Talk:Quantum logic

Introduction needs clarification

Particularly, I think the statements q = "the particle is in the interval [-1,1]" and r = "the particle is not in the interval [-1,1]" need to be clarified. What does it mean the particle is in interval [-1,1]? Does it mean that the amplitude of the wavefunction squared integrates to non-zero over this interval? Or does it mean that we have measured whether the particle is there or not and found it is in [-1,1]?

The other thing is that p = "the particle is moving to the right" is also not clear. I imagine that this means that the particle is in a momentum eigenstate such that it is moving to the right.

And one last thing, if (p and q) or (p and r) = false because "they assert tighter restrictions on simultaneous values of position and momentum than is allowed by the uncertainty principle" then it is clearly not true that the particle is in the particular momentum eigenstate and so p = false, which means p and (q or r) = p = false and thus (q or r) = (p and q) or (p and r). So distributive law is OK. Meznaric (talk) 12:26, 25 August 2009 (UTC)

I agree with the above. The example author's logic was flawed. He seemed to assume that in the p and (q or r) case, that or was exclusive, and yet for (p and q) or (p and r), it was suddenly inclusive. Therefore I deleted, as there was no justification to keep it in there. 153.104.46.187 (talk) 21:48, 9 October 2009 (UTC)

The introduction's counterexample is completely wrong. For example q = "the particle is in the interval [−1, 1]" does not make sense when talking about physics: 1) On a classical sense: it allows the particle to be in a set of measure zero. 2) On a quantum mechanical sense: By the Heisenberg principle, the position can only known up to an error (variance) given by the theorem. A correct statement would be : q = "the particle is in an open sub-interval of the interval [−1, 1]", in which case there is no contradiction and the distributive law is (of course ) OK. Where does that counterexample come from??

Why I renamed it "Quantum Logic"

I've read a lot about this subject and it's always called "quantum logic", never "quantum reason". For evidence, try some of the papers and books in the References for this article, starting with the Birkhoff--von Neumann paper cited at the beginning. As the article notes, they called this subject "quantum logic" --- not "quantum reason". --John Baez 11:01, 5 September 2007 (UTC)

Revert to older version

Something screwed up and I had to revert to an older version.

I removed the category: theorems since quantum logic isn't a theorem. user: Gene Ward Smith

Disambiguation

"Quantum logic" first makes me think of the processing elements needed for Quantum computers. David R. Ingham 05:48, 24 October 2005 (UTC)

As used in the article, it's standard. --CSTAR 13:11, 24 October 2005 (UTC)

Article history

What happened to the revision history of this article? --CSTAR 15:39, 12 October 2007 (UTC)

This appears to be the last version with the full edit history [1]. Subsequently, the edit history got obliterated by what seems (to me at least) to be a copy-and-paste rename rather than a move.--CSTAR 16:40, 12 October 2007 (UTC)

Whoever is deleting the material on the limitations of quantum logics concerned with super-geometry and non-commutative geometry please stop! If you don't understand the comment, contact me: dedwards@math.uga.edu

Failure to consider all the possibilities

When someone wrote…

   q = "the particle is to the left of the origin"
   r = "the particle is to the right of the origin"
   then the proposition "q or r" is true, so

… [s]he ignored the possibility that the particle is _at_ the origin, at least in the left/right sense of "at". —Preceding unsigned comment added by 66.114.69.71 (talk) 20:15, 20 July 2008 (UTC)

This is true but it doesn't really matter, since the chance of being "at" the origin is zero. More technically, in the lattice of projections on L^2(R), the projection onto functions supported at the origin is the same as the zero projection, so we identify the corresponding propositions.

In short, we should only change what's written if it seems likely to be confusing to most nonexperts, and if so, we should change it in some minimal way, e.g. by saying r = "the particle is at or to the right of the origin". John Baez (talk) 20:22, 1 June 2009 (UTC)

Can the Rules of Quantum Logic be Listed Simply?

The article begins

In mathematical physics and quantum mechanics, quantum logic is a set of rules for reasoning about propositions which takes the principles of quantum theory into account.

and then

Quantum logic can be formulated as a modified version of propositional logic.

And then an example of one of its properties is given. OK, so rule one:

1. p and (q or r) ≠ (p and q) or (p and r)

2. ...

