Orientation and diagnostic

1 / 5 · Learn

How a lesson runs

About ten minutes: what to expect, five words you will meet constantly, how quantum states are written, and a quick check of the math that the lessons use without teaching it.

This app makes you work. Most of your time goes on answering questions, and old questions come back days later, mixed in with new ones. That feels slower and less satisfying than reading a clear explanation, and the feeling is a poor guide. In a controlled comparison in a university physics course, students taught actively scored higher on a test but believed they had learned less than students given polished lectures (Deslauriers et al., PNAS 2019). Recalling an idea, getting it wrong and being corrected takes effort, and that effort is a large part of what makes it last.

Check yourself

Halfway through a lesson you feel slower and less sure of yourself than when you simply read a clear explanation. What is the best reading of that feeling?

AYou are learning less than you would by reading. Skip the questions and read the cards.
BThe lesson is too hard for you, so you should go back and start again.
CIt is a normal side effect of active practice, and a poor guide to how much you are learning. Results on the mastery check, and on reviews days later, are better evidence.

Show the answer

Answer: C. It is a normal side effect of active practice, and a poor guide to how much you are learning. Results on the mastery check, and on reviews days later, are better evidence.

Feeling of learning and actual learning come apart under active practice. Judge your progress by what you can do without help, especially in the reviews that come back days later.

Why not the others

A. A very common idea is that smooth reading means learning. In the study above, the students who felt they learned less actually scored higher. Ease while reading measures how familiar the words feel, not whether you can use the idea later. Check that with what you can do unaided days later, which is what the reviews measure.
B. Effort alone is not a sign that a lesson is too hard. Mistakes during practice are expected: each one gets targeted feedback and a second try. If the mastery check at the end shows a real gap, it links you to the exact segment to revisit.

Inside a lesson the steps always come in the same order:

Guess first. One or two questions on the core idea before any teaching, marked Guess, not scored. When the answer comes up, usually later in the same lesson, your guess is shown again.
Short cards, each followed by a question. Feedback aimed at your particular mistake, hints after a first try, and the full solution after a second.
Predict, observe, explain. A simulation stays locked until you commit to a prediction.
Worked examples that hand you the last steps, then a problem on your own.
Explain it in your own words, then compare with a model answer.

Check yourself

A lesson opens with a question about something it has not taught yet. What is that question for?

AIt is a placement test: a wrong guess means you should go back a lesson.
BGuessing first, even wrongly, makes the explanation that answers it stick better. The guess is not scored.
CIt shows you how much you still have to learn.

Show the answer

Answer: B. Guessing first, even wrongly, makes the explanation that answers it stick better. The guess is not scored.

A prequestion helps you learn the material it asks about, provided you see the answer later (St. Hilaire et al., 2024). The lesson flags that moment with Remember your guess?

Why not the others

A. It is natural to read any question as a test. These ones are not scored and decide nothing; most people get some wrong. Their job is to make you think about the question before the lesson answers it.
C. A wrong guess can feel like a verdict, but that is not what the question is for, and most people get some wrong. Its job is to make you think about the question first: an answer you have already wondered about is easier to keep when it comes.

After a lesson. A mastery check gives one try per item, with no hints, and a retake uses new numbers. Already know a lesson’s material? Show me what you know, at the top of its first page, takes you straight to the check. Then spaced review: the lesson’s key questions come back days later, mixed with other lessons and with fresh numbers.

Three buttons. You submit each answer with Guessing, Fairly sure or Certain. They don’t change your score; they change what happens next. A wrong answer you were Certain about offers the full solution straight away, because confident errors are the ones people correct best once they see why (Butterfield & Metcalfe, 2001). A right answer marked Guessing comes back in review sooner.

Check yourself

You press Certain and your answer turns out to be wrong. What happens, and why?

ANothing different from any other wrong answer.
BYou can open the full solution straight away, without a second try first, because a confident error is the best moment to fix an idea.
CYou lose more points than you would for a wrong guess.

