A Verbose Solution with commentary of Problem (3-4)
from the Feynman and Hibbs text book, with some comments
concerning FCCR, the discretation of space and time, and
a multiplicity of formal concepts of time.

NB: This is very much a verbatim transcript of my notes of something I did about 20 years ago, and that was transcribed over 10 years ago, so please don't expect me to have the interminable details in my head.

NB2: What feynman is telling us with this "student excercise" is that the path integral technique is powerful enough to connect classical physics with relativistic quantum physics - but that the way is not necessarily the easiest way. Feynman loved throwing things like this at his students to separate the men from the boys who just wanted to play PhD. He had undoubtedly already done the calculations himself, of course, but I have never even seen a reference to them.

[Feynman 1965], pp, 34-36

PROBLEM 2-6:

A One dimensional relativistic problem: Suppose that a particle moving in one dimension can go only forward or backward at the speed of light. For convenience, define the units so that the speed of light, the mass of the particle and Planck's constant are all unity. Then in the xt-plane all trajectories shuttle back and forth with slopes of (± π/4). The amplitude for such a path can be defined as follows: suppose time is divided into small equal steps of length ε (epsilon).

Suppose that reversals of path can only occur only at the boundaries of these steps, i. e. at

t = t_0 + nε, with n integral.

For this relativistic problem, the amplitude to go along such a path P is correctly defined to be

φ(P, R) :=  (i ε)^R  =  exp( i (π/2) R ) ε^R

=  exp( i (π/2) R ) exp( R ln( ε ) )

=  exp( i ( π/2 -i ln( ε ) ) R )

where R = R(P) is the number of reversals (or corners) of the path P.
NB: For the symmetry trick below it is important to remember that the verticies of the path corners are points taken on the x-axis.

As the problem, calculate the Kernel K(a, b), by adding together the contributions of 1 corner, 2, corners, ..., etc to determine

K(a, b)  =  Σ N(R) (i ε)^R
R

=  Σ N(R) exp( R ln( ε ) ) exp( i (π/2) R )
R

=  Σ N(R) exp( i( π/2 -i ln( ε ) ) R )
R

where N(R) (an R-degeneracy function) is the number of paths possible with R corners, and the arguments (a, b) of the kernel K are points of closure of the interior lightcone centered at (0, 0) in the 2 dimensional spacetime with coördinates x and t.

Next, suppose the unit of time is (ħ/mc²), [the reduced Compton wavelength/c]. If the time interval is long and the average velocity is small, show that the resulting kernel is correct for a relativistic free particle moving in one dimension. The result is equivalent to the Dirac equation for that case.

The problem is first to construct the path integral (sum) = probability amplitude for a one dimensional relativistic particle starting from the concept of the Zitterbewegung.

Suppose a particle moving in one dimension can go only forward or backward at the velocity of light. In the x-t plane, the particle shuttles back and forth only in lightlike directions. "The amplitude for such a path can be defined as follows: Suppose time is divided into small equal steps of length ε. Suppose that reversals of path direction can occur only at the boundaries of these steps, i.e., at t = t_a + nε, where n is an integer."

For this relativistic problem the amplitude to go along such a path is different from the amplitude defined in terms of the classical action for a nonrelativistic problem. For the classical nonrelativistic case,

φ[x(t)]  =  const exp( (i/ħ) S[x(t)] )

where the action integral over a path

⌠
S[x(t)]  =  ┃ L(dx/dt, x, t) dt
⌡

and x(t) is the particle position on the on the x-axis
parametrized by t.

is the classical action computed for the path, and L() is the Lagrangian. Then the amplitude or kernel to go from point a to point b is given by

K(x_b, t_b; x_a, t_a)  =  Σ φ[x(t)]

where the summation is taken over all kinematically possible paths from

a to b.

The correct definition of the contributing amplitude for a path P for the present relativistic case is given as

φ  =  (iε)^R'

where R' is the number of reversals or corners along the path: bounces off the lighcone.

There are multiple paths with a given number of reversals; simply consider all the permutations of the flight segments of every path.

NB: the elements i^k are those in the strong operator topology, have a limit that is the QM q-p Fourier transform.

As a problem, calculate the kernel K(b, a) by adding together the contribution for the paths of one corner, two corners, etc. Thus determine

for a < b, (Minkowski precedence)

?
K(b, a)  =  Σ n(R) (iε)^R
R=0

where n(R) is the number of paths possible with R corners.
R' → R reduces R' to a number count, R = 0, 1, 2, ...?
and n(R), a corner density: how many paths have R corners.

F&H suggest that it is advisable to calculate four separate K's, namely the amplitude K_++(b, a) of starting at a with positive velocity and coming into point b with positive velocity, the amplitude K_+-(b, a), of starting at the point a with a negative velocity and coming into point b with a negative velocity, and the amplitudes K_-+(b, a), and K_--(b, a) similarly.

Second part of the problem:

Suppose the time quantum is given by the ((1/c)reduced Compton wavelength).

ε :=  ħ/(mc²).

If the total time interval is long

(t_b - t_a)  >> ε  :=  ħ/mc²

and the average velocity is small

(x_b - x_a)  << c (t_b - t_a)

show that the resulting kernel is approximately the same as that for a nonrelativistic free particle

K_nr(b, a)  :=  [2 π i ħ (t_b - t_a)/m]^(-1/2)   X

exp[ i m (x_b - x_a)²/(2 ħ) (t_b - t_a) ]

but for a factor

exp[ (-i m c²/ħ)(t_b - t_a) ]  =

exp[ (-i/ε)(t_b - t_a) ]

The definition given for the amplitude and the resulting kernel is claimed to be correct for a relativistic theory of a free particle moving in one dimension and is further claimed to be equivalent to the Dirac equation for that case.

