Does QM unequivocally violate the law of bivalence?

Question

I had heard that QM violates the law of bivalence.

Does anyone claim that?

Answer 1

The words are very dangerous as the "principle of bivalence" is unavoidable in any science or system using logic – any thinking that is "rational" in this sense.

However, I would agree with the statement when carefully interpreted. The correct interpretation of the statement is that quantum mechanics forbids one to assume that either $P$ or $¬P$ is true at an intermediate moment, in the absence of an actual measurement that decides which of the options is realized, and one may always compute the probabilities of histories by summing the probabilities of the histories that include $P$ or $¬P$ at the intermediate moment.

To be specific, calculate the probability that $A_j$ at the beginning evolves to $B_k$ at the end. In the middle of the time interval, we may classically have $P$ or $¬P$. Classically, we could assume $$ P(A_j\to B_k) = P(A_j\to P \to B_k) +P (A_j\to ¬ P \to B_k)=\\ =P(A_j\to P)P(P \to B_k) +P (A_j\to ¬ P)P(¬ P \to B_k) $$ You may imagine that the experiment is the double slit experiment, $A_j$ is the particle emitted from a "cannon" while $B_k$ is the particle detected somewhere on the photographic plate. $P$ is the particle going through the left slit and its negation is the right slit.

However, quantum mechanically, this reasoning – which follows from the "framework of classical physics" and the corresponding "principle of bivalence"– is incorrect.

In particular, if $P$ and $¬P$ are two possible intermediate states of the physical system (let's assume it is a qubit, and the full Hilbert space is a tensor product of this qubit and something else), the correct law is actually similar to equation above but it uses the probability amplitudes $c$. We must first sum the complex probability amplitudes and then square the result to obtain the probability. And that's why the double slit experiment produces an interference pattern, among other things.

When we square this equation with the probability amplitudes and realize that probabilities $P=|c|^2$ by the Born rule, we will "almost" get the classical formula (my only displayed formula so far) but there will be extra mixed terms on the right hand side.

If I simplify, the classical reasoning with the "principle of bivalence" would lead us to write the total probability of the transition as $$P_P + P_{¬P} = |c_P|^2 + |c_{¬P}|^2$$ where the subscript indicates whether $P$ or $¬P$ was right in the intermediate state. But the correct actual quantum mechanical formula is actually $$|c_P + c_{¬P}|^2=\dots$$ which may also be written as $$|c_P|^2 + |c_{¬P}|^2 + 2\,{\rm Re}(c_P^* c_{¬P})$$ The last term is the mixed term i.e. the interference term. And if one is very careful about the interpretation of the statement, one could say that this mixed term exactly quantifies how much Nature deviates from the principle of bivalence.

However, to say such a thing safely, one must really understand quantum mechanics correctly – and also know why quantum mechanics basically agrees with classical physics in ordinary situations, and so on. If the statement about the "violation of the principle of bivalence" is used in the context of the people who don't really think quantum mechanically, they will unavoidably misinterpret it in a way that isn't sensible at all. And if they will believe the violation, they will abandon the good old rule of logic and keep everything else – and this will lead them to wrong conclusions and an inconsistent set of axioms. But the only correct way is to replace the whole classical reasoning about Nature by a different one, the quantum mechanical one, not just to "damage" some isolated rules.

Quantum mechanics isn't some inconsistent injured crazification of classical physics. It is (at least) an equally consistent but inequivalent framework as classical physics and its rules and axioms are as careful as those in classical physics, just different in content. A straightforward denial of the principle of bivalence in the classical reasoning wouldn't be "equally careful".

As a minimum, I would clarify the statement by saying that "quantum mechanics refutes the principle of bivalence for statements about observable quantities that were/are actually not being measured by any observer". When the measurements are made, it is always possible to assume the principle of bivalence for the possible results of such measurements.

P.S.: The equation with the sum of two probabilities that I referred to as a consequence of the "law of bivalence" may be given other names in the probability calculus. In their famous 1936 paper "The Logic of Quantum Mechanics" which has over 2,000 citations, Birkhoff and von Neumann discuss the violation of a closely related identity of mathematical logic in quantum mechanics and call it a violation of the distributive law in logic. But the source of the novelty of quantum mechanics relatively to the classical expectations about logic (expressed for probabilities) is always the same, namely the interference of the probability amplitudes.

Answer 2

There have been attempts to make sense of quantum mechanics by developing a so-called "quantum logic"; this goes back to von Neumann in the thirties.

These approaches are discussed in Quantum Logic and Its Role in Interpreting Quantum Theory (Tabia, 2010).

Two-valued logic is being questioned when one tries to formulate logical propositions involving non-commuting observables. This is clearly impossible since complementary observables do not have values defined at the same time.

An example from the above paper is a particle with a spin measured as $up$ along $z$ which could be said to have a spin either $left$ or $right$ along $x$, but at the same time could not be said to have either spin $up$ and $right$ or $up$ and $left$ (since once measured along $x$, the $z$ spin gets undefined and can be $up$ or $down$).

We end up with ($up$ and ($left$ or $right$)) being different from (($up$ and $left$) or ($up$ and $right$)))

Edit: As pointed out in the comments, once it is seen that ($up$ and $left$) is not an observable the question is not so much about how to logically articulate propositions but better about which propositions actually relate to anything real.