Re: Hawton Position Operator. (Was: EM field of photon)

I have no idea which meaning could be assigned to matrix-valued
probabilities; neither is it clear to me why Hawton's approach
should yield such probabilities; the blog does not seem to give details.
...
On the other hand, if the probabilities for being in some region
of space are ordinary numbers (as I think they should be)


The post is rather clear on this. Rather than take traces, a
mathematically equivalent method of defining probabilities
for pure quantum states is to take them as the real multiples
of primitive idempotents (states). By doing this, one keeps
one's formalism entirely in matrix form. Probabilities are not
matrices, but ratios of matrices.

Perhaps this isn't sufficiently explicit. Working in a finite
Hilbert space, let B be a primitive idempotent. Then
BB = B and trace(B) = 1.

Let M be some matrix. Then BMB is also a matrix, and by
the properties of primitive idempotents, it will be a complex
multiple of B. Therefore, assign <M>_B to be this complex
multiple. This allows one to define quantum averages without
need of specifying "trace". Instead, trace is built into the
definition of "primitive idempotent".

In the formula for Pr(H) a few paragraphs before, the trace
is already missing after each equal sign. But the trace is necessary
(and it is indeed mentioned in the English context).
With the trace, the probabilities discussed there become
ordinary numbers; without the trace, the formulas simply become wrong:
The third equality would not be correct since, as the context states,
it uses the fact that  tr AB = tr BA.

You don't have to define probabilties with "trace" in order to
have the equivalent of tr(AB) = tr(BA). However, since
the alternative definition of probabilities gives the same
numbers as the trace, one can move back and forth between
the definitions however one wishes. Similarly, when one
is working in the density matrix formalism, it is frequently
very convenient to take results well known in spinor
formalism and use them.

ABA is a complex (real for Hermitian) multiple of A,
and BAB is a complex multiple of B. The statement
that tr(AB) = tr(BA) amounts to the observation that
the "complex multiples" involved (i.e. ABA / A and
BAB / B), are the same.

To prove this, without reference to trace is easy and
I will leave it as an exercise for the reader with one
hint: It might be useful to prove that the matrix product
of two primitive idempotents AB is either zero, or a
complex multiple of a (not necessarily Hermitian)
primitive idempotent. What do you suppose that
complex multiple would be?

Somewhere in here I should mention that density matrices
are particularly handy in QFT in that virtual particles are
density matrices. When one writes the formalism in density
matrix form, one unifies the formalism for the virtual and real
states.

I think the most useful applications of density matrices is
in deep bound states. In analyzing these, one wants to
have the initial and final states both be bound states.

The usual QFT makes the initial and final states free, but
for a deeply bound state they should instead be virtual states.
In other words, one needs to analyze deeply bound states by
ignoring the state vectors that one usually sandwhiches the
virtual states between, but instead one looks at the algebra
of the virtual states among themselves.

To do all this, it helps to look at things from a point of view
where the virtual particles are at the foundation and the real
particles are just approximations of virtual particles. And
that is why density matrices are useful to understand.

On the other hand, if you are dealing in the usual formalism
where the state vectors are fundamental, then it makes perfect
sense to define average values by <a| M | a>. And that naturally
turns into traces in the density matrix formalism. From the spinor
point of view, defining probabilities as ratios of matrices is rather
contrived, but from a density matrix point of view, the trace is
also a rather contrived object.

As is usual in QM, the different formalisms give the same
result. It's just easier in one formalism or another to set
up the problem.

.

Relevant Pages