I've explained the inevitability of quantum mechanics and the incorrectness of all the criticisms from many angles.
The main reason why the wrong opinions keep on spreading is that the "natural selection" that used to eliminate the people who are totally wrong about fundamental issues has ceased to exist. The inkspillers who write complete nonsense about quantum mechanics should really be "terminated" as scientists – for their life (that's how it still worked e.g. in the times of Hugh Everett) – but instead, we live in an epoch in which they find a few thousand dumb readers who will say that the "realism" or a related stupidity is cool and the authors may often keep on calling themselves "physicists" although they have just proven to be incompetent.
Quantum mechanics is not only a very useful and precise theory. The general framework of quantum mechanics is also a set of principles that are pretty much totally guaranteed not to change in the future at all – because their unavoidability may be almost rigorously proven. Aside from a few thousand of "real experts" in quantum mechanics, almost no one gets this point. Almost no one understands quantum mechanics. The problem is that almost all people think that it's OK to consider QM to be just a "bunch of random counterintuitive equations" that may be deformed just like some unreadable incomprehensible classical equations. But that's not the case at all.
Quantum mechanics is all about the postulates that are robust and natural. Quantum mechanics says that the evolution is determined by unitary transformations acting on the density matrix (which transforms as the adjoint representation of a unitary group) and probabilities of any statements are given by linear functions of the density matrix (Born's rule). That's it.
It's very natural for probabilities to be "linear functions" of some mathematical variables; and it's very natural for transformations and evolution to be mathematically expressed by unitary transformations. We might reasonably argue that this basic framework of quantum mechanics is more natural than the framework of classical physics and I am going to do so now, too.
In classical physics, the evolution is encoded by a deterministic evolution that may be integrated over some time and the original values of the coordinates and momenta get transformed to the final values:\[
E: \,x_j(t_{\rm initial}),p_j(t_{\rm initial})
\mapsto
x_j(t_{\rm final}),p_j(t_{\rm final})
\] This function that calculates the final values of the coordinates and momenta from the initial ones is continuous (whenever the time and the differential equations are continuous); and it is a symplectic transformation that preserves the symplectic form on the phase space, i.e. the Poisson brackets\[
\{x_i,p_j\} = \delta_{ij}.
\] Ignore all the Poisson symplectic comments if you have never encountered those things. A more general, simple point is that the evolution map \(E\) may be classified as a permutation of the points on the phase space! There are infinitely many (and uncountably many) points on the phase space but if you generalize the notion of a permutation to such sets (or if you truncate and discretize the phase space and approximate it by a finite set), the evolution is a permutation.
This permutation character of the classical deterministic evolution holds for the probabilistic description of classical physics, too. If you don't know the initial state precisely, you may express your knowledge by a classical probabilistic distribution on the phase space\[
\rho(x_i,p_j).
\] Again, if you imagine that the phase space is a finite set – because you discretized and truncated the phase space – the values of \(\rho\) at individual points of the phase space may be replaced with finitely many (or countably many) probabilities \[
P_k, \quad k\approx (x_i,p_j).
\] The index \(k\) remembers the equivalent of the values of all coordinates and momenta – \(k\) labels the point on the phase space. The evolution in time means that the probabilities \(P_k\) are permuted with each other – by the same permutation that was scrambling the original points in the phase space (or by the inverse one, depending on whether it's an active or passive transformation). So\[
P_k \mapsto P_{\Pi(k)}
\] where \(\Pi(k)\) encodes the permutation of the points on the phase space. Note that the total probability of all points on the phase space is one:\[
1 = \sum_k P_k \quad {\rm or} \quad 1 = \int d^N x\, d^N p\, \rho(x_j,p_j).
\] It's nice except that the the "set of all permutations that are also continuous and/or symplectic transformations" of the phase space isn't a terribly natural set. The groups of permutations of finite sets or countable sets aren't even "continuous" manifolds. When the phase spaces are continuous i.e. the number of the phase space points becomes uncountable, the group of permutations becomes a "manifold" with some topology (definition of continuity) on it. But it's still not "too" continuous.
