Complex conjugate and |z|²
In Track 1 you measured qubits and watched histograms build. The rule was: probability equals amplitude squared. This lesson is where that rule comes from — and why the "squared" actually means multiplying by the complex conjugate, not squaring directly.
Probability must be real. It must be non-negative. $z^2$ satisfies neither.
In Track 1 you knew that probability equals the amplitude squared — $P = |\alpha|^2$. You saw it in the measurement histograms: prepare a qubit, measure it many times, watch the bars converge to the predicted values. The rule worked. But there was a hidden problem we didn't need to address yet: what exactly does "squared" mean for a complex number?
The naive answer is: just square it. $z^2 = (a + bi)^2$. Let's see what happens.
Take $z = 1 + i$. Then $z^2 = (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$. The result is purely imaginary. A probability of $2i$ is meaningless — probabilities must be real numbers between 0 and 1. Squaring a complex number almost always produces another complex number. That can never represent a probability.
The problem is clear: ordinary squaring keeps the imaginary part alive. What we need is an operation that kills the imaginary part reliably — turning any complex number into a guaranteed non-negative real number. The complex conjugate is precisely that operation.
In lessons L06 and L09, you measured qubits and saw probabilities emerge from amplitudes. The rule "$P = |\alpha|^2$" was stated as a fact. Now you'll see exactly why it must be written as $\alpha \cdot \alpha^*$ rather than $\alpha^2$ — and why this choice is the only one that guarantees real, non-negative probabilities for every possible complex amplitude.
The conjugate is the mirror image — flip the imaginary part, keep the real part.
Given a complex number $z = a + bi$, its complex conjugate is written $z^*$ (pronounced "z star") and defined as:
$$z = a + bi \qquad \Longrightarrow \qquad z^* = a - bi$$
One rule: keep $a$ exactly as it is; flip the sign of $b$. That's all. The conjugate is the reflection of $z$ across the real axis on the complex plane. If $z$ sits above the axis, $z^*$ sits symmetrically below it — same horizontal distance from the origin, same vertical distance, but on the opposite side.
The real axis is a horizontal mirror. The conjugate $z^*$ is the reflection of $z$ in that mirror. Both $z$ and $z^*$ have the same distance from the origin — the same magnitude $|z|$. Their angles are equal and opposite: if $z$ makes angle $\theta$ with the positive real axis, then $z^*$ makes angle $-\theta$. The conjugate never moves the point left or right — it only flips it up or down.
Three examples worth checking directly:
$z = 3 + 4i \;\Rightarrow\; z^* = 3 - 4i$. The real part stays; the $+4i$ flips to $-4i$.
$z = -2 + 0i \;\Rightarrow\; z^* = -2 - 0i = -2$. A real number is its own conjugate — flipping $0$ does nothing.
$z = i \;\Rightarrow\; z^* = -i$. Pure imaginary numbers flip sign entirely.
Multiply $z$ by $z^*$ and watch the imaginary parts vanish — leaving only $a^2 + b^2$.
Why $z^2$ is not useful
First, let's see the failure mode in full algebra. Take $z = a + bi$:
$$z^2 = (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 + 2abi + b^2 i^2$$
Since $i^2 = -1$:
$$z^2 = a^2 - b^2 + 2abi$$
The result has real part $(a^2 - b^2)$ and imaginary part $2ab$. The imaginary part survives. This is a complex number — not usable as a probability.
Computing $z \cdot z^*$ step by step
Now multiply $z = a + bi$ by its conjugate $z^* = a - bi$:
$$z \cdot z^* = (a + bi)(a - bi)$$
Multiply each term in the first factor by each term in the second:
$$= a \cdot a + a \cdot (-bi) + bi \cdot a + bi \cdot (-bi)$$
$$= a^2 - abi + abi - b^2 i^2$$
Notice: $-abi + abi = 0$. The two cross terms are exactly opposite. They cancel completely, regardless of the values of $a$ and $b$. This is why conjugate multiplication works — the cross terms are guaranteed to cancel because one is the negative of the other.
$$= a^2 - b^2 i^2$$
$$= a^2 - b^2(-1) = a^2 + b^2$$
$$\boxed{z \cdot z^* = a^2 + b^2 = |z|^2}$$
Since $a^2 \geq 0$ and $b^2 \geq 0$ for all real $a, b$, the result $a^2 + b^2$ is always a real number and always greater than or equal to zero. It equals zero only when $a = 0$ and $b = 0$ simultaneously — i.e., only when $z = 0$ itself.
Why the cross terms always cancel — explained plainly
The cross terms are $+abi$ (from $a \times (-bi)$, with a sign flip giving $-abi$, then $bi \times a = +abi$). The reason they cancel: the conjugate $z^*$ is constructed precisely to introduce the negative sign on $b$. When you multiply $(a + bi)(a - bi)$, you are multiplying a number by its mirror image — and the "upward" and "downward" imaginary components destroy each other by design.
Geometrically: $z$ and $z^*$ have equal and opposite imaginary components. When their product is computed, those equal-and-opposite contributions add to zero — the same mechanism as destructive interference, applied to imaginary parts.