An important aspect of the problem to bear in mind is that it is for a 2-D spacetime, and not for the usual 4-D Minkwoski space. Thus, the 24 = 16 dimensional Dirac-Clifford algebra does not come into play. If the context was a 3-D Minkowski space, we would have the Pauli-Clifford algebra of 23 = 8 dimensions.
As it stands the algebra is that of 1x1 matrices, and so there is only one equation, so it will not look like the usual Dirac equation, especially in an approximating form.

Thinking classically, this is a problem of a single particle in a 1-D space, with a disconnected 1-D local newtonian time.

For a free particle of any type, the Lagrangian is the same as the Hamiltonian.

XYZZY Calculate the path integral for a free relativistic particle.

Aside question:
Can the number of corners in a path be associated with some entropic measure?

Solution of the Main Problem:

The first part of the problem is another way of looking at a standard random flight problem:

Consider the particle as above with velocity c = 1, that executes an N step path that ends M steps from its starting place "the origin". If we know how many ways this can happen, we can calculate the probability of arriving at M after N steps. Let the probability of a forward step be p+ and the probability of a backward step be p-. Assume that at each step the probabilities are independent of previous or future choices. Any path of N steps will have M+ forward steps and M- backward steps so that

N  =  M+ + M-        and      M  =  M+ - M-

where we have also assumed that the arrival point is positive. Notice that M can be even or odd only as N is even or odd. M+ and M- can characterize a path just as N and M and

M+  =  (1/2)(N + M)  and      M-  =  (1/2)(N - M)

(N+M and N-M are always even)

For a SINGLE path with M+ and M- the probability is then

(p+)^M+ (p-)^M-

The values of M+ and M- do not, however, fix the path. For the arrival, we must consider all paths with fixed M+ and M- so the arrival probability can be expressed as

W(M+, M-)  =  #(M+, M-) (p+)^M+ (p-)^M-

The possible paths are ennumerated by the number of ways that M+ indistinguishable unit steps can be distributed over N segments, i.e.,

#(M+, M-)  =  (N M+)  =  (N M-)  =  ((M+ + M-)  M+)
=  N!/(M+! M-!)
=  N!/( [(1/2)(N + M)]! [(1/2)(N - M)]! )

a binomial coefficient.  Then

W(N, M+)  =  (N  M+) (p+)^M+ (p-)^(N-M+)

so from the binomial theorem

N
Σ W(N, M+)  =  (p+ + p-)^N  =  1
M+=0

showing that the probability distribution W(N, M+) is normalized.

In terms of N and M

W(N, M)  =

N!/( [(1/2)(N + M)]! [(1/2)(N - M)]! )   X
(p+)^((1/2)(M + N)) (p-)^((1/2)(M - N))

where p+ + p-  = 1.

For the unbiased random flight  p+ = p- = 1/2.  This is an
expression of microtime reversibility.  Then

W(N, M)  =   N!/( [(1/2)(N + M)]! [(1/2)(N - M)]! ) (1/2)^N

is a Bernoullian distribution.  This is the standard random
flight problem in its simplest form.

For long waiting times, N very large; and for low velocities,
M << N, by using Stirling's approximation

log N!  ≃  (N + 1/2)log N - N + (1/2)log(2π)
+ O(N^(-1), as N → ∞

and the approximation for M/N << 1

log(1 (± (M/N))  ≃  (±)(M/N) - (1/2)(M/N)²

We have the asymptotic behavior of a Bernoullian distribution

W(N, M)  → (2/(πN))^(1/2) exp( -(M²/(2N)) )

Cf. also, the De Moivre-Laplace theorem [Wilks 1962], p. 274.

Binom(np, npq) -> N(np, npq), asymptotically.

An aside:

NM  =  M+² - M-²

M² + N²  =  M+² + M-²

(N + M)(N - M)  = N² - M²  =  4M+ M-

1 + (M/N)
M+/M-  =   ━━━━━━━━━━━
1 - (M/N)

Take
the final displacement x  :=  c ε M
and
elapsed time           t  :=  ε N

define the average macro velocity

v  :=  x/t
Then

M/N    =  v/c
M+/N   =  (1/2)(1 + v/c)
M-/N   =  (1/2)(1 - v/c)

M+ / M-  =  (1 + v/c) / (1 - v/c)

2 c ε M+  =  ct + x
2 c ε M-  =  ct - x

4 c² ε² M+ M-  =  c² ε² (N² - M²)

=  (c² t² - x²)