Show the answer

Answer: B. You can open the full solution straight away, without a second try first, because a confident error is the best moment to fix an idea.

Confident errors are corrected well once you see the reasoning (the hypercorrection effect). So press Certain only when you are, and Guessing when you are guessing. The home page shows how often you are right at each level.

Why not the others

A. A confident error is treated differently: the full solution is offered at once instead of after a second try, because that is when you are most likely to correct the idea for good.
C. There are no points to lose: confidence never changes a score. It decides what you see next, and how soon an idea comes back.

2 / 5 · Learn

Five words to know

Five words carry most of what follows. Here is one plain sentence for each; the lessons fill them in.

Amplitude. A qubit’s state gives a complex number for each basis state, and the probability of getting that outcome is its squared magnitude: $\abs{\alpha}^2$ for an amplitude $\alpha$ .
Basis. A pair of reference states, such as the two states a measurement reports as 0 and 1, in terms of which any qubit state can be written; every measurement is made in some basis.
Normalization. A state’s squared magnitudes add up to 1, because some outcome must occur.
Measurement. It returns one outcome at random, with those probabilities, and leaves the qubit in the basis state that was observed.
Gate. A reversible operation that changes the amplitudes in a fixed way, without reading anything out.

Check yourself

Match each description to its word.

Match each of these…

Turns a state into a single random outcome, and leaves the qubit in the state for that outcome.
Changes the amplitudes, can always be undone, and reveals nothing about the state.
Its squared magnitude is the chance of one particular outcome.
The requirement that all the probabilities add up to 1.
The set of reference states that a measurement sorts the qubit into.

…with one of these

Measurement
Basis
Amplitude
Gate
Normalization

Hint

Two of these are things you do to a qubit. Which one gives you a result you can write down?

Show the answer

Turns a state into a single random outcome, and leaves the qubit in the state for that outcome.	Measurement
Changes the amplitudes, can always be undone, and reveals nothing about the state.	Gate
Its squared magnitude is the chance of one particular outcome.	Amplitude
The requirement that all the probabilities add up to 1.	Normalization
The set of reference states that a measurement sorts the qubit into.	Basis

Measurement gives an outcome and changes the state; a gate changes the state and gives no outcome. An amplitude’s squared magnitude is a probability, normalization makes those probabilities add up to 1, and a basis lists the possible outcomes.

3 / 5 · Learn

Reading the notation

Quantum states are written as kets: a label between a bar and an angle bracket, such as $\ket0$ , read “ket zero”. The label is only a name. A qubit’s two basis states are $\ket0$ and $\ket1$ , and each stands for a column vector:

\ket0 = \begin{bmatrix}1\\0\end{bmatrix},\qquad \ket1 = \begin{bmatrix}0\\1\end{bmatrix}

Any state is a combination of the two, with amplitudes usually called $\alpha$ (alpha) and $\beta$ (beta). The amplitude of $\ket0$ goes on top:

\alpha\ket0 + \beta\ket1 = \begin{bmatrix}\alpha\\ \beta\end{bmatrix}

Two combinations get names of their own, $\ket+$ (“ket plus”) and $\ket-$ (“ket minus”). The factor $\tfrac{1}{\sqrt2}$ makes their squared amplitudes add up to 1:

\begin{aligned}\ket+ &= \tfrac{1}{\sqrt2}(\ket0 + \ket1) \\ \ket- &= \tfrac{1}{\sqrt2}(\ket0 - \ket1)\end{aligned}

A Greek letter in a ket, as in $\ket\psi$ (“ket psi”), names a state in general.

Check yourself

Which column vector is $\ket0$ ?

A $\begin{bmatrix}0\\ 0\end{bmatrix}$
B $\begin{bmatrix}0\\ 1\end{bmatrix}$
C $\begin{bmatrix}1\\ 0\end{bmatrix}$

Show the answer

Answer: C. $\begin{bmatrix}1\\ 0\end{bmatrix}$

$\ket0$ is the first basis vector: a 1 in the top slot, the slot that belongs to $\ket0$ . Its label is a name, not the number 0.

