quantization - Equivalence of classical and quantized equation of motion for a free field

Suppose a classical free field $\phi$ has a dynamic given in Poisson bracket form by $\partial_o\phi=\{H, \phi\}$. If we promote this field to an operator field, the dynamic after canonical quantization is given by $\partial_o\phi=i[H, \phi]$.

How do we prove the equivalence of these two equation of motions? Does the procedure break down for interacting fields?

Edit: The following is my understanding of how the question is answered. I consider a scalar field for simplicity.

1/ Because the field is free we can write the operator $\phi$ as a superposition of plane-wave operators:

$$\phi(x)=K\int (d^3p) a_p e^{-ipx}+a_p^{\dagger}e^{ipx}$$

Where K is a normalisation constant. The problem is now to find what equation governs the time evolution of $a_p$

2/As an operator, $a_p$ evolves according to $\frac{d}{dt}a_p=i[H, a_p]$

3/ Because the field is free we can write $H$ in the form:

$$H=\int(d^3p)\omega a_p^{\dagger}a_p$$ 4/Inserting into the commutator and using commutation relations this gives $$\frac{d}{dt}a_p=i\omega a_p$$ $$\frac{d^2}{dt^2}a_p=-\omega^2 a_p$$ 5/These equations for operators are the same as the classical equations, which justifies the use of the classical Euler-Lagrange equations for the operators.

Is this correct?

Answer

Let's assume that there are no obstructions to quantization, ordering issues, etc. This is perfectly fine in most physical cases and I think this makes the answer more understandable.

The answer has two parts:

1. Given that the quantum Hamiltonian is nothing more than the classical Hamiltonian with hats in the fields and momenta $$\hat H=H_{cl}\, (\hat \Pi , \hat \Phi)$$ and that he Dirac prescription holds $$[\cdot, \square]=i\hbar \{\cdot , \square\}$$ with the dot and the square any field or momentum, then it is clear that the classical and the quantum equations of motion in the Heisenberg picture are formally the same.



2. If the equations of motion are linear in the fields, then the previous formal equivalence is additionally "real", namely: the expectation values of the fields evolve like the classical fields. This is Ehrenfest theorem.

Example: For simplicity, considerer the following quantum mechanical problem (the generalization to QFT is immediate):

$$H_{cl} (P,Q)= {P^2\over 2}+{Q^2\over 2}+g{Q^3\over 3}$$ with the standard Poisson brackets. Note that this is the harmonic oscillator (in some convenient units where the mass and frequency are set to 1) plus an interaction term. The classical equation of motion are obtained as you very well know (taking Poisson brackets with the Hamiltonian) and in its second order form is: $$\ddot Q + Q + gQ^2=0$$ So far everything is classical. Now, the quantum Hamiltonian is simply: $$\hat H= H_{cl}\, (\hat P , \hat Q)= {\hat P^2\over 2}+{\hat Q^2\over 2}+g{\hat Q^3\over 3}$$ with the canonical commutation relations obtained from the Dirac prescription. We are in the Heisenberg picture. As you should verify (taking commutators with the Hamiltonian) the quantum equation of motion is: $$\ddot {\hat Q} +\hat Q + g\hat Q^2=0$$
This is the previous first part; as you see both equations are formally the same.

However, physically —rather than formally— we are interested in the evolution of the expectation values of observables instead of in the evolution of the own operators $\hat Q$ in the Heisenberg picture (these operators do not depend on time in the Schrödinger picture and physics cannot depend on the picture humans decide to use). So, we can take the expectation value of the previous equation in a generic state $|\Psi \rangle$

$$\frac{d^2}{dt^2} {\langle \Psi |\hat Q|\Psi\rangle }+\langle \Psi |\hat Q|\Psi\rangle + g\langle \Psi |\hat Q^2|\Psi\rangle =0$$ Note that since we are in the Heisenberg picture, physical states do not evolve in time and we are allow to write the time derivatives out of the expectation value. The big question is: Does this equation imply that the expectation value of the quantum observable (the physical thing) evolve classically? And the answer is rotundly negative because: $$\langle \Psi |\hat Q^2|\Psi\rangle \neq\langle \Psi |\hat Q|\Psi\rangle ^2$$ that is: in genera, the expectation value of the square is not the square of the expectation value (the difference is the square of the standard deviation or indetermination).

In the $g=0$ case the last term is absent, the equation is linear and the evolution of the quantum expectation value is classical. That is, calling $q\equiv \langle \Psi |\hat Q|\Psi\rangle$:

$$\ddot q + q=0$$ which is the classical equation (with $g=0$).