With the definitions of N and M, the asymptotic form of the relativistic kernel for the second part of the problem is supposed to be

exp( -i (2(M+M-)/(M+ + M-) ) ) exp( -i(M+ + M-)/2 )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
c ε sqrt(2 π i) sqrt(N)

exp( i (1/2N)(M² - N²) ) exp( -iN/2 )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
c ε sqrt(2 π i) sqrt(N)

exp( i (M²/2N) ) exp( -iN )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
c ε sqrt(2 π i) sqrt(N)

exp( i (M² - 2N²)/2N )
━━━━━━━━━━━━━━━━━━━━━━━━━━  =
c ε sqrt(2 π i) sqrt(N)

exp( i (M² - N² - N²)/2N )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
c ε sqrt(2 π i) sqrt(N)

exp( i ((M² - N²)/2N - N²/2N) )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
c ε sqrt(2 π i) sqrt(N)

exp( i ((-1/c²ε²)(x² - c²t²)/2N - N/2 ))
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
c ε sqrt(2 π i) sqrt(N)

exp( -i ( (x² - c²t²)/N - εc²N ) )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
2 c(^3/2) ε sqrt(2 π i) sqrt(N)

exp( -i ( (x² - c²t²)/(t/ε) - εc²(t/ε)²/(t/ε) ) )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
2 c(^3/2) ε sqrt(2 π i) sqrt(t/ε)

exp( -i ( (x² - c²t²) - ε(ct/ε)² ) )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
2 c(^3/2) ε sqrt(2 π i) (t/ε)^3/2

√ε exp( -i ( (x² - c²t²) - (ct)²/ε ) )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
2 √(2 π i) (ct)^3/2

√ε exp( -i ( (x² - (1+(1/ε))c²t²) ) )
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━  =
2 √(2 π i) (ct)^3/2

The kernel has the physical dimension of [L^(-1)], as it should.

BUT we want to partition the (N M+) paths for fixed N and M+ into subsets each characterized by the number R of reversals in the path. Generally N > M+, and so the minimum number of reversals is

R_min  =  1.

The maximum number of reversals happens assuming M is positive, (M+ > M-) Crudely, every negative step is associated with two reversals.

R_max  ≃  2M-  =  (N - M).

Generally for M of either sign

R_max  ≃  2 min[ M+, M- ]

Then we also know that for M positive

R_max
Σ  n(R)  =  (N M+)
R=1

But, in considering paths of R corners there are distinctions to be made for R even and R odd. It is not too hard to see that with regard to the sign of the initial and terminal velocities:

R even      start +   end +             R = 2r
start -   end -

R odd       start +   end -             R = 2r - 1
start -   end +

exhibits the possibilities.

Thus for paths beginning and ending with the same sign of velocity, an even number of reversals is required; for opposite signs, an odd number of reversals is required.

Reconsider R_max more precisely: For + -> + paths all the negatively directed steps are associated with exactly 2 reversals. Therefore for (M+ > M-)

R_max  =  2 M-  =  M- + M-

In this case R is even and R = 2r, so

r_max  =  M-

For - -> - paths all the negatively directed steps are associated with 2 reversals except for the start and end termini. Therefore for (M+ > M-)

R_max  =  2 M- - 2  =  (M- - 1) + (M- - 1)

Again, R is even and R = 2r, so

r_max  =  (M- - 1)

For + -> - and - -> + paths all the negatively directed steps are associated with 2 reversals but for a single terminus. Therefore for (M+ > M-)

R_max  =  2 M- - 1  =  (M- - 1) + M-
=  2 M- - 1  =  M- + (M- - 1)

These two cases cover odd R = 2r - 1.  So then

r_max  =  M-

Let n_++(R), n_+-(R), n_-+(R), n_--(R) be the numbers of paths distinguished by the F&H suggestion for the kernels, so that,

n(R)  =  n_++(R) + n_+-(R) + n_-+(R) + n_--(R)

Then, also from simple symmetries (reverse the sign for each flight segment in a path):

n_++(2r)  =  n_--(2r)  not=  0
n_+-(2r)  =  n_-+(2r)  =  0

n_++(2r-1)  =  n_--(2r-1)  =  0
n_+-(2r-1)  =  n_-+(2r-1)  not=  0

So
n(2r)  =  2 n_++(2r)
n(2r-1)  =  2 n_+-(2r-1)

We need only determine n_++(2r), and n_+-(2r-1).

Consider a path contributing to n_++(2r).

Number of corners or reversals = 2r
Number of flight steps = 2r + 1
Number of positively directed flight steps = r + 1
Number of negatively directed flight steps = r

while for a path contributing to n_--(2r).

Number of corners or reversals = 2r
Number of flight steps = 2r + 1
Number of positively directed flight steps = r
Number of negatively directed flight steps = r + 1

Let m_+k be the length of the kth positively directed flight segment where k = 1, 2, ..., r+1, and let m_-k, where k = 1, 2, ..., r, be the length of the kth negatively directed flight segment of some path. Then with M+ and M- defined as for the usual random flight path ennumeration,

r+1
Σ m_+k  =  M+
k=1

r
Σ m_-k  =  M-
k=1

Now consider a path contributing to n_+-(2r - 1) or n_-+(2r - 1).

Number of corners or reversals = 2r - 1
Number of flight steps = 2r
Number of positively directed flight steps = r
Number of negatively directed flight steps = r

Let m_+k be the length in steps of the kth positively directed flight segment where k = 1, 2, ..., r, and let m_-k, where k = 1, 2, ..., r, be the length of the kth negatively directed flight segment of some path. Then with M+ and M- defined as for the usual random flight path ennumeration,

r
Σ m_+k  =  M+
k=1

r
Σ m_-k  =  M-
k=1

A distinct flight path with R corners and M+ and M- fixed is determined by giving the ordered sequences of the m_±k, subject to the constraints of their sums. Distribute the M+ and M- flight segments separately.

