The purpose of this post will be to review a standard method for passing from the classical to the quantum world (or more poetically, turning \(\hbar=0\) on to \(\hbar=1\) in natural units). This procedure is called canonical quantization.
It is first worth emphasizing a conceptual point. Morally, it is quantum mechanics, and not classical mechanics, that represents ground truth. Thus, the whole notion of trying to “pass” from the classical world to the quantum world (i.e. the whole notion of “quantizing” a classical theory) is a bit fallacious; from a logical perspective, it makes much more sense to start with quantum mechanics and pass the other way into the classical world by taking the limit \(\lim_{\hbar\to 0}\). For instance, the fact that a particle’s position and momentum along some direction \(\hat{\textbf n}\) can never be simultaneously known with certainty should be thought of as something very normal, the way things just are and have always been, and that one’s apparent ability to evade this principle in classical mechanics should be thought of as a strange, unusual luxury that deviates from the quantum norm. Although as classical creatures it is the Newtonian physics of the classical world that is certainly more intuitive, as a serious physicist it is essential to think clearly and get one’s priorities straight.
First Quantization (Nonrelativistic Quantum Mechanics)
Mathematically, canonical quantization \(f\mapsto\hat f\) is a Lie algebra representation that maps functions \(f\in C^{\infty}(\textbf R^2\to\textbf R)\) on classical phase space \(\textbf R^2\) to operators \(\hat f\in \frak u\)\((\mathcal H)\) on quantum state space \(\mathcal H\) (all this discussion is easily generalizable to a higher-dimensional phase space like \(\textbf R^{6N}\)). In the former, the Lie bracket is provided by the Poisson bracket \(\{,\}\) while in the latter the Lie bracket is the commutator \([,]\), and one requires canonical quantization to preserve this structure:
\[[\hat f,\hat g]=\widehat{\{f,g\}}\]
Thanks to Taylor’s theorem, a countably infinite basis of \(C^{\infty}(\textbf R^2\to\textbf R)\) is given by monomial functions on phase space of the form \((x,p)\mapsto 1,x,p,x^2,xp,p^2,…\), but it turns out there is good reason to stop at quadratic order. Specifically, it is worth noticing (using standard abuse of notation of writing e.g. \(x\) to mean the function \((x,p)\mapsto x\)) that the subspace \(\text{span}_{\textbf R}(1,x,p,x^2,xp,p^2)\subset C^{\infty}(\textbf R^2\to\textbf R)\) is in fact a \(6\)-dimensional Lie subalgebra of the Lie algebra \(C^{\infty}(\textbf R^2\to\textbf R)\):
In fact, the \(3\times 3\) yellow box highlighted above defines yet an even smaller Lie subalgebra \(\text{span}_{\textbf R}(1,x,p)\subset C^{\infty}(\textbf R^2\to\textbf R)\), called the Heisenberg Lie algebra. One can then check that the canonical quantization map, defined on the basis functions by:
\[\hat 1:=-i\hbar 1\]
\[\hat x:=-iX\]
\[\hat p:=-iP\]
\[\widehat{x^2}:=\frac{X^2}{i\hbar}\]
\[\widehat{xp}:=\frac{1}{i\hbar}\frac{XP+PX}{2}\]
\[\widehat{p^2}:=\frac{P^2}{i\hbar}\]
is a legitimate Lie algebra representation when restricted to either the first-order Heisenberg subalgebra \(\text{span}_{\textbf R}(1,x,p)\) or the quadratic subalgebra \(\text{span}_{\textbf R}(1,x,p,x^2,xp,p^2)\).