Why not the others

A. It is natural to read the 0 as a value, so that $\ket0$ means “nothing”. But the label inside a ket is only a name: $\ket0$ is the first basis state, with a 1 in the top slot. The all-zero vector is not a state at all, since its squared entries add up to 0, not 1.
B. That is $\ket1$ , the second basis state, with its 1 in the bottom slot. $\ket0$ has its 1 on top, because the top slot belongs to $\ket0$ .

Check yourself

Which column vector is $\ket-$ ?

A $\begin{bmatrix}1/\sqrt2\\ 1/\sqrt2\end{bmatrix}$
B $\begin{bmatrix}1/\sqrt2\\ -1/\sqrt2\end{bmatrix}$
C $\begin{bmatrix}1\\ -1\end{bmatrix}$
DThe single number $-1$

Show the answer

Answer: B. $\begin{bmatrix}1/\sqrt2\\ -1/\sqrt2\end{bmatrix}$

$\ket- = \tfrac{1}{\sqrt2}\ket0 - \tfrac{1}{\sqrt2}\ket1$ : the $\ket0$ amplitude $\tfrac{1}{\sqrt2}$ goes on top and the $\ket1$ amplitude $-\tfrac{1}{\sqrt2}$ below.

Why not the others

A. That is $\ket+$ . The two differ only in the sign of the $\ket1$ term: $\ket- = \tfrac{1}{\sqrt2}(\ket0 - \ket1)$ , so its bottom entry is $-\tfrac{1}{\sqrt2}$ .
C. The factor $\tfrac{1}{\sqrt2}$ multiplies both terms, so each entry carries it. Without it the squared entries add up to 2, and a state’s squared amplitudes must add up to 1.
D. It is tempting to read the minus sign as a value. But the label of a ket is a name, and a qubit’s ket always stands for a column vector with two entries: $\ket-$ names $\tfrac{1}{\sqrt2}(\ket0 - \ket1)$ .

4 / 5 · Diagnostic

Math check

Eight short items on the math the lessons use without teaching it. Pencil and paper help. Type answers such as 7, 1/sqrt(3) or 5-2i. Anything you miss points to a three-minute refresher, and none of it is needed until the lesson that uses it.

Check yourself

What is $\abs{1 - 3i}^2$ , the squared magnitude of $1 - 3i$ ?

Show the answer

Answer: $10$

For $z = a + bi$ , $\abs{z}^2 = a^2 + b^2 = 1 + 9 = 10$ .

Write $z = 1 - 3i$ , so the real part is $1$ and the imaginary part is $-3$ . Then $\abs{z}^2 = 1^2 + (-3)^2 = 10$ . The same number comes out of $z\,z^* = (1 - 3i)(1 + 3i)$ , because the cross terms cancel. Probabilities in quantum mechanics are squared magnitudes like this one. Refresher: Complex numbers.

Common slips

$3.162$ : That is $\abs{1 - 3i} = \sqrt{10}$ , the magnitude itself. The question asks for its square: $\abs{z}^2 = a^2 + b^2 = 10$ .
$-8-6i$ : That is $(1 - 3i)^2$ , the number multiplied by itself. The squared magnitude is $\abs{z}^2 = z\,z^*$ , which is always real and never negative: here $1 + 9 = 10$ .
$-8$ : That brings $i^2 = -1$ into the magnitude and gives $a^2 - b^2$ . The magnitude squared adds the squares of the real and imaginary parts as plain numbers: $1^2 + 3^2 = 10$ .
$16$ : That squares the sum of the two parts. Square each part separately, then add: $1^2 + 3^2 = 10$ .

Check yourself

What is $\abs{e^{i\pi/3}}$ , the magnitude of $e^{i\pi/3}$ ?

Show the answer

Answer: $1$

$e^{i\varphi} = \cos\varphi + i\sin\varphi$ lies on the unit circle, so $\abs{e^{i\varphi}} = 1$ for every real $\varphi$ .