For R = 2r even, consider the distribution of the M+ unit steps. This is equivalent to the problem of distributing M+ indistinguishable unit steps in r+1 labeled segments (distinguishable and ordered), such that NO BOX IS EMPTY. This is a classic combinatorial problem [Percus 1971], p. 34, 67. The integer valued 's' will, for our purposes, take on the values r and r+1.

#(M_±, s)  =  ((M_± - 1) (s - 1))  =
=  ((M_± - 1) (M_± - s))
So

#(M, r)  =  (r/M) (M r)
and
#(M, r+1)  =  ((M - r)/r) (M r)
then
#(M, r+1)  =  ((M - r)/r) #(M, r)
=  (M/r - 1) #(M, r)

Also,

#(M_±, s)  =

s-1
Σ (-1)^k (s k) ((M_± + s - 1 - k) M_±)
k=0

Terms with k > s-1 will vanish identically from the poles of the factorial function. Then also the following summation holds as an identity:

s
Σ (s k) #(M_±, s-k)  =
k=0

s
Σ (s k) #(M_±, k)  =
k=0

s
Σ (s s-k) #(M_±, s-k)  =
k=0

((M_± + s - 1) M_±)

or

s   k
Σ   Σ (-1)^j (s k) (k j) ((M_± - 1 + k - j) M_±)
k=0 j=0

=  ((M_± + s - 1) M_±)

From standard binomial formulas

((M+ - 1) r-1) ((M- - 1) r)  =

[r(M- - r)]/[M+ M-] (M+ r) (M- r)

((M+ - 1) r) ((M- - 1) r-1)  =

[r(M+ - r)]/[M+ M-] (M+ r) (M- r)

((M+ - 1) r) ((M- - 1) r)  =

[M+M- - r(M+ - M-) + r²]/[M+ M-] (M+ r) (M- r)

Then,

((M+ - 1) r-1) ((M- - 1) r) + ((M+ - 1) r) ((M- - 1) r-1) =

[r(M+ - M-) - r²]/[M+ M-] (M+ r) (M- r)

and

((M+ - 1) r-1) ((M- - 1) r) + ((M+ - 1) r) ((M- - 1) r-1)
+ 2((M+ - 1) r) ((M+ - 1) r)  =

[2M+ M- - r(M+ - M-)]/[M+ M-] (M+ r) (M- r)  =

2[ 1 - (2rN)/(N² - M²) ] (M+ r) (M- r)

M_±
Σ ((M_± + s - 1) M_±)  =  ((M_± + M_±) M_±)
k=0

=  (N M_±)  =  (N M_±)

M-1
Σ #(M, k)  =  2^M-1
k=1

M
Σ #(M, k) z^k =  z(1 + z)^(M-1)
k=0

M+
Σ (M+ k) (M- k) =  (N M+)
k=0
(k partitions the full set of paths)

(M+ k)(M- k)  =  (M+-1 k-1)(M--1 k-1)
(M+-1  k )(M--1 k-1)
(M+-1 k-1)(M--1  k )
(M+-1  k) (M--1  k )

(M+-1 k-1)(M--1 k-1)  =
k²/(M+M-) (M+ k)(M- k)

(M+-1  k )(M--1 k-1)  =
(M+ - k)(M- - k)/(M+M-) (M+ k)(M- k)

(M+-1 k-1)(M--1  k )  =
k(M- - k)/(M+M-) (M+ k)(M- k)

(M+-1  k) (M--1  k )  =
k(M+ - k)/(M+M-) (M+ k)(M- k)

Check: the sum of the coefficients of the (M+ k)(M- k) does = 1.

The path counting functions are then

n_++(2r)  =  #(M+, r+1) #(M-, r)
n_--(2r)  =  #(M-, r+1) #(M+, r)
n_+-(2r-1)  =  #(M+, r) #(M-, r)
n_-+(2r-1)  =  #(M-, r) #(M+, r)

n_+-(2r+1)  =  #(M+, r+1) #(M-, r+1)
n_-+(2r+1)  =  #(M-, r+1) #(M+, r+1)

n(2r)  =  n_++(2r) + n_--(2r)
=  #(M-, r+1) #(M+, r) + #(M+, r+1) #(M-, r)

n(2r-1)  =  n_+-(2r) + n_-+(2r)
=  #(M-, r) #(M+, r) + #(M+, r) #(M-, r)
=  2 #(M-, r) #(M+, r)

Some Special Cases:

#(M+, 0)  =  0
#(M+, 1)  =  1
#(M+, 2)  =  M+ - 1
#(M+, 3)  =  (M+ - 1)(M+ - 2)/2
......
#(M+, (M+ - 1))  =  M+ - 1
#(M+, M+)  =  1

The separate kernels with pretty much the same structure are: (See above for maximum values in summation for r.)