As an aside, the standard physicist’s convention with regards to the meaning of a Lie algebra representation is to instead impose the rule:
\[[\hat f,\hat g]=i\hbar\widehat{\{f,g\}}\]
with the extra factor of \(i\hbar\) compared with the mathematician’s convention. This makes it so that the canonical quantization map corresponds to “just putting hats on everything” (the exception is for the function \(xp\) below where a symmetric combination is used to maintain Hermiticity):
\[\hat 1:=1\]
\[\hat x:=X\]
\[\hat p:=P\]
\[\widehat{x^2}:=X^2\]
\[\widehat{xp}:=\frac{XP+PX}{2}\]
\[\widehat{p^2}:=P^2\]
Despite the hopes of Dirac and others that such a simple prescription could extend to a Lie algebra representation over all functions \(f\in C^{\infty}(\textbf R^2\to\textbf R)\) on phase space, it was subsequently shown in work of Groenewold and Van Hove that this is impossible; i.e. that there does not exist any “reasonable” quantization map that would continue to remain a Lie algebra representation beyond the subalgebra \(\text{span}_{\textbf R}(1,x,p,x^2,xp,p^2)\) of quadratic polynomials on phase space. One way to see this is that, continuing to cubic order, one could reasonably demand, in analogy to \(\widehat{xp}:=\frac{XP+PX}{2}\), that for instance \(\widehat{x^2p}:=\frac{X^2P+XPX+PX^2}{3}\), or more generally that canonical quantization should proceed by the Weyl transform:
\[\hat f=\frac{1}{(2\pi\hbar)^2}\int_{\textbf R^2}dx’dp’e^{i(p’X+x’P)/\hbar}\int_{\textbf R^2}dxdpf(x,p)e^{-i(p’x+x’p)/\hbar}\]
But if this were so, then one can see that:
\[\{x^3,p^2\}=6x^2p\to 2i\hbar(X^2P+XPX+PX^2)\]
whereas, using (ironically) the identity \([f(X),P]=i\hbar f'(X)\) in light of the obvious analog \(\{f(x),p\}=f'(x)\):
\[[X^3,P^2]=P[X^3,P]+[X^3,P]P=3i\hbar(X^2P+PX^2)\]
So these disagree, at least if one were to implement canonical quantization using the Weyl transform as above (thus, this is not a rigorous proof that there isn’t some other clever way to quantize that preserves the Lie algebra structure consistently, but just an intuitive argument for why such a scheme can’t exist, and that at its heart the reason is due to the non-abelian nature of operators).
Second Quantization (Quantum Field Theory)
Just as one can implement canonical quantization to pass from classical particle mechanics on phase space to quantum particle mechanics on state space, one can perform an analogous procedure to pass from classical field theory to quantum field theory. More precisely, it will be most convenient to work in the framework of Hamiltonian classical field theory (if one were to start with Lagrangian classical field theory, then in lieu of canonical quantization one would instead utilize path integral quantization but fundamentally all of these are equivalent as they must be).
Recall that Lorentz invariance is often manifest in Lagrangian classical field theories because one can just check that all indices are suitably contracted in the theory’s Lagrangian density \(\mathcal L\) to check whether or not the theory’s action \(S=\int dtd^3\textbf x\mathcal L\) is Lorentz-invariant. By contrast, the Legendre transform to an equivalent Hamiltonian formulation of the same classical field theory doesn’t break Lorentz invariance, but obscures it, since defining the conjugate momentum field \(\pi^j(X):=\frac{\partial\mathcal L}{\partial\dot{\phi}_j}\) picks out a preferred time coordinate \(ct=x^0\) and this is further emphasized in the Hamiltonian density itself:
\[\mathcal H=\pi^i\dot{\phi}_i-\mathcal L\]
In the Schrodinger picture, a quantum field is considered to be an operator-valued function \(\phi(\textbf x)\) on space \(\textbf x\in\textbf R^3\) (not spacetime because in the Schrodinger picture one takes operators to be \(t\)-independent) which satisfies the canonical commutation relations (i.e. the commutation relations of the Heisenberg algebra that canonical quantization would aim to preserve):
\[[\phi_i(\textbf x),\phi_j(\textbf x’)]=[\pi^i(\textbf x),\pi^j(\textbf x’)]=0\]
\[[\phi_i(\textbf x),\pi^j(\textbf x’)]=i\hbar\delta_i^j\delta^3(\textbf x-\textbf x’)\]
In the Schrodinger picture, all the \(t\)-dependence lies in the state \(|\psi\rangle=|\psi(t)\rangle\) which evolves by the usual Schrodinger equation \(i\hbar\dot{|\psi\rangle}=H|\psi\rangle\) where \(H=\int d^3\textbf x\mathcal H\). However, note that \(|\psi\rangle\) is more like a “wavefunctional” than a “wavefunction” because it takes in some field \(\phi(\textbf x)\) as an input and outputs a complex number. Thus, to emphasize again, Lorentz invariance was already broken from the get-go, but here it’s just getting even worse since only space \(\textbf x\) appears in the field \(\phi(\textbf x)\) whereas only time \(t\) appears in the Schrodinger equation.