Are the intensities of waves equal when they are represented in real vs complex notation?

Question

The intensities should be equal no matter how a wave is represented. So clearly I think i'm making some elementary mistake, it seems they are not same :

$$ \Phi(x,t) = A_0 \cos{(kx-wt)} \\ \Phi(x,t) = A_o e^{i(kx-wt)} \qquad \text{complex notation} $$

So in the above it is understood that the physical part of the wave is the real part, and the imaginary part is to simply calculations by avoiding complicated trigonometric calculations of the former representation.

But then intensity is given by square of the amplitude and modulus square of the wavefunction respectively for the two forms. Then, $$ I = \Phi(x,t)^2 = A_0^2 \cos^2{(kx-wt)} \hspace{2cm} \text{for the 'real' representation} \\ \text{while,} \hspace{1cm} I = |\Phi(x,t)|^2 = \Phi^*\Phi = A_0^2.\hspace{2cm} \text{for the complex representation} $$

These are different...What's the error that I am making? My prof couldn't clear it up, gave some explanation I couldn't understand.

Thank you for any answers.

Answer 1

The error you are making is in the notation. If $\Phi = A_0 cos(\phi)$ then $\Phi \ne A_0 e^{\mathfrak j \phi}$. Instead you should write, say, $\tilde \Phi = A_0 e^{\mathfrak j \phi}$. Then notice that $\Phi ^2 \ne |\tilde \Phi|^2$ so your definition of the intensity associated with $\tilde \Phi$ should be something else.

In fact, the magnitude square operation on the complex sinusoid is a form of averaging over a cycle. This you can see by calculating $\langle \Phi^2\rangle = \frac{1}{T}\int_0^T A_0^2cos^2(\frac{2\pi}{T} u)du = \frac{1}{2}A_0^2$, which is equal to $\frac{1}{2}|\tilde \Phi|^2$. Now if you define the power (intensity) associated with the complex envelope $A_0$ to be $ \frac{1}{2}A_0^2$ you get the correct average power (intensity) for both cases.

You could also view this complex wave $\tilde \Phi = A_0 e^{\mathfrak j \phi}=A_0 cos(\phi) + \mathfrak j A_0 sin (\phi)$ as having two sinusoids each of intensity $\frac{1}{2}A_0^2$ in it in quadrature, so the total intensity of the two is $A_0^2$. Naturally, then only the half of the latter is the real thing.

Answer 2

As long as you manipulate the complex representation of the wave by linear operations (e.g. adding complex waves, multiplying by a scalar, integrating or differentiating), you can always recover the "real" wave by taking the real part. However, as you have noticed, the square or square modulus operations are not linear and do not have the same property: more generally for a complex number $A+iB$, $$\text{Re}\left\{(A+iB)^2\right\}=A^2-B^2\ne A^2,\ \ \ \ \text{Re}\left\{|A+iB|^2\right\}=A^2+B^2\ne A^2.$$

The square modulus of the complex representation does have a meaning when the time variation is harmonic (i.e. sinusoidal), as in this case. Then, we can represent the wave as a phasor by dropping $e^{-i\omega t}$, the factor describing the time variation. $$\Phi(x, t)=A_0\cos(kx-\omega t)\ \ \ \rightarrow\ \ \ \Phi_P(x)=A_0e^{ikx}.$$ While you don't really have to drop the time factor, there is usually no value in keeping it because it is implied and doesn't affect calculations. $|\Phi_P(x)|^2/2$ gives the time average of $\Phi^2(x,t)$. This is true for more complicated functions $\Phi(x,t)$ as well, as long as they have sinusoidal time variation. Some applications where this arises are complex power and the complex Poynting vector.