Exercises and Problems

Practice the central calculations of Core Formalism with nine short problems and worked solutions. Try each problem before opening its solution. All systems here are finite dimensional; the qubit basis $\lvert0\rangle,\lvert1\rangle$ is orthonormal, and tensor products are ordered with subsystem $A$ first.

Practice by topic

Start with Problems 1–3 if you are learning the notation. Problems 4–6 combine operator algebra with measurement and dynamics. Problems 7–9 test subsystem states, positivity, and the meaning of the probability rule. Follow the preparation links when a step is unfamiliar.

Problem	What you will practice	Preparation
1. Basis components	Change amplitudes without changing the state	Change of Basis
2. Two measurement bases	Find where relative phase affects probabilities	Born Rule
3. Mean and variance	Separate possible outcomes from their average	Variance and Standard Deviation
4. Uncertainty	Compare a lower bound with actual spreads	General Uncertainty Relations
5. Degenerate measurement	Normalize a conditional state and preserve unresolved coherence	Lüders Rule
6. Time evolution	Turn an energy difference into an oscillating probability	Unitary Time Evolution
7. Entanglement	Compare joint states with identical local states	Reduced States
8. Positivity	Find which matrices represent possible states	Density Operators
9. Meaning of the postulates	Distinguish predictions from interpretive assumptions	What the Postulates Do Not Say

Exercises

1. Basis Components of a Qubit

Let

$$\lvert\psi\rangle = \frac{\lvert0\rangle+i\lvert1\rangle}{\sqrt2}, \qquad \lvert+\rangle = \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2}, \qquad \lvert-\rangle = \frac{\lvert0\rangle-\lvert1\rangle}{\sqrt2}.$$

Compute the components of $\lvert\psi\rangle$ in the $\{\lvert+\rangle,\lvert-\rangle\}$ basis and check normalization.

Solution

The new components are inner products with the new basis:

$$c_+ = \langle+\vert\psi\rangle = \frac{1+i}{2}, \