# Why is there a negative sign in front of the optical wave?

In undergrad I lost (a lot) of marks in my optics class for writing:

$$A(t) = \exp(i(\omega t + \phi))$$

$$A(t) = \exp(i(-\omega t + \phi))$$

In a derivation where I must have needed a plane wave. At the time I thought the TA was being pedantic. Both forms represent a plane wave and are mathematically equivalent within some aesthetic changes (namely $\omega -> -\omega$). Now I realize this has huge implications if we need to take the derivative of the signal.

Physically why is it negative omega? Why do I see a negative sign pop up in every time it involves $t$, but the more standard for of the Fourier series is used otherwise?

I don't think it's very likely, but one other thing I can think of: when the sign before the $\omega$ is a minus then the wave represents a wave travelling to the "right" - positive x direction - and maybe your TA wants only waves travelling to the right. ^^

In case you don't know why the minus sign represents a wave travelling in the positive x direction:

Normally I write a plane wave like this (I consider only the x component here):

$$exp[i(k_xx\pm\omega t)]$$

The phase of the plane wave is obviously the term in the exponential. Let's consider a point of constant phase, let's say zero. The term in the brackets must then give

$$k_xx\pm\omega t=0$$

When the wave and hence this point of constant phase is travelling to the right then $x$ increases. Since time also increases then the only possibility to obtain a wave travelling in the positive x-direction is to put a minus sign before the omega, such that

$$k_xx-\omega t=0$$

for all times for this point of constant phase.

This convention of course depends on the way he writes the plane wave. If your write the plane wave as

$$exp[i(\omega t\pm k_xx)]$$

Then there would be no minus sign befor the omega.

• Your TA wants waves travelling to the right was pretty much the first thing I thought on reading the question. +1 – 299792458 Apr 7 '15 at 12:30
• This sounds like it is THE answer, although, as you say, not very clear from how the OP posed the question. – WetSavannaAnimal Apr 7 '15 at 13:32
• To be honest the way the OP stated his question it's also not very clear to me what the TA wants... But since you, @WetSavannaAnimalakaRodVance, covered the "conventions"- and "fourier series" aspect I thought that "travelling right" could be another possibility. – user42076 Apr 7 '15 at 13:48
• I took optics ten years ago, I forgot a few details in the meantime. So I can't honestly say if there was anything explicitly in the question about direction of travel. – Lenzuola Apr 7 '15 at 17:58

You may have lost marks for leaving out the imaginary unit $i$. If not, then understand that either $e^{\pm i\,\omega\,t}$ can be used to represent the real signal with positive frequency $\omega$ - it's wholly a question of convention. But once you have made the choice you must stick with it and the choice has implications throughout all the equations of physics you may use. This may have been why your prof took marks off: if you learn equations one way, you must use the corresponding sign in $e^{\pm i\,\omega\,t}$ to make them work. For instance, if we choose $e^{-i\,\omega\,t}$ to represent positive frequencies, then Faraday's law is $\nabla \times \vec{E} = +i\,\omega\,\vec{B}$ and Ampère's law $\nabla\times \vec{H}=\vec{J}-i\,\omega\,\vec{D}$. They become $\nabla \times \vec{E} = -i\,\omega\,\vec{B}$, $\nabla\times \vec{H}=\vec{J}+i\,\omega\,\vec{D}$ if we use $e^{+i\,\omega\,t}$ for positive frequencies. Also, the Fourier transform changes of course: if you want to find the superposition weights $F(k)$ of $e^{-i\,\omega\,t}$ representing the signal $f(t)$ then you use $F(k)= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{+i\,\omega\,t}\,f(t)\,\mathrm{d}t$ and the sign in the exponent swaps if instead your convention calls on you to find superposition weights of $e^{+i\,\omega\,t}$.

Lastly, the convention $e^{-i\,\omega\,t}$ is often chosen in optics to make it consistent with quantum mechanics, in particular the Schrödinger equation, which is $i\,\hbar\,\partial_t \psi = \hat{H}\,\psi$. In this convention, a positive energy eigenstate at energy eigenvalue $E_0$ has time phase variation defined by $i\,\hbar\,\mathrm{d}_t \psi = E_0\,\psi$. This of course is $\exp\left(-i\,\frac{E_0}{\hbar}\,t\right)$.

