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 easy. He 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 +&- pi/4.
The amplitude for such a path can be defined as follows: suppose time is
divided into small equal steps of lenth epsilon.
Suppose that reversals of path can only occur only at the bpoundaries of
these steps, i. e. at t = t0 + (n epsilon), with n integral.
For this relativistic problem, the amplitude to go along such a path
is correctly defined to be
phi = (i epsilon)^R
= exp( i (pi/2) R ) exp( R ln( epsilon ) )
= exp( i ( pi/2 -i ln( epsilon ) ) R )
where R is the number of reversals (or corners) of the path.
As the problem, calculate the Kernal K(a, b), by adding together
the contributions of 1 corner, 2, corners, ..., etc to determine
K(a, b) = SIGMA N(R) (i epsilon)^R
R
= SIGMA N(R) exp( R ln( epsilon ) ) exp( i (pi/2) R )
R
= SIGMA N(R) exp( i( pi/2 -i ln( epsilon ) ) R )
R
where N(R) (an R-degeneracy function) is the number of paths possible
with R corners.
Next, suppose the unit of time is (h-bar/mc^2), [the Compton wavelength].
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
epsilon. Suppose that reversals of path direction can occur only at the
boundaries of these steps, i.e., at t = t_a + (n epsilon), 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,
phi[x(t)] = const exp( (i/h-bar ) S[x(t)] )
where
S[x(t)] = INTEGRAL L(dx/dt, x, t) dt
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(b, a) = SIGMA phi[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 for the
present relativistic case is
phi = (i epsilon)^R
where R is the number of reversals or corners along the path.
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
K(b, a) = SIGMA n(R) (i epsilon)^R
R
where n(R) is the number of paths possible with 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
epsilon = (h-bar/mc^2)
If the total time interval is long
(t_b - t_a) >> (h-bar/mc^2)
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 pi i h-bar (t_b - t_a)/m]^(-1/2) X
exp[ i m (x_b - x_a)^2/(2 h-bar) (t_b - t_a) ]
except for a factor
exp[ (-i m c^2/h-bar )(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.
--------------------------------------------------------------------------
Aside question:
Can the number of corners in a path be associated with some entropic measure?
--------------------------------------------------------------------------
Solution:
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
SIGMA 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. 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.
For long waiting times, N very large; and for low velocities,
M << N, by using Stirling's approximation
log N! approx= (N + 1/2)log N - N + (1/2)log( 2 pi )
+ O(N^(-1), as N -> infinity
and the approximation for M/N << 1
log(1 (+|-) (M/N)) approx= (+|-)(M/N) - (1/2)(M/N)^2
We have the asymptotic behavior
W(N, M) -> (2/(pi N))^(1/2) exp( -(M^2/(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
M^2 + N^2 = M+^2 + M-^2
(N + M)(N - M) = N^2 - M^2 = 4M+ M-
1 + (M/N)
M+/M- = ------------
1 - (M/N)
Take
the displacement x := c epsilon M
and
elapsed time t := epsilon 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 epsilon M+ = ct + x
2 c epsilon M- = ct - x
4 c^2 epsilon^2 M+ M- =
c^2 epsilon^2 (N^2 - M^2) =
(c^2 t^2 - x^2)
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 epsilon sqrt(2 pi i) sqrt(N)
exp( i (1/2N)(M^2 - N^2) ) exp( -iN/2 )
--------------------------------------- =
c epsilon sqrt(2 pi i) sqrt(N)
exp( i (M^2/2N) ) exp( -iN )
------------------------------ =
c epsilon sqrt(2 pi i) sqrt(N)
exp( i (M^2 - 2N^2)/2N )
------------------------------
c epsilon sqrt(2 pi i) sqrt(N)
Kernel then has the physical dimension of [L^(-1)]
----------------------------------------------------------------------
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 approx= 2M- = (N - M).
