Self-Assessment Quiz: The Qubit

Twenty questions on superposition, the Bloch sphere, phase, normalization, and what a qubit actually is. Answer each before opening the key. Aim for 16+.


Question 1

A general single-qubit state is $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$. The normalization condition is:

A) $\alpha + \beta = 1$ B) $|\alpha| + |\beta| = 1$ C) $|\alpha|^2 + |\beta|^2 = 1$ D) $\alpha^2 + \beta^2 = 1$

Question 2

For $|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle$, the probability of measuring $0$ in the computational basis is:

A) $\tfrac{1}{\sqrt 2}$ B) $\tfrac12$ C) $-\tfrac12$ D) $1$

Question 3

The state in Question 2 is conventionally written:

A) $|+\rangle$ B) $|-\rangle$ C) $|i\rangle$ D) $|1\rangle$

Question 4

In the Bloch sphere parametrization $|\psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\varphi}\sin\frac{\theta}{2}|1\rangle$, the state $|+\rangle$ sits at:

A) $\theta = 0$ B) $\theta = \pi/2,\ \varphi = 0$ C) $\theta = \pi/2,\ \varphi = \pi/2$ D) $\theta = \pi$

Question 5

Why does the Bloch parametrization use $\theta/2$ rather than $\theta$?

A) To make the math simpler B) Because orthogonal quantum states ($|0\rangle$, $|1\rangle$) must sit at antipodal points, 180° apart C) Because $\theta$ would exceed $2\pi$ D) It is an arbitrary convention with no physical content

Question 6

Multiplying an entire state by $e^{i\gamma}$ (a global phase):

A) Rotates the Bloch vector by $\gamma$ B) Has no observable consequence whatsoever C) Changes measurement probabilities in the $X$ basis only D) Makes the state unnormalized

Question 7

The relative phase between $|0\rangle$ and $|1\rangle$ components:

A) Is also unobservable B) Is observable — it determines the $\varphi$ coordinate and affects measurements in bases other than $Z$ C) Only matters for two or more qubits D) Is always zero for a physical qubit

Question 8

Measuring $|+\rangle$ in the computational ($Z$) basis yields:

A) $0$ always B) $1$ always C) $0$ or $1$ with equal probability D) $|+\rangle$ again

Question 9

Measuring $|+\rangle$ in the $X$ basis yields:

A) $+$ with certainty B) $-$ with certainty C) $\pm$ equally likely D) Undefined

Question 10

A classical probabilistic bit that is $0$ with probability ½ and $1$ with probability ½ differs from $|+\rangle$ because:

A) It does not — they are the same thing B) $|+\rangle$ has a definite value in the $X$ basis; the probabilistic bit does not C) The probabilistic bit cannot be measured D) $|+\rangle$ is not normalized

Question 11

How many real parameters specify a pure single-qubit state, after removing global phase and enforcing normalization?

A) 1 B) 2 C) 3 D) 4

Question 12

A point strictly inside the Bloch sphere represents:

A) An unnormalized state B) A mixed state C) An entangled state D) An impossible state

Question 13

The Bloch vector of $|1\rangle$ points:

A) $+z$ B) $-z$ C) $+x$ D) $-x$

Question 14

Applying $H$ to $|0\rangle$ produces:

A) $|1\rangle$ B) $|+\rangle$ C) $|-\rangle$ D) $|0\rangle$

Question 15

$T|+\rangle$ equals:

A) $\frac{1}{\sqrt2}(|0\rangle + e^{i\pi/4}|1\rangle)$ B) $\frac{1}{\sqrt2}(|0\rangle + |1\rangle)$ C) $|+\rangle$ rotated by $\pi/4$ about $x$ D) $|1\rangle$

Question 16

True or false: A qubit "stores infinitely much information" because $\alpha$ and $\beta$ are continuous.

Question 17

True or false: Two states differing only by global phase are physically indistinguishable by any measurement.

Question 18

True or false: The Bloch sphere picture extends straightforwardly to two qubits as a pair of spheres.

Question 19

Short answer. A colleague says "a qubit is a bit that's 0 and 1 at the same time." Give a one-sentence correction that a non-physicist can follow.

Question 20

Short answer. You are handed a device that always outputs $0$ and $1$ with 50/50 frequency in the $Z$ basis. Describe one experiment that distinguishes $|+\rangle$ from a classical coin flip.


Answer Key

Q Ans Note
1 C Amplitudes are complex; the squared moduli are probabilities and must sum to 1 (Born rule).
2 B $|-1/\sqrt2|^2 = 1/2$. The minus sign is a relative phase and does not affect $Z$-basis probabilities.
3 B $|-\rangle = (|0\rangle - |1\rangle)/\sqrt2$.
4 B $\cos(\pi/4)|0\rangle + \sin(\pi/4)|1\rangle = |+\rangle$, on the $+x$ axis.
5 B The half-angle maps the 180° Hilbert-space orthogonality onto 180° of Bloch-sphere separation. Without it, $|0\rangle$ and $|1\rangle$ would be 90° apart on the sphere.
6 B Global phase cancels in $|\langle\phi|\psi\rangle|^2$ for every measurement. It is pure gauge.
7 B Relative phase is the entire content of the $\varphi$ coordinate, and it is what interference exploits.
8 C $|\pm 1/\sqrt2|^2 = 1/2$ each.
9 A $|+\rangle$ is the $+1$ eigenstate of $X$; measuring $X$ on it is deterministic.
10 B Same $Z$ statistics, completely different $X$ statistics. Superposition is not ignorance.
11 B Two: $\theta$ and $\varphi$ — hence a 2-sphere.
12 B The interior is the mixed states; the surface is the pure states; the center is maximally mixed.
13 B $|0\rangle$ is $+z$, $|1\rangle$ is $-z$ — antipodal, as orthogonality requires.
14 B $H|0\rangle = (|0\rangle+|1\rangle)/\sqrt2 = |+\rangle$.
15 A $T$ applies phase $e^{i\pi/4}$ to the $|1\rangle$ component only.
16 False You may only extract one classical bit per measurement, and measurement destroys the state. Holevo's bound caps the accessible information at one bit per qubit.
17 True This is why global phase is discarded when we draw the Bloch sphere.
18 False Two qubits need 6 real parameters and cannot be drawn as two independent spheres — entanglement is exactly the information the two-sphere picture loses.
19 Better: "a qubit's state is a direction in a continuous space; when you measure it along an axis you always get one bit, but the direction it pointed determines the odds — and unlike a coin, some directions give a certain answer."
20 Apply $H$ before measuring (equivalently, measure in the $X$ basis). $|+\rangle$ then yields $0$ every single time; a genuine coin flip still yields 50/50. Interference distinguishes superposition from ignorance.