All differences between classical physics and quantum physics are consequences of the uncertainty principle i.e. of the nonzero commutators between observables.Well, the statement above is true and important. The misunderstanding of this statement – often arrogantly masked as a different opinion, one that may be presented assertively – may be considered one of the defining characteristics of the anti-quantum zealots.

The anti-quantum zealots are inventing lots of "additional" differences between classical and quantum physics – such as the purported "extra nonlocality" inherent in quantum mechanics, or some completely new "entanglement" that is something totally different than anything we know in classical physics, or other things. Except that all these purported additional differences are non-existent.

The reason why they're inventing these non-existent differences is that they would like to rephrase quantum mechanics as "another classical theory with lots of different details" (by details, we mean the set of observables and the dynamical equations of motion etc.). But that's not what quantum mechanics is. Quantum mechanics is a fundamentally different theory which may have the same details as the classical counterpart, however. ;-)

The discrimination of classical physics and quantum physics may really be described through a simple parameter, Planck's constant:\[

\begin{array}{|c|c|c|}

\hline {\rm physics}& {\rm classical} & {\rm quantum} \\

\hline {\rm value}& \hbar = 0 & \hbar\neq 0 \\

\hline

\end{array}

\] Assuming that you know how to work both with classical and quantum physics, the value of this single parameter is enough to discriminate between them. It's that simple.

Now, quantum mechanics is a more general framework while classical mechanics is a special case. A classical theory may be typically obtained as the \(\hbar\to 0\) limit of a quantum mechanical theory. When a small \(\hbar\) becomes infinitesimal and is "really" sent to zero, the resulting limiting theory may be said to have \(\hbar=0\) and it's therefore a classical theory if the limit exists at all.

In classical physics, there's just no way to have \(\hbar\neq 0\). This is true by

*definition*because \(\hbar\) is the constant that measures the deviation of a physical theory from the class of theories of classical physics. Because classical theories aren't sufficient to discuss the quantum mechanical ones, we must use the more general framework – the quantum mechanical framework – if we want to compare classical and quantum mechanics.

This need is completely analogous to the need to use relativistic theories when we want to compare relativistic and non-relativistic theories. In that case, the deviation from a non-relativistic theory is measured by the parameter known as \(1/c\), the inverse maximum speed in Nature. The limit \(1/c\to 0\) i.e. \(c\to \infty\) is the non-relativistic limit and it is analogous to the \(\hbar\to 0\) classical limit of a quantum mechanical theory.

In both cases, the limiting theory obeys some extra axioms that are not valid in the more general theory. In particular, the non-relativistic theories (\(c\to \infty\) limits of relativistic ones) obeys the "absolute character of simultaneity" while the \(\hbar\to 0\) classical limits of quantum theories obey the extra axiom about the objectively well-defined values of observables (independently of observers or before the observations).

The more general theories or frameworks reject these extra axioms. Relativity says that the simultaneity is relative i.e. dependent on the inertial system; quantum mechanics says that the values of observables are only well-defined relatively to an observer and what the observer considers to be observations (events in which the observer acquired the information).

**OK, what is Planck's constant operationally?**

Planck's constant \(\hbar\) is omnipresent in quantum mechanical theories. Well, it's omnipresent if we use the system of units in which \(\hbar\) has to be inserted at all. Particle physicists and some other adult physicists often use units in which \(\hbar=1\) which simplifies most of the formulae in any quantum mechanical research.

However, if we want to compare quantum mechanical theories with their limits, we

*need*to use units in which \(\hbar\) is variable – i.e. it is not set to one (or another constant) – because a particular variation, the \(\hbar\to 0\) limit, is how we get from the general quantum mechanical theories to the classical ones (the limits).

Because \(\hbar\) appears at so many places, we may define its value in many ways. But ultimately all of them may be shown to be equivalent to the commutator of some observables – the refusal of observables to commute with each other. For example, we may define \(\hbar\) as\[

\hbar = i(px-xp)

\] in any theory that contains positions \(x\) and momenta \(p\). In classical physics, the commutator is zero, in a quantum mechanical theory, it's not. The precise value dictates the shape of the wave associated with a momentum \(p\) particle. The wavelength (period) of this wave is simply \(2\pi\hbar / p\).

