Editing Quantum logic (section)

== Differences with classical logic ==
The structure of ''Q'' immediately points to a difference with the partial order structure of a classical proposition system.  In the classical case, given a proposition ''p'', the equations
: ⊤ = ''p''&or;''q'' and
: ⊥ = ''p''&and;''q''
have exactly one solution, namely the set-theoretic complement of ''p''.  In the case of the lattice of projections there are infinitely many solutions to the above equations (any closed, algebraic complement of ''p'' solves it; it need not be the orthocomplement).

More generally, [[Valuation (logic)|propositional valuation]] has unusual properties in quantum logic.  An orthocomplemented lattice admitting a [[Total function|total]] [[Lattice Homomorphism|lattice homomorphism]] to {{mset|⊥,⊤}} must be Boolean.  A standard workaround is to study maximal partial homomorphisms ''q'' with a filtering property:
: if ''a''≤''b'' and ''q''(''a'') = ⊤, then ''q''(''b'') = ⊤.{{sfn|Bacciagaluppi|2009}}

=== Failure of distributivity ===
Expressions in quantum logic describe observables using a syntax that resembles classical logic.  However, unlike classical logic, the distributive law ''a'' ∧ (''b'' ∨ ''c'') = (''a'' ∧ ''b'') ∨ (''a'' ∧ ''c'') fails when dealing with [[Observable#Incompatibility of observables in quantum mechanics|noncommuting observables]], such as position and momentum.  This occurs because measurement affects the system, and measurement of whether a disjunction holds does not measure which of the disjuncts is true.

For example, consider a simple one-dimensional particle with position denoted by ''x'' and momentum by ''p'', and define observables:
* ''a'' — |''p''| ≤ 1 (in some units)
* ''b'' — x ≤ 0
* ''c'' — x ≥ 0

Now, position and momentum are Fourier transforms of each other, and the [[Fourier transform]] of a [[square-integrable]] nonzero function with a [[compact support]] is [[Entire function|entire]] and hence does not have non-isolated zeroes.  Therefore, there is no wave function that is both [[Normalizable wave function|normalizable]] in momentum space and vanishes on precisely ''x'' ≥ 0.  Thus, ''a'' ∧ ''b'' and similarly ''a'' ∧ ''c'' are false, so (''a'' ∧ ''b'') ∨ (''a'' ∧ ''c'') is false.  However, ''a'' ∧ (''b'' ∨ ''c'') equals ''a'', which is certainly not false (there are states for which it is a viable [[quantum measurement|measurement outcome]]).  Moreover: if the relevant Hilbert space for the particle's dynamics only admits momenta no greater than 1, then ''a'' is true.

To understand more, let ''p''<sub>1</sub> and ''p''<sub>2</sub> be the momentum functions (Fourier transforms) for the projections of the particle wave function to ''x'' ≤ 0 and ''x'' ≥ 0 respectively.  Let |''p''<sub>i</sub>|↾<sub>≥1</sub> be the restriction of ''p''<sub>i</sub> to momenta that are (in absolute value) ≥1.

(''a'' ∧ ''b'') ∨ (''a'' ∧ ''c'') corresponds to states with |''p''<sub>1</sub>|↾<sub>≥1</sub> = |''p''<sub>2</sub>|↾<sub>≥1</sub> = 0 (this holds even if we defined ''p'' differently so as to make such states possible; also, ''a'' ∧ ''b'' corresponds to |''p''<sub>1</sub>|↾<sub>≥1</sub>=0 and ''p''<sub>2</sub>=0). Meanwhile, ''a'' corresponds to states with |''p''|↾<sub>≥1</sub> = 0. As an operator, ''p'' = ''p''<sub>1</sub> + ''p''<sub>2</sub>, and nonzero |''p''<sub>1</sub>|↾<sub>≥1</sub> and |''p''<sub>2</sub>|↾<sub>≥1</sub> might interfere to produce zero |''p''|↾<sub>≥1</sub>.  Such interference is key to the richness of quantum logic and quantum mechanics.