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%% subsection 2.1 Basic Formalism [slac-pub-7056-0-0-2-1 in slac-pub-7056-0-0-2: ^slac-pub-7056-0-0-2 >slac-pub-7056-0-0-2-2]
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\subsection{\usemenu{slac-pub-7056::context::slac-pub-7056-0-0-2-1}{ Basic Formalism }}\label{subsection::slac-pub-7056-0-0-2-1}
LCQ is formally similar to equal-time quantization (ETQ) apart from
the choice of initial-value surface. In ETQ one chooses a surface
of constant time in some Lorentz frame on which to specify initial
values for the fields. In quantum field theory this corresponds to
specifying commutation relations among the fields at some fixed
time. The equations of motion, or the Heisenberg equations in the
quantum theory, are then used to evolve this initial data in time,
filling out the solution at all spacetime points.
In LCQ one chooses instead a hyperplane tangent to the light
cone---properly called a null plane or light front---as the
initial-value surface. To be specific we introduce LC coordinates
\begin{equation}
x^\pm \equiv x^0\pm x^3
\end{equation}
(and analogously for all other four-vectors). The selection of the
3 direction in this definition is of course arbitrary. Transverse
coordinates will be referred to collectively as $x_\perp = (x^1,
x^2)$. A null plane is a surface of constant $x^+$ or $x^-.$ It is
conventional to take $x^+$ to be the evolution parameter and choose
as the initial-value surface the null plane $x^+=0$.
In terms of LC coordinates, a contraction of four-vectors decomposes
as
\begin{equation}
p\cdot x = \ha(p^+x^-+p^-x^+)-p_\perp \cdot x_\perp\; ,
\end{equation}
from which we see that the momentum ``conjugate'' to $x^+$ is $p^-$.
Thus the operator $P^-$ plays the role of the Hamiltonian in this
scheme, generating evolution in $x^+$ according to an equation of
the form (in the Heisenberg picture)
\begin{equation}
[\p,P^-] = 2i{\del\p\over\del x^+}\; .
\end{equation}
What is the effect of this new choice of initial-value surface,
apart from the change of coordinates? The main point is that it
represents a change of {\em representation}, that is, of the Fock
basis used to represent the Hilbert space of a field theory. The
creation and annihilation operators obtained by projecting fields
onto a null plane create and destroy different states than do the
corresponding operators projected out at equal time. Furthermore,
the relationship between the LC and ET Fock states is complicated in
an interacting field theory---complicated enough to perhaps be
useful. A simple way to appreciate this is to imagine starting with
a theory formulated at $t=0$ and solving for the LC Fock states. To
do this one would evolve the fields to the surface $x^+=0$ and
project out its Fourier modes there. Because this requires evolving
the fields in time, however, this requires knowing the full solution
of the theory. Thus the relationship between the two bases is
highly nontrivial, involving the full dynamics of the theory at hand
\cite{20}.
There are two main reasons why the LC representation might be useful
in the context of diagonalizing Hamiltonians for quantum field
theories. First, it can be shown that in LCQ a maximal number of
Poincar\'e generators are kinematic, that is, independent of the
interaction \cite{1,21}. In ETQ six generators are
kinematical (the momentum and angular momentum operators) and four
are dynamical (the Hamiltonian $H$ and boost generators
${\overrightarrow K}$). The fact that the boost operators contain
interactions is a serious difficulty, however. For imagine that we
could actually diagonalize $H$ in some approximation to obtain the
wavefunction for, say, a proton in its rest frame. Boosting the
state to obtain a moving proton, for use in, e.g., a scattering
calculation, would be quite difficult. The state transforms as
\begin{equation}
|\psi^\prime\rangle=e^{-i\a K_3}|\psi\rangle
\end{equation}
for a finite boost in the 3-direction, and since $K_3$ is a
complicated operator (as complicated as the Hamiltonian) calculation
of the exponential is difficult. What is worse is that the
interactions in $K_3$ change particle number. The boost will
therefore take us out of the truncated space in which we are
working, and a suitable effective boost operator, which acts in the
truncated space, must be constructed. This may be expected to be as
difficult as that of determining the effective Hamiltonian;
furthermore, there is no reason to expect that the approximations
used to obtain $P^-_{\rm eff}$ will also be appropriate for
constructing the effective boost operator.
