# How do we know that there isn't a classical solution to the measurement problem/Quantum Mechanical uncertainty?

It was mentioned to me that it can be shown that there is no classical explanation for the uncertainty in Quantum Mechanics -- i.e. that there are no hidden workings that we have just not yet seen, which could be explained classically and would explain the probabilistic nature of Quantum events in a 'deterministic' fashion.

Can someone explain how this is known please?

-
See physics.stackexchange.com/q/35516. Esp comment by Lubos Motl. If that doesnt answer this maybe question not clear. – user12811 Apr 20 '13 at 2:32

While quantum mechanics might not be as weird as we used to think (classical wave-particle systems exhibit many quantum-like properties - see eg this article), there's a fundamental disconnect between quantum and classical theories and various no-go theorems that go along with it (Bell, Kochen-Specker, Greenberger–Horne–Zeilinger are probably the most famous ones).

Basically, we cannot accomodate entanglement and incompatible observables in a classical theory, and quantum logic is the attempt to abstract and formalize this in a mathematically precise way.

-
Agree! I mentioned you in my answer but I guess @Christoph links don't work across answers... – FrankH May 20 '13 at 19:55

Classical Mechanics is an abstraction. Simply put, in the reality out there there is only one kind of object, it is not the point-with-momentum of classical mechanics, it is the (for lack of a better word) Quantum.

A Quantum is described by a QM-wave function (as opposed to a particle described by two vectors). The QM-wave function is so called because it somewhat resembles the wave equation.

Wave-equation:

$$\frac{\partial^2 u(\mathbf{x}, t)}{\partial t^2} = c^2 \nabla^2 u(\mathbf{x}, t)$$

Where $u$ is a scalar field of potential energy, $\mathbf{x}$ is a vector, $t$ is time, $\nabla^2$ is the Laplacian operator and $c$ is a constant.

Schrödingers equation (QM-wave equation, single non-relativistic quantum, no potential energy):

$$i \hbar \frac{\partial \Psi(\mathbf{x}, t)}{\partial t} = \frac{-\hbar^2}{2 m} \nabla^2 \Psi(\mathbf{x},t)$$ Where $\Psi$ is the complex-numbered amplitude field, $m$ is the mass of the quantum, $i$ is the imaginary unit and $\hbar$ is Planck's constant.

Notice how the structure is similar, barring the order of the partial differentiation. The thing is you can take the squared-modulus of the QM-wave function:

$$| z |^2 = z \bar z \quad \bar z = \Re z - \Im z$$

This gives you a positively-real-numbered field, which looks even more like the wave equation's behaviour.

Now, the thing to realize is that a wave doesn't consistenly have a "position" and a "momentum". A plane wave only has momentum, while a wave packet has a position. Minutephysics video on the topic.

So you cannot "cheat" the Heisenberg-principle. And you shouldn't even try, because there literally isn't any "particles" with "momenta" out there in reality.

-

The no-go theorems (Bell, free-will, etc.) do not exclude deterministic theories. They require the existence of at least some objects with non-deterministic behavior, leaving the deterministic hidden-variable theories outside of their scope. The authors of these theorems admit this fact.

Once this assumption is rejected, no incompatibility between QM and classical determinism remains.

-

The final and undeniable proof that there cannot be a classical physics explanation of quantum mechanics will be when a real quantum computer has been constructed and is able to implement and compute a quantum algorithm in a time faster than would be possible for a classical computer. An example of such an algorithm is Grover's Algorithm:

Grover's algorithm is a quantum algorithm for searching an unsorted database with N entries in $O(N^{1/2})$ time and using $O(log N)$ storage space ... In models of classical computation, searching an unsorted database cannot be done in less than linear time (so merely searching through every item is optimal). Grover's algorithm illustrates that in the quantum model searching can be done faster than this; in fact its time complexity $O(N^{1/2})$ is asymptotically the fastest possible for searching an unsorted database in the quantum model. It provides a quadratic speedup, unlike other quantum algorithms, which may provide exponential speedup over their classical counterparts. However, even quadratic speedup is considerable when N is large.

This demonstration of quantum mechanics will be far more convincing than the other interference and entanglement experiments and no-go theorems that Christoph mentions in his answer:

...there's a fundamental disconnect between quantum and classical theories and various no-go theorems that go along with it (Bell, Kochen-Specker, Greenberger–Horne–Zeilinger are probably the most famous ones).

Current quantum computers are very primitive and have not been able to do any significant computation and certainly cannot currently beat our very fast classical computers since the quantum computers have very few quantum bits and have very slow cycle times. However a quantum computer with a very slow cycle time implementing Grover's algorithm can beat a faster classical computer if you have enough quantum bits and give it a large enough $O(N)$ problem.

Currently there are many technological problems with building a quantum computer. If we are able to overcome all these known problems and we build a quantum computer that we are sure should work and we find that it does not work, then we will know that there is something fundamentally wrong with quantum mechanics and will have the fun of finding the new theory that explains why quantum computers won't work and still gives all the other well tested results that quantum mechanics predicts.

-
I think there are several problems with this answer. (1) It's an open problem whether factorization is outside the P complexity class. Although it's suspected that it is, there is no proof. (2) You can't prove by experiments what complexity class a problem is in, because the complexity class depends on the limiting behavior. (3) Quantum computers have already been built. (4) Inability to build more complex quantum computers could be due to practical problems, not the failure of this interpretation of quantum mechanics. – Ben Crowell May 20 '13 at 20:39
@BenCrowell I made a mistake in using the factoring problem. I should have used any of the problems where there is a known classical computer complexity class. The current quantum computers are very trivial - factoring 15 for example. A real quantum computer that does a non trivial problem faster than current classical computers will be non controversial and very convincing. I will leave another comment when I have updated the answer and perhaps you will take away the down vote. – FrankH May 20 '13 at 22:24

The answer is simple: classical mechanics is macroscopic (Planck's constant zero) and quantum mechanics is microscopic (Planck's constant not zero). It is a myth that any hidden variable theory that underpins quantum mechanics must be classical. They cannot be. Local realism uses non-classical ideas.

Locality is not a quantum notion of course but is defined as:

Locality: when two particles have separated beyond the range of any of the four forces, their states form a product

Reality is a microscopic property:

Reality: a microscopic system is described only by pure states and all the pure states are simultaneously dispersion free.

I am about to submit a paper to Phys Rev A on the reconciliation of the EPR paradox, By making a small, but physically quite different, change to the definition of spin when it is isolated, then a simulation using this local realistic spin accounts for the EPR correlation without persistence of entanglement (without non-locality)

http://quantummechanics.mchmultimedia.com/

So major points that people tend to forget:

quantum mechanics is a theory of measurement: spin changes when it is measured.

Entanglement is a property of quantum mechanics but not of Nature

Local Hidden Variables are not classical.

-

## protected by Qmechanic♦Apr 20 '15 at 0:25

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.