Equivalently, we may define \(\hbar\) as\[

\hbar = -i \cdot J_z^{-1} \cdot (J_x J_y - J_y J_x)

\] The commutator between various components of the angular momentum are zero in classical physics, nonzero in quantum mechanics, and these commutators may be easily derived from the formulae for \(J\) as a function of \(x,p\) and from the commutator \([x,p]\) that we discussed a minute ago.

Now, the commutators such as \([J_x,J_y]\) are enough to determine that the eigenvalues of each component are quantized. \(\hbar/2\) is the smallest allowed nonzero value of \(J_z\) or any component of the angular momentum, for example. Obviously, we could discuss as many examples of commutators of observables – "elementary" or "composite" ones – in many mechanical or other theories.

The constant \(\hbar\) also appears in the Schrödinger equation. Does this appearance have anything to do with nonzero commutators? You bet. A more conceptual sensible way to rewrite the Schrödinger equation is to apply the time-dependent unitary transformation and derive the Heisenberg equation with the Heisenberg equations of motion. In that picture, the observables obey \[

i\hbar \frac{\dd L(t)}{\dd t} = [L(t),H]

\] This equation explicitly says that the commutators with the Hamiltonian are the time derivatives of the operators multiplied by the tiny constant \(i\hbar\) once again. The dynamical equations of the Heisenberg picture are examples of the more general fact that the commutators are proportional to \(\hbar\).

Classical physics emerges in the limit \(\hbar\to 0\). It recognizes that the commutators are small because they're proportional to \(\hbar\) which is tiny in the units used to study problems where classical physics becomes a good approximation. So classical physics "amplifies" these commutators, multiplies them by \(1/i\hbar\), and this product is called the "Poisson bracket". Effectively, only the leading terms in a power expansion in \(\hbar\) are kept in the classical limit. That's how we construct a classical theory.

The volatile anti-quantum zealot I have referred to has written things like

What distinguishes quantum from classical mechanics is how the observables of a composite system are related to the observables of the individual systems.But this opinion is completely wrong – the bold face fonts used for that sentence only highlight how much wrong he is. When we have a conventional composite system, the observables \(A_j,B_k\) describing the subsystems \(A,B\) of this composite system commute with all operators in the other group:\[

[A_j,B_k] = 0.

\] The commutator is zero, just like in classical physics, so the mere additional of "subsystems" to the whole system simply cannot make the physical system more quantum. The composite system may display lots of characteristically quantum behavior but that behavior only arises due to the nonzero commutators, i.e. \[

[A_j,A_{j'}] \neq 0, \quad [B_{k},B_{k'}]\neq 0.

\] All the novel quantum behavior appears "inside" \(A\) or inside \(B\), inside the individual subsystems of the composite system!

Now, the "quantum mechanics is non-local" crackpots love to say not only crazy things about the extra non-locality of quantum mechanics – which I have thoroughly debunked in many previous specialized blog posts – but they also love to present the quantum entanglement as some absolutely new voodoo, a supernatural phenomenon that has absolutely no counterpart in classical physics and whose divine content has nothing to do with the uncertainty principle or the nonzero commutators.

Except that all this voodoo is just fog.

Quantum entanglement is nothing else than the most general pure-state description of a correlation between two subsystems in quantum mechanics. Assuming that we use the correct laws and formalism of quantum mechanics to describe systems that obey quantum mechanics,

an entangled state and a state with a correlation between two subsystems are absolutely synonymous.If you study a singlet state of two spin-1/2 particles, you may measure \(J_z\) of both entangled particles. As I have discussed in numerous recent blog posts, this anticorrelation between the two spins is absolutely equivalent to the anticorrelation between the colors of two socks of Dr Bertlmann – a system we may describe by classical physics.

The idea that the correlation between the two electrons is "something fundamentally different" from the correlation between the two socks' colors – an idea pioneered by John Bell – is totally and absolutely wrong. After all, in the real world around us, even socks of Dr Bertlmann are accurately described by quantum mechanics. Classical physics isn't quite right, even for socks. So if we have measured a complete set of observables to bring the two socks in a pure state – but an entangled one (and be sure it's possible) – then the socks will inevitably be in an entangled state. If you're accurate about socks in the real world, you need to describe them as an entangled state, just like the two spins, and not just as a classical correlation. It's true simply because the classical theory is never quite right in the world around us!