M-
K_++(b, a)  =    Σ #(M+, r+1) #(M-, r) (i ε)^(2r)
r=0

(M- - 1)
K_--(b, a)  =    Σ #(M-, r+1) #(M+, r) (i ε)^(2r)
r=0

M-
K_+-(b, a)  =    Σ #(M+, r) #(M-, r) (i ε)^(2r+1)
r=0

M-
K_-+(b, a)  =    Σ #(M-, r) #(M+, r) (i ε)^(2r+1)
r=0

with

#(M+, r+1) #(M-, r)  =  ((M+ - 1) r)  ((M- - 1) (r - 1))
=  ((M+ - r)/M-)  (M+ r)(M- r)
#(M-, r+1) #(M+, r)  =  ((M- - 1) r)  ((M+ - 1) (r - 1))
=  ((M- - r)/M+)  (M+ r)(M- r)

#(M+, r) #(M-, r)  =
((M- - 1) (r - 1))  ((M+ - 1) (r - 1))

=  #(M+, r) #(M-, r)
=  (1/M+M-) r² (M+ r)(M- r)

#(M+, r+1) #(M-, r) + #(M-, r+1) #(M+, r)  =

M+² + M-² - r(M+ + M-)
────────────────────── (M+ r) (M- r)
M+ M-

N² + M² - rN
=       ────────────── (M+ r) (M- r)
(1/4)(N² - M²)

N² + M²      N
=  4 [ ─────── - ─────── r] (M+ r) (M- r)
N² - M²   N² - M²

Cf. [Wilks 1962], p. 147 eq 6.6.11, p. 149, eq. 6.6.20 The p(u) distribution on p. 149 eq. 6.6.20 is (N M+) n(R) for R even and R odd if the following replacements are made:

n_1  ->  M+
n_2  ->  M-
n    ->  N
u    ->  R
Then

N
Σ n(R)  =  (N M+)
R=0

The full kernel is expressed as a sum of the four kernels:
(The k = M- term has been added to the sum in K_--. This is OK since this term contains a factor #(M-, M- + 1) which is identically zero.)

K(b, a)  =

M- 2 (r² - i (ε/2)(M+ + M-)r + i (ε/2)(M+² + M-²))
Σ ─────────────────────────────────── (M+ r)(M- r)(i ε)^(2r)
r=0         i ε      M+ M-

=

M- 2 (r² - (iε/2)Nr + (iε/2)(N² + M²))
Σ ───────────────────────────────── (M+ r)(M- r)(i ε)^(2r)
r=0 i ε    (1/4) (N² - M²)

ASIDE:

Is it true that using a n(R), K(b, a) can be written in the uniform form?

n(r)  =  n(N-r)  ?

M+
Σ n(r)  =  ?
r=0

K(b, a) =

M-
Σ n(r)  =  (1/2)(N M+)  ?
r=0

What we would really like to have is a closed expression for

M-
Σ r^d (M+ r) (M- r) z^(2r)
r=0

END ASIDE

where the exponent d takes the three values d = 0, 1, 2. If we have the result for d = 0, then the others can be had by differentiation.

Find the asymptotic form of M+! M-!:

M+! M-!  =  N!/(N M+)

From the asymptotics of the random walk problem for large N and M << N,

we know that

(N M+)  →  2^N (2/(π N))^(1/2) exp( -M²/2N )

From Stirling's approximation of the factorial,

N!  →  (2 π N)^(1/2) N^N exp(-N)

Then

M+! M-!  →   π N (N/2)^(N) exp( M²/2N - N )

Now get the asymptotic expression for (M+ - r)! (M- - r)!

log[ (M+ - r)! (M- - r)! ]

=  log[(M+ - r)!] + log[(M- - r)!]

=  [M+ - r + 1/2] log(M+ - r) - (M+ - r) + (1/2)log 2 π
+ [M- - r + 1/2] log(M- - r) - (M- - r) + (1/2)log 2 π

=  [M+ - r + 1/2] log[(N/2)(2(M+ - r)/N) - (M+ - r) + (1/2)log 2 π
+ [M- - r + 1/2] log[(N/2)(2(M- - r)/N) - (M- - r) + (1/2)log 2 π

=  [M+ - r + 1/2] log[(N/2)(2(M+ - r)/N) - (N - 2r) + log 2 π
+ [M- - r + 1/2] log[(N/2)(2(M- - r)/N)

=  [N - r + 1/2] log[(N/2)] - (M - 2r) + log 2 π
+ [M+ - r + 1/2] log[(2(M+ - r)/N)
+ [M- - r + 1/2] log[(2(M- - r)/N)

=  [N - r + 1/2] log[(N/2)] - (M - 2r) + log 2 π
+ [M+ - r + 1/2] log[(N + M - 2r)/N]
+ [M- - r + 1/2] log[(N - M - 2r)/N]

=  [N - r + 1/2] log[(N/2)] - (M - 2r) + log 2 π
+ [M+ - r + 1/2] log[1 + (M - 2r)/N]
+ [M- - r + 1/2] log[1 - (M + 2r)/N]

The summation variable ranges from r = 1 to
r = M- = (1/2)(N - M).

At r = 1

(M - 2r)/N  << 1  and (M + 2r)/N  << 1

At 2r = (N - M)

(M - N + M)/N  = 2M/N - 1
1 + (M - N + M)/N  = 2M/N  < 1

[We have assumed that M << N and therefore M/N << 1.
The absolute value of (2M/N - 1) must be less than 1]

and

(M + N - M)/N  = -1
1 - (M + N - M)/N  =  1 - 1  =  0

which is a singularity of the log function. Trust to the damping of contributions to paths at this value and perform the approximations

log[1 + (M - 2r)/N]  ≃

+(M - 2r)/N - (1/2)[ (M - 2r)/N ]²

log[1 - (M + 2r)/N]  ≃

-(M + 2r)/N - (1/2)[ (M + 2r)/N ]²

Then,

log[(M+ - r)! (M- - r)!]  =

log[(M+ - r)!] + log[(M- - r)!]  =

[N - r + 1/2] log[(N/2)] - (M - 2r) + log 2 π
+ [M+ - r + 1/2] [ +(M - 2r)/N - (1/2)(M - 2r)²/N² ]
+ [M- - r + 1/2] [ -(M + 2r)/N - (1/2)(M + 2r)²/N² ]

Some Points of interest, along the way:

In the approximation leading to the (x-t) Dirac kernel, the limit as ε → ∞ is specifically not taken. The limit is one of long waiting times and low velocities (many or heavily weighted moderately wiggly paths). Paths with the most reversals are few in number; similarly, paths with the fewest reversals. The paths with the most reversals are the paths that are most quantum mechanical in nature. The more numerous paths with a moderate number of reversals contribute to an averaging by phase cancellation in the summation defining the kernel in analogy to the Riemann-Lebesgue lemma.

