69 )
21Tnfl
These are the correct energy levels in the discrete version. this we notice that the free particle Hamiltonian is given byH=----
To see
~
fi2
d2
2m dx2
Since the SchI1.5dinger wave function must be periodic, the eigenfunctions are given by 21Tnx] o/n(x) = exp [ 1. -Land the corresponding eigenvalues precisely reproduce the above result(5.69)
The second example is concerned with a field theory. of periodic solutions: are given by (cf. (5.70)~(x,t)
In the
discussion of the sine-Gordon model we found an interesting family The bions (or breathers). In a slightly w, they changed notation, where we emphasize the cyclic frequency (4.50): n Sinwt ] with Arctan [ Cosh(nw~
4 =~
n
= ---w--
/].l_w 2
and
O
313generates a unique tensor of rank 1, which we also denote vp'
II
* is a linear map Tp(M) ~ R, since the bracket is bilinear. Therefore vp
...
Exercise 6.9.1
Problem: Conslder a tangent vector vp ....Show that the components of vp as a tangent vector and the components of vp as a tensor of rank 1 are iaentical.
.
...
...
Ylhen we have tensors of an arbitrary t:'pe at our disposal w" can ge-
neralize the contractions. Let, for instance, Tp be a tensor of type
(1,2). It is then characterized by its components
n with respect to some coordinates (x1, ... ,x ). Here the index a transforms contravariantly and the indices band c transform covariantly. But then we may contract the indices a and b obtalltingthe quantity:
As we sum over the index a, this quantity is characterized by only one index
The important point is now that the quantity Sc transforms covariantly! This follows from the computation j a S Ta av Ti ax axk (Toc (21 ac axi C!l jk aya a~/
a j where we have used that (~) and (~) are reciprocal matrices. But axl. aya if Sc transforms covariantly, we may regard it as the components of acotencor Sp' We say that Sp is generated from Tp by contraction in the first two variables. This is obviously a general rule: Whenever you
contract an upper index with a Zower index, the resuZting quantitytransforms as the components of a tensor. (where the degree of course is lowered by two!). The only trouble with contractions is that they are almost impossible to write in a coordinate free manner! As contractions play an important role in the applications, we will therefore write down many equations involving tensors using component notation.
314
II
In the following table l'/e have summarized the most important properties of mixed tensors:
A mixed tensor Tp of type termined by its componentsComponents
(k,~)
is completely de'
T
i
l .. i
k
JI'''~
.
. = T (dxP
i
l
; ... ; dX1k;~. ; ... ;~. )JIJ~
with respect to a coordinate system. The indices i l , ... ,i transform contr8:variantly , and the indices k jl, .. ,j~ transform covariantly.
Linear structure
If Sp andTp are mixed tensors of the same type and A 1S a real scalar, then you can form the mixed tensors Sp+Tp and ATp of the same type (k,~). Furthermore these mixed tensors are characterized by the components(k,~),
(6.71)
Tensor product(6.72)
If Sp and Tp are mixed tensors Of type (kl'~I) and (k 2 ,P-2), you can form the mixed tensor S1' ~ Tp of type (k l + k2 '~l +P-2). The tensor product is characterized by the componentsS
il .. i k
il ... i1. TJI"Jt 1
k
jl"j~2
2
Contraction
If Sp is a mixed tensor of type (k,~), you can form a mixed tensor Tp of type (k-l ,~-1) by contracting a contravariant and a covariant index. The contraction is characterized by the components
(6.73)
In what follows we are going to deal a lot with tensor fieZds.
To con-
struct a tensor field T of rank 3 we attach to each point P in our man nifold a tensor Tp of rank 3! Let us introduce coordinates (xl, ... ,x ) n on M. Then the tensor at the point P(xl, .. ,x ) is characterized by its components T abc (x 1 , ... ,x n )
315 \7e say that the tensor field T is a smooth field if the components
II
Tabc(X', ... ,xn) are smooth functions of the coordinates. If nothing else is stated we will always assume the tensor fields to be smooth.
Illustrative example:
The unit-tensor field.