• I didn't forget the imaginary unit - it's not a plane wave without it. I remember asking the TA specifically about the negative sign, and being told the negative had to be there. – Lenzuola Apr 7 '15 at 11:42
• @Lenzuola Try asking him/ her whether the objection is ultimately owing to the use of the "wrong" sign together with another equation of physics. As in my example, the sign does bear on the signs of quantities in time-harmonic Maxwell equations as well as FT and many others. – WetSavannaAnimal Apr 7 '15 at 12:06
• I took optics ten years ago. I asked the question because I was reading a paper that wrote $U = U_{o}exp(-j\omega t)$ for vacancy concentration. I'm implementing the paper's model in a program, and that negative sign makes a huge difference (and is welcome). But I can't just add a negative sign in just because I like it. I wanted to know the implications of the negative sign so I can figure out where in my own model I have implicitly assumed my that concentration wave is $U=U_{o}exp(-j\omega t)$ – Lenzuola Apr 7 '15 at 18:06
• @Lenzuola I think that NoEigenvalue's answer seems the most plausible for the TA incident: there must have been a direction of travel in the question. I'm sure the negative sign does make a difference to your software: I think that you're almost certainly using it implicitly in several places (through the different equations of physics and the FT). I have had this same problem with software many times and it is a gigantic headache. Make sure you understand the sign convention for your FT libraries, as well as sign conventions for the branches of square roots (if you need to do things like .... – WetSavannaAnimal Apr 7 '15 at 22:43
• ... work out $\sqrt{k^2 - k_x^2-k_y^2}$). Another possibility is the workings of the function $\mathrm{atan2}: \mathbb{R}\times\mathbb{R}\to \mathbb{R}$ and branches of complex logarithms. I don't think any of these things is standardized by e.g. ANSII C++. Certainly Microsoft Visual Studio, gcc, Code Warrior and C++ Builder have slight differences in the branch conventions of all these things. – WetSavannaAnimal Apr 7 '15 at 22:47

I think that you are missing some $i$ or $j$ (complex $\sqrt{-1}$ in your expression, like in $exp[j(\omega t-\phi)]$.

consult this page of the book: The Light Fantastic: A Modern Introduction to Classical and Quantum Optics and read the equation 5.10 and the side note: The choice of the complex form ... is made for convenience.

The frequency $\omega$ is not negative. The variable time $t$ can be positive or negative depending on the choice of the origin. Idem for the phase ($\pm\phi$) .

As long as your formulas and derivations are consistent and correct no points should be subtracted.

The real exponential you wrote (always increasing or decreasing) is not the same as the complex exponential: real part (cos..) + imaginary part (j.sin..)

In electronics we are used to express quantities like current and tension as complex quantities but those quantities are indeed real valued. It is a mathematical treatment to simplify the description: For instance the fact that tension and current are out of phase for a non-resistive circuit (inductive or capacitive) is easy to describe with complex notation. The tension, or current, is indeed described by a sin function or equally well by a cos (actually by an infinite sum of such functions - see Fourier transform).
The $\omega=2\pi\ f$ is an angular frequency (it reads 'the number of times' that..flips..per unit time) and I can only understand it as non negative.

The time variable $t$ can have any real value which means that it is irrelevant if you use $+t$ or $-t$ in the expression. Of course when you represent the time derivative (the temporal rate of change) the signal must correspond to the correct one. By convention negative values of time represent past moments irt to the $t=0$. Usually $+t$ is used in the equations.
Now look at the equation 5.10 as I suggested above. The $kz$ part represents that the wave appears to be in motion thru space, where $k=\omega/c$ . See Wave Equation $u(x, t) = F(x - c t) + G(x + c t)$

In other words, solutions of the 1D wave equation are sums of a right traveling function F and a left traveling function G. "Traveling" means that the shape of these individual arbitrary functions with respect to x stays constant, however the functions are translated left and right with time at the speed $c$

In physics it is common an erroneous interpretation of the complex plane as been a fundamental quality of the nature, notably in Quantum Mechanics. Indeed it is only a mere convenient mathematical description, the same as in the electronics domain. QM under the Geometric Algebra (GA) perspective do not use the complex plane at all (find Hestenes).

The time is source of misinterpretations. The fact that one can use negative values, or even represent the description of a phenomenon backwards like presenting the evolution of the Solar System from the present to the past make some to forget the nature of time: a notation to describe the evolution of events. And events by definition can not be undone: One can describe it but the past is always inaccessible.

• Missing $i$ < corrected. Thank you for your other comments. Although, for unsymmetrical signals (I'm speaking generally, not specifically about optics), isn't the sign of the $\omega$ relevant? (Since $exp(jx) = cos(x) + jsin(x)$ and $sin(-x) = - sin(x)$) – Lenzuola Apr 7 '15 at 18:09
• @Lenzuola I've added a long story to my answer. – Helder Velez Apr 7 '15 at 23:55