The quantum of time is inversely proportional to the mass of the particle, or more precisely to the rest energy or "proper energy" of the particle. If we invert the definition of ε, so

m  =  ħ/(ε c²)

The smallest unit of time possible takes the Planck time,

τ_0  =  (ħ G_0/c^5)^(1/2)

=  5.391 x 10^(-44) sec

with G_0 the gravitational constant. Intervals of time are naturally measured in integral multiples of τ_0.

Let ε(n) := n τ_0.

Then there is a discrete mass spectrum

m(n) :=  ħ /(ε(n) c²)
=  (1/n) (ħ c/G_0)^(1/2)
=  (1/n) m_0

where m_0 is the Planck mass. We note that the Planck mass is notoriously too big as a fundamental mass quantum since it is approximately 2.177 x 10^(-5) g, while the mass of the electron is approximately 9.1095 x 10^(-28) g. Feynman pointed out that the Planck mass is about the mass of a flea. So, a flea or a Planck mass converted to energy would release about

1.956 x 10^(16) ergs

The mass spectrum m(n) determines a smallest mass by a largest value of n. If the smallest mass is of the order of the mass of the electron, then

n  ≃  m_0/m(n)  ≃  0.20 x 10^(23)

Muons and tauons, however, have masses that are already smaller than that of electrons and positrons.

In any case, This inversrse linear mass spectrum, for sufficiently large n, reaches down into the range of real elementary particle masses.

With regard to FCCR, n is the dimension of the Hilbert space that accurately covers particle behavior contained within the known universe. This is a Hilbert space (C*-algebra) of process whose intrinsic time is of the order of 10^(-21) sec. A Hilbert space that is already big enough quantitatively and dimensionally to approximate those of quantum mechanics well.

It is curious that the relativistic limit should be one for low velocities, since the standard relativistic regime specifically embraces velocities close to the speed of light in a vacuum.

Modifications to a standard relativistic quantum theory then occur for short times or high velocities.

The relativistically invariant quantity

c² t² - x²  =  c² ε² (N² - M²)

is what replaces the action integral in the nonrelativistic formulation. It is the invariant path length integral that is varied in the derivation of the Einstein equations by variational methods.

⌠
δ ┃  √(-g) ds  =  0
⌡

From the more primitive viewpoint wherein special relativity (the Lorentz group) can be seen as a type of thermodynamic limit, the only velocity is that of c. The fundamental group is then not the Lorentz group but rather the symmetry group of the light cone, E(2).

[Feynman 1965], p. 42. For a nonrelativistic free particle where the Lagrangian is given by the kinetic energy.

In one time and one spatial dimension, for a free
particle, the Lagrangian is

L  =  (m/2)(dx/dt)²  =  p²/(2m)

For one particle in a k dimensional real space,
with a one dimensional real common time parameter

k                          k
L(k)  =  (m/2) Σ (dx_j/dt)²  =  (1/(2m))  Σ  (p_j)²
j=1                        j=1

Due to the usual assumed isotropy (direction
independence) of a Newtonian space, the Cartesian
components, (x_j) are independent and separable.

The propagation kernel then becomes

φ[x_1(t) ... x_k(t)]  =

const exp( (i/ħ) S[x_1(t) ... x_k(t)] )

The action integral over a path in k-space is

t_b
k   ⌠
S[x(t)]  =   (m/2) Σ   ┃  (dx_j/dt)²  dt
j=1  ⌡
t_a

and x_j(t) is the particle position on the
(x_j)-axis parametrized by t.

For a particle in a k dimensional space, the
action integral for a free particle is a sum
of 1 dimensional free particle action integrals.

XYZZY - all below
1.
with the 1-dim Dirac equation.  Show that the combinatorial
kernel satisfies the 1 dimensional Dirac equation.

2.
Show that the k dimensional Dirac equation is satisfied by
the k dimensional Dirac kernel that pulls in the Clifford
algebra of 2^k dimensions as a matter of factoring a bilinear
form.

3.
when defining the velocity operator.

4.
Make the transition for FCCR "dynamics".

Feynman's 1-dimensional nonrelativistic kernel for a free particle is defined by the limit (The order of the two limits matters) of the (N-1)-fold path integral:

K_C(b, a)  =

lim    lim   (2 π i ħ ε/m)^(-N/2)  X
N → ∞  ε → 0

+∞     +∞
⌠     ⌠
┃ ... ┃     X
⌡     ⌡
-∞     -∞

N                    N-1
exp[ (i m ħ ε)/2   Σ (x_j - x_(j-1))² ]  Π dx_j
j=1                   j=1

With
N ε  =  t_b - t_a,  ε  =  t_(j+1) - t_j
t_0  =  t_a,  t_N  =  t_b
x_0  =  x_a,  x_N  =  x_b

From the successive application of the definite integral formula [Feynman 1965], p. 43 equation (3-4); (See proof below.)