Is rule number two?

p and (q or r) ≠ (q or r) and p

What are the other rules please? I mean for fuzzy logic, for example, instead of a proposition just being true or false and having a value 1 or 0, it can have any value between 1 and 0, so to AND you take the smallest value of two propositions, and to OR you take the largest, and to NOT you just take the value of the proposition away from one. Simple. And that's all you need to know to start playing around with fuzzy logic.

So how do you do quantum logic?

--Vibritannia (talk) 12:23, 17 August 2008 (UTC)

deep inference

I added a sentence about application of System BV (a deep inference logic) to quantum evolution, since I came across it while reading about proof theory and it sounded sort of related to this article, but I actually don't understand it at all, so feel free to revert it if it's insufficiently relevant. 207.241.239.70 (talk) 06:43, 16 November 2008 (UTC)

I think it's only marginally related. --CSTAR (talk)

errors in lead

The physics of the example given in the lead seems to have been oversimplified to the point where it's incorrect. There's nothing in the Heisenberg uncertainty principle that prohibits us from knowing both p (the particle is moving to the right) and q (the particle is to the left of the origin). We can know p while still having an arbitrarily large uncertainty in momentum. We can know q while still having an arbitrarily large uncertainty in position. It's certainly possible to have the product greater than Planck's constant. The argument in the lead could probably be reformulated correctly, but as it stands it's just plain wrong.

I think the lead also oversimplifies the logical issues. Even if p and q were reformulated correctly so that it would violate the Heisenberg uncertainty principle for them both to be simultaneously verified as true with 100% probability, it's certainly possible to have a wavefunction for which the p-and-q operator has a value of true, with some probability. So I don't think it's correct to say that p-and-q is false. The operator p-and-q just doesn't have eigenstates whose eigenvalues are true and false.--76.167.77.165 (talk) 19:03, 21 November 2008 (UTC)

I believe a valid version of the example is as follows: Consider a particle on the real axis. Let q be the proposition that the position of the particle is in the bounded interval [-1,1] and the proposition p is as before: i.e. the particle moves to the left. Then for any wave function ψ the proposition q is true precisely when ψ is supported in [-1,1]. By one version of the Paley-Wiener theorem, this implies that the Fourier transform of ψ (the wave function in momentum space) is real analytic and hence cannot vanish in any non-trivial interval; in particular it cannot vanish on either the set of positive or the set of negative numbers. Therefore neither the proposition p nor its negation, not p is true. The rest of the example remains as it s currently.

Does this seem reasonable to you? --CSTAR (talk) 16:33, 23 November 2008 (UTC)

PS. What follows is a formal justification for the claim made earlier concerning violation of the distributive law:

1. p and q = 0
2. not p and q = 0
3. p or not p = 1

where propositions are considered as orthogonal projections on L²(R).

So this example definitely shows that the distributive law fails.

Though the claims 1) and 2) are showm by a form of the Paley-Wiener theorem, I think it is fair to say it is a form of the Heisenberg uncertainty principle.

--CSTAR (talk) 17:25, 23 November 2008 (UTC)

PPS. To say a proposition p about a quantum system (ie a projection) is true in a state ψ means

$\langle \mathbf {p} \psi \mid \psi \rangle =1$

--CSTAR (talk) 19:58, 23 November 2008 (UTC)

One thing that this leaves me unclear about is whether truth and falsehood are then only defined on a per-wavefunction basis. The example in the lead doesn't start by assuming any particular state, it starts by assuming a particular space of states that are solutions to a particular wave equation.--76.167.77.165 (talk) 20:44, 23 November 2008 (UTC)

Another thing that isn't said explicitly in this definition is whether p=false is defined as $\langle \mathbf {p} \psi \mid \psi \rangle =0$, or as $\langle \mathbf {p} \psi \mid \psi \rangle \neq 1$. I think you intend the latter, which treats true and false asymmetrically, but preserves the law of the excluded middle.--76.167.77.165 (talk) 21:07, 23 November 2008 (UTC)

Hi -- Thanks for the reply. Your reformulation seems intuitively reasonable to me (I haven't dug into Paley-Wiener), but IMO it would be preferable to have an example that a reader can understand without having to know any more than the Heisenberg uncertainty principle. How about this. Let a particle be confined in a one-dimensional box with infinitely hard walls, i.e., V(x)=0 in [-1,1], infinity elsewhere. Let p be the proposition that the particle is in the ground state, let q be the proposition that the particle is in [-1,0], and let r be the proposition that it's in [0,1]. Then (q or r) is true, because the particle has unit probability of lying in [-1,1]. But 1 isn't an eigenvalue of (p and q), since such an eigenstate would violate the uncertainty principle. (The ground state has velocities lying within a certain range. That range is too narrow to be consistent with a position-uncertainty of 1.)