By Euler's formula $e^{i\pi/3} = \cos(\pi/3) + i\sin(\pi/3)$ , so $\abs{e^{i\pi/3}}^2 = \cos^2(\pi/3) + \sin^2(\pi/3) = 1$ and the magnitude is 1. A factor $e^{i\varphi}$ changes a phase and never a magnitude, which is why it never changes a probability on its own. Refresher: Complex numbers.

Common slips

$\frac{1}{2}+\frac{\sqrt{3}}{2}i$ : That is the complex number $e^{i\pi/3} = \cos(\pi/3) + i\sin(\pi/3)$ itself. Its magnitude is a single non-negative number, its distance from 0: $\sqrt{\cos^2(\pi/3) + \sin^2(\pi/3)} = 1$ .
$\frac{1}{2}$ : That is $\cos(\pi/3)$ , the real part only. The magnitude uses both parts: $\sqrt{\cos^2(\pi/3) + \sin^2(\pi/3)} = 1$ .
$\frac{\sqrt{3}}{2}$ : That is $\sin(\pi/3)$ , the imaginary part only. The magnitude uses both parts: $\sqrt{\cos^2(\pi/3) + \sin^2(\pi/3)} = 1$ .
$1.047$ : That is the angle $\pi/3$ itself, the phase of the number. The magnitude is the distance from 0, and $e^{i\varphi}$ always sits on the unit circle.
$2.85$ : That is $e^{\pi/3}$ , with the $i$ dropped. With the $i$ , $e^{i\varphi} = \cos\varphi + i\sin\varphi$ lies on the unit circle, so its magnitude is 1.
$-1$ : A magnitude is a distance from 0, so it is never negative. $e^{i\varphi}$ lies on the unit circle, at distance 1 from 0.

Check yourself

What is the complex conjugate of $5 + 3i$ ? (Type it like 5-2i.)

Show the answer

Answer: $5-3i$

$(a + bi)^* = a - bi$ : reflect across the real axis. So $(5 + 3i)^* = 5 - 3i$ .

The conjugate $z^*$ keeps the real part and flips the sign of the imaginary part, so $(5 + 3i)^* = 5 - 3i$ . Comparing two quantum states (an inner product) conjugates the amplitudes of one of them, so this step comes up constantly. Refresher: Complex numbers.

Common slips

$5+3i$ : That is $5 + 3i$ unchanged. Conjugating flips the sign of the imaginary part only: $(a+bi)^* = a - bi$ .
$-5-3i$ : That flips both signs, which gives $-z$ , not $z^*$ . The conjugate keeps the real part as it is and flips only the imaginary part.
$-5+3i$ : That flips the sign of the real part instead of the imaginary part. The conjugate keeps the real part ( $5$ ) and flips only the sign of the imaginary part.
$34$ : That is $z\,z^* = \abs{z}^2$ , a real number. The question asks for $z^*$ itself, which is still complex.

Refresher: Complex numbers If any of these three felt shaky: magnitudes, phases and conjugates in about three minutes.

Check yourself

Normalize the vector $(1, 3)$ : rescale it so that the squares of its entries add up to 1. Enter the two entries.

Show the answer

Answer: $v_1 = 0.316,\quad v_2 = 0.949$

The length is $\sqrt{1^2 + 3^2} = \sqrt{10}$ , so the normalized vector is $(\tfrac{1}{\sqrt{10}}, \tfrac{3}{\sqrt{10}})$ .

Divide every entry by the vector's length $\sqrt{1^2 + 3^2} = \sqrt{10}$ . That gives $(\tfrac{1}{\sqrt{10}}, \tfrac{3}{\sqrt{10}})$ , and the squares of the entries now add up to 1. A qubit state must be normalized like this, because its squared amplitudes are probabilities. Refresher: Vectors and normalization.

Refresher: Vectors and normalization Normalizing, and why a state’s squared entries must add up to 1.