Let M be an arbitrary manifold. Then we construct a mixed tensor of type (1,1) in the following way. At each point P we consider the bilinear map
The corresponding components arei .... To( dx ie.) J '" < ax i
" l e.> J
i.e. the Kronecker-delta! It is a remarkable fact that the components of this tensor have the same values inall coordinate systems. As the components are constant throughout the manifold, they obviously depend smoothly upon the underlying coordinates! We therefore conclude:
The Kronecker-delta oi. are the components of a smooth tensor field on M, called the unit tensor field of type (1,1).J
Exercise 6.9.2Problem: Show that the Christoffel fieldsrV
as
=
..,g
l.
v].! [a/>:].l(:t ax
--S+-a --ax ax].!
a/>:S].!
g a aS ]
are not the components of a mixed tensor field. (Hint: Show that they do not transform homogenously under a coordinate transformation. )
Exercise 6.9.3n Problem: Consider a point P on our manifold. To each coordinate system (x', ... ,x ) around P we attach a quantity TlJk with two upper indices and one lower index. Apriori the upper indices need not transform contravariantly, and the lower index need not transform covariantly. Show the following:
If the quantity U~ given.byQ, " RU k - T1-J S
-
k
ij
are the components of a mixed tensor of type (1,1) .Ulhenever S1.ij are the eomponents of a mixed tensor of type (l j 2), then p1-J k are the components of a mixed tensor of type (2,1)
316
II
The method outlined in exercise 6.9.3 is very useful when you want to show that a given quantity transforms like a tensor. Clearly it can be generalized to mixed tensors of arbitrary type. Suppose now that we have attached a
metria g to our manifold M. Then
we have previously shown (Section 6.8) how to identify tangent vectors and covectors using the bijective linear map
I:T;(M) - Tp(M)
,
generated by the metric. If we write it out in components it is given by
Exercise 6.9.4Problem: Show that I(
axk )
is characterized by the components gki
11/e can now in a similar way identify all tensor spaces of the same rank. For simplicity we sketch the idea using tensors of rank 2. Let T be a cotensor of rank 2 characterized by the covariant components T ij . Then we identify T with the following tensor of rank 2: II (T)
(~;4)d~f.T(I(~) ;In two. o.f the indices must always co.inside (since an index a can i o.nly take the values l, . ,n), whence F aj a vanishes! So. in the k case o.f k-fo.rms we do. not have an infinite family o.f tenso.r spaces! We will use the convention that co.tensors of rank 1 are co.unted as l-fo.rms. Of ~o.urse, it has no. meaning to. say that a co.vecto.r is skewsymmetric, but it is useful to. include them amo.ng the fo.rms. In a similar way it is useful to. treat scalars as O-fo.rms. Co.nsequently the who.le family o.f fo.rms lo.o.ks as fo.llo.ws:A~ (M) =R;A~ (M) =Tp (M) lJ\.~ (M) l lAp (M) lAp
*
n
n+l
(M) ={O} lAp
n+2
(M) ={O};
Wo.rking with o.rdinary co.tenso.rs we have previo.usly intro.duced the tensor pro.duct: If F and G are arbitrary co.tensors then the tenso.r pro.duct F II G def ined by
is a tria
co.tenso.r o.f rank k+m. But if we restrict o.urselves to. skewsymmeco.tenso.rs this co.mpo.sitio.n is no. lo.nger relevant because F ~ G1.
is no.t necessarily skewsymmetric.(If yo.u interchange V. and say no.thing about what happens to. F(Vji ;vk)G(Uji .. therefo.re try to. mOdify this compo.sitio.n:
;rtm.
rt. J
yo.u can
We will
If Waj a is a quantity with indices aj, ,a k then we can co.nk struct a skewsymmetric quantity in the fo.llo.wing way (7.1)w[ aj a 1 k
1 = k!