I'm not really clear on what you're saying in your P.S., which addresses the second point I originally made. It seems to me that we're talking about the issue of how to connect quantum-mechanical states and operators with logical propositions. Later in the article, we have "Measurement of f yields a value in the interval [a, b] for some real numbers a, b," which is less technical in its formulation, but seems a little ambiguous to me. Suppose we have operators A, B, and C. Suppose that every state is an eigenstate of A with eigenvalue 1, every state is an eigenstate of B with an eigenvalue of 0, and C has some eigenvalues that are 0 and some that are 1. Then clearly A=1 is true, and B=1 is false. By the definition used in this article, it sounds to me like C=1 is also considered false. Is that correct? If so, then maybe this article should be broadened a little to discuss other nonclassical logics that have been applied to quantum mechanics. E.g., An Introduction to Non-Classical Logic, by Graham Priest, p. 125, mentions that "Other examples of truth-value gluts that have been suggested include ... certain statements about micro-objects in quantum mechanics." Here a truth-value glut refers to a proposition that is both true and false, violating the law of excluded middle. This is in contrast to a truth-value gap, in which neither P nor not-P holds. To me, an operator like C sounds like an example that would most naturally be described using a truth-value glut; simply calling C false seems like a strangely asymmetric way of treating truth and falsehood, but maybe the compensating advantage of such a definition is that it preserves the law of the excluded middle?

I'm also wondering how the distributive law relates to other aspects of nonclassical logic. The WP article distributive law doesn't specifically mention the case of Boolean logic, and Classical logic doesn't discuss the distributive law. The index of Priest doesn't have an entry for "distributive law." I wonder whether violation of distributivity is more often referred to by some other term, or whether violation of distributivity is strictly equivalent to some other logical rule, with the other rule being the one more commonly referred to in the literature...?--76.167.77.165 (talk) 20:08, 23 November 2008 (UTC)

It looks like we were both editing at the same time, and your PPS came in at the same time as my edit. To me, it sounds like your PPS confirms my interpretation phrased in terms of the operators A, B, and C...?--76.167.77.165 (talk) 20:14, 23 November 2008 (UTC)

BTW, Priest also has some interesting remarks about non-classical logic as applied to the Bohr model. This is the kind of thing that I think would be valuable to mention in a WP article for a general readership, even though a specialist might just pooh-pooh the Bohr model.--76.167.77.165 (talk) 20:16, 23 November 2008 (UTC)

Reply I'll have to think more carefully about your other comments, but in regard to the complexity of the example, I think it is sufficient to refer to the Heisenberg uncertainty principle. My reference to Paley-Wiener was to prove conclusively to myself that p and q is false, but this is intuitively true (at least to me) just by relying on Heisenberg. I don't think it would be a good idea to put to put so many details in an example the lead.

I disagree. IMO the lead is simply incorrect as it stands. There's nothing in the example as stated that violates the uncertainty principle.--76.167.77.165 (talk) 04:29, 27 November 2008 (UTC)

In regard to this

Suppose we have operators A, B, and C. Suppose that every state is an eigenstate of A with eigenvalue 1, every state is an eigenstate of B with an eigenvalue of 0, and C has some eigenvalues that are 0 and some that are 1. Then clearly A=1 is true, and B=1 is false.

Your operator C is not identically false: in some states it is false and in some it is true and yet others it has a probability between 0 and 1. --CSTAR (talk) 00:01, 24 November 2008 (UTC)

Hmm...so you're talking about a logic that has truth-value gluts, and doesn't obey the law of excluded middle? That really isn't clear in the article.--76.167.77.165 (talk) 04:29, 27 November 2008 (UTC)

Reply I think this is a misunderstanding of the truth-value semantics of quantum logic. Even in classical propositional calculus some formulas are true regardless of truth value assignments to its atomic constituents (where such a truth value assigment can be regarded as a state of the world so to speak), some formulas are false regardless of the truth value assigment and some fall in between. One notable difference (among many) is that in quantum logic, for each state of the world every "proposition" has a numerical probability. This is definitely not the case in classical propositional logic. Probabilities can be assigned but not in any canonical way.

