To quantize a scalar field theory with the action:

$$S=\int \mathrm d^Dx\mathscr{L}(\phi,\partial_\mu\phi)=\int \mathrm dx^0L(\phi,\partial_0\phi)$$

we promote $\phi(\vec{x})$ and $\pi=\frac{\delta L}{\delta(\partial_0\phi)}=\pi(\vec{x})$ to be field operator at fixed time $x^0$ (with the canonical commutation relation $[\phi(\vec{x}),\pi(\vec{y})]=\delta^{D-1}(\vec{x}-\vec{y})$), then we use the classical equation of motion to find $\phi$ and $\pi$ at all time.

What principles lie behind this? Why can we assume that the field operators satisfy the classical equation of motion?

By example if we have a classical theory with $S=\int \,\mathrm dtL(\phi,\partial_t\phi)$, momentum $\pi=\frac{\delta L}{\delta(\partial_t\phi)}$ hamiltonian $H=\pi\phi-L$, the classical equation of motion is:

$$\frac{\mathrm d\phi}{\mathrm dt}=\frac{\delta L}{\delta(\partial_t\phi)}$$

When we do the canonical quantization, we promote $[\phi,\pi]=\delta(\vec{x}-\vec{y})$ and the equation of motion for $\phi$ is $$\frac{\mathrm d\phi}{\mathrm dt}=i[H,\phi]$$

Why should $i[H,\phi]$ has the same form as $\frac{\delta L}{\delta(\partial_t\phi)}$? This fact comes from where?

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    $\begingroup$ The r.h.s. of your classical e.o.m. should have $H$, not $L$. Why don't you write the r.h.s. of the classical e.o.m. as the Poisson bracket $\{\phi,H\} = \frac{\delta H}{\delta \pi}$. Then canonical quantization gives that the equation holds in the quantum theory with $\mathrm{i}[H,\phi]$ instead of $\{\phi,H\}$ by definition of canonical quantization. I don't see what you think there is to show. $\endgroup$ – ACuriousMind Feb 9 '16 at 16:11
  • $\begingroup$ I think I don't really understand canonical quantization to the level that I can see clearly why $i[H,\phi]$ can always replace $\{\phi,H\}$... I'm trying to do this by a concrete example I guess $\endgroup$ – quangtu123 Feb 9 '16 at 16:13
  • $\begingroup$ @ACuriousMind By example here: en.wikipedia.org/wiki/… It's not always true that we can replace the Lie bracket by the commutator... $\endgroup$ – quangtu123 Feb 9 '16 at 16:27
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    $\begingroup$ Of course. Canonical quantization is not the only quantization procedure, and it is actually a theorem of Groenewold and van Howe that it doesn't work as desired. But nevertheless, canonical quantization is what one usually does first in QFT. I added the path integral version to my answer to show this is not a peculiar feature of the canonical approach, but a property of the QFT. $\endgroup$ – ACuriousMind Feb 9 '16 at 16:37

How to see it in canonical quantization: All operators $\mathcal{O}$ in a quantum theory fulfill the Heisenberg equations of motion $$ \frac{\mathrm{d}}{\mathrm{d}t}\mathcal{O}(t) = \mathrm{i}[H,\mathcal{O}(t)]$$ where $H$ is the Hamiltonian density and which is exactly the quantum version of the classical Hamiltonian equations of motion. So the fields, as they are operators, indeed must obey the classical equations of motion.

How to see it in path integral quantization: Write $\phi'(x) = \phi(x) + \epsilon\delta(y-x)$ and observe the path integral measure is invariant under this. Expand the integrand to first order in $\epsilon$, and deduce $$ \int\left(\frac{\delta S}{\delta\phi(x)} + J(x)\right)\mathrm{e}^{\mathrm{i}S[\phi]+J\phi}\mathcal{D}\phi = 0$$ which is known as the Schwinger-Dyson equation. Setting $J=0$ gives $\delta S/\delta\phi = 0$ (inside the path integral, which is the PI version of "as an operator equation"), which is exactly the classical equation of motion.

