Using the EoM in the canonical quantization of EM field

Question

Starting from the classical electromagnetic field, there two approaches to quantization that I want to compare. The problem arises when I write the classical fields in terms of $a$ and $a^*$, which after quantization will "become" the usual creation and annihilation operators. A brief description of the two approaches is the following

1. Starting from the Maxwell Lagrangian, I derive the corresponding Hamiltonian. Now, I perform a spatial Fourier expansion of $A(\vec{r},t)$ then making the proper substitutions I can rewrite it in terms of $a$ and $a^*$;
2. Starting from the Maxwell Lagrangian, I write energy (so I'm still working with lagrangian variables) and I write the solution to the EoM in terms of Fourier modes (in other words, I use $\omega(k)=ck$) and then I write energy in terms of $a$ and $a^*$, obtaining the same functional form as in point 1. but having used the EoM to get a critical cancellation.

My question is: why do I need to use the EoM in the second approach? Writing in terms of $a$ and $a^*$ shouldn't need EoM: consider e.g. a simple harmonic oscillator, where we'd only make a substitution (ignoring constants) $$\begin{cases}a\sim q+ip \\ a^\star\sim q-ip\end{cases}$$ we're not using the EoM to make this substitution (or rather, we're not using them in the Hamiltonian). In case I am wrong, where does my argument break? In other words, it seems that in one case we use the EoM and in the other we postulate.

Approach 1

Considering the electromagnetic potential $(\phi,\vec{A})$ in the quantization volume $V$, I expand in a Fourier series (I'm assuming Coulomb gauge) $$\phi(\vec{r},t)=\frac{1}{\sqrt{V}}\sum_{\vec{k}}\phi_{\vec{k}}(t)e^{i\vec{k}\cdot\vec{r}}\qquad \vec{A}(\vec{r},t)=\frac{1}{\sqrt{V}}\sum_{\vec{k},\sigma}q_{\vec{k},\sigma}(t)\vec{\epsilon}_{\vec{k},\sigma}e^{i\vec{k}\cdot\vec{r}} \tag{1.1}\label{1.1}$$ I haven't used the EoM (Maxwell equations), which after the transformation are HO equations for the fourier modes. Note that the reality condition amounts to $$\phi^\ast_{\vec{k}}=\phi_{-\vec{k}}\qquad q^\ast_{\vec{k},\sigma}=q_{-\vec{k},\sigma}\tag{1.1.1}\label{1.1.1}$$ I now consider the Maxwell Lagrangian $$L=\frac{1}{2}\int d^3\vec{r}\left\{\bigg\lvert-\frac{1}{c}\frac{\partial\vec{A}}{\partial t}-\nabla\phi\bigg\rvert^2-\lvert\nabla\times\vec{A}\rvert^2\right\}\tag{1.2}\label{1.2}$$ I use \eqref{1.1} inside \eqref{1.2} $$L=\frac{1}{2}\sum_{\vec{k},\sigma}\frac{1}{c^2}|\dot{q}_{\vec{k},\sigma}|^2-k^2|q_{\vec{k},\sigma}|^2\tag{1.2.1}\label{1.2.1}$$ (I omitted the calculations of this part but it's really a bunch of fourier integrals and vector calculus). We may now consider the canonical momentum $$p_{\vec{k},\sigma}=\frac{\partial L}{\partial\dot{q}_{\vec{k},\sigma}}=\frac{1}{c^2}\dot{q}_{\vec{-k},\sigma}\implies \dot{q}_{\vec{-k},\sigma}=c^2p_{\vec{k},\sigma}\tag{1.2.2}\label{1.2.2}.$$ As we expressed the $\dot{q}$ in terms of the momenta we can now obtain the Hamiltonian by means of a Legendre transformation $$H=\frac{1}{2}\sum_{\vec{k},\sigma}c^2|p_{\vec{k},\sigma}|^2+k^2|q_{\vec{k},\sigma}|^2\tag{1.3}\label{1.3}$$ Finally, definining $$q_{\vec{k},\sigma}=:\sqrt{\frac{\hbar c}{2k}}(a_{\vec{k},\sigma}+a^{\ast}_{-\vec{k},\sigma})\qquad p_{\vec{k},\sigma}=:\frac{1}{i}\sqrt{\frac{\hbar k}{2c}}(a_{-\vec{k},\sigma}-a^{\ast}_{\vec{k},\sigma})\tag{1.4}\label{1.4}$$ It is easy to see that $$H=\sum_{\vec{k},\sigma}\hbar cka^*_{\vec{k}}a_{\vec{k},\sigma}\tag{1.5}\label{1.5}$$

Approach 2

To make things easier I'll consider the radiation gauge $\phi=0, \nabla\cdot\vec{A}=0$.

In this case, in addition to the spatial fourier expansion, I also use the EoM - which amount to $\omega(k)=ck$ - to write the Fourier expansion (in a different fashion) as $$\vec{A}(\vec{r},t)=\frac{1}{\sqrt{V}}\sum_{\vec{k},\sigma}\left\{A_{\vec{k},\sigma}(t)\vec{\epsilon}_{\vec{k},\sigma}e^{i\vec{k}\cdot\vec{r}-ickt}+A^{\ast}_{\vec{k},\sigma}(t)\vec{\epsilon}_{\vec{k},\sigma}e^{-i\vec{k}\cdot\vec{r}+ickt}\right\}\tag{2.1}\label{2.1}$$

Using \eqref{1.2}, I write the corresponding energy $$E=\frac{1}{2}\int d^3\vec{r}\left\{\frac{1}{c^2}\bigg\lvert\frac{\partial\vec{A}}{\partial t}\bigg\rvert^2+\lvert\nabla\times\vec{A}\rvert^2\right\}\tag{2.2}\label{2.2}$$

Using \eqref{2.1} inside \eqref{2.2} leads to $$E\propto\sum_{\vec{k},\sigma}k^2(A^{\ast}_{\vec{k},\sigma})A_{\vec{k},\sigma}$$ which can be recast as \eqref{1.5} defining $a$ and $a^*$ by means of a proper rescaling of $A$ and $A^*$ If we didn't use the EoM, a critical cancellation leading to this form wouldn't happen (see Ref.1). The reason why it happens is because we have that $\dot{a}(t)\propto ck a(t).$

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References

1. J.J. Sakurai, Jim Napolitano. Modern Quantum Mechanics, Third edition. Section 7.8.1, pages 466-467.

Answer 1

I think you can use the equation of motion for the simple harmonic oscillator too. The Euler Lagrange equation for the position operator $X$ is

$$\frac{d^2X}{dt^2}+\omega ^2 X=0$$

so we get:

$$X=ae^{-i\omega t}+a^{\dagger} e^{i\omega t}$$

and a similar solution for $P$

If you substitute $t=0$, you recover the usual substitutions : $X=a+a^{\dagger}$, $P=a-a^{\dagger}$ (upto constants).

Since energy is conserved, it doesn't matter whether you used the time-dependent solutions obtained from the equations of motion, or you use the time independent variable subsitution obtained by subsituting $t=0$ in the general solution. Either way, you obtain the same Hamiltonian written in terms of $a$ and $a^{\dagger}$. The time dependence cancels out to give you a conserved energy.