Compatibility in Lattice of Propositions
John Forkosh
space H. As per [B&C, page 98 and following]
[B&C] The Logic of Quantum Mechanics, E.Beltrametti and
G.Cassinelli, Addison-Wesley 1981, ISBN 0-201-13514-0
(subsequently Cambridge U.P., ISBN 0-521-30235-8)
L is atomistic, with atoms A the one-dimensional subspaces of H,
such that \forall p \in L : p = V{ a \in A | a <= p }
where q <= r means q is a subspace of r, and V denotes join
which is the spanned subspace. My question concerns the
operational meaning of join, but let me frame it by first
reviewing the operational meaning of meet.
For p,q \in L, the meet p^q = q^p is the collection of
vectors in both p and in q, which is always a subspace of H.
So for any p,q \in L, we always have p^q = q^p \in L, too.
Physically, however, p^q is interpreted as the simultaneous
knowledge of p and q measurements, only possible when the
corresponding projection operators pq=qp commute.
And, of course, that happens iff p and q are simultaneously
diagonalizable, i.e., not always. So how do we operationally
interpret p^q \in L when p,q don't commute?
The answer to this question is explained in [B&C, page 188],
and if you notice the *footnote on that page, we'll discuss the
explanation provided by [Jauch, pages 75 and 38]
[Jauch] Foundations of Quantum Mechanics, J.M.Jauch,
Addison-Wesley 1968, 1973, ISBN 0-201-03298-8
We always operationally interpret p^q, whether or not pq=qp
commute, as the alternating sequence of filters
incident identically p-filter q-filter p-filter q-filter etc...
prepared test systems +--+ +--+ +--+ +--+
----------------------> | | | | | | | |
| | | | | | | |
----------------------> | | | | | | | | ......
| | | | | | | |
----------------------> | | | | | | | |
+--+ +--+ +--+ +--+
That is, algebraically, p^q = \lim_{n-->\infty} (pq)^n
(where, sorry, ^n on the rhs denotes exponent rather than meet).
If pq=qp happen to commute, you can rearrange terms so that
(pq)^n = (p^n)(q^n) = pq, reducing to the customary operational
interpretation. So that's all fine, i.e., we now understand what
p^q \in L means regardless of whether or not p,q commute.
Now I'd like to ask about the join. For example, when [B&C] say
\forall p \in L : p = V{ a \in A | a <= p }, how are we supposed
to operationally interpret pvq when p,q don't commute? To clarify
the operational intent of the question, suppose I'm in my lab with
apparatus corresponding to a filter that tests for (measures)
proposition p, and other apparatus that measures q. If someone
asks, "measure p^q", [Jauch] has told me how to do it -- construct
an alternating sequence of the apparatuses I already have. But if
they ask, "measure pvq", then exactly what am I supposed to
construct with the apparatus I already have when p,q don't commute?
--
John Forkosh ( mailto: j...@f.com where j=john and f=forkosh )
a student
> Now I'd like to ask about the join. For example, when [B&C] say
> \forall p \in L : p = V{ a \in A | a <= p }, how are we supposed
> to operationally interpret pvq when p,q don't commute? To clarify
> the operational intent of the question, suppose I'm in my lab with
> apparatus corresponding to a filter that tests for (measures)
> proposition p, and other apparatus that measures q. If someone
> asks, "measure p^q", [Jauch] has told me how to do it -- construct
> an alternating sequence of the apparatuses I already have. But if
> they ask, "measure pvq", then exactly what am I supposed to
> construct with the apparatus I already have when p,q don't commute?
You could use de Morgans rule (which holds
for quantum logics), i.e.,
a v b = ~( ~a ^ ~b ).
Hence, operationally measure the meet of
~a and ~b, and reverse the yes/no result.