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 : (b+1) / (b) / \ 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 : x / = 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 formalism) 3. What if we try to make a fermion which is kink-invariant? \ \ \ = / / \ / / 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 conditions: 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 Send Me Email
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