Deriving the Duo-Dimensional Discrete Dirac Dynamics is Delightful

(or Definitely a Dastardly over-Dose of D's)

There's an amazing way to derive the Dirac equation (in 2d) which everyone should know. Louis Kauffman and Richard Feynman came very close to finding it (and probably knew it, but never published it).

(First see the traditional Dirac way to do this, if you don't know it yet). Or, jump to newest additions

Let L and R be the chiral fermion components.  Let dl and dr be the
light-cone derivatives.  The (massive) Dirac Equation in 2d is :

    dr R = -i L
    dl L = -i R

In units where hbar = c = m = 1 .  
(BTW the massless Eq. is trivial : L moves l and R moves r).

We discretize the l/r on a lattice ala the "Feynman Checkerboard".
If up is time, and side is space, the lattice is like:


I won't actually play checkers like Feynman and Kauffman do, but the
lattice is useful for conversing.  We want to derive the Dirac Eq
we just stated.  First, what is R(a,b) (or L) ?  We normalize so
that R(0,0) = 1 : a fermion was emitted at 0,0 and now R(a,b) is
the amplitude to find it at light-cone lattice site (a,b).  This
simple observation provides us with a consistency condition to
derive the Dirac equation.

The component R always moves instantaneously to the right.  This
means the amplitude to find R at (a,b) is a sum on paths that
travel only to the right.  (I am not using path integrals, simply
the fact that quantum mechanics is a sum on amplitudes).  So,
we can derive dr(R).  In our units, the time step is (h/mc^2) = 1,
so dr(R) = R(a,b+1) - R(a,b) .  We need to know how the amplitude
to go (0,0) -> (a,b) changes when we increment b by one.  It's
useful to draw a picture now (boys and girls); we need only count
the new paths can get to (a,b+1) which did not get to (a,b).
The answer is simple : only paths that arrived at (a,b) from the
left.  (remeber that R(a,b) takes contribution from all paths that
arrive at (a,b) from the right = all paths from R(a,b-1) + a jog,
so R(a,b+1) = R(a,b) + paths that hit (a,b) from the left).  I'll
try to draw something :

          /   \
        R       L

We have found :

    R(a,b+1) - R(a,b) = dr R = L(a,b) * C

Where C is the amplitude to turn a corner.  To find C we need to
draw some pictures.  We demand that our fermions result in Fermi
statistics.  To get Fermi statistics it is necessary and sufficient
for the exchange amplitude for two paths to be -1.  So :

     / /         / /
     \ \         \/
     / /    = -  /\
    / /         / /

(the jogging in the first picture is so the lines start and end in
the same place).  Or :

    C^4 = - C^2

So, C^2 = -1 .  Thus C = +/- i .  We may choose either one, it's
equivalent to an r<->l handedness convention. 

I note in passing that this means C^4 = 1 ; this seems to be
a direct constraint, like Rotate(2pi) = 1 , but it is hard to
derive rigorously.  If we relax our idea of the lattice, we
see something like

    |_              |
     _|     C^4 =   |   1
    |               |

    \               \
     \/\             \
        \             \
        /   C^5 =     / C
     /\/             /
    /               /

These seem nice, but both violates our idea of the lattice in one way
or another; the first one is on an x-t lattice, not a lightcone lattice,
and the second one involves steps backwards in time.  However, both
of these graphical relations are correct.  We might try to justify these
relations by some sort of analytic continuation of the lattice structure,
but that seems difficult.

Anyhoo, we choose C = -i, so:

    R(a,b+1) - R(a,b) = dr R = L(a,b) * C

    dr R = - i L

Which is indeed the Dirac equation !!  We see that this equation states
only two things, and is simply a relation forced upon us by consistency
of the amplitude picture :

    L is the amplitude to arrive from the left (and R ...)
    they may turn into eachother with amplitude -i

From this we can go on to derive the Feynman Checkerboard as Kauffman
does, and from that we can derive the ordinary path integral in the
limit of t >> x (that is, (r+l) >> (r-l)).  

Some amusing notes : obviously ,  dldr R/L = - R/L which is klein-gordon.
This means that Phi = R +/- L is a (very small subset) special case solution
of the massive boson equation.  If we re-insert the 'm' but keep c = h = 1,
the dirac equation is :

    dr R = - i m L

and the amplitude to corner is (-i m).  We can understand this very
nicely in terms of the Lagrangian which creates our equation :

    L = R dr R + L dl L - i m L R

If we take a field theory viewpoint and call the first two terms the
"free part" we find the propagator moves R in r and L in l at the speed
of light.  The "vertex" (-i m L R) is simply the amplitude for L to
become R and vice-versa, which is exactly our cornering amplitude (-im).
The instantaneous velocities are always 'c' ; the mass only 'slows down'
the particle by making it turn around and go backwards once in a while.
The massive fermion travels very much in a random walk (though the
probability to go left and right are both one, and corners contribute
an amplitude (-i)).