Connection to geometry: $|z|^2$ is the squared distance from the origin
On the complex plane, the distance from the origin to the point $(a, b)$ is $\sqrt{a^2 + b^2}$ by the Pythagorean theorem. So $|z|^2 = a^2 + b^2$ is the squared distance — also called the squared modulus. This gives a completely geometric interpretation: the probability of measuring an outcome is the square of the distance of the amplitude from the origin on the complex plane.
$$P(\text{outcome}) = |\alpha|^2 = \alpha \cdot \alpha^* = \text{(distance from origin)}^2$$
Drag a point on the complex plane and watch its conjugate mirror and $|z|^2$ update live.
The canvas shows the complex plane. The blue point is $z = a + bi$ — drag it anywhere. The rose point is $z^* = a - bi$ — always the reflection of $z$ across the real axis. The cyan bar shows $|z|^2 = a^2 + b^2$, the probability this amplitude would contribute to a measurement outcome.
Goal 2 — $|z|^2$ is distance squared: Drag $z$ away from the origin. As the distance grows, $|z|^2$ grows as the square. Notice that moving along a circle centered at the origin keeps $|z|^2$ constant.
Goal 3 — Unit circle check: Drag $z$ to the unit circle (distance = 1 from origin). Confirm $|z|^2 = 1$. Any amplitude on the unit circle represents a state with 100% probability for that single outcome.
Micro practice — three conceptual checks
Think before reading: what would $z^2$ give for $z = i$? $(i)^2 = -1$. A probability of $-1$ is impossible — probabilities are never negative. More generally, $z^2 = a^2 - b^2 + 2abi$: it has a surviving imaginary part whenever $a \neq 0$ and $b \neq 0$. Using $z \cdot z^* = a^2 + b^2$ instead eliminates both problems in one move — no imaginary part, no negative values.
Use the formula directly: $|z|^2 = a^2 + b^2 = 3^2 + 4^2 = 9 + 16 = 25$. Notice: $|z| = \sqrt{25} = 5$ — the familiar 3-4-5 right triangle. The complex plane makes Pythagoras unavoidable: the modulus is always the hypotenuse, and $|z|^2$ is always the sum of the squares of the two legs.
$a^2 \geq 0$ and $b^2 \geq 0$ for every real $a$ and $b$, so $a^2 + b^2 \geq 0$ always. This is a consequence of how squaring works on real numbers — you cannot square a real number and get a negative result. Since $a$ (the real part) and $b$ (the imaginary part) are both real numbers, their squares are both non-negative, and their sum is non-negative. The conjugate multiplication is engineered to reduce to exactly this sum — guaranteeing non-negativity for every possible complex amplitude.
Four things you now own — and why they matter for every quantum calculation.
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The conjugate $z^*$ flips the imaginary sign — nothing else
$z = a + bi \Rightarrow z^* = a - bi$. Geometrically, it's the reflection across the real axis. The magnitude $|z^*| = |z|$ — the mirror image has the same distance from the origin. Real numbers are their own conjugates: if $b = 0$, then $z^* = z$.
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$z \cdot z^* = a^2 + b^2$ — always real, always non-negative
The cross terms $-abi$ and $+abi$ cancel exactly because the conjugate introduces a sign flip on $b$. What remains is $a^2 + b^2$ — a sum of squares of real numbers, guaranteed non-negative. This is the complete proof that $|z|^2$ is a valid probability measure.
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$|z|^2$ is the squared distance from the origin — Pythagorean
By the Pythagorean theorem, the distance from $(0,0)$ to $(a,b)$ is $\sqrt{a^2+b^2}$. So $|z|^2 = a^2 + b^2$ is literally distance-squared on the complex plane. Probability is geometric: it's how far the amplitude sits from the origin, squared.
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This is why Track 1's histograms converged to $|\alpha|^2$
In L06 you measured qubits 50 times and watched bars converge. The rule $P = |\alpha|^2$ was stated without proof. Now you have it: probability is $\alpha \cdot \alpha^* = a^2 + b^2$. Every histogram you built in Track 1 was physically implementing this formula — sampling from a distribution whose probabilities are moduli squared.
You now know $|z|^2 = z \cdot z^*$.
The next step: a quantum state is a vector of amplitudes.
Probability for the whole state means summing $|\alpha_j|^2$ — an inner product.
- Nielsen, M. A. & Chuang, I. L. — Quantum Computation and Quantum Information, Cambridge, 2000. §2.1.1: The state space postulate; the inner product and probability rule.
- Preskill, J. — Lecture Notes for Physics 229, Caltech, 1998. Chapter 2: Hilbert space, inner products, and the probability interpretation. Available online
- Needham, T. — Visual Complex Analysis, Oxford, 1997. Chapter 1 §2: The modulus, argument, and conjugate — geometric interpretation.
- Feynman, R. P., Leighton, R. B., Hibbs, A. R. — Quantum Mechanics and Path Integrals, McGraw-Hill, 1965. Chapter 1: The probability amplitude and why $|A|^2$ gives probability.