Vectors as quantum states —
the column picture
In Track 1 you met $\alpha$ and $\beta$ — the two amplitudes that describe a qubit. You knew $|\alpha|^2+|\beta|^2=1$. But there was one thing left unnamed: when you stack those two numbers vertically, you get something mathematicians call a vector — and that object is exactly what a quantum state is.
You've been holding a vector the whole time — you just didn't know its name.
In Track 1 you learned that a qubit has two amplitudes: $\alpha$ for $|0\rangle$ and $\beta$ for $|1\rangle$. You knew $|\alpha|^2+|\beta|^2=1$.
Here's the realization that unlocks everything in Track 2:
$$|\psi\rangle=\begin{bmatrix}\alpha\\\beta\end{bmatrix}$$
The vector is not describing the state. The vector IS the state.
Stack $\alpha$ on top, $\beta$ on the bottom. That two-entry column is the complete physical object. Nothing is hidden behind it.
That's the whole lesson. What follows is making this feel natural — so you can read, write, and manipulate quantum states fluently.
Once states are vectors, gates become matrices and measurements become inner products. Every quantum algorithm becomes readable algebra. Nielsen & Chuang open Chapter 2 with exactly this (N&C §2.1, p. 63). This is where that vector space finally has a face you can calculate with.
Two slots. Fixed order. That's it.
A quantum state has exactly two slots. Slot 1 holds the amplitude for $|0\rangle$. Slot 2 holds the amplitude for $|1\rangle$. Stack them vertically and you have the complete state. Swap them and you have a completely different state.
That's the whole rule. The rest is learning to read and write it fluently.
$$|\psi\rangle=\begin{bmatrix}\alpha\\\beta\end{bmatrix},\quad\alpha=\text{amplitude for }|0\rangle,\quad\beta=\text{amplitude for }|1\rangle$$
The top entry always belongs to $|0\rangle$, the bottom always to $|1\rangle$. Swap them and you have a completely different state. That is the entire rule.
The four key states in column form
In Track 1 you met $|0\rangle$, $|1\rangle$, $|+\rangle$, and $|-\rangle$. Here they are as explicit columns — every entry derived from the definition:
$|+\rangle$ and $|-\rangle$ have identical Z-basis probabilities ($|\pm 1/\sqrt{2}|^2=1/2$). But that minus sign on $\beta$ makes them physically distinct — they interfere differently when gates act on them. The vector captures this difference exactly. A “picture” of the state never could.
The transpose ${}^\top$ — column and row are the same data
You'll often see states written as $|\psi\rangle=[\alpha,\beta]^\top$. The superscript ${}^\top$ means transpose: rotate the arrangement from horizontal to vertical. The entries stay the same; only the layout changes. Writing $[\alpha,\beta]^\top$ is shorthand for “a column.” The ${}^\top$ reminds you to stand it upright.
Write $\frac{1}{2}|0\rangle+\frac{\sqrt{3}}{2}|1\rangle$ as a column. What are $\alpha$ and $\beta$? Is this state normalized?
The formal definition — and why every symbol is necessary.
Definition: the qubit state vector
A qubit state is a unit vector in the complex vector space $\mathbb{C}^2$: a column of two complex numbers whose squared moduli sum to 1.
$$|\psi\rangle=\begin{bmatrix}\alpha\\\beta\end{bmatrix},\quad\alpha,\beta\in\mathbb{C},\quad|\alpha|^2+|\beta|^2=1$$
$|\psi\rangle$ — “ket psi.” Dirac notation for a quantum state. Think of it as a label for the column; full Dirac notation arrives in §4.
$\alpha$ — amplitude for outcome $|0\rangle$. A complex number $a+bi$. Probability of measuring 0 is $|\alpha|^2=a^2+b^2$.
$\beta$ — amplitude for outcome $|1\rangle$. Also complex: $c+di$. Probability of outcome 1 is $|\beta|^2=c^2+d^2$.
$\mathbb{C}^2$ — the set of all columns $\begin{bmatrix}z_1\\z_2\end{bmatrix}$ where $z_1,z_2\in\mathbb{C}$. An $n$-qubit system lives in $\mathbb{C}^{2^n}$.
$|\alpha|^2+|\beta|^2=1$ — the Born rule in disguise: probabilities must sum to 1.
The Bloch sphere parameterization
The most general valid qubit state is:
$$|\psi\rangle=\cos\!\left(\frac{\theta}{2}\right)|0\rangle+e^{i\varphi}\sin\!\left(\frac{\theta}{2}\right)|1\rangle=\begin{bmatrix}\cos(\theta/2)\\e^{i\varphi}\sin(\theta/2)\end{bmatrix}$$
$\theta\in[0,\pi]$ = polar angle, $\varphi\in[0,2\pi)$ = azimuthal angle on the Bloch sphere.
With $\alpha=\cos(\theta/2)$ and $\beta=e^{i\varphi}\sin(\theta/2)$:
$|\alpha|^2=\cos^2(\theta/2)$ (real, so modulus squared is just the square)
$|\beta|^2=|e^{i\varphi}|^2\cdot\sin^2(\theta/2)=1\cdot\sin^2(\theta/2)=\sin^2(\theta/2)$
Note: $|e^{i\varphi}|^2=e^{i\varphi}\cdot e^{-i\varphi}=1$ — any unit complex number has modulus 1.
