Requirements for theory where Hamilton is the generator for time evolution

Question

Introduction

It is well known, that the Hamilton operator is the generator for the time evolution. In Heisenberg picture $$ -i\ \partial_t \phi(x, t) = [H, \phi(x, t)]. $$ For a Quantum Field Theory, we usually start with the Lagrangian $\mathcal{L}(\phi(x, t), \partial_\mu \phi(x, t))$, and then construct the Hamilton from there $$ H(t) := \int d^3x\ \Big( \underbrace{\frac{\partial\mathcal{L}}{\partial (\partial_0 \phi)}}_{\pi(x, t)}\ \partial_0 \phi(x, t) -\mathcal{L} \Big). $$ The other axioms besides how the Lagrangian looks like are the Canonical Commutator Relations for $\phi(x, t)$, $\pi(x, t)$.

Question

It is not at all obvious (at least for me), that given a Lagrangian, and the Canonical Commutation Relations, that the constructed Hamilton will be the time evolution generator.
Is there any simple check given a Lagrangian to see whether it's constructed Hamilton will behave as the generator? What's one of the most accepted, fundamental requirement for Lagrangian for this property?

Or we set Hamilton as the generator as an axiom, and then get the Canonical Commutation Relations?

I can even give counter examples, where this is not the case. For example, some probably not valid theories, like $\mathcal{L} \propto \phi (\partial_\mu \phi) (\partial^\mu \phi)$. This example is a Lorentz scalar, so in theory it can even be a good candidate, however, if the Canonical Commutator Relations hold, the Hamilton is not the time evolution generator.

Notes

• This question has been asked numerous times, but I have not seen satisfactory answers, only this one https://physics.stackexchange.com/a/360077/254794, but it does not answer why commutation holds like that. For other Noether Charges (momentum, internal symmetry charges), his/her answer explains this, but not for energy.
• I do not consider using the Poisson bracket $\rightarrow$ Commutator Relations quantization here as a good starting point. There are several quantum theories as far as I know, which do not have classical correspondence like that.

Related

• Conserved charges and generators (I do not want to use Poisson Bracket as a starting point, not every quantum theory corresponds to a classical one.)
• Why does the classical Noether charge become the quantum symmetry generator? (Also uses Poisson brackets, and one example Lagrangian to try to prove the general case.)
• Connection between conserved charge and the generator of a symmetry (Still just partial explanation as far as I saw.)
• How a symmetry transformation acts on quantum fields (Related, but the answer does not answer my specific question.)

Answer 1

This might not be the most general formulation, but maybe it's general enough to answer the question. I'll use some abbreviations that are hopefully clear, like omitting the time-argument and writing $\partial\phi$ instead of $\partial_\mu\phi$. I'll write $\dot\phi$ for the time-derivative of $\phi$.

