What does the complex electric field show?

Question

We have a complex electric field. Is there any definition for absolute and imaginary part of a complex electric field? What do they stand for?

Answer 1

Jonas's answer shows one way wherein complex numbers are useful in representing sinusoidally varying with time quantities. The quantity $e^{i\,(\omega\,t + \delta)}$ when it replaces $\cos(\omega\,t+\delta)$ in a linear equation (or $\sin(\omega\,t+\delta)$ if one "favours the imaginary part" in Jonas's words) is called a phasor. The phasor method is applied widely throughout physics, not only to electric fields.

However, in the particular case of Maxwell's equations, there is a radically different way to bring in complex equivalents of the electromagnetic field that has a neat interpretation in terms of polarization. In practice, it ends up being used in a way very like the phasor method, even though its grounding is altogether different.

This is the idea of diagonalizing the Maxwell curl equations (Faraday and Ampère laws) with the Riemann-Silberstein fields which are:

$$\vec{F}_\pm = \sqrt{\epsilon_0} \,\vec{E} \pm i\,\sqrt{\mu_0} \,\vec{H}\quad\quad\tag 1$$

and which decouple the Maxwell curl equations into the following form:

$$i\, \partial_t \vec{F}_\pm = \pm c\,\nabla \wedge \vec{F}_\pm\quad\quad\tag2 \label{eq:2}$$

Note that by taking the divergence of both sides of $\eqref{eq:2}$ we get $i\, \partial_t \vec{F}_\pm =0$, so that if the fields are time varying and have no DC (zero frequency) component (i.e. $\partial_t$ is invertible), $\eqref{eq:2}$ also implies the Gauss' laws $\nabla\cdot\vec{F}_\pm=0$ too.

Now one could simply sit with real electric and magnetic fields and one would need only one complex Riemann-Silberstein vector (either of $\vec{F}_\pm$ will do just as well as the other) to stand in the stead of two real fields and then the real valued curl equations are replaced by one complex-valued one. So one would interpret the real part as the field $\sqrt{\epsilon_0} \,\vec{E}$ and the imaginary part as $\pm\sqrt{\mu_0} \,\vec{H}$ (depending on whether $\vec{F}_\pm$ were used) at the end of the calculation.

However, it turns out to be more physically meaningful to keep both vectors, but to throw away their negative frequency parts and keep the positive frequency parts alone of both vectors. What's really neat about this second approach is that if the light is right circularly polarized, only $\vec{F}_+$ is nonzero; if left, only $\vec{F}_-$ is non-zero. So the positive frequency parts of the electromagnetic fields are decoupled precisely by splitting them into left and right circularly polarized components.

Now to restore a field's negative frequency part from the positive frequency part alone, one adds the complex conjugate, i.e. we're still effectively taking the real part of the $\vec{F}_\pm$ fields at the end of the calculation, so the practicalities are rather like the phasor method. But now we take:

$$\begin{array}{lcl} \vec{E} &=&\operatorname{Re}\left(\displaystyle \frac{\vec{F}_+ + \vec{F}_-}{2\,\epsilon_0}\right)\\ \vec{H} &=&\operatorname{Re}\left(\displaystyle \frac{\vec{F}_+ - \vec{F}_-}{2\,i\,\,\mu_0}\right)=\operatorname{Im}\left(\displaystyle \frac{\vec{F}_+ - \vec{F}_-}{2\,\mu_0}\right) \end{array} \tag 3$$

to get our "physical" fields at the end of the calculation. But, given the physical, manifestly Lorentz covariant interpretation of the Riemann-Silberstein vectors I talk about below (see "more advanced material" below), one might just as well say that $\vec{F}_\pm$ are the physical fields (even though they're not what you would measure with a vector voltmeter or magnetometer). In this framework of thought, a quantity's being real or imaginary has a geometric meaning as whether it is bivector or a Hodge dual thereof in the Clifford algebra $C\ell_3(\mathbb{R})$ wherein the now "spinor" $\mathbf{F}_\pm$ live and the entity $i$ is now the unit pseudoscalar in this algebra. Bivectors and their Hodge duals mix and transform differently under the Lorentz transformation $\eqref{eq:8}$, so, if you like, you can very soundly take this difference as the meaning of real and imaginary parts.

Lastly, since now $\eqref{eq:2}$ is confined to two equations in positive frequency (therefore positive energy) we can now interpret $\eqref{eq:2}$ as the time evolution, i.e. Schrödinger equation for the quantum state of a first quantized photon. See:

• I. Bialynicki-Birula, "Photon wave function" in Progress in Optics 36 V (1996), pp. 245-294 also downloadable from arXiv:quant-ph/0508202*

for more details.