!(_l)TI w TI TI
(all TI (a ) k
329 where we sum over all permutations is the sign of the permutation, i.e.(-1)TI TI
II
of the indices al, ... ,ak,and
(-If
(-1) TI
= =
+1 -1
if if
11 11
is an even permutation is an odd permutation
For instance we get W[ab] and W[abc]1 = 2![Wab
- wba]
=
1 3! [w abc + wbca + wcab - wacb - wcba - wbac ]
Observe that w[a
The quantity w[al .... ~] is called the skewsymmetrization of wal ak . ] is completely skewsymmetric, and that if ~ wal .... ak is born skewsymmetric, then
I
1 (This is, of course, the reason why we have included the factor k!') If we introduce the abbreviation
(7.2)
sgn[bl .. b k al ... ak
]= { -1
+l if (b l ... bk) is an even permutation of (al"
.ak)
if (bl ... bk) is an odd permutation of(al ... ak)
o otherwise
we can write down the skewsymmetrization as an explicit summation W = 1 sgn[bl ... b k ] W [al" .ak ] k! al'" o
VtP Divergence: 'i/ a~ ~
dtP
a itP1 ai(lgAi ) n
..... +' ..... "
- SAdA
Curl:
vx;:Laplacian: lltP
.rg E:ijkaJ A1 .
k
- &H
~d i Vial.,)
The preceding discussion of the dual map, the scalar product, the wedge product, and the exterior derivative should convince you that exterior algebra is capable of doing almost anything you can do in conventional vector analysis. But exterior algebra is not just another way of saying the same thing. It is a much more powerful machinery for at least two reasons I 1. 2. It works in any coordinate system, i.e. it is a covariant formalism. It works in any number of dimensions.
372
II
(A)l.
d2 = 0If 'We apply
(A)
to a scalar fieldd(d.p) = 0
.p.
'We get
(*)0 .... '" ....()
.
f
M
~l ~k
We shall refer to this inner product as the Hilbert product. compact, then the Hilbert product is always well-defined. If compact, the integral is not necessarily convergent.
If M is M is not
421 has compact support.Consider a Riemannian manifold. with compact support.
II
It is, however, always well defined if one of the differential formsLet Ak(M) denote the vector space of k-forms o The Hilbert product defines a positive definite metric in is a pre-Hilbert space. We can
this infinite dimensional vector space.
Thus Ak(M) o complete it and obtain a conventional Hilbert space
~(M)called the Hilbert space of square-integra?le k-forms . An element of the form 1 l.l l.k T = k' T. (x) d x A A d x l.l .. l.k with measurable components
~(M) is on
is convergent. For a pseudo-Riemannian manifold things are slightly different. Here the scalar product (Tlu) is indefinite and therefore the Hilbert product is indefinite too. If we complete Ak(M) o we therefore get a Hilbert space with indefinite metric.
In the following we shall always assume that all the differential forms we are considering have a well defined Hilbert product. preserves the inner product between two differential forms: Observe first that up to a sign the dual map is a unitary operator, i.e. it
Theorem ? (a) The dua'l map * is a unitary operator on a Riemannian manifo'ld: = < T I u > (b) The dual map * is an anti-unitary operator on a manifold with Minkowski metric: = -< T I u > Proof:This follows immediately from the corresponding local property (Theorem 9, section 7.5):< *T
I
*u > =
t
(*T
I
*u)'
Then we finally arrive at a most important relationship between the exterior derivative -form, U
d
and the codifferential
5.
Let
T
be a
(k-l)
a k-form and consider the inner product:
=
I*UM
A dT
This can be rearranged using a "partial integration":
422 d(*UAT)
II
=
d*U A T + (_l)n-k *U A dT
From exercise (7.6.1) we know that
d * UFurthermore
=
(_l)n-k+l * IS U
J d(*u Meither because M
A
T) =
JaM
*U
A
T = 0*U A T Therefore we obtain: i .e . < UI d T > =
is compact without boundary or because
"vanishes sufficiently fast at infinity".
(8.31)
J * U A dT = J *d.
& U A T
Thus we see that due to our sign convention, 6 operator of
is simply the adjoint
It should be emphasized that this holds both for Let
Riemannian manifolds and manifolds with a Minkowski metric. This has important consequences for the Laplacian operator.
M be a Riemannian manifold.Laplacian as
Using (8.31) we can rearrange the
where
&'
...denotes the adjoint operator. Consequently -A is a
positivQk-forms.
B~rmitian operator.