$|\alpha|^2+|\beta|^2=\cos^2(\theta/2)+\sin^2(\theta/2)=1$ ✓ (Pythagorean identity)
Why $\theta/2$ instead of $\theta$?
Short answer: two states that are orthogonal as vectors (zero overlap, proved in M16) must sit at opposite poles of the Bloch sphere. If we used $\theta$ directly, orthogonal states would only be 90° apart on the sphere instead of 180°. Halving the angle stretches the geometry so opposite poles correctly correspond to orthogonal states. Check it: $|0\rangle$ is at $\theta=0$ (north pole) and $|1\rangle$ at $\theta=\pi$ (south pole) — exactly opposite, as you'd expect from two perfectly distinguishable states.
$|0\rangle$: set $\theta=0$ → $\begin{bmatrix}\cos 0\\e^{i\varphi}\sin 0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$ ✓. $|1\rangle$: set $\theta=\pi$ → $\begin{bmatrix}\cos(\pi/2)\\e^{i\varphi}\sin(\pi/2)\end{bmatrix}=\begin{bmatrix}0\\e^{i\varphi}\end{bmatrix}$. With global phase $e^{i\varphi}=1$ this gives $\begin{bmatrix}0\\1\end{bmatrix}$ ✓.
Column Vector Builder — state, sphere, and probabilities in sync.
Discover for yourself that changing the vector entries changes the state — not the description of the state. Watch the probability bars respond instantly. Every number you drag is a real change to a real quantum state.
Step 1 — Start at $|0\rangle$. Click the $|0\rangle$ preset. Notice the vector shows $[1, 0]^\top$ and P(0) = 100%. This is the north pole. Nothing surprising.
Step 2 — Predict first. Before touching anything: if you drag $\alpha$ down from 1 toward 0, what do you expect P(0) to do? Commit your prediction mentally.
Step 3 — Test it. Drag the $\alpha$ slider slowly downward. Watch the probability bar change. Was your prediction right?
Step 4 — Connect the dots. The number you moved is the state. You didn't change a “setting” — you changed the physical quantum state itself.
Step 5 — Mini-challenge. Create a state where P(0) = P(1) = exactly 50%. Try $|+\rangle$ or $|-\rangle$. What do those vectors look like? Why are the probabilities identical even though the vectors are different?
Every slider move changed the state itself — not a setting, not a view, not a representation. The column vector in the display is the complete physical state of the qubit. When $\alpha$ went up and P(0) went up: that's not a display updating. That's the state changing. The vector is the state.
- The column picture: $|\psi\rangle=[\alpha,\beta]^\top$
A qubit state is a pair of complex numbers arranged vertically. Top entry $\alpha$ = amplitude for $|0\rangle$; bottom $\beta$ = amplitude for $|1\rangle$. Not a metaphor — the definition. N&C §2.1.1, p. 66.
- Normalization: $|\alpha|^2+|\beta|^2=1$
Not every pair of complex numbers is a valid state — only those satisfying this constraint. It is the Born rule in vector form: probabilities must sum to 1.
- Bloch sphere: $[\cos(\theta/2),\,e^{i\varphi}\sin(\theta/2)]^\top$
Every normalized state maps to a unique point on the sphere. The $\theta/2$ factor ensures orthogonal states (zero overlap) correspond to opposite poles.
- Relative phase: invisible to Z-measurement, visible to gates
Changing $\varphi$ (imaginary part of $\beta$) moves the sphere without changing $|\alpha|^2$ or $|\beta|^2$. Undetectable by Z-basis measurement, but critical the moment a gate mixes the amplitudes.
- Transpose ${}^\top$: same entries, different shape
$[\alpha,\beta]^\top$ is shorthand for a column. The symbol tells you to stand the row upright. Values don't change; layout does. You'll see this notation throughout all of quantum computing.
State = column vector:
$|\psi\rangle = \begin{bmatrix}\alpha\\\beta\end{bmatrix}$
$\alpha$ → amplitude of $|0\rangle$ // top slot, always
$\beta$ → amplitude of $|1\rangle$ // bottom slot, always
$|\alpha|^2$ → P(outcome = 0)
$|\beta|^2$ → P(outcome = 1)
$|\alpha|^2 + |\beta|^2 = 1$ // always, no exceptions
The vector is not describing the state. The vector IS the state. Say it once more, then move on.
- Nielsen, M. A. & Chuang, I. L. — Quantum Computation and Quantum Information, Cambridge, 2000. §2.1.1 pp. 63–67: Quantum states as vectors in Hilbert space.
- Preskill, J. — Lecture Notes for Physics 219/Computer Science 219, Caltech. §2.1–2.2: State vectors and the Bloch sphere parameterization. Available online
- MIT OpenCourseWare — 8.370/18.435 Quantum Computation. §1: Qubit state representation and Bloch sphere. ocw.mit.edu
- IBM Quantum Learning — Basics of quantum information: Single-qubit gates. learning.quantum.ibm.com