Well, quantum mechanics replaces the group of continuous/symplectic transformations of the phase space by the unitary group (possibly an infinitedimensional group). And unitary groups are much more natural than permutation groups. In particular, the unitary group \(U(M)\) contains the permutation group \(S_M\) as a small finite subgroup. If you can replace a theory whose permutation group \(S_M\) is enhanced to a whole \(U(M)\), this enhancement is surely an "improvement". Greater symmetries are better. And more tolerant theories are more general and less arbitrarily constrained. The enhancement of the permutation group \(S_M\) to \(U(M)\) is also how Matrix theory guarantees the symmetry or antisymmetry of wave functions for \(M\) identical particles – that symmetry or antisymmetry follows from the \(U(M)\) gauge symmetry of the matrix model. The experts in Matrix theory would especially agree that it's very natural to enhance \(S_M\) to \(U(M)\) – and the fathers of quantum mechanics did it 70 years before the fathers of the BFSS Matrix theory.
These comments of mine may be dismissed as some "aesthetic" judgments that you don't need to agree with. Except that people who have a good physical intuition do agree with me. And they agree for a good reason. These statements may be made much more precise and justified.
In quantum mechanics, you may be given a mixed state. And there are infinitely many "Yes/No" questions in quantum mechanics. Every such question is associated with a linear Hermitian projection operator\[
P = P^\dagger = P^2
\] and the probability that the answer is "Yes" in the state given by the density matrix \(\rho\) is simply\[
{\rm Prob} = {\rm Tr} (P \rho).
\] Now, the evolution evolves \(\rho\) in a unitary way assuming that \(\rho\) transforms in the adjoint representation\[
\rho \to U\cdot \rho \cdot U^{1}, \quad U^{1} = U^\dagger.
\] When you learn that some question \(P\) had the answer "Yes", you replace \(\rho\) by \[
\rho \mapsto \frac{ P\cdot \rho \cdot P }{{\rm Tr}(P\cdot \rho \cdot P)}.
\] This replacement is nothing else than the quantum counterpart of the Bayesian inference. We added \(P\) on both sides to keep \(\rho\) Hermitian. The new \(\rho\) has the property that \(\rho_{\rm new}P = \rho_{\rm new}=P\rho_{\rm new}\) which is just a mathematical way to express that the answer to the question \(P\) was "Yes" ("\(P=1\)").
The denominator was added there just to keep \({\rm Tr}\rho=1\). You may omit one \(P\) in the denominator because of the cyclic property of the trace and the projection character of \(P=P^2\). Also, don't worry that you may divide by zero. The denominator can't be zero because we previously said it was the probability of getting the "Yes" answer for \(P\) so if that denominator is zero, the probability of that "division by zero" is also zero, so it cannot happen! ;)
But the simple rules above define pretty much the whole framework of quantum mechanics. They're extremely natural and there is nothing to change. We may view the rules of quantum mechanics for the pure states \(\ket\psi\) to be just a corollary of our treatment assuming the mixed state i.e. the density matrix \(\rho\). Every \(\rho\) may be written as a sum of objects \(\ket{\psi_i}\bra{\psi_i}\). If you take just one object \(\ket{\psi_i}\bra{\psi_i}\), all the probabilities will depend on \(\ket\psi\) instead of \(\rho\) – and they will depend on \(\psi\) in the sesquilinear way (quadratic dependence with one complex conjugation). That's the \(\ket\psi\) version of Born's rule. But we don't need to add anything new for pure states. The treatment involving the pure states is just a special case of the apparatus for the mixed states – just like the probabilities of the form\[
P_k = (0,0,1,0,0,0, \dots , 0,0,0)
\] are a special subclass of the probabilistic distributions on the phase space in classical physics. So it's really just the quantum mechanical rules for the density matrix \(\rho\) that need to be defined. In that formalism, all the probabilities are simply linear in \(\rho\) which is as natural as in classical physics. Probabilities have to evolve linearly – both in classical physics and quantum mechanics – because a 5050 mixture of two initial states has to evolve into the 5050 mixture of the corresponding final states!
In quantum mechanics, it means that the evolution equation for \(\rho\) has to be linear in \(\rho\) as well. And the most general transformations that quantum mechanics allows involves unitary transformations of \(\rho\) instead of the "permutations of points that are also continuous/symplectic maps on the phase space" which was the rule in classical physics.