The above argument applies to arbitrary
In the special case of scalarfields it is well-known from
elementary quantum mechanics, where the Hamiltonian, H
=-
i'J2
2mA ,is a
positive Hermitian operator reflecting the positivity of energy! We can also use (8.31) to deduce an important property of harmonic forms on a compact manifold.
Theorem 8 Let M be a compact Riemannian manifold.A T
A k-form
T
is harmonic
i f and only if it is primitively harmonic, i.e.
=0
iff;d T
= &T =
0
Proof:Let us first observe that = + = + If T is harmonic, i.e. AT 0, then the left-hand-side vanishes
automatically. But the right-hand-side consists of two non-negative terms: Hence they must both vanish: < & T I & T > = = 0
423
II
But as the inner product is positive definite, this implies
&T
=d
T
=0
0i. e. a smooth function, _m2]From this we immediately deduce
o
dS =- de: I e:=O
428
II
( H ere we have used that 1jJ vanishes on the boundary to throwaway the boundary term coming from the partial integration). But this is only consistent if ~ satisfies the equation (8.40) -Sd~ = m2~ , which is the geometrical form of the Klein-Gordon equation: This is a one-form A , and the action is based The Max~eZZ fieZd: upon the Lagrangian density (3.50) which leads to a covariant action given by
This is rearranged as(8.41)
Then we perform a variation,
Awhere
A+E:U:2
uS(g)
is a one-form, which vanishes on the boundary of-~
-~
- - :f
From
this we immediately getdS o = cr---= - = -
I=o
( Here we have used that U vanishes the boundary of n to throw away the boundary term coming from the partial integration). But this is only consistent if(8.42 )
Le.
-OF
o
which is nothing but the Maxwell equations!Exercise 8.6.4Problem:a)Consider the massive vector field by
A
Show that the action (3.51) is given
(8.43)b)Perform the variation tion A ~ A+:U and deduce the following equations of mo-
(8.44)c)Show that they are equivalent to
(8.45)
2 cA= m A
-6A=O
We may summarize the preceding discussion in the following scheme:
429
II
FIELD(8.39)
Action S
Equation of motion-6d~ = m2~
Klein-Gordon~
= f~~(*d~Ad~)-~m2(*~A~)=
(8.40)(8.41)
Maxwell A Massive vectorA
S
-Is, ~ (*dA) AdAn
-odA
=
0
( 8.42) (8.43) (8.44)
S
=J
-~(*dAAdA)-~m2(*AAA) -odA
= m2 A
8.7 INTEGRAL CALCULUS AND ELECTROMAGNETISMAs an other example of how to apply the integral calculus we will use it re-express in a geometrical form various electromagnetic quantities like the electric and magnetic flux through a surface and the electric charge contained in a 3-dimensional regular domain. We start out peacefully in 3-space to get some feeling for the new formalism. Remember that tially B E is a I-form, but B is a 2-form. Essenis the dual of the conventional magnetic field. Now let
n
be a 3-dimensional regular domain. From the discussion of fluxintegrals (8.24) we get (8.46) (8.47) (8.48)
fan
B
The magnetic flux through the closed surface The electric flux through the closed surface The electric charge contained in
an . an .
Jan *Ef;p
n
We can now use the integral calculus to deduce some wellknown elementary properties:Example 1 If n~ke
c~ntains
no singularities, then the magnetic flux
~
through
closed surface
an
vanishes.
This follows from an application of Stoke's theorem:
~(due to (7.76.
J an B = JndB =
0
0
430Example 2
II
(Gauss' theorem)
The electric flux through the closed surface electric charge contained in
an
is
n.(7.78)we get
This follows from an application of theorem 5 (Corollary to Stokes' theorem). From the Maxwell equation;
o
x
[the charge]
=
lJ *p0
n
=
-J n*QES be a sur-
Example 5:
This time we consider a static situation. Let face with boundaryto
r . Then the flux of current through S is equal
oc 2 times the circulation of the magnetic field along Observe first that the Maxwell equa1 oc2 J
r.
