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Why does the distributive law fail in quantum logic?

Recently, I learned that the distributivity law of logic does not always hold for physical systems, and on Wikipedia; I found the proof: https://en.wikipedia.org/wiki/Quantum_logic

We have: $p$ = "the particle has momentum in the interval $[0, +1/6]$"

$q$ = "the particle is in the interval $[−1, 1]$"

$r$ = "the particle is in the interval $[1, 3]$"

(using a system of units where the reduced planks constant is 1)

In this case, we might observe that

$$p\land(q\lor r)=T$$

(where T stands for true)

In other words, the particle's momentum is between 0 and +1/6, and its position is between -1 and +3.

However, the propositions $p\land q$ and $p\land r$ are both false. They assert tighter restrictions on simultaneous values of position and momentum than is allowed by the uncertainty principal.

So $(p\land q)\lor(p\land r)=\bot$

Thus the distributive law fails.


After reading this proof, I cannot see a reason why it is necessary for distributivity to fail. For instance; if it is insisted that the particle cannot have momentum in the interval $[0, +1/6]$, and be in $[−1, 1]$, then we may "and" our first proposition with this insistence ($\ref{1}$). Similarly, the same can be done for the interval $[1, 3]$.

$p\land (q\lor r)\land [(p\land q)\leftrightarrow \bot]\land[(p\land r)\leftrightarrow \bot]\label{1}\tag{1}$

$p\land q\land(\lnot r)\tag{2}$

($\ref{1}$) simplifies to ($\ref{2}$), which reads:

The particle has momentum in the interval [0,1/6], it is within [-1,1] and it is not in the interval [1,3]. $\label{2}\tag{2}$

($\ref{2}$) simplifies again to ($\ref{3}$):

The particle has momentum in the interval [0,1/6], it is in [-1,1) . $\label{3}\tag{3}$

($\ref{3}$) is a sentence without contradiction to empirical results. It was not necessary for distributivity to fail in ($\ref{3}$)'s case.

So here I ask; why is it necessary that the distributive law of logic fails in quantum logic? What have I misunderstood?

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    $\begingroup$ I can't get over the ambiguity of these statements. If "the particle has a momentum between [0,1/6]", have we made a measurement of absolute precision in momentum space? If not, do we assert we have a wavefunction which is precisely zero in momentum space outside this region? If that's the case, then both restrictions on position are false. You can get lots of contradictions and nonsense if you assume that a measurement has a result without actually performing the measurement, and it seems like that's what this calculation is touching on. $\endgroup$
    – user12029
    Feb 24, 2017 at 9:38
  • $\begingroup$ That particular statement is a quote from the page I got the proof from. So I am unsure exactly how that property of the particle is known. $\endgroup$
    – user400188
    Feb 24, 2017 at 9:44
  • $\begingroup$ @NeuroFuzzy I must ask though, why does a wave function which is zero in momentum space outside the region of [0,1/6] make the restrictions on position false? $\endgroup$
    – user400188
    Feb 24, 2017 at 9:46
  • $\begingroup$ it looks like some of the logical statement depends on an exact position of the particle which isnt possible to narrow down because of heisenberg uncertainty principle... $\endgroup$
    – vzn
    Feb 25, 2017 at 1:29
  • $\begingroup$ We are not assigning locations to the particle; we are assigning regions the particle may be in. For instance; it would be fine to say a particle is within the location (the entire universe). Similarly we can say it is somewhere within a location of [here, to here] so long as that location is larger than the uncertainty in position of the particle. $\endgroup$
    – user400188
    Feb 25, 2017 at 1:39

2 Answers 2

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You're not keeping your definition of "is in" consistent.

When you're evaluating $p \land (q \lor r)$, you're treating $q$ like "the position might be in [-1, 1]". But when you're evaluating $q \land r$ you're treating $q$ like "the position must be in [-1, 1]".

Note that your "contradiction" doesn't require the system to be quantum. It works just as well with a plain old probability distribution. Simply assert that the variable $X$ is a uniform random variable over [-1, 3]. Now define $q =$ "$X$ is in $[-1, 1]$" and $r = $ "$X$ is in $[1, 3]$". It seems like we have $q \lor r = \text{True}$, and yet clearly $q$ can't be right because $X$'s range spans over 4 units but $q$'s range only spans two units. The same for $r$. And so $q = \text{False}$ and $r = \text{False}$ and therefore $q \lor r = \text{False}$. Contradiction. But this apparent contradiction is simply because we failed to keep the meaning of "is in" consistent. Pick a definition and stick with it!

The error is a lot more obvious when using a probability distribution. I bet that "yet clearly $q$ can't be right because $X$'s range spans over 4 units but $q$'s range only spans two units" sounded a bit off, right? When you're more familiar with quantum mechanics, the leap from "the uncertainty principle says the standard deviation is wider than the range of $q$" to $\lnot q$ sounds just as strange.

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The problem has nothing to do with imprecise language or consistency of verbiage "is in", but turns on quantum logic refusing to honor the distributivity of classical logic.

For example, in fact, (p∧q∨r) = ((p∧q)∨(p∧r)), which is a tautology and theorem.

Using the modal logic model checker named Meth8/VŁ4, your equations are mapped as follows, where the designated proof value is T, and 16-value truth tables are row-major and horizontal to save space. Also, & is And, + is Or, = is Equivalent, @ is Not Equivalent, (p=p) is T, and (p@p) is F for contradiction.

(p&(q+r))=(p=p)     ; FFFT FTFT FFFT FTFT (0.1)
((p&q)+(p&r))=(p@p) ; TTTF TFTF TTTF TFTF (0.2)
(p&(q+r))&(((p&q)=(p@p))&((p&r)=(p@p))) 
                    ; FFFF FFFF FFFF FFFF (1)
(p&q)&~r            ; FFFT FFFF FFFT FFFF (2)
((p&(q+r))&(((p&q)=(p@p))&((p&r)=(p@p))))=((p&q)&~r)    
                    ; TTTF TTTT TTTF TTTT to mean that
                      (1) is not equivalent to (2), but almost.
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