As regards your comment on the uncertainty principle, at least in my formulation using the Paley Wiener theorem it is arguably a form of the uncertainty principle. --CSTAR (talk) 05:12, 27 November 2008 (UTC)

PS I don't agree that the lead is as bad (or wrong) as you suggest. Actually I don't know for sure that the example currently there is actually wrong, but the modification I suggested does work at least as a counterexample to the distributive law; we seem to disagree whether it's because of something related to HUP. I realize now that you are making some other suggestions about the differences between classical and quantum logic which should be in the article.--CSTAR (talk) 06:38, 27 November 2008 (UTC)

I think in general what's probably going on is that we have a couple of very large fields of study (nonclassical logic and the interpretation of quantum mechanics) that have been studied for generations, attacked from many different angles, and discussed within a variety of foundational frameworks. The article seems to be describing one very specific approach within a much larger field of quantum logic and nonclassical logic. The difficulties this raises are, I think, that (1) the article doesn't acknowledge its narrow focus, and (2) because of its narrow focus, the article makes foundational assumptions that will not be understood by people who aren't acquainted with the particular, specific approach it describes. E.g., I have a PhD in physics, and have a general acquaintaince with nonclassical logic from reading the book by Priest, but the hidden assumptions behind some of the article's casual use of terminology aren't at all clear to me. I can't tell whether it's assuming the law of the excluded middle. I can't tell whether it assumes truth-value gluts. I can't tell whether the definition of p=true as $\langle \mathbf {p} \psi \mid \psi \rangle =1$ is meant to imply the definition of p=false as $\langle \mathbf {p} \psi \mid \psi \rangle =0$, or as $\langle \mathbf {p} \psi \mid \psi \rangle \neq 1$. In the lead, we have On the other hand, the propositions "p and q" and "p and r" are both false, since they assert tighter restrictions on simultaneous values of position and momentum than is allowed by the uncertainty principle. Now the premise stated in the second clause of this sentence is literally incorrect, although you claim that it's correct in spirit, but I think the conclusion stated in the first clause is also another example of the lack of appropriate attention to foundational issues. I think it's true that there is no wavefunction for which $\langle (\mathbf {p} {\text{and}}\mathbf {q} )\psi \mid \psi \rangle =1$. On the other hand, there are many wavefunctions for which $\langle (\mathbf {p} {\text{and}}\mathbf {q} )\psi \mid \psi \rangle$ is greater than 0 and less than 1. There may be some specific foundational framework in which this leads to "(p and q)=false", but the article doesn't spell out what that framework is, and doesn't show an awareness that the choice of such a framework is a nontrivial one.--76.167.77.165 (talk) 18:42, 27 November 2008 (UTC)

Reply Re

On the other hand, there are many wavefunctions for which $\langle (\mathbf {p} {\text{and}}\mathbf {q} )\psi \mid \psi \rangle$ is greater than 0 and less than 1.

As a matter of fact, for the example I gave of p and q there are no such vectors ψ. The range of the projections p and q have only the vector 0 in their intersection. That's the reason I invoked the Paley Wiener theorem, to rigorously support my claim: The Fourier transform of any distribution of compact support has an entire extension (defined and analytic on all of C). In particular its zero set of the Fourier transform cannot have any accumulation points. In particular, in my version of Baez' example the set of points in momentum space where the wave function vanishes has no accumulation points.

As to what is comonly called "quantum logic" I think the article takes a fairly conventional approach, using standard terminology from well-known sources such as works by von Neumann, Mackey, Vardarajan, Piron, d'Espagnat and others.--CSTAR (talk) 19:22, 27 November 2008 (UTC)

I think the introduction is more or less fine as written, since it gives an example that correctly illustrates the nondistributivity of 'and' over 'or' which is characteristic of quantum logic. There are indeed no states for which both the wavefunction and its Fourier transform are nonzero only on one half of the real line, thanks to Paley--Wiener. John Baez (talk) 20:26, 1 June 2009 (UTC)

Please take my point of view as someone who knows nothing about quantum logic, but something about propositional, predicate and modal logics. Also I do not know nothing about quantum mechanics.

