Complex Numbers and Vectors

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Complex Numbers and Vectors

The complex-number and vector-space toolkit every later lesson leans on, i, modulus, inner products, and norms.

Quantum amplitudes are complex numbers, not just probabilities, and quantum states are vectors built from them. Everything in this course, gates, superposition, interference, eigenvalues, is linear algebra over the complex numbers. This lesson is the toolkit; the next four build on it one piece at a time before the course touches a single line of Qiskit.

Complex Numbers

A complex number is $z = a + bi$ , where $a$ (the real part) and $b$ (the imaginary part) are ordinary real numbers and $i$ satisfies $i^2 = -1$ . There's no "solving for $i$ "; it's simply defined into existence because $x^2 = -1$ has no real solution, and it turns out to be exactly the right extra ingredient quantum mechanics needs.

Two operations matter most for this course:

Complex conjugate: $z^* = a - bi$ , flip the sign of the imaginary part.
Modulus (absolute value): $|z| = \sqrt{a^2 + b^2}$ , and conveniently $|z|^2 = z z^*$ , since $(a+bi)(a-bi) = a^2 + b^2$ .

That second identity is the one you'll use constantly: whenever a lesson says "the probability of measuring this outcome is $|\alpha|^2$ ," this is exactly what's being computed.

Polar Form and Phase

Every complex number can also be written as $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$ (Euler's formula), where $r = |z|$ is the modulus and $\theta$ is the phase, the angle $z$ makes on the complex plane. Multiplying two complex numbers in this form just adds their phases and multiplies their moduli: $r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 \, e^{i(\theta_1+\theta_2)}$ .

This is why quantum computing talks about "phase" so much: a gate that only changes an amplitude's phase (leaves $r$ alone) doesn't change any measurement probability at all, since $|re^{i\theta}|^2 = r^2$ regardless of $\theta$ , but phase differences between amplitudes are exactly what produce interference once you combine them, which is the entire mechanism later lessons (starting with superposition and interference) are built on.

Play with a complex number

z

Drag the sliders to change $z$ 's modulus $r$ and phase $\theta$ , and watch its position on the complex plane along with its conjugate $z^*$ and squared modulus $|z|^2$ .

Complex plane

$Modulus r$

1.00

$Phase θ$

45°

$z = a + bi$

0.71 + 0.71i

$∣ z ∣$

1.000

$∣ z ∣^{2}$

1.000

$z^{*}$

0.71 - 0.71i

Drag either slider: $r$ sets how far $z$ sits from the origin, $\theta$ sets its angle. If $z$ were a qubit amplitude, $|z|^2$ (bottom row) is exactly the measurement probability it contributes, regardless of what $\theta$ is, only $r$ matters for that number.

Vectors

A qubit's state is a column vector with complex entries, living in $\mathbb{C}^2$ :

$|\psi\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}, \qquad \alpha, \beta \in \mathbb{C}$

$n$ qubits together live in $\mathbb{C}^{2^n}$ (the Tensor Products lesson later in this track covers exactly how that space is built up). The ket notation $|\psi\rangle$ is just a label for "this is a vector"; $|0\rangle = \begin{pmatrix}1\\0\end{pmatrix}$ and $|1\rangle = \begin{pmatrix}0\\1\end{pmatrix}$ are the two vectors you'll see the most.

Inner Product and Norm

The inner product of two vectors $|\psi\rangle = \begin{pmatrix}\alpha_1\\\alpha_2\end{pmatrix}$ and $|\phi\rangle = \begin{pmatrix}\beta_1\\\beta_2\end{pmatrix}$ is

$\langle\psi|\phi\rangle = \alpha_1^*\beta_1 + \alpha_2^*\beta_2$

The bra $\langle\psi|$ is $|\psi\rangle$ 's entries, conjugated and laid out as a row instead of a column; $\langle\psi|\phi\rangle$ is literally that row times $|\phi\rangle$ 's column, ordinary matrix multiplication. Conjugating the first vector is what makes $\langle\psi|\psi\rangle$ come out real and non-negative for every vector, which is exactly what makes the next definition work:

$\||\psi\rangle\| = \sqrt{\langle\psi|\psi\rangle} = \sqrt{|\alpha|^2 + |\beta|^2}$

Two vectors are orthogonal when $\langle\psi|\phi\rangle = 0$ . Check $|0\rangle$ and $|1\rangle$ : $\langle 0|1\rangle = 1^*\cdot 0 + 0^*\cdot 1 = 0$ . They're orthogonal, and each has norm 1, together making them orthonormal, exactly what you want from the two basis states of a bit.

Check your understanding

A qubit's amplitude for

|1\rangle

\beta = \frac{1}{\sqrt2}i

(purely imaginary). What is the probability of measuring

|1\rangle

\frac{1}{\sqrt2}

, since that's the value of

\beta

0

, since imaginary numbers can't contribute to a real probability.

0.5

, since

|\beta|^2 = \beta\beta^* = \frac{1}{\sqrt2}i \cdot \left(-\frac{1}{\sqrt2}i\right) = \frac{1}{2}

Undefined, since

\beta

isn't a real number.

Check your understanding

Are

|+\rangle = \frac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}

and

|-\rangle = \frac{1}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}

orthonormal?

Yes: each has norm 1, and

\langle+|-\rangle = \frac{1}{2}(1\cdot1 + 1\cdot(-1)) = 0

No, they have norm 1 but aren't orthogonal.

No, they're orthogonal but don't have norm 1.

No to both properties.