---

More advanced material

The Riemann-Silbertein vectors are actually the electromagnetic (Maxwell) tensor $F^{\mu\nu}$ in disguise. We can write Maxwell's equations in a quaternion form:

$$\begin{array}{lcl} \left(c^{-1}\partial_t + \sigma_1 \partial_x + \sigma_2 \partial_y + \sigma_3 \partial_z\right) \,\mathbf{F}_+ &=& {\bf 0}\\ \left(c^{-1}\partial_t - \sigma_1 \partial_x - \sigma_2 \partial_y - \sigma_3 \partial_z\right) \,\mathbf{F}_- &=& {\bf 0}\end{array}\tag 4$$

where $\sigma_j$ are the Pauli spin matrices and the electromagnetic field components are:

$$\begin{array}{lcl}\dfrac{1}{\sqrt{\epsilon_0}}\mathbf{F}_\pm &=& \left(\begin{array}{cc}E_z & E_x - i E_y\\E_x + i E_y & -E_z\end{array}\right) \pm i \,c\,\left(\begin{array}{cc}B_z & B_x - i B_y\\B_x + i B_y & -B_z\end{array}\right)\\ & =& E_x \sigma_1 + E_y \sigma_2+E_z\sigma_3 \pm \mathbb{i}\,c\,\left(B_x \sigma_1 + B_y \sigma_2+B_z\sigma_3\right)\end{array}\tag 5$$

The Pauli spin matrices are simply Hamilton's imaginary quaternion units reordered and where $\mathbb{i}=\sigma_1\,\sigma_2\,\sigma_3$ so that $\mathbb{i}^2 = -1$. When inertial reference frames are shifted by a proper Lorentz transformation:

$$L = \exp\left(\frac{1}{2}W\right)\tag 6$$

where:

$$W = \left(\eta^1 + i\,\theta\, \gamma^1\right) \sigma_1 + \left(\eta^2 + i\,\theta\, \gamma^2\right) \sigma_2 + \left(\eta^3 + i\,\theta \gamma ^3\right)\,\sigma_3\tag7$$

encodes the transformation's rotation angle $\theta$, the direction cosines of $\gamma^j$ of its rotation axes and its rapidities $\eta^j$, the entities $\mathbf{F}_\pm$ undergo the spinor map:

$${\bf F} \mapsto L {\bf F} L^\dagger. \tag 8 \label{eq:8}$$

Here, we're actually dealing with the double cover $PSL(2,\mathbb{C})$ of the identity-connected component of the Lorentz group $SO(3,1)$, so we have spinor maps representing Lorentz transformations, just as we must use spinor maps to make a quaternion impart its represented rotation on a vector.

Answer 2

Actually, electric fields are real. Using complex exponentials bears no advantage other than convenient calculation. The interpretation usually is that the imaginary part is discarded and only the real part is taken to be, well, real. Of course, this is completely arbitrary. One could also favor the imaginary part and take it to represent the physics. This works because we usually only add electric fields or multiply them by scalars (linear operations). If the scalar is complex as well, we can use this to represent phase shifts which is also a convenient thing to do. One has to pay attention, though, when taking higher-order functions of those complexified quantities. E.g. the electric field energy density which would usually be proportional to $|\vec E|^2$ (where the vertical bars refer to the euclidian vector norm rather than the complex mod) has to be replaced by $(Re(\vec E))^2$.

Another common thing is to write something like

$\vec E(\vec x,t)=\vec E_0\,(e^{i(k\vec x-\omega t)} + c.c.).$

c.c. means complex conjugate, thus, you take the complex conjugate of the first term and add it up such that the result will be twice the real part. Therefore, in this notation, the electric field is real but we can still work with complex exponentials and we don't have to write out the complex conjugate.

Answer 3

The real and imaginary parts of an electric field are similar to the position and velocity of a pendulum. If you only know the pendulum's position at a given moment, you can't predict its future motion. You also need to know its velocity. In this analogy, the imaginary part of the electric field corresponds to velocity.

For instance, if you know the pendulum is $0.2 \ \text{m}$ to the left of its equilibrium point (the real part), you can't tell whether it will keep moving left or start swinging back to the right without knowing its velocity.

However, you can capture both the position and velocity in the real part if you express it in polar form, which reveals how the position evolves over time.

$$x(t) = A \cos(\omega t + \phi)$$

where:

• $A$ is the amplitude of oscillation.
• $\omega$ is the angular frequency.
• $\phi$ is the phase, which determines the pendulum's starting point in its swing.

But to determine this polar form, you need both the real and imaginary parts.

You can rewrite this equation using complex notation, just like you would for the electric field:

$$x(t) = \text{Re} \left\{ A e^{\displaystyle j(\omega t + \phi)} \right\}.$$

In this form, $ A e^{\displaystyle j\phi} $ is a complex number that encodes both the pendulum's position (real part) and its velocity (imaginary part) at any given time.