tion (7.79) reduces to&B
=
for a static configuration. The proposition then follows from an application of theorem 5 (Corollary to Stokes' theorem) :
Fig. 156We conclude the discussion of electromagnetism putation of two important integrals: a) Consider the spherically symmetric monopole field. Let 52 be the closed surface of a sphere with radius r . Then polar coordinates (r,8,~) are adapted to the sphere, and we can choose (8,~) to parametrize it! We Can now compute the magnetic flux through the closed surface 5 First we observe that the monopole field B is gi ven by (cf. (7. 89) ) :B
in 3-space with the explicit com-
= t,;- 5in8d8AiAp2Tf Tf
Then we immediately get:
(8.49)
J B = tL f J 5in8d8d~ = g 52 ~O 8=0Tf
Fig. 157
431b) This time we consider the magnetic field around a wire with current j. Let r be a circle around the wire with radius p. Then the cylinder coordinates (p,~,z) are adapted to the circle, and we can choose ~ to parametrize the circle. We can now evaluate the line integral of the magnetic field along the closed curve r. For this purpose we must find the dual form *B which is the one-form representing the magnetic field. This has been done previously (cf. section 7.8) *B Thus we get- 21TE: o c 2
II
z
-L2'TT
dip-t
J
fr *B = 21T~ c 2 fd~ oo
=
E7 n0
x
Fig. 158
Then we proceed to consider electromagnetism in Minkowski space, i.e. 4-dimensional space-time. Here it is more complicated to express suitable quantities, so we shall adopt the following terminology: Suppose we have chosen an inertial frame S. Let (xO,X 1 ,X 2 ,X 3 ) denote the corresponding inertial coordinates. We say that the three-dimensional submanifoldF,; is a space slice if it is a subset on the form F,; = {XEMlxo=tO} F,;
One dimension
suppressed~
i.e.
points at a specific time
consists of all the spatial to . Ob-
serve that the spatial coordinates (X 1 ,X 2 ,X 3 ) are adapted to F,; The fundamental quantities describing the properties of the electromagnetic field are the field strengths A , and the current F and *F Fig. 159 ,the Maxwell field S Then we
J . Now let
n
be a 3-dimensional regular domain
contained in a space slice relative to the inertial frame
can form the following integrals which we want to interpret:
Ian F
Ian*F
and dxo
fn*Jto
Observe first that the restriction of grands. But(8.50~
n
vanishes. Conse-
quently the integrals involve only the space-components of the inte-
F
and
*F
are decomposed as and
F
=
[*]
*F
[*J
432 So the restriction of (respectively striction to intn*JI~
II *F) to an is given by B
F
(respectively
*E). Similarly the dual current has the following re-
..
=
(*J)
123
dX ' Adx 2 Adx 3
= -JOdX ' Adx 2 Adx'
We can now generalize the results obtained in (8.46)-(8.48) to the followingLemma 1 Let Then(8.51)
n
be a J-dimensional regular domain contained in a space slice.
JanF
The magnetic flux through The electric flux through
an an n
(8.52)
Jan
r
*F
(8. [, J)
-In*J
=
The electric charge contained in
Exercise 8.7.1 Introduction: Let n be a. 3-dimensional regular domain obtained in a space slice and let F be smooth throughout n. Problem: Use the 4-dimensional integral formalism to re-examine the following well known results: a) The magnetic flux through an is zero. b) The electric flux through an is equal to the electric charge contained in n (As usual we have put EO = c = 1 ).