Check yourself

Multiply: $\begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix}\begin{bmatrix}a\\ b\end{bmatrix} = \;?$

A $\begin{bmatrix}-b\\ a\end{bmatrix}$
B $\begin{bmatrix}a\\ b\end{bmatrix}$
C $\begin{bmatrix}a\\ -b\end{bmatrix}$

Show the answer

Answer: C. $\begin{bmatrix}a\\ -b\end{bmatrix}$

Each output entry is a row dotted with the column: row 1, $(1, 0)$ , gives $a$ , and row 2, $(0, -1)$ , gives $-b$ .

Take the rows one at a time. Row 1, $(1, 0)$ , dotted with $(a, b)$ gives $a$ ; row 2, $(0, -1)$ , gives $-b$ . So the product is $\begin{bmatrix}a\\ -b\end{bmatrix}$ . Every gate in these lessons acts on a state exactly like this. Refresher: Matrices acting on states.

Why not the others

A. Right entries, wrong order. Entry 1 comes from row 1: row 1, $(1, 0)$ , dotted with $(a, b)$ gives $1\cdot a + 0\cdot b = a$ .
B. That leaves the vector unchanged, which only the identity matrix does. Each entry is a row of the matrix dotted with the vector: row 2, $(0, -1)$ , dotted with $(a, b)$ gives $0\cdot a + (-1)\cdot b = -b$ .

Refresher: Matrices acting on states Matrix times vector, and reading a gate’s columns.

Check yourself

Compute the tensor product $v = (0, 1) \tp (\tfrac{3}{5}, \tfrac{4}{5})$ , a vector with four entries.

Show the answer

Answer: $v_1 = 0,\quad v_2 = 0,\quad v_3 = \frac{3}{5},\quad v_4 = \frac{4}{5}$

$(u_1, u_2) \tp (w_1, w_2) = (u_1 w_1,\ u_1 w_2,\ u_2 w_1,\ u_2 w_2) = (0, 0, \tfrac{3}{5}, \tfrac{4}{5})$ .

Take the first entry of the first vector times the whole second vector, then its second entry times the whole second vector: $(u_1 w_1, u_1 w_2, u_2 w_1, u_2 w_2) = (0, 0, \tfrac{3}{5}, \tfrac{4}{5})$ . In ket language, two qubits have four basis states, $\ket{00}, \ket{01}, \ket{10}, \ket{11}$ , where $\ket{01}$ means the first qubit is in $\ket0$ and the second in $\ket1$ . As vectors, $\ket0 \tp \ket1 = (1, 0) \tp (0, 1) = (0, 1, 0, 0)$ , the second of the four. Refresher: Tensor products.

Refresher: Tensor products How two vectors combine into four entries, in the standard order.

Check yourself

What is $0 \oplus 1 \oplus 1$ ? ( $\oplus$ is XOR, addition mod 2.)

Show the answer

Answer: $0$

XOR is addition mod 2: $0 + 1 + 1 = 2$ , and 2 mod 2 is 0.

Add the bits as ordinary numbers, $0 + 1 + 1 = 2$ , then keep the remainder mod 2: $0$ . Equivalently, the result is 1 exactly when an odd number of the bits are 1. XOR turns up throughout the later lessons, from the CNOT gate of Lesson 6 to the algorithms of Lesson 9. Refresher: Bits, XOR and bit strings.

Common slips

$2$ : That is the ordinary sum, $0 + 1 + 1 = 2$ . $\oplus$ is addition mod 2: keep only the remainder after dividing by 2, so the answer is $0$ .
$1$ : That is what OR gives. XOR is 1 exactly when an odd number of the bits are 1; here 2 of them are, so the answer is $0$ .

Check yourself

For bit strings, $x \cdot z = x_1 z_1 + x_2 z_2 + \dots + x_n z_n \pmod 2$ . What is ${1100 \cdot 0110}$ ?

Show the answer

Answer: $1$

$1\cdot 0 + 1\cdot 1 + 0\cdot 1 + 0\cdot 0 = 1$ , and 1 mod 2 is 1.

