"Igor Khavkine" <email@example.com> wrote in message
> On Jun 15, 3:01 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
.. . .
> The operator m is a scalar (of the form (real number)*I_oo, as you've
> put it above). That, by itself, is plenty sufficient to show that
> [O,m^k]=0, for *any* operator O and any power k of m.
>> [...snip introduction of Dirac equation...]
> Let me repeat this: m is a scalar, [O,m] = 0 for any operator O.
> Introducing anything else into the picture is merely muddying the
I agree completely. So, I have linked a 2.5 page calculation at:
It is simple and to the point. Start with [O,m] = 0 for any operator O
including O=x_u. Take Dirac's equation to be true. Take the canonical
[x_i,p_j]=i hbar eta_ij i,j=1,2,3 (1)
to be true, where diag (eta_ij) = (1,1,1) are the space components of
the Minkowski metric tensor. From this, one may directly deduce that:
[x_k,p_0] psi = -i alpha_k psi (2)
where alpha_k = gamma^0 gamma_k, and the gamma^u are the Dirac matrices.
Equation (2) suggests that the [x_k,p_0] commutators times i=sqrt(-1)
must be equal to the eigenvalues of the Dirac alpha_k, which alpha_k, by
the way, also sit along the 0k components of the
-2i sigma_uv = [gamma_u, gamma_v]
and that these [x_k,p_0] commutators can therefore only be considered in
relation to their associated eigenstate vectors.
Just as Dirac's equation reveals some features that cannot be seen
strictly from the Klein Gordon equation, the calculation here seems to
reveal some features about the canonical commutators that the usual
calculation based on [O,m^2]=0 and m=p^u.p_u cannot, by itself, reveal.
I would like some input on whether this is on the right track or whether
I am overlooking something.