ILLUSTRATIVE EXAHPLE:
MAGNETIC STRINGS IN A SUPERCONDUCTOR
We have earlier been discussing some of the features of superconductivity, especially the flux quantization (see section 2.12). Recall that in the superconducting state of a metal, the electrons will generate Cooper pairs. These Cooper pairs act as bosons and we can'therefore characterize the superconduc~ing state by a macroscopic wave function the order parameter,1~12,
~
called the order parameter of the superconducting metal. The square of represents the density of the Cooper pairs. We want now to study equilibrium states in a superconductor. Consider a static configuration ~(~), where ~(~) is a slowly varying spatial function. In the Ginzburg-Landau theory one assumes that the static energy density is given on the form:(8.54)H
433
II
Here a is a temperature dependent constant(8.55)
a
=
aCT)
=
a ~cT
(with a positiv)
c
where Tc is the socalled critical temperature. The constant y is just inserted to normalize H to be zero at its global minima. The equilibrium states are found by minimizing the static energy. This leads to the Ginzburg-Landau equation(8.56)
Now consider the potential(8.57)
It has the well-known shape shown on figure 160. Above Tc the vacuum
T>T
c
TOx:TW-WaIT Ua8 (x)dA
1
dA d x
2,
Within the framework of distributions we therefore have
(9.28)We summarize the main properties of the singular form following lemma Lemma 1 (Dirac's lemma) The singular form 1)(9.29 )(9.30)
S
in the
S
has the following properties:
It vanishes outside the string.dS
2)3)
=
*K
fns
=-gM
for any cZosed surface
n
surrounding the mono-
pole. Proof:
(1) (2)
If
x
lies outside the sheet(9.2~
L,u~e
then the
o-function
o'(x-X(A 1 ,A 2
automatically vanish. we that it is equivalent to
To check the relation_1_
a
(Fg a8)
;=ga
s
= KS
(chain-rule)
(Stoke's theorem) = (3) Finally we mustCCIlIpUte
th= flux through a
closed surface sur-
rounding the monopole.
But a closed surface
n
surrounding
the monopole is the boundary of a regular domain
W con-
481 taining the monopole. Consequently we get using (9.29) Ins =
II
laws
=
Iw ds
=
-lw*K
=-gM
D
Worked Problem:
e~ercise
9.2.2 Prove (9.30)by an explicit computation of the integral.
Remark:
We can also reformulate (9.29 ) as
5*S = - K Within the framework of distributions the boundary operation coincides with the codifferential and we therefore get &*S = -g~l:
= -gMal: = ~rM
= - K
Using Dirac I s lenuna we can now "cure the diseases" of therefore had the properties (9.24) dF=-*K InF=gM
F.
It was
generated by a magnetic monopole and an electrically charged particle and
We now choose an arbitrary string extending from the monopole to infinity. Associated with this string we have a weak form properties: (9.29,30) S with the
dS
= *KF + S is exact, although singular, and we which generate
Consequently we see that The gauge potential A
therefore can find a global gauge potential A,
F + S F ,
will, of course, be singular too. It will be singu-
lar at the position of the monopole, reflecting the singularity of and it will be singular at the string, reflecting the singularity of S Formally S represents a concentrated magnetic flux flowing towards the monopole. Hence we may formally interpret F + S as a concentrated magnetic flux flowing towards the monopole position along the string and then spreading out to produce the monopole field. However, it should be emphasized that S has no physical meaning. The position of the string can be chosen completely arbitrarily and the introduction of S is a purely formal F! trick, which cures the diseases of
Fig. 176
482
II
9.3 DIRAC'S LAGRANGIAN PRINCIPLE FOR MAGNETIC MONOPOLESWe are now in a position where we can state the Lagrangian principle of
Dirac~)Dirac'sS
action consists of three pieces:
(9.32)
= SpARTICLES
+ SINTERACTION + SpIELD
where SpARTICLES
SINTERACTION
q
Jr
A dxpa dA
a dA
q
SpIELD
=
Notice that the interaction term only contains a coupling between the electrically charged particle and the field! However, if we look a little closer at(9.33)
SpIELDP
we see that it contains information about the-aA-S v ~ ~v
monopole trajectory because~v
=aA~
v
Hence when you vary the trajectory of the monopole, you will have to vary the sheet which terminates on the trajectory. But that will forceS~v to vary, and thus SpIELD contains the coupling between the monopole and the field. That the above action in fact gives the expected
equations is the contents of the following famous theorem:Theorem 1 (Dirac's theorem) All equations of motion for a system consisting of monopoles, electrically charged particles and the electromagnetic field can be derived from Dirac's action, provided you respect Dirac's veto, i.e. netic strings are never cally charged particle.Proof: The proof is long and technically complicated and you may skip it in a first reading. First we list the equations of motion which we are going to derive d2 x a dx ~ dx v a dx S e + ra e e ] _ qP 6--e(1) ( 3) me [ --;rrzdF =-* K ~ v liT dTallo~ed
the mag-
to cross the worldl