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    $\begingroup$ How exactly can we go from $\frac{\mathrm{d}}{\mathrm{d}t}\mathcal{\phi}(t) = \mathrm{i}[H,\mathcal{\phi}(t)]$ to $\frac{d\phi}{dt}=\frac{\delta H}{\delta\pi}$? It can derive it for particular cases but for a general case, can we prove that the classical equation of motions is always satisfied? $\endgroup$ – quangtu123 Feb 9 '16 at 15:39
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    $\begingroup$ @Mr.T: There is nothing to derive. The classical e.o.m. is $\mathrm{d}/\mathrm{d}t\mathcal{O} = \{\mathcal{O},H\}$ and canonical quantization means that $\{-,-\} \mapsto -\mathrm{i}[-,-]$, i.e. the Poisson bracket becomes the commutator, so it is by design that this is exactly the classical equation of motion. If you don't like the canonical quantization argument, you can very easily derive the Schwinder-Dyson equation in the path integral approach. $\endgroup$ – ACuriousMind Feb 9 '16 at 15:49
  • $\begingroup$ I will add more details to my questions so that maybe you can make it more clear for me. $\endgroup$ – quangtu123 Feb 9 '16 at 15:53

Suppose first that you deal with free field theories.

You start from canonical commutation or anticommutation relations for coordinate $q(x, t)$ and momentum $p(x, t)$: $$ \tag 0 [q_{i}(\mathbf x , t), p_{j}(\mathbf y , t)]_{\pm} = i\delta (\mathbf x - \mathbf y)\delta_{ij}, \quad [q_{i}(\mathbf x , t),q_{j}(\mathbf y , t)]_{\pm} = [p_{i}(\mathbf x , t), p_{j}(\mathbf y , t)]_{\pm} = 0 $$ This is in some sense quantum analog of Poisson brackets.

Such relations reault in important variational identities (since here operators are coordinate dependent, we need to use variational calculus instead of differential one) for bosonic operator $F \equiv F(q, p)$: $$ \frac{\delta F[q(t),p(t)]}{\delta q_{n}(\mathbf x , t)} \equiv i[p_{n}(\mathbf x , t),F[q(t), p(t)]], \quad \frac{\delta F[q(t),p(t)]}{\delta p_{n}(\mathbf x , t)} \equiv i[F[q(t), p(t)], q_{n}(\mathbf x , t)] $$ So that for arbitrary $c$-valued functional $F[q(t), p(t)]$ its variation is $$ \tag 1 \delta F [q(t), p(t)] = $$ $$ \sum_{n}\int d^{d}\mathbf r \left( \delta q_{n}(\mathbf r , t)i[p_{n}(\mathbf r , t),F[q(t), p(t)]] + \delta p_{n}(\mathbf r , t)i[F[q(t), p(t)], q_{n}(\mathbf r , t)]\right) $$ From the mathematical sense of $\hat{H}_{0}$ (it is time translation operator) follows that $$ \tag 2 q_{n}(\mathbf r, t) = e^{iH_{0}t}q_{n}(\mathbf r , 0)e^{-iH_{0}t}, \quad p_{n}(\mathbf r, t)e^{iH_{0}t}p_{n}(\mathbf r , 0)e^{-iH_{0}t} $$ From $(1)$ and $(2)$ follows canonical Heisenberg EOM: $$ \tag 2\frac{dF[q(t), p(t)]}{dt} = i[F, H_{0}] $$

Now, let's assume case of interacting theory. We may interact $Q(\mathbf x , t), P(\mathbf x , t)$ from free field coordinates and momentum by using Heisenberg picture: $$ \tag 3 Q(\mathbf r , t) = e^{iHt}q(\mathbf r , 0)e^{-iHt}, \quad P(\mathbf r , t) = e^{iHt}p(\mathbf r , 0)e^{-iHt}, $$ where $H$ is the full hamiltonian. Since this is similarity transformation (and it commutes with $H$), then the full hamiltonian is a functional of $Q, P$ as well as it was functional of $q, p$: $$ H[Q, P] = e^{iHt}H[q, p]e^{-iHt} = H[q, p] $$ Next, due to the same reason $Q, P$ obeys the same canonical relations $(0)$ and EOM $(2)$ as free theory $q, p$, but with $H$ instead of $H_{0}$.

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