Start with any Lagrangian density $L(x)$ that can be expressed in terms of a field $\phi$ and its first derivatives $\partial\phi$, without any higher derivatives. Define the canonical conjugate$^\dagger$ $$ \newcommand{\pl}{\partial} \pi(x) := \frac{\delta }{\delta\dot \phi(x)}\int dy\ L(y) \tag{1} $$ and the Hamiltonian $$ H = \int dx\ \big(\pi(x)\dot\phi(x) - L(x)\big). \tag{2} $$ To define the quantum theory, impose the canonical equal-time commutation relations \begin{gather} [\phi(x),\,\pi(y)]=i\delta(x-y) \\ [\phi(x),\,\phi(y)]=0 \\ [\pi(x),\,\pi(y)]=0. \tag{3} \end{gather} Equation (2) gives $$ [H,\phi(x)]=-i\dot\phi(x) +\int dy\ \big(\pi(y)[\dot\phi(y),\phi(x)] -[L(y),\phi(x)]\big). \tag{4} $$ The definition of $\pi(x)$ implies that last two terms cancel each other. To see this, use the identity $$ \int dy\ [L(y),\phi(x)] = \int dy\ \frac{\delta L(y)}{\delta\, \pl\phi(y)}[\pl\phi(y),\phi(x)] + \int dy\ \frac{\delta L(y)}{\delta \phi(y)}[\phi(y),\phi(x)]. \tag{5} $$ (The derivatives of $L$ with respect to $\phi$ and $\pl\phi$ are formal, because $\phi$ and $\dot\phi$ are operators, but we can think of these derivatives as convenient abbreviations for a proper calculation that keeps track of the ordering of the operators.) The last term in (5) is zero. In the second-to-last term in (5), the commutator with the spatial derivatives of $\phi$ is zero, and the commutator with the time-derivative of $\phi$ cancels the second-to-last term in (4). Altogether, this leaves $$ [H,\phi(x)]=-i\dot\phi(x), \tag{6} $$ which says that $H$ generates time-evolution. (The sign-difference compared to the OP might be due to different sign-conventions.) We can also check $[H,\pi(x)]$: $$ [H,\pi(x)] = \int dy\ \big(\pi(y) [\dot\phi(y),\pi(x)] - [L(y),\pi(x)]\big). \tag{7} $$ As in (5), use the identity $$ \int dy\ [L(y),\pi(x)] = \int dy\ \frac{\delta L(y)}{\delta\, \pl\phi(y)}[\pl\phi(y),\pi(x)] + \int dy\ \frac{\delta L(y)}{\delta \phi(y)}[\phi(y),\pi(x)]. \tag{8} $$ The time-derivative part of the second-to-last term in (8) cancels the second-to-last term in (7), and the remaining terms together with (3) and Euler-Lagrange equation of motion reduce equation (8) to $$ [H,\pi(x)]=-i\dot\pi(x). \tag{9} $$ Altogether, this confirms that $H$ generates time-evolution.

The $L\sim \phi(\pl\phi)(\pl\phi)$ example that was mentioned in the question satisfies the assumption that $L$ can be expressed in terms of $\phi$ and $\pl\phi$ alone, so the results (6) and (9) do hold in that example.

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$^\dagger$ Qmechanic's answer correctly points out that the definition (1) is potentially ambiguous because it doesn't specify the order of the product (if any) of field operators on the right-hand side. The right-hand sides of equations (5) and (8) are abbreviations for more explicit identities that keep track of the order of the products of field operators, and since I didn't show the more explicit versions (which would be hard to do without specifying $L$), I didn't actually prove the cancellations that were claimed in the narration. So this answer is just an outline, not a complete proof.

Answer 2

1. OP's question about existence of quantization is very broad and is literally the topic of whole books and current research. For a general classical system, there is no proof that a consistent quantization exists. Although many partial results in e.g. geometric quantization and deformation quantization have been obtained.

2. The quantization procedure is not unique. When quantizing a classical model, there are often operator ordering ambiguities.

3. OP's example: OP considers the classical Lagrangian density $${\cal L}~=~ \frac{1}{2}\phi \dot{\phi}^2-{\cal V}, \qquad {\cal V}~=~\frac{1}{2}\phi(\nabla\phi)^2. \tag{A}$$ The momentum becomes $$ \pi~=~\frac{\partial{\cal L}}{\partial\dot{\phi}}~=~\phi \dot{\phi} .\tag{B}$$ It is not difficult to find the corresponding classical Hamiltonian density $$ {\cal H}~=~ \frac{1}{2}\phi^{-1} \pi^2+{\cal V}, \qquad H~=~\int\!d^3x~{\cal H}. \tag{C}$$ The issue is now: How should we order the quantum Hamiltonian density? One possibility is $$ \hat{\cal H}~=~\frac{1}{2}\hat{\phi}^{-1/2} \hat{\pi}^2\hat{\phi}^{-1/2}. \tag{D}$$ This is Hermitian. The time-evolution is $$\frac{d\hat{\phi}}{dt}~=~\frac{1}{i\hbar}[\hat{\phi}, \hat{H} ]~=~\hat{\phi}^{-1/2}\hat{\pi}\hat{\phi}^{-1/2},\tag{E}$$ or equivalently, $$\hat{\pi}~=~\hat{\phi}^{1/2}\frac{d\hat{\phi}}{dt}\hat{\phi}^{1/2}. \tag{F}$$