As was first shown by Dirac \cite{1}, on the LC seven of the
ten Poincar\'e generators become kinematical, the maximum number
possible. The most important point is that these include Lorentz
boosts. Thus in the LC representation boosting states is
trivial---the generators are diagonal in the Fock representation so
that computing the necessary exponential is simple. One result of
this is that the LC theory can be formulated in a manifestly
frame-independent way, yielding wavefunctions that depend only on
momentum fractions and which are valid in any Lorentz frame. This
advantage is somewhat compensated for, however, in that certain
rotations become nontrivial in LCQ. Thus rotational invariance will
not be manifest in this approach.
The second advantage of going to the LC is even more striking: the
vacuum state seems to be much simpler in the LC representation than
in ETQ. Indeed, it is sometimes claimed that the vacuum is
``trivial.'' We shall discuss below to what extent this can really
be true, but for the moment let us give a simple kinematical
argument for the triviality of the vacuum. We begin by noting that
the longitudinal momentum $p^+$ is conserved in interactions. For
particles, however, this quantity is strictly positive,
\begin{equation}
p^+=\left(p_3^2+p_\perp^2+m^2\right)^\ha + p^3 > 0\; .
\end{equation}
Thus the Fock vacuum is the only state in the theory with $p^+=0$,
and so it must be an exact eigenstate of the full interacting
Hamiltonian. Stated more dramatically, the Fock vacuum in the LC
representation is the {\em physical} vacuum state.
To the extent that this is really true, it represents a tremendous
simplification, as attempts to compute the spectrum and
wavefunctions of some physical state are not complicated by the need
to recreate a ground state in which processes occur at unrelated
locations and energy scales. Furthermore, it immediately gives a
constituent picture; all the quanta in a hadron's wavefunction are
directly connected to that hadron. This allows a precise definition
of the partonic content of hadrons and makes interpretation of the
LC wavefunctions unambiguous. It also raises the question, however,
of whether LC field theory can be equivalent in all respects to
field theories quantized at equal times, where nonperturbative
effects often lead to nontrivial vacuum structure. In QCD, for
example, there is an infinity of possible vacua labelled by a
continuous parameter $\theta$, and chiral symmetry is spontaneously
broken. The question is how it is possible to identify and
incorporate such phenomena into a formalism in which the vacuum
state is apparently simple.
One clue as to how the physics associated with the vacuum can
coexist with a simple vacuum state is provided by the following
series of observations \cite{22,12,19}. In LC
coordinates the free-particle dispersion relation takes the form
\begin{equation}
p^- = {p_\perp^2+m^2\over p^+}\; ,
\label{disprel}
\end{equation}
from which we see that particle states that can combine to give a
complicated vacuum (i.e., that have $p^+\sim0$) are {\em
high-energy} states.\footnote{We ignore for the moment quanta for
which the dispersion relation (\docLink{slac-pub-7056-0-0-2.tcx}[disprel]{8}) does not hold, i.e.,
massless particles with $p_\perp=0$.} Thus an effective Hamiltonian
approach is natural. For example, we can introduce an explicit
cutoff on longitudinal momentum for particles:
\begin{equation}
p^+>\l\; .
\end{equation}
This immediately gives a trivial vacuum and the corresponding
constituent picture. Since the states thus eliminated are
high-energy states, their effects may be incorporated in effective
interactions in the Hamiltonian; the effective interactions they
mediate will be local in (LC) time, so that they can be expressed as
the integral of some Hamiltonian density over the initial-value
surface.\footnote{It is instructive to contrast this with the
situation in ET field theory. Here, many of the states that are
kinematically allowed to mix with the bare vacuum are low-energy
states, so that a description of vacuum physics in terms of
effective Hamiltonians is not practicable.} In this approach one
can consider the problem of the vacuum as part of the
renormalization problem, that is, the problem of removing dependence
on the cutoff $\l$ from the theory.
Quanta that do not obey Eq. (\docLink{slac-pub-7056-0-0-2.tcx}[disprel]{8}) can simultaneously have
$p^+=0$ and low LC energies, and these may give rise to nontrivial
vacuum structure that cannot be expressed in the form of effective
interactions. Experience with model field theories, however,
suggests that even in this case the physical vacuum state has a
significantly simpler structure than in ETQ \cite{23}. In
addition, these states constitute a set of measure zero in a
(3+1)-dimensional theory. We shall elaborate on this somewhat when
we return to the vacuum problem below.
\subsubsectionInput{slac-pub-7056-0-0-2}{slac-pub-7056-0-0-2-1-1}{Connection to the Constituent Quark Model}%