Natural Science Forum / Physics / Research / June 2008

At this point you already depart from quantum mechanics, relativistic
or not. There is no x^v operator when v is the time index. But, say
that you take this relation as an axiom (*).

OK, here you take on another axiom (**), [m^2,x^v] = 0.

Logically, both (4) and (5) are correct starting from axioms (*) and
(**). Congratulations, you've just proven that 0 = 1 (subtract (4)
from (5) and contract with p_v).

Such inconsistencies are not present when you stick with usual QM
axioms. Basically, this little calculation shows how much care must be
taken when arbitrarily departing from them.

Igor

Etc.

Normally in the quantum mechanics (or the classical
mechanics) of a relativistic particle the relation between
rest mass and 4-momentum is treated as a *constraint*,
not as an identity. Then these sorts of issues and associated
paradoxical computations do not arise.

For a thorough treatment of constrained Hamiltonian systems
and their quantum formulation, see, "Quantization of Gauge
Systems", by Henneaux and Teitelboim.

charlie torre

Jay,

One thing to note is that in relativistic QM of single partikl m^2 is
the identity operator
times the number m^2  which commutes with all operators on that
sector. When you
promote M^2 to many particle theory, its no longer the identity, so it
may not commute
with x's. Will kommute with the p's though!

So,..Mr Yablon proved that

[x_m, p_n] = eta_{mn}                     (theres ihbar but that dont
matter for us here)

[M^2, x_n] = 0

and  p^2 = M^2

           with m,n ranging 0,1,2,3

is a logically inconsistent set of statements.
(what makes this go wrong compared to the usual picture,
is that in the usual picture m,n range from 123 in [x_m, p_n] =
eta_{mn}  ,
while the indices range from 0123
in p^2 = M^2 )

Now if what he wants to know is some kind of fix to this, that is
consistent with the usual picture,

just "assume the usual picture"

[x_m, p_n] = eta_{mn}   for m,n=123  --------leave m,n=4 and
combination of 4 with 123 undetermined

[M^2, x_n] = 0     for n=123

p^2 = sum_k  p_k^2 eta_{kk}  = M^2   where the k=0123

then we ask  M^2 x_n = Sum_{j \= n, 0} p_j^2 x_n + p_0^2 x_n + p_n^2
x_n

(Here make the added assumption that [M^2, x_0] = 0)

Now the above becomes   x_n M^2 =  Sum_{j \= n, 0} x_n p_j^2 +
eta_{00} p_0^2 x_n + x_n p_n^2  + 2p_n

Then, whatever freedom is left are basically your choices for the
operator x_0