Note that 'i' here is the precursor to spin : the quaternions in 4d
which we usually call spin (Pauli matrices) are just the algebra of
i's extended by adding j's and k's.  In this sense, "i" can be thought
of as the "x" quaternion (one of the Pauli matrices).  It is nice to
interpret each quaternion as the instruction "rotate by 90" , so
that xy = z , and x^4 = 1, so x^2 = -1 , etc.  This works perfectly
with our interpretation of i as the "corner" amplitude : a corner is
just a rotate by 90 operator.  Presumably in the 4d version, there
would be 3 different corner amplitudes : x,y, and z, being the three
quaternions (Pauli Matrices acting on the two-component spinor).

I recommend reading Kauffman's paper hep-th/9603202 .
Feynman's remarks are in the classic "Feynman & Hibbs", page 35,
but Feynman considers the problem too trivial to solve, and
assigns it as homework (!?! , in chess notation)

It's difficult to extend this kind of construction beyond 2d, for
several reasons.  First of all, you must go to four components, and
explicitly separating the fermions becomes non-trivial.  Furthermore,
beyond 2d you have spin and a little group : that is, there is a
continuous symmetry group, and a continuous degree of freedom, so it is
not clear how to discretize and count paths properly.

It's also very difficult to extend this to bosons.  In fact, I think I
can prove that you cannot find any solutions to Klein-Gordon this way,
except for the fermion solution itself.  (that is, Phi = R +/- L)

There is something rather mysterious here.  If we
consider the most general (chiral) reduction relation
allowed by causality/locality :

    /    = A  x + B x
          /          \

That is, amplitude to hit (a,b+1) from the left is
some combo of amplitudes to hit (a,b) from either
direction.  (you can add terms in like the amp. to
hit (a,b+1) from the left, but by parity symmetry
these reduce back into this form).  If we now
demand that this reduction relation lead to a
solution of the graphical Klein-Gordon equation :

     x       /\
    / \  -  x  x =  - m^2   / \
    \x/      \/             \x/

with m = 1 :

     x       /\
    / \ =   x  x  - / \
    \x/      \/     \x/

You find A = 1 and B = +/- i

But that is nothing other than the graphical
Dirac reduction relation :

     x         /       /
    /    =    x -   i x
             /         \

Which we understood in terms of the cornering amplitude!

I cannot find any reduction relation which satisfies
Klein-Gordon other than the Dirac solution!  How can
we interpret this?  Perhaps it is simply because the
bosons cannot be put on a light-cone lattice like this?
They do not instantaneously have the velocity of light?

(BTW there is one exception to this : if A is not = 1 ,
then this reduction relation is the solution to the
Dirac equation in the presence of a maxwell field).



These are various comments which need to be added to
my note on the 2d (1+1) Dirac equation.  They come
primarily from thoughts inspired in conversation with
Louis Kauffman.

1. First, it appears Feynman knew of these results,
 though he never published them.  You may note in
 Feynman & Hibbs, and also in the classic
 "A Space-Time Approach to NRQM" (section 11) Feynman
 makes comments like "there is another way to derive
 the Dirac equation".  This way is found in Feynman's
 manuscripts, as demonstrated by Silvan Schweber's
 biographical review : "Feynman and the Visualization
 of Space-Time Processes" .  Here Feynman basically
 goes through what I reported last time.  It appears
 that Feynman never published because he was troubled
 by what I am going to say in part 2.

2. Our derivation that the corner-amplitude , c = -i
 appears to be inconsistent.  This is because the
 amplitudes are *not* topological, so that simply
 requiring that one pair of paths cross once, and the
 others do not cross does not specify them.  In other
 words, the choice of which pair of pairs to compare
 is ambiguous and leads to different answers.  In the
 last note, we compared :

      / /             / /
     / /             / /
     \ \    = -      \/
     / /             /\
    / /             / /

    c^4 = - c^2

 to conclude c = +/- i
 However, we can introduce a "kink" at any point :

    \           \
     \  = c^2   /
     /          \
    /           /

 Which changes the power of c in any graph by two.
 Thus, our choice of which two graphs to compare
 to find "fermi statistics" is ambiguous because
 of these kinks.  (one can show that all other
 topo-moves on graphs to do not introduce ambiguity (see part 4);
 if there is something special about these kinks, it
 is unknown to me, and it seems also to Feynman).