Generally for M of either sign
R_max approx= 2 min[ M+, M- ]
Then we also know that for M positive
Rmax
SIGMA 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
SIGMA m_+k = M+
k=1
r
SIGMA 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
SIGMA m_+k = M+
k=1
r
SIGMA 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,
s-1
#(M_+|-, s) = SIGMA (-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
SIGMA (s k) #(M_+|-, s-k) =
k=0
s
SIGMA (s k) #(M_+|-, k) =
k=0
s
SIGMA (s s-k) #(M_+|-, s-k) = ((M_+|- + s - 1) M_+|-)
k=0
or
s k
SIGMA SIGMA (-1)^j (s k) (k j) ((M_+|- - 1 + k - j) M_+|-)
k=0 j=0
= ((M_+|- + s - 1) M_+|-)
From standard binom 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-) + r^2]/[M+ M-] (M+ r) (M- r)
Then
((M+ - 1) r-1) ((M- - 1) r) + ((M+ - 1) r) ((M- - 1) r-1) =
[r(M+ - M-) - r^2]/[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)/(N^2 - M^2) ] (M+ r) (M- r)
-----------------------------------------------------------------
M_-|+
SIGMA ((M_+|- + s - 1) M_+|-) = ((M_+|- + M_-|+) M_+|-)
k=0
= (N M_-|+) = (N M_+|-)
M-1
SIGMA #(M, k) = 2^M-1
k=1
M
SIGMA #(M, k) z^k = z(1 + z)^(M-1)
k=0
M+
SIGMA (M+ k) (M- k) = (N M+) (k partitions the full set of paths)
k=0
(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) = k^2/(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)
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) = SIGMA #(M+, r+1) #(M-, r) (i epsilon)^(2r)
r=0
(M- - 1)
K_--(b, a) = SIGMA #(M-, r+1) #(M+, r) (i epsilon)^(2r)
r=0
M-
K_+-(b, a) = SIGMA #(M+, r) #(M-, r) (i epsilon)^(2r+1)
r=0
M-
K_-+(b, a) = SIGMA #(M-, r) #(M+, r) (i epsilon)^(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-) r^2 (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-
N^2 + M^2 - rN
= ---------------- (M+ r) (M- r)
(1/4)(N^2 - M^2)
N^2 + M^2 N
= 4[ --------- - --------- r] (M+ r) (M- r)
N^2 - M^2 N^2 - M^2
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
SIGMA 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 (r^2 - i epsilon(1/2)(M+ + M-)r + i epsilon(1/2)(M+^2 + M-^2))
SIGMA ----------------------------------- (M+ r) (M- r) (i epsilon)^(2r) =
r=0 i epsilon M+ M-
M- 2 (r^2 - i epsilon(N/2)r + i epsilon(1/2)(N^2 + M^2))
SIGMA --------------------------------- (M+ r) (M- r) (i epsilon)^(2r) =
r=0 i epsilon (1/4) (N^2 - M^2)
Is it true that
n(R) = n(N-R) ?
M+
SIGMA n(R) = ?
R=0
M-
= SIGMA n(R) = (1/2)(N M+) ?
R=0
What we really want is closed expression for
M-
SIGMA r^d (M+ r) (M- r) z^(2r)
r=0
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/(pi N))^(1/2) exp( -M^2/2N )
From Stirling's approximation of the factorial,
N! -> (2 pi N)^(1/2) N^N exp(-N)
Then
M+! M-! -> pi N (N/2)^(N) exp( M^2/2N - 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 pi
+ [M- - r + 1/2] log(M- - r) - (M- - r) + (1/2)log 2 pi
= [M+ - r + 1/2] log[(N/2)(2(M+ - r)/N) - (M+ - r) + (1/2)log 2 pi
+ [M- - r + 1/2] log[(N/2)(2(M- - r)/N) - (M- - r) + (1/2)log 2 pi
= [M+ - r + 1/2] log[(N/2)(2(M+ - r)/N) - (N - 2r) + log 2 pi
+ [M- - r + 1/2] log[(N/2)(2(M- - r)/N)
= [N - r + 1/2] log[(N/2)] - (M - 2r) + log 2 pi
+ [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 pi
+ [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 pi
+ [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] approx= +(M - 2r)/N - (1/2)[ (M - 2r)/N ]^2
log[1 - (M + 2r)/N] approx= -(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 pi
+ [M+ - r + 1/2] [ +(M - 2r)/N - (1/2)(M - 2r)^2/N^2 ]
+ [M- - r + 1/2] [ -(M + 2r)/N - (1/2)(M + 2r)^2/N^2 ]
--------------------------------------------------------------------------
Some Points of interest:
In the approximation leading to the Dirac kernel, the limit as
epsilon -> infinity 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. They 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 epsilon, so
m = h-bar / (epsilon c^2)
The smallest unit of time possible take as the Planck time,
tau_0 = (h-bar 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 tau_0.
