As far as I know (or I thought I knew), if we have an electric field $$\mathbf{E}=\mathbf{E_0}\cos(\omega t - kx),$$ we can define it as the real part of $$\mathbf{E}=Re(\mathbf{E_0}e^{i(\omega t - kx)}).$$

Introducing imaginary components to the electric field is only a matter of making the maths easier, in particular in the case of complex exponentials which are eigenfunctions of the differential operators.


In a Ph.D. thesis I was reading, about lasers and cavity misalignment (which doesn't really matter, my question does not require any knowledge of this), I came across something along the lines of:

The electric field to first order misalignment can be written as $$\mid E\rangle = | 00 \rangle + i | 01 \rangle, $$ where the two kets on the LHS are the fundamental and 1st excited Hermite-Gaussian modes.

But the physical part of the electric field is surely its real part? What purpose can the imaginary component above serve?

  • $\begingroup$ In what sense the electric field is a vector of the Hilbert space? Is it an eigenvector of the operator $\hat{E}$? If it is so, it seems plausible that it is a complex superposition of other states. $\endgroup$
    – yuggib
    Sep 12, 2014 at 13:59
  • $\begingroup$ The Hermite-Gaussian $|nm>$ modes form a complete orthonormal basis $\endgroup$
    – SuperCiocia
    Sep 12, 2014 at 14:02
  • 1
    $\begingroup$ Ok, but it is the meaning of your $\lvert E\rangle$ that is not so clear (to me). In QM, $E$ is usually an observable (self-adjoint operator), not a vector. A vector of the Hilbert space is usually complex valued, while an observable is real-valued. An eigenvector of the observable $E$ is complex-valued, but corresponds to a real eigenvalue. $\endgroup$
    – yuggib
    Sep 12, 2014 at 14:06
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/80028/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 12, 2014 at 14:44
  • $\begingroup$ Note that there are multiple ways of getting a real value out of a complex value. You're correct that you can just take the real part. But you can also take the modulus: $\sqrt{\phi^{*}\phi}$, or take the imaginary part, or a variety of other things. What convention is the author using? $\endgroup$ Sep 12, 2014 at 16:17

2 Answers 2


The $i$ in front of the $|01\rangle$ tells you that the first order mode is $90^\circ$ out of phase with the fundamental mode. The absolute phase generally has no meaning so you could just as well have put $-i$ in front of the fundamental mode.

The phase difference between the two has physical meaning. When you take the projection onto the real axis one will be acting as a cosine and the other as a sine which means that one will be maximum when the other is zero and vice versa. This phase difference is important e.g. in detecting the misalignment between a laser and an optical resonator.

It might help to think about it in the rotating phasor picture. In this picture you think about arrows rotating at the frequency of the optical field, and the physical portion is the projection onto the real axis. Figure 3 on this page has a good example of rotating phasors which are $\sim 30^\circ$ out of phase.


I agree with Chris, the imaginary part here is just the mathematical way of showing a rotation. In this case, it happens to be that the second ket is 90 degrees out of phase with the first. You can think of the imaginary axis as a direction orthogonal to any real axis, thus it is similar to saying the imaginary part is at some angle with respect to the real part [though mathematicians may scold me for my somewhat careless use of terms].

In the first part, if either $\omega$ or $\kappa$ have imaginary parts, those have very important physical consequences. $\Im \left[ \omega \right] \neq 0$ can imply that there is a temporal dependence in the field amplitude. $\Im \left[ \kappa \right] \neq 0$ can imply that there is a spatial dependence in the field amplitude.


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