Re: EM field of photon

On Jan 4, 12:36 am, Arnold Neumaier wrote

This is strange terminology which she uses repeatedly. Her position
operator is a 3x3 matrix, not a vector as one would reasonably demand?
On closer inspection it turns out that it is a 3-vector with 3x3 matrix
components.

I took position to imply a vector, in the same way that velocity
implies a vector. I guess I should have been more explicit and
referred to the components of the position 3-vector as matrices, but I
didn't anticipate this misinterpretation. I think the explicit
expression for the position operator, r=i I Del+kxS/|k|+ ... is clear.
In our papers r, k and S are bold to denote vectors and I is the 3x3
unit matrix.

the position vector must satisfy the standard commutation rules (3) with
the angular momentum, which is violated, as she says explicitly on p.5.

I don't agree that the standard commutation relations must be obeyed.
The position operator determines the localized basis states. These
localized states have definite total angular momentum in some
specified direction. But the coefficient of these basis vectors in the
expression for the wave function can modify this to give the correct
angular momentum for any physical state. The absolute value squared of
the coefficient of the localized state is the probability density to
detect a photon at r with the z-component of angular momentum of the
localized state. This is not so different from the case of the
electron where spin + or -1/2 is relative to some specified axis.

I suspect that the photon position operator will be a more
natural object as a density matrix / density operator
density.

There is a very nice blog that discusses our position operator at
http://carlbrannen.wordpress.com/2008/01/14/consistent-histories-and-density-operator-formalism/

margaret.hawton@xxxxxxxxxxxx
http://physics.lakeheadu.ca/facNstaff/hawton/hawton.html

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