Let epsilon(n) = n tau_0.
Then there is a mass spectrum
m(n) = h-bar /(epsilon(n) c^(2)
= (1/n) (h-bar c/G_0)^(1/2
= (1/n) mu_0
where mu_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) gm., while the mass of the
electron is approximately 9.1095 x 10^(-28) gm.
The mass spectrum m(n) determines a smallest mass by a largest n. If the
smallest mass is of the order of the mass of the electron, then
n approx= mu_0/m(n) approx= 0.20 x 10^(23)
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 of process whose intrinsic time is of the order of 10^(-21) sec.
A Hilbert space that big already quantitatively approximates 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.
Modifications to a standard relativistic quantum theory then
occur for short times or high velocities.
The relativistically invariant quantity
c^2 t^2 - x^2 = c^2 epsilon^2 (N^2 - M^2)
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.
delta INTEGRAL 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
L = (m/2)(dx/dt)^2
Feynman's nonrelativistic kernel is defined by the limit of the (N-1)-fold
integral:
K_C(b, a) =
lim lim (2 pi i h-bar epsilon/m)^(-N/2) X
N->infinity epsilon->0
+infinity +infinity
INTEGRAL ... INTEGRAL X
-infinity -infinity
N N-1
exp[ (i m/2 h-bar epsilon) SIGMA (x_j - x_(j-1))^2 ] PI dx_j
j=1 j=1
With
N epsilon = t_b - t_a, epsilon = 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.)
+infinity
(alpha/sqrt(epsilon)) INTEGRAL X
-infinity
exp[ (alpha^2/epsilon) ( (x-a)^(2 + x-b)^2 ) ] dx =
(alpha/sqrt(2 epsilon)) exp[ (alpha^2/(2 epsilon)) (a-b)^2 ]
the integrals of the kernel can be be performed for any N so that
K_C(b, a) =
lim lim (2 pi i h-bar epsilon N/m)^(-1/2) X
N->infinity epsilon->0
exp[ (m/(i2h-bar) N epsilon) (x_b - x_a)^2 ] =
lim lim (2 pi i h-bar (t_b - t_a)/m)^(-1/2) X
N->infinity epsilon->0
exp[ (m/(i2h-bar) ) (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 h-bar epsilon)) SIGMA
(x_j - x_(j-1))^2 - c^2 (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 kernel is then
K_R(b, a) =
lim lim (2 pi i h-bar (t_b - t_a)/m)^(-1/2) X
N->infinity epsilon->0
exp( m/(2 i h-bar) ) [(x_b - x_a)^(2/2(t_b - t_a) - c^(2(t_b - t_a)]
= lim lim exp (mc^(2/2ih-bar ) [(t_b - t_a)] K_C(b, a)
N->infinity epsilon->0
Which 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/(2h-bar epsilon) SIGMA (x_j - x_(j-1))^2 - 2 c^2 epsilon^2 ]
Then the kernel is
K_R(b, a) =
lim lim exp( mc^2/(ih-bar) ) [(t_b - t_a)] K_C(b, a)
N->infinity epsilon->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/h-bar epsilon) SIGMA sqrt(c^2 epsilon^2 - (x_j - x_(j-1))^2 ) ]
=
exp[ (-imc^2/h-bar ) SIGMA sqrt( 1 - (x_j - x_(j-1))^2/(c^2 epsilon^2) ) ]
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
INTEGRAL exp[ (+|-)i alpha (sqrt(1 - (x-a)^2/beta^2) ] dx
-z
+z
INTEGRAL exp[ (+|-)i alpha X
-z
( sqrt(1 - (x-a)^2/beta^2) + sqrt(1 - (x-a)^2/beta^2) ) ] 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
(N^2 - M^2)^(1/2) =
= N (1 - (M^2/N^2))^(1/2)
and with M << N [low average velocities]
approx= N (1 - (1/2)(M^2/N^2))
= - (M^2/(2N) - N )
and that for timelike paths:
(N^2 - M^2) > 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 (N^(2 - M^2)^(1/2) )
------------------------------
c epsilon sqrt(2 pii) sqrt(N)
then approximates the said low velocity Dirac kernel. Note that this kernel
form is in no way derived; it is merely plunked down as approximating the
low velocity Dirac kernel.