Multiply the strings position by position and add: $1\cdot 0 + 1\cdot 1 + 0\cdot 1 + 0\cdot 0 = 1$ (the positions where both have a 1). Then reduce mod 2: 1 mod 2 is $1$ . This bitwise dot product appears in the Deutsch–Jozsa algorithm as the sign $(-1)^{x\cdot z}$ . Refresher: Bits, XOR and bit strings.

Common slips

$100$ : 0100 is the bitwise AND, the positions where both strings have a 1. The dot product goes one step further: count those 1s (1) and reduce mod 2, giving 1.

Refresher: Bits, XOR and bit strings XOR and bit strings, as used from the CNOT gate on.

5 / 5 · Recap

Recap

From memory

Before reading the summary, write down what you remember. Pulling it back from memory is what makes it last.

Name the five words from Five words to know and say what each one means.
Show the key points
- Amplitude: a complex number for each basis state; its squared magnitude is a probability
- Basis: the reference states a measurement sorts into
- Normalization: the squared magnitudes add up to 1
- Measurement: gives one random outcome and leaves the qubit in the matching basis state
- Gate: changes the amplitudes reversibly and reads nothing out
What do $\ket0$ , $\ket1$ , $\ket+$ and $\ket-$ stand for, and how is $\alpha\ket0 + \beta\ket1$ written as a column vector?
Show the key points
- $\ket0$ and $\ket1$ are the basis vectors $(1, 0)$ and $(0, 1)$ , written as columns; the label in a ket is a name, not a value
- $\alpha\ket0 + \beta\ket1$ is the column with $\alpha$ on top and $\beta$ below
- $\ket\pm = \tfrac{1}{\sqrt2}(\ket0 \pm \ket1)$ , so $\ket-$ has entries $\tfrac{1}{\sqrt2}$ and $-\tfrac{1}{\sqrt2}$
What should a lesson here feel like, and what do the three confidence buttons change?
Show the key points
- Slower and more effortful than reading, which does not mean you are learning less
- Guess first; short cards, each followed by a question; a mastery check with one try per item; reviews days later
- A wrong Certain offers the full solution at once; a right Guessing comes back sooner; the buttons never change a score

Key ideas

Judge your progress by what you can do unaided, especially in reviews days later, not by how smooth a lesson felt.
Answer honestly with Guessing, Fairly sure or Certain: it decides what you see next and when an idea comes back.
A ket names a column vector: $\ket0$ and $\ket1$ are the two basis vectors, $\alpha\ket0 + \beta\ket1$ has $\alpha$ on top, and $\ket\pm = \tfrac{1}{\sqrt2}(\ket0 \pm \ket1)$ .
If the math check flagged something, the refreshers take a few minutes each: complex numbers, vectors, matrices, tensor products, bits and XOR. The lessons also link each one where it is first needed.

Go further

Free resources for when you want more than this lesson covers.

Essence of linear algebra3Blue1Brown · Video series
Short animated videos that build the geometry of vectors, matrices and bases behind the math check.
Algebra and Trigonometry 2e: Complex numbersOpenStax · Free book
Worked examples and exercises on adding, multiplying and dividing complex numbers, for more practice than the check gives.
Introduction to Classical and Quantum ComputingThomas G. Wong · Free book
A whole beginner’s course in one free PDF that assumes only trigonometry and builds from classical bits to quantum algorithms.

Math refreshers

The toolkit topics this lesson leans on.

Complex numbersMagnitude, phase, and why a phase factor has size 1
Inner products and bases⟨φ|ψ⟩, orthogonality, and |⟨φ|ψ⟩|² as a probability
Vectors and normalizationKets are column vectors; probabilities must add to 1
Matrices acting on statesGates are matrices; circuit order vs matrix order
Tensor productsCombining qubits: |a⟩⊗|b⟩ as a 4-entry vector
Bits, XOR and bit stringsx ⊕ y, x · z mod 2, sums over {0,1}ⁿ, and signs that depend on f(x)