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 $\phi(x, t)$, $\pi(x, t)$.


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.


  • 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.


  • 3
    $\begingroup$ Are you sure your counterexample really is a counterexample? Can you show the calculation that made you think it's a counterexample? This may help potential answerers see what you might be missing. $\endgroup$ Commented Apr 18, 2020 at 17:18
  • $\begingroup$ Yes, I will outline the calculation steps here. $\pi = \frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)} = 2 \phi \partial_0 \phi$, so $\partial_0 \phi = 1/2 \phi^{-1} \pi$. Insert this form in $H$. We need to find $[H, \phi]$. Because $[\partial_i \phi, \phi] = 0$, $i \in \{1, 2, 3\}$, and from Commutator Relations $[\phi(y, t), \phi(x, t)^{-1}] = 0$, $[\pi(x, t), \phi(y, t)^{-1}] = i \delta^3(x - y)$, we can say (after some calculations), that $[H(t), \phi(y, t)] = i \phi(y, t) \partial_0 \phi(y, t) - 1/4 \delta^3(0)$. Correct me, if something is wrong. $\endgroup$
    – Gabor
    Commented Apr 18, 2020 at 17:32
  • $\begingroup$ A little bit more information about my calculation. As I mentioned, $[\partial_i \phi, \phi] = 0$, so $[Q(t), \phi(y, t)] = \int d^3 x [\phi(x, t) \partial_0 \phi(x, t) \partial^0 \phi(x, t), \phi(y, t)]$. Now, we insert $1/2 \phi^{-1} \pi$ for $\partial_0 \phi$ to get $\int d^3 x 1/4 [\phi(x)\phi(x)^{-1}\pi(x)\phi(x)^{-1}\pi(x), \phi(y)]$. Even if I made some error in my calculation, the correct result will still not give the generator for the Hamilton I think. $\endgroup$
    – Gabor
    Commented Apr 18, 2020 at 17:45
  • $\begingroup$ I made some minor mistakes in the calculation, sorry. The correct result is $[H(t), \phi(y, t)] = -i \partial_t \phi(y, t) + 1/4 \delta^3(0)$. So, my statement still holds even if I made some mistakes in my first calculation. $\endgroup$
    – Gabor
    Commented Apr 18, 2020 at 18:22
  • 1
    $\begingroup$ Have you tried a different ordering in attempt to remove that extra delta function? $\endgroup$
    – JF132
    Commented Apr 18, 2020 at 18:26

2 Answers 2


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.

$^\dagger$ 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.

  • $\begingroup$ Could you expand more why (5) is true? What does it really mean? Why $\delta L$ needs to be written there instead of $\partial L$? Also, could you specify when do you use Lagrange density, and when the full Lagrangian? $\endgroup$
    – Gabor
    Commented Apr 18, 2020 at 19:14
  • $\begingroup$ @Gabor Equation (5) is just a concise way of expressing the consequences of the identity $[AB,C]=A[B,C]+[A,C]B$. If $L=AB$, then $\delta L/\delta A = B$, and so on. I wrote $\delta L$ instead of $\partial L$ only to avoid expressions like $\partial L/\partial\partial\phi$, which would be superficially ambiguous because the to $\partial$'s in the demoninator have different meanings. And oops - you're right. I was careless about mixing the lagrange density and its integrals. I'll come back soon and clean that up. $\endgroup$ Commented Apr 18, 2020 at 19:33
  • $\begingroup$ @Gabor I edited the answer so that $L$ consistently denotes the lagrange density, with no implicit integrals over space or time. $\endgroup$ Commented Apr 18, 2020 at 20:30
  • $\begingroup$ Thanks you. Ok, I understand this answer intuitively, but still trying to check how can this be true a little more rigorously. $\endgroup$
    – Gabor
    Commented Apr 19, 2020 at 5:42
  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, $$\hat{\pi}~=~\hat{\phi}^{1/2}\frac{d\hat{\phi}}{dt}\hat{\phi}^{1/2}. \tag{F}$$

  • $\begingroup$ This does not answer the general question I think, only hints how we may get rid of problems in general, if possible. I'm assuming the ordering problem comes from the $\delta L \propto \frac{\partial \mathcal{L}}{\partial A} \delta A$ terms, because $\delta A$ may be on the other side, or inside $\mathcal{L}$. Am I right? Is it proven, that you can always find an ordering for which $H$ is the time evolution generator? For which kind of physical theories does this hold? $\endgroup$
    – Gabor
    Commented Apr 19, 2020 at 5:41
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Apr 19, 2020 at 6:06

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