Regarding FCCR, we have associated <psi|G(n)|psi> with the
action of a process. Consider states decomposed on the q-eigenbasis.
(n-1)
|psi> = SIGMA alpha_k |q(n, k)>
k=0
Any |psi> 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 "additive quantum number", in that
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
process |psi> is supposedly given by
<psi| Q(n) |psi>
while the time path length (as the number of vicissitudes) is given by n.
A process |psi> 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 |alpha_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
---------------------
The 4 formal notions of time:
Relativistic proper time
<psi|G(n)|psi> = <psi|psi> - n|<psi|n, n-1>|^2
Modified expectation value of G(n)
<psi|G(n)|psi>/<psi|psi> = 1 - n|<psi|n, n-1>|^2/<psi|psi>
If it vanishes, then
<psi|psi> = n|<psi|n, n-1>|^2
<psi|n, n-1> = (1/sqrt(n)) exp(i theta(|psi>)) <psi|psi>
Minkowskian coordinate time
c^2 t^2 = <psi|Q^2(n)|psi> + <psi|G(n)|psi>
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 time operators
t_+(n) and t_-(n)
probably best understood as operators for phase or some sort of "microtime".
--------------------------------------------------------------------------
Relativistic FCCR kernel:
n = number of proper time quanta per lightlike segment. (A lightcone measure)
|q(n, k)> are the possibilities for a lightlike flight segment.
|psi> 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.
{|psi_k>} represents a Q-statistical path (polychotomic flight)
where each subsegment is Q-weighted but indeterminate.
SIGMA p_k |psi_k><psi_k|
k
is a density matrix with classical statistical weighting by probabilities p_k
of a sequence of Q-flight segments.
<psi_k|G(n)|psi_k> = the expectation value of the invariant
distance squared "s^2" for the k-th flight segment.
Relativistic FCCR kernel phase
exp[ (i m/(2h-bar) tau_0) SIGMA <psi_k|G(n)|psi_k> ]
k
Then the actual QM kernel should be something like
INTEGRAL exp[ (im/(2h-bar) tau_0) SIGMA <psi_k|G(n)|psi_k> ] dmu(psi)
k
where mu(psi) is a measure on the Hilbert space Hilb(n).
--------------------------------------------------------------------------
--------------------------------------------------------------------------
Proof of F&H equation (3-4):
Evaluate the integral
+infinity (x-a)^2 + (x-b)^2
INTEGRAL exp[ (+|-)i ----------------- ] dx
-infinity epsilon
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 is then equal to
(a-b)^2 +infinity
exp[ +|-i --------- ] INTEGRAL exp[ (+|-)(i 2/epsilon) (x - (a+b)/2)^2 ] dx
2 epsilon -infinity
For epsilon > 0 and real, let
u = (2/epsilon)^(1/2)(x - (a+b)/2)
so
du = (2/epsilon)^(1/2) dx or dx = (epsilon/2)^(1/2) du
Then the integral becomes, by variable substitution
(a-b)^2 +infinity
exp[ +|-i ------ ] (epsilon/2)^(1/2) INTEGRAL exp[ (+|-)i u^2 ] du
epsilon -infinity
and
+infinity +infinity
INTEGRAL exp[ +|-i u^2 ] du = INTEGREL (cos u^2 + i sin u^2) du
-infinity -infinity
Which is the sum of two double sided Fresnel integrals.
infinity infinity
INTEGRAL sin u^2 du = INTEGRAL cos u^2 du = (1/2) (pi/2)^(1/2)
0 0
From symmetry
+infinity
INTEGRAL exp[ (+|-)i u^2 ] du = 2(1 (+|-) i) (1/2) (pi/2)^(1/2)
-infinity
But
sqrt((+|-)i) = (1 +|- i)/sqrt(2)
So the integral evaluates as
+infinity (x-a)^2 + (x-b)^2
INTEGRAL exp[ (+|-)i ----------------- ] dx =
-infinity epsilon
(a-b)^2
sqrt(+|-i pi epsilon/2) exp[ (+|-)i --------- ]
2 epsilon
Making the replacement
epsilon -> 2 epsilon h-bar / m
and choice of the minus sign, F&H equation (3-4) follows exactly.
QED
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Created 1996, transcribed from work done sometime before 1986.
Last updated: August 21, 2007
bhammel@graham.main.nc.us
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