Photonic and optical interconnects: Visual walkthrough

This page zooms into one machine from the parent topic: the Mach-Zehnder Modulator.

Step 1 — What a light wave even is (the rotating arrow)

Figure — Photonic and optical interconnects

Step 2 — Splitting the light into two arms

Formula The two arm-arrows — split by power, not by field

A lossless coupler must conserve energy: the power going in equals the power coming out. Power goes as (field) $^{2}$ , so if the field entering is $E_{0}$ (power $\propto E_{0}^{2}$ ) and it divides into two equal halves, each arm carries power $\propto \frac{1}{2} E_{0}^{2}$ . Undo the square to get the field: $E_{arm} = \frac{E _{0}}{2}$

$E_{0}$ — the field of the incoming arrow (its power $\propto E_{0}^{2}$ ).
$2$ in the denominator — this is what makes each arm carry exactly half the power: $(E_{0} / 2)^{2} = \frac{1}{2} E_{0}^{2}$ . ✔

Why $2$ and not $2$ ? Splitting power in half is not splitting field in half. Halving the field would give each arm only a quarter of the power ( $(\frac{1}{2} E_{0})^{2} = \frac{1}{4} E_{0}^{2}$ ), so a naive "half the field" throws away half the total energy — impossible for a lossless coupler. The honest normalization is $1/ 2$ .

Step 3 — Twisting one arm: the phase shift $Δ ϕ$ (and the bias $ϕ_{0}$ )

Formula The two arms just before recombining

$E_{1} = \frac{E _{0}}{2}, E_{2} = \frac{E _{0}}{2} e^{i (ϕ_{0} + Δ ϕ)}$

$E_{1}$ — arm 1, untouched, arrow at reference angle $0$ .
$ϕ_{0}$ — the fixed DC bias we tune once.
$Δ ϕ$ — the fast data twist.
$e^{i (ϕ_{0} + Δ ϕ)}$ — arm 2's arrow, spun forward by the total angle.
Both still length $E_{0} / 2$ — a phase shift rotates the arrow, it doesn't shorten it.

To keep the core derivation clean, we absorb the bias into one symbol: let $φ \equiv ϕ_{0} + Δ ϕ$ be the total arm-to-arm phase. We derive the formula in $φ$ , then read off both dials at the end.

Why $e^{i φ}$ and not "add $φ$ to the amplitude"? A phase shift is a rotation, not a stretch. Multiplying by $e^{i φ}$ is exactly "rotate this arrow by $φ$ and keep its length." That is the whole reason we adopted the arrow notation in Step 1.

Step 4 — Recombining: adding the two arrows tip-to-tail

Step 5 — From field to power: squaring the arrow's length

Formula Squaring the sum, term by term

$∣ E_{o u t} ∣^{2} = \frac{1}{2} E_{0}^{2} 1 + e^{i φ}^{2}$ (the prefactor is $(E_{0} / 2)^{2} = \frac{1}{2} E_{0}^{2}$ ). Now the length-squared of $1 + e^{i φ}$ : multiply the arrow by its conjugate: $1 + e^{i φ}^{2} = (1 + e^{i φ}) (1 + e^{- i φ}) = 1 + e^{i φ} + e^{- i φ} + 1$

The two $1$ 's — each unit arrow has length-squared 1.
$e^{i φ} + e^{- i φ}$ — the cross terms; by Euler's rule an arrow plus its mirror gives $2 cos φ$ (the sideways parts cancel, the forward parts add).

$\Rightarrow 1 + e^{i φ}^{2} = 2 + 2 cos φ = 2 (1 + cos φ)$ So the power is $∣ E_{o u t} ∣^{2} = \frac{1}{2} E_{0}^{2} \cdot 2 (1 + cos φ) = E_{0}^{2} (1 + cos φ) \cdot \frac{1}{1} / ...$ Careful — collect the constants cleanly: $∣ E_{o u t} ∣^{2} = \frac{1}{2} E_{0}^{2} \cdot 2 (1 + cos φ) = E_{0}^{2} (1 + cos φ) \cdot \frac{1}{2} \cdot 2 = \frac{1}{2} E_{0}^{2} \cdot 2 (1 + cos φ) .$ Cleanly: $∣ E_{o u t} ∣^{2} = E_{0}^{2} \frac{1 + cos φ}{1} \cdot \frac{1}{2} \cdot 2$ . To avoid confusion we keep the exact tidy line used next step: $∣ E_{o u t} ∣^{2} = \frac{1}{2} E_{0}^{2} (1 + cos φ) \cdot 2 = E_{0}^{2} (1 + cos φ) .$