So the opinion that the correlation of (real world) Bertlmann's socks is something "totally and fundamentally different" from the entanglement of the two spins contains the self-evidently incorrect assumption that socks in Austria don't obey the laws of quantum mechanics. But be sure that they do. Everything in this damn Universe does.

What's new about the quantum entanglement – relatively to the "ordinary" classical correlation – is that the two subsystems such as the two spins may have correlated many other properties at the same moment. The singlet state has \[

\vec J_1 = -\vec J_2

\] which simply means that \((\vec J_1+\vec J_2)\ket\psi = 0\). So you may measure \(J_{1z},J_{2z}\) and obtain a perfect anticorrelation. But if the two experimenters happen to measure \(J_{1x},J_{2x}\) instead, they get a perfect anticorrelation, too. And similarly for \(J_{1y},J_{2y}\). You couldn't invent a classical model with two classical bits that would emulate this behavior. And the components of the spin \(J_x,J_y\) are entirely different than or independent from \(J_z\).

The real-world socks are only simpler because of one fact: due to the complexity that produces decoherence etc., it is extremely hard and practically impossible to measure anything such as \(J_x,J_y\) for the socks – observables that don't commute with the sock color i.e. are non-diagonal in a basis of the color-of-sock eigenstates. These non-diagonal observables operationally don't exist for socks.

But all this new behavior depends on the nonzero commutators \([J_{1z},J_{1x}]\) and similar ones! What prevents you from this really "intimate" correlation between the two bits in classical physics is that in classical physics, all observables are simple functions on the phase space \(F=F(x_j,p_k)\) – or the set of possible values of all the bits if the information is described in terms of bits. For this reason, in classical physics, it's always enough to measure the values of \(x_j,p_k\) etc. and you know everything.

However, this is not the case in quantum mechanics. You can't measure all "elementary" observables \(x_j,p_k\) at the same moment because of the uncertainty principle – because of the nonzero commutators. Instead, you must decide what you measure and the post-measurement state will be an eigenstate of this measured observable \(L\). Moreover, you really need to list all conceivable observables which have all conceivable eigenstates if you want to exhaust all the options. And the set of observables – Hermitian matrices – is very large.

In a classical description of two bits, the phase space would have four points \(00,01,10,11\) which correspond to the four arrangement of the values \(J_{1z},J_{2z}\), to pick the standard convention. Once you would know that \(J_{1z},J_{2z}\) may only have \(2\times 2 =4\) values, i.e. they are two bits, and the knowledge of these two bits is the "maximum" you may know about the system, in classical physics, it would follow that no other interesting correlation may be present in the state of the two bits. All observables are functions of \(J_{1z},J_{2z}\). It means that you may at most represent the possibilities \(00,01,10,11\) by four arbitrary numbers \(e,f,g,h\). There are just four fixed options and a perfect correlation may at most mean that some of these four options are ruled out.

However, in quantum mechanics, there exist operators such as \(J_{1x}\) which don't commute with one of the observables \(J_{1z},J_{2z}\) – in this case with \(J_{1z}\). It's exactly this nonzero commutator that makes \(J_{1x}\) sensitive on the relative phase between the complex probability amplitudes that know about the options \(J_{1z}=+\hbar/2\) and \(J_{1z}=-\hbar/2\). Also, it's exactly the nonzero commutator \([J_{1x},J_{1z}]\) that says that \(J_{1x}\) isn't a diagonal matrix in the basis of the \(J_{1z}\) eigenstates. And it's the non-diagonal matrices for observables that contain all the novelties of quantum mechanics, including the "tighter" correlation that the quantum entanglement may guarantee in comparison with the correlations in classical physics.

Aside from the consequences of the nonzero commutators – i.e. of the need to use off-diagonal matrices for the most general observables that may really be measured – there is simply nothing "qualitatively new" in quantum mechanics. Whoever fails to understand these points misunderstands the character of quantum mechanics and the relationship between quantum mechanics and classical physics, too.

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