+∞
⌠
(α/sqrt(ε)) ┃ exp [ (α²/ε) ( (x-a)² + (x-b)² ) ] dx  =
⌡
-∞

(α/sqrt(2 ε)) exp [ (α²/(2 ε)) (a-b)² ]

The integrals of the kernel can be be performed for any finite
N so that the Classical Kernel,

K_C(b, a)  =

lim    lim    (2 π i ħ ε N/m)^(-1/2)   X
N → ∞  ε → 0

exp[ (m/(i2ħ) N ε) (x_b - x_a)² ]  =

lim    lim  (2 π i ħ (t_b - t_a)/m)^(-1/2)    X
N → ∞  ε → 0

exp[ (m/(i2ħ) ) (x_b - x_a)²/(t_b - t_a) ]

It appears, that if the kernel phase is instead given by the relativistically invariant quantity

exp [ (i m/(2 ħ ε)) Σ (x_j - x_(j-1))² - c² (t_j - t_(j-1))² ]

where m is taken as the rest mass, that then the indefinite integral formula is still useful since the extra term in the exponential is not a function of the variable of integration and simply factors out of each integral. The resulting Relativistic Kernel is then:

K_R(b, a)  =

lim    lim   (2 π i ħ (t_b - t_a)/m)^(-1/2)   X
N → ∞  ε → 0

exp( m/(2 i ħ) ) [(x_b - x_a)² / 2(t_b - t_a) - c² (t_b - t_a)]

=   lim    lim  exp (mc²/2iħ ) [(t_b - t_a)] K_C(b, a)
N → ∞  ε → 0

This is almost the form claimed by F&H to be the correct Dirac kernel. But I think what is meant is that it is the correct Dirac kernel in the low velocity limit.

The phase that gives the kernel claimed to be correct is

=  exp[ (i (m/(2ħ ε)) Σ (x_j - x_(j-1))² - 2 c² ε² ]

Then the kernel is

K_R(b, a)  =

lim    lim  exp( mc²/(iħ) ) [(t_b - t_a)] K_C(b, a)
N → ∞  ε → 0

The difference being the correct absence of the factor of 2 in the denominator of the phase.

In both nonrelativistic and relativistic cases, the kernel form is correct whether or not the infinitely fine grained limit is taken. The model of a grained spacetime gives the same formal result as that of one based on the continuum. Observe, however, that the kernel integrations allow that each variable of integration x_j has a doubly infinite continuous range.

Taken over to the context of FCCR with only convex combinations of |q(n, k)> allowed, the integration limits should be q_min = -q_max to q_max. (q_max)² is approximately proportional to n.

But this is arrived at by a low velocity limit, and it appears that the rigorously correct phase should be

exp[ (-imc/ħ ε) Σ sqrt(c² ε² - (x_j - x_(j-1))² ) ]

=

exp[ (-imc²/ħ ) Σ sqrt( 1 - (x_j - x_(j-1))²/(c² ε²) ) ]

since the integrand of the action principle is the path length and not the path length squared. The integration is now rougher. We need to evaluate integrals of the form

+z
⌠
┃ exp [ (±)i α (sqrt(1 - (x-a)²/β²) ] dx
⌡
-z

+z
⌠
┃ exp [ (±)i α    X
⌡
-z

( sqrt(1 - (x-a)²/β²) + sqrt(1 - (x-a)²/β²) ) ] dx

On the other hand, if action is taken as the sum of squares of the individual invariant lengths of the flight subsegments of the paths, then the successive integrations of the kernel can be performed using the F&H equation (3-4), and it appears that specified low velocity approximate kernel of F&H problem (6-2) is the result. We have not derived this form from the rigorously correct relativistic kernel. Explain this.

Connection with Hausdorff/fractal measure?

The low velocity limit as follows:

With regard to the exponent of the Dirac kernel, note that

(N² - M²)^(1/2)  =

=  N (1 - (M²/N²))^(1/2)

and with M << N [low average velocities]

≃   N (1 - (1/2)(M²/N²))

=  - (M²/(2N) - N )

and that for timelike paths:

(N² - M²)  >  0

is associated with the action of the path. (Feynman's relativistic path summation by definition, sums only over lightlike paths.) A kernel of the form

exp( i (N² - M²)^(1/2) )
──────────────────────────
c ε sqrt(2 π) sqrt(N)

then approximates the said low velocity Dirac kernel. Note that this kernel form is in no way derived; it is ultimately plunked down as approximating the low velocity Dirac kernel.

Regarding FCCR, we have associated <ψ|G(n)|ψ> with the action of a process. Consider states decomposed on the q-eigenbasis.

(n-1)
|ψ>   =     Σ α_k  |q(n, k)>
k=0

Any |ψ> is a weighted complex superposition of lightlike

( <q(n, k)|G(n)|q(n, k)>  =  0 )

transitions of spatial displacements q(n, k).

( <q(n, k)|Q(n)|q(n, k)>  =  q(n, k) )

where for large n, q(n, k) becomes an

for all k

q(n, k+1) - q(n, k)

approximates a constant that depends only on n,
and decreases with increasing n.

Convex combinations of the |q(n, k)> are timelike or spacelike processes.