At the beginning article I do not see why two names were chosen for an assertion and it's negation, instead of:

$p\land (q\lor r)\equiv (p\land q)\lor (p\land r)$

you could write

$p\land (q\lor \neg q)\equiv (p\land q)\lor (p\land \neg q)$

What about the time in this example? a particle is moving always to right and may be or not within the $[-1,1]$ interval? other interpretation is: after moving to right it is or not within the $[-1,1]$ interval.

If the last is the intended meaning, why it is not stated in some temporal logic.

This part is not clear to someone with my background.

I know that fuzzy logic may be seen as a multivalued logic.

I know that different true value distributions exists, it seems the connection with quantum logic is that the true distribution depends on the probability function of the particle position, but I can't see why a new logic is needed, if a fuzzy logic is enough.

I think that it would be more clear to start with the difference between our common experience and the quantum world.

Why conventional logic can not model the quantum world, giving just an informal narrated example (no equations yet, please.)

Then you can show how Propositional Logic may be extended to support the behavior of the quantum world.

What the introduced operators mean, and the rules of the logic.

Then a more formal support for this may be presented.

after that, the multivalued version may be explained in the same style.

As I told you, I am not an expert in the subject, I am actually a reader searching about quantum logic. As such kind of reader I think that the suggested style would be more easy to read for people like me or even with less background. —Preceding unsigned comment added by Elias (talk • contribs) 03:51, 7 June 2009 (UTC)

Reply to Elias: Indeed, instead of writing

$p\land (q\lor r)\equiv (p\land q)\lor (p\land r)$

in the first example, we could write

$p\land (q\lor \neg q)\equiv (p\land q)\lor (p\land \neg q)$

Whatever makes it clearest! We're talking about a failure of the distributive law here, and the first equation is precisely the distributive law, so that's the advantage of using that. But we're doing an example where $r$ is $\neg q$.

As for your question "What about the time in this example?", time plays no real role here: the propositions refer to a particle at a particular but unspecified ***instant*** of time, just like the proposition "it is raining" in ordinary logic. Note that in quantum mechanics, a particle can have a specific momentum even at an instant of time: we don't need to wait and see how it moves.

I agree that it would be nice to write a more leisurely, expository article. There are many issues that need to be explained. I don't have time for this myself, alas.

It's probably wise to assume some familiarity with quantum mechanics in this article, since quantum logic is a rather advanced topic in quantum mechanics, at least in the usual scheme of things. John Baez (talk) 17:13, 20 July 2009 (UTC)

Do quantum mechanics actually break the distributive law? An argument against it.

Considering the illustrative example - how the distributive law fails, I've found that there is another logical solution: The distributive law might be not violated.

In constructive logic disjunction "q or r" is not a logical tautology. To recieve evidence, that it's true, we should have at our disposal a procedure to verify "q" or "r". From the quantum mechanical point of view this procedure is nothing more than a position meter. From the constructive point of view we aren't allowed to substitute this verifing device for nothing (as we did it in classical logic). But the uncertainty principle states, that if we have a position meter in an experimental scheme, then we cannot accurately measure a momentum of the object, i.e. "p" cannot be unambiguously verified. Thus, conjunction "p and (q or r)" can never be resolved in constructive logic owing to uncertainty principle: To resolve a conjunction we should resolve both its parts. For the same reason, "p and q" and "p and r" are both unresolvable.

So, constructive logic, which respects distributive law, treats this example as NOT contradictory.

Eugepros (talk) 13:02, 29 July 2010 (UTC)

Quantum logic has some properties which clearly distinguish it from classical logic, most notably, the failure of the distributive law of propositional logic:

p and (q or r) = (p and q) or (p and r),

where the symbols p, q and r are propositional variables. To illustrate why the distributive law fails, consider a particle moving on a line and let

p = "the particle is moving to the right"

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

r = "the particle is not in the interval [-1,1]"

then the proposition "q or r" is true, so

p and (q or r) = true

On the other hand, the propositions "p and q" and "p and r" are both false, since they assert tighter restrictions on simultaneous values of position and momentum than is allowed by the uncertainty principle. So,

(p and q) or (p and r) = false

Thus the distributive law fails.

In this fragment there are confused spaces of events and probabilities. True is a probability, which is equal 1. False is a probability, which is equal 0. If you write the example correctly you'll see that the distributive law doesn't fail. — Preceding unsigned comment added by 76.10.183.181 (talk) 18:52, 24 February 2013 (UTC)