 It seems that this may be salvageable in some way,
 because it is hard to think of a consistent theory
 in which c^4 = 1 is not satisfied.  The trouble is
 in finding the postulate to justify this.
  (for example, the Fermion loop condition 
        /\ = -1
   seems to work, (here we are considering the product of
   two different ways to go straight in time, each
   way goes like c) but is hard to justify in this

3. What if we try to make a fermion which is

    \       \
     \  =   /
     /      \
    /       /

  We can try this by letting corner-to-the-left and
  corner-to-the-right have different amplitudes : l and r
  respectively.  Then the kink invariance says
    l = lrl
  or lr = rl = 1
  (let me briefly note : if we used our Fermi requirement
  in this generalized case, we would get lr = -1 .  This can
  be satisfied by l = r = i , the ordinary formulation,
  or via l = 1, r = -1 , (modulo overall -1 factors).  These
  simply correpond to using i*sigma_x (l=r=i) or i*sigma_y (l=-r=1)
  and produce the same complex structure (lr = -1))
  So, we have dr R = r L , dl L = l R , using my old notation.
  Thus dl dr R = r dl L = rl R , or (box - rl) R = 0
  This is Klein-Gordon with mass = (- rl) .  In the usual case,
  rl = -1, we have mass = 1 in our units.  In a kink-invariant
  theory, rl = 1, so mass = -1, is a tachyon !
  It's interesting that the tachyon in 2d actually seems to
  be a topological theory; you seem to run into trouble
  with constructing a conserved probability current.
  (in the usual case, it's just Jr = R* R , Jl = L* L , so
    d*J = dr Jr + dl Jl ~ R dr R + L dl L = R r L + L l R 
        = c ( RL + LR ) = 0
   when r=l=c is violated, this is ruined).

4. The original thoughts on this problem, for the record :

At 07:47 PM 4/9/98 , the esteemed Louis Kauffman wrote:
>Further thought on the Fermi Statistics. How do you know which trajectories
>to compare? ...  [for example]
>(rrrr, llll) crossing in the middle, versus  (rlrl, lrlr). 

Well, this is c^6 = -1 , which is not inconsistent, it's simply a slightly
weaker condition.  We can take the usual view in quantum and demand that it
work the same way regardless of the specific paths taken.  
I think that if we demand that any pair of trajectories which differ by a
crossing have the opposite sign, we should find a whole family of different 
conditions which are related to c^2 = -1 by multiplying either side by c^4 . 
With this more general view of the Fermi condition, it would be helpful to
know c^4 = 1, but I think we can proceed by simply requiring all of the

c^4 = - c^2
c^6 = - 1
c^10 = - c^6

be satisfied simultaneously.  The trick would be to show that it's impossible
to find any paths other than these, and there's trouble here.
The trouble is basically that we can introduce a wiggle :

         /      /
        /       \
        \       /
         \      \

The first of these has one corner, the next has 3 (it's llrr vs. lrlr),
they differ by a c^2 , but are topologically indistinguishable.  Thus
any condition like "crossing" .  We can save the idea if we demand that
this wiggle move only be done to both fermions simultaneously.  If
we don't demand that, then we run into the trouble :

    \  /        \  /
     \/          \ \
     /\          / /
    /  \        /  \

    1   = - c^4 

is no good.  (this is llll+rrrr vs. rrll + rlrl).  Again, this graph
can be saved only if we demand the fermions be symmetric, so that
(rrll + rlrl) is replaced by (rrll+llrr) or (rlrl+lrlr)

If we introduce the light-cone Reidemiester moves:

0.  \       \
     \  =   /
     /      \
     \       \
    (lrll) = (llrl)

0*. \           \
     \  = c^2   /
     /          \
    /           /
    (rrll) = cc (rlrl)

2.  \ /         \ /
    /^\ = c^4   / \
    \ /         \ /
    /^\         / \
    (a crossing is is indicated by a caret mark)
    (llrr+rrll) = c^4 (lrlr+rlrl)

S.  \   /           \   /
     \ /             \ /
     /^\    = c^2    / \
    /   \           /   \
    (re-attach crossing)

the "S" move is the Skein for the "Feynman Polynomial" to
be a solution of the Dirac equation.  Apparently we can
deform any graph into any other by application of these
moves  (BTW crossings here are not oriented, overcrossing
or undercrossings are equivalent; this might be inconsistent
if c^4 != 1) , as long as they have the same endpoints.
Moves 0 and 2 are clearly topological (once we figure out
that c^4 = 1).  It is move 0* that seems to be a problem.
This move is a problem for the "fermi statistics" way of figuring
out the value of 'c' , because move 0* can introduce arbitrary
powers of c^2 into the graph without changing the topology
(number of crossings).  Only if we demand symmetrical use
of move 0* can we maintain the "fermi" argument.

Charles Bloom / cb at my domain
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