The unitary transformations acting on fundamental vectors \(\ket\psi\) are those that preserve the sesquilinear norm \(\bra\psi \psi\rangle\). Equivalently, if \(\rho\) transforms as the adjoint representation under the unitary transformations, \[
\rho \mapsto U \cdot \rho \cdot U^{1},
\] then the unitary condition simply means that the trace of \(\rho\) is preserved. The trace plays exactly the same role as \(\sum P_k\) or \(\int d^N x\,d^N p\,\rho\) in classical physics, and has the meaning of the total probability of all mutually exclusive options. Now, the unitary group is exactly what you need because the preservation of the "total probability" is the only general condition that the evolution laws for the probabilities have to obey. In special situations, you know something more precise about the transformations \(U\), of course, so the transformations may be known to be elements of a smaller subgroup of \(U(M)\).
Again, what quantum mechanics has changed relatively to classical physics is that the group of permutation of points on the phase space \(S_M\) was enhanced to a nice and full \(U(M)\) although the value of \(M\) was reduced from an infinite and uncountable value a little bit because the phase space cells of volume \((2\pi\hbar)^N\) correspond to basis vectors in quantum mechanics.
But the extension of the allowed transformations from \(S_M\) to \(U(M)\) is very natural and obviously equivalent to the statement that the observables in quantum mechanics generically refuse to commute. That's also true for the special observables \(P\), the projection operators that encode "Yes/No" questions. Those refuse to commute in quantum mechanics, too.
The whole quantum mechanical description is therefore determined. There has to be an object that remembers all the probabilities, \(\rho\). All the probabilities have to be linear in it and by the basic probability calculus that works not only in classical physics but generally, the evolution has to be linear in this \(\rho\). The allowed transformations have to preserve the invariant corresponding to the "total probability". And the only invariant that may be written as a sum of a corresponding number of probabilities is the sesquilinear invariant of the unitary group. So the unitary group is the right group from which you should choose the transformations acting on \(\rho\). The only freedom of the model builder is to pick the dimension of the Hilbert space \(M\), which determines the group \(U(M)\) – it's usually infinite – and the embedding of the algebra of observables into this group.
The fact that the quantum mechanical framework cannot be "deformed" really boils down to the statement that there is no nearby group similar to \(U(M)\). Groups in mathematics are separated from each other. The linearity in probabilities is dictated by logic which is why you can't allow things like "quantum groups". At most, you may imagine that you add new degrees of freedom. Indeed, a quantum field theory may have a greater number of fields and massive particles that we haven't produced yet. But the relevant effective quantum mechanical theory is basically dictated by the experiment. In particular, the heat capacity of an atom is comparable to \(k_B\), Boltzmann's constant, which means that an atom effectively carries \(O(1)\) qunats (quantum nats) of information only. So you simply have a rather small number – extracted from the low heat capacity of the atoms – of distinguishable states \(M\) to describe an atom. As I argued, the evolution of the probabilities has to be a linear unitary map in \(U(M)\) and the only question is what it exactly is i.e. what are the matrix entries of the Hamiltonian.
You can't really "deform" the linearity condition on the evolution because the linearity of evolution in probabilities follows from general laws of logic or probability calculus such as\[
P(A\,{\rm or} \,B) = P(A)+P(B)
\] for mutually exclusive \(A,B\). Also, you can't really deform the condition that requires the evolution to be unitary because the "total probability" (equal to one) has to be an invariant of the evolution. The only thing that you can do is to restrict the unitary group \(U(M)\) further to a subgroup, such as \(S_M\) from classical physics. But this is in no way a "small perturbation" of the quantum mechanical theory because \(S_M\) is "very far" from \(U(M)\). We have "experimentally proven" that \(S_M\) isn't enough because the different observables really don't commute with each other.
That's why \(S_M\) as a group of allowed transformations on the phase space – transformations acting on the probabilities \(\rho\) or \(\rho(x_j,p_j)\) – has been experimentally refuted and \(U(M)\) is the next surviving group. Once you know that the probabilities evolve in a linear and unitary way, it's really quantum mechanics. There is nothing to change. You may only identify the relevant observables and e.g. the Hamiltonian operator that governs their time evolution. In other words, you may pick a particular quantum mechanical model. But you won't be able to deform anything about the general postulates of quantum mechanics because they boil down to linearity and unitarity and these conditions follow from the universal probability calculus – i.e. from mathematical logic applied in the context of probabilities.
Monday, May 20, 2019 ... //
Why the framework of quantum mechanics cannot be deformed
Vystavil
Luboš Motl
v
10:09 AM



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