The quantum statistical spatial displacement associated with a process |ψ> is supposedly given by

<ψ| Q(n) |ψ>

while the time path length (as the number of vicissitudes) is given by n.

A process |ψ> is associated with a statistical distribution of lightlike transitions, i. e., a collection of complex weighted lightlike path subsegments. The subsegments are labeled by size and complex weight but are considered to be unordered (commutativity of vector addition). The weighted subsegments are added as interfering alternatives. We are really dealing with a polychotomic random flight problem.

The |α_k|² is the probability of executing a lightlike transition of spatial displacement q(n, k) at any vicissitude. The Hilbert space dimension n therefore limits the transition lengths, consequently the total flight time.

So n is a measure of the total flight time in terms of the number of vicissitudes OR the number of "time quanta" in a waiting time. Vicissitude may be thought of as an uncertainty or hesitation within a time quantum. The actual displacement can be apparent only after the passage of a time quantum, while the possibilities beforehand are the q(n, k).

Virtual (total) positive displacement

Virtual (total) negative displacement

Four formal notions of time:

Relativistic proper time
<ψ|G(n)|ψ>  =  <ψ|ψ> - n|<ψ|n, n-1>|²

Modified expectation value of G(n)
<ψ|G(n)|ψ>/<ψ|ψ>  =  1 - n|<ψ|n, n-1>|²/<ψ|ψ>

If it vanishes, then
<ψ|ψ>  =   n|<ψ|n, n-1>|²
<ψ|n, n-1>  =  (1/√n) exp(i ф(|ψ>) ) <ψ|ψ>

Minkowskian coördinate time in SR or SR tangent space in GR,

c² t²  =  <ψ|Q²(n)|ψ> + <ψ|G(n)|ψ>

Time direction in Hilb(n)

|n, n-1>

A vector in the Hilbert space is associated with a ST point; so |n, n-1>
should be associated with a single timelike event unless vector magnitude
acquires a meaning.

Oscillator forward and backward time operators,
Fourier Transforms

t_+(n) and t_-(n)

probably best understood as operators for phase or some sort of "microtime".
C.f.,  Origins of Newtonian Time

Relativistic FCCR kernel:
Density Matrices

n = number of proper time quanta per lightlike segment.  (A lightcone measure)
|q(n, k)> are the possibilities for a lightlike flight segment.
|ψ> is a weighting of the set of possibilities as interfering
alternatives for a single segment or process.
A sequence of any number of process vectors.
{|ψ_k>} represents a Q-statistical path (polychotomic flight)
where each subsegment is Q-weighted but indeterminate.

Σ p_k |ψ_k><ψ_k|
k

is a density matrix with classical statistical weighting by probabilities p_k
of a sequence of Q-flight segments.

<ψ_k|G(n)|ψ_k>  =  the expectation value of the invariant
distance squared "s²" for the k-th flight segment.

Relativistic FCCR kernel phase

exp[ (i m/(2ħ) τ_0) Σ <ψ_k|G(n)|ψ_k> ]
k

Then the actual QM kernel should be something like

⌠
┃ exp[ (im/(2ħ) τ_0) Σ <ψ_k|G(n)|ψ_k> ] dμ(ψ)
⌡                    k

where μ(ψ) is a measure on the Hilbert space Hilb(n), which for
any finite n is a complex n-dimensional Lebesgue measure.

Proof of F&H equation (3-4):

Evaluate the integral

+∞
⌠           (x-a)² + (x-b)²
┃ exp[ (±i) ─────────────── ] dx
⌡                  ε
-∞

Expanding the quadratics of the exponential, collecting
terms in x and completing the square for the terms in x,
obtain

(x-a)² + (x-b)²  =  2(x - (a+b)/2)² + (1/2)(a-b)²

The integral then becomes

+∞
(a-b)²   ⌠
exp[ ±i ────── ] ┃ exp[ (±i 2/ε) (x - (a+b)/2)² ] dx
2 ε     ⌡
-∞

Integrating by substitution, for ε > 0 and real, let

u  :=  (2/ε)^(1/2)(x - (a+b)/2)
so
du  =   (2/ε)^(1/2) dx , or dx  =  (ε/2)^(1/2) du.

Then, the integral becomes, by variable substitution

+∞
(a-b)²                ⌠
exp[ ±i ────── ]  (ε/2)^(1/2) ┃ exp[ (±)i u² ] du
ε                   ⌡
-∞

and

+∞                 +∞
⌠                  ⌠
┃ exp[±i u²] du =  ┃ (cos u² + i sin u²) du
⌡                  ⌡
-∞                 -∞

Which is the sum of two double sided Fresnel integrals.

∞                   ∞
⌠                   ⌠
┃ sin u² du  =      ┃ cos u² du  =
⌡                   ⌡
0                   0

(1/2) (π/2)^(1/2)

From symmetry

+∞
⌠
┃ exp[±i u² ] du  =  (1 ± i) (π/2)^(1/2)
⌡
-∞

But
sqrt(±i)  =  (1 ± i)/sqrt(2)

So the integral evaluates as

+∞
⌠         (x-a)² + (x-b)²
┃ exp[ ±i ─────────────── ] dx  =
⌡                ε
-∞

(a-b)²
sqrt( ±i π ε/2 ) exp[ ±i ────── ]
2 ε

Making the replacement

ε  ━→  2 ε ħ / m

and choice of the minus sign, F&H equation (3-4)
follows exactly.

QED

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