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I'm currently studying Quantum Mechanics, and I have just been presented Schrödinger's (time dependent) equation. Of course, the first solution to said equation I've been taught is that of a (complex) plane wave: $\Psi(\vec{r}, t)=A\cdot e^{i[\vec{k}\cdot \vec{r}-\omega t]}$. I also understand Schrödinger's equation does not accept real functions as solutions, such as that of a travelling wave, which we typically get from solving Maxwell's equations. Nonetheless, I'm wondering whether the opposite is also true. Could we find a solution to the equation:

$$\nabla^2 \vec{E} = \frac{1}{c^2}\cdot \frac{\partial^2 E}{\partial t^2}$$

which is not real but in fact complex? Again, the complex form of a plane wave is a solution for this equation, not only its real part. That is to say, we often introduce $\vec{E} = \vec{E}_0\cdot ex^{i[\vec{k}\cdot \vec{r}-\omega t]}$ in the classical wave equation for the electric field and then take its real part, but why couldn't we take the whole complex expression as a solution, as we do in QM? Would this imply the electric field is also a complex function, just as a quantum wave function? If so, would the electric field lose its physical meaning, just as a wave function is not considered a physical entity but rather a mathematical object associated to a particle?

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  • $\begingroup$ "I also understand Schrödinger's equation does not accept real functions as solutions," --> This is not correct. For example, for a free particle, $e^{i k x}$ and $e^{-ikx}$ are solutions. Since the Schrodginer equation is linear, superpositions are also solutions, so $\cos(kx) = \frac{1}{2}\left(e^{ikx} + e^{-ikx}\right)$ is also a solution. $\endgroup$
    – Andrew
    Commented Feb 4 at 17:00

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Mathematically, you can consider complex valued solutions of Maxwell's equations.

Physically, it doesn't make sense to consider a complex electric field. Consider the equation for the Lorentz force

$$ \vec{F} = q(\vec{E} + \vec{v}\times \vec{B}) $$

If $\vec{E}$ could be complex, then the force $\vec{F}$ could be complex, which would mean that the acceleration of a charged particle could be complex. However, the acceleration of a particle is always a real-valued quantity. So the field can't be complex in a physical situation.

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Remember, the electric field strength $\vec{E}$ is defined by the force $\vec{F}$ experienced by a charge $q$: $$\vec{F}(\vec{r},t)=q\vec{E}(\vec{r},t)$$ Both charge $q$ and force $\vec{F}$ are real measurable quantities. Hence the electric field $\vec{E}$ must be real too.
The fact that we (for mathematical convenience) sometimes use a complex $\vec{E}$ in intermediate calculational steps doesn't matter, because at the end we throw away its imaginary part and take only its real part as the final result.

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