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. |