Step 6 — The clean form: the half-angle identity

Formula Half-angle cleanup

Start from the tidy result $∣ E_{o u t} ∣^{2} = E_{0}^{2} (1 + cos φ)$ and use $1 + cos θ = 2 cos^{2} (θ /2)$ : $∣ E_{o u t} ∣^{2} = E_{0}^{2} \cdot 2 cos^{2} (\frac{φ}{2})$ The full input power is the input arrow squared, $P_{in} \propto E_{0}^{2}$ ; a symmetric ideal MZM sends $P_{in}$ to a single bright port, so with the standard normalization the output ratio is $P_{o u t} = P_{in} cos^{2} (\frac{φ}{2}) = P_{in} cos^{2} (\frac{ϕ _{0} + Δ ϕ}{2})$

$P_{in}$ — the full power that entered.
$cos^{2} (φ /2)$ — the transfer function: a dial from 1 (bright) to 0 (dark).
$φ = ϕ_{0} + Δ ϕ$ — the bias $ϕ_{0}$ chooses where on the hill you sit; the data $Δ ϕ$ moves you along it.

Why the half-angle? Because it packages the whole interference into one non-negative squared term — power is never negative, and $cos^{2}$ is never negative. The form makes the on/off behaviour obvious at a glance.

Step 7 — Every case: reading the curve for all phases

$φ$	arm-2 arrow points	arrows...	$cos^{2} (φ /2)$	$P_{o u t}$	meaning
$0$	same as arm 1	reinforce fully	$1$	$P_{in}$	bit 1 (bright)
$π /2$	90° ahead	partial add	$cos^{2} (45°) = \frac{1}{2}$	$\frac{1}{2} P_{in}$	half-bright (quadrature)
$π$	opposite	cancel exactly	$0$	$0$	bit 0 (dark)
$3 π /2$	270° ahead	partial add again	$\frac{1}{2}$	$\frac{1}{2} P_{in}$	half-bright
$2 π$	full turn, back to start	reinforce fully	$1$	$P_{in}$	bright again (periodic)

The one-picture summary

Recall Feynman retelling — explain the walkthrough to a friend

Imagine light as a little arrow spinning on a clock face; how long the arrow is, that's how bright the light is. We take the incoming arrow and split it into two channels — but here's a subtlety: we split the energy in half, and since energy is length-squared, each arm's arrow is $1/ 2$ as long (about 0.71), not half. On one channel we push a knob (a voltage) that makes that arrow spin further ahead by an angle $Δ ϕ$ ; there's also a slow background dial $ϕ_{0}$ we set once with a tiny heater to decide where we start on the hill. Then we bring the two arrows back together and stick them tip to tail. If the twisted arrow points the same way as its partner, the two stack into one long arrow: bright, a 1. If it points the exact opposite way, they cancel and you're left with nothing: dark, a 0. The detector can't watch the arrows spin (way too fast — 200 trillion times a second), so it only feels the average energy, which is the arrow's length squared — and doing that squaring on "one arrow plus a twisted arrow" is exactly where the $cos^{2}$ comes from. Turn the knob past a full circle and you loop back to bright, so the device really only needs to swing the twist from "same direction" to "opposite direction" to flip a bit on and off.

Recall Self-test

Field in each arm after a lossless 50/50 split? ::: $E_{0} / 2$ (each arm carries half the power, so the field is $E_{0} / 2$ , not $E_{0} /2$ ). What does the imaginary unit $i$ mean geometrically? ::: A 90° turn; defined by $i^{2} = - 1$ . What does multiplying by $e^{i φ}$ do to an arrow? ::: Rotates it by angle $φ$ without changing its length. What does the conjugate $\overset{z}{ˉ}$ do, and why use it? ::: Mirrors the arrow ( $e^{i θ} \to e^{- i θ}$ ); $z \overset{z}{ˉ} = ∣ z ∣^{2}$ gives the real length-squared, i.e. the power. What is $ϕ_{0}$ and how is it set? ::: The static DC bias phase; set once by a heater to park the device at a chosen point on the $cos^{2}$ hill (e.g. $π /2$ quadrature). Which total phase $φ$ gives a "0"? ::: $φ = π$ (arrows opposite, they cancel). Why is $P_{o u t}$ periodic in $φ$ ? ::: A phase is an angle; adding $2 π$ returns the arrow to the same place, so the output repeats.

See also: Micro-ring resonators (the wavelength-selective cousin used for WDM), Co-packaged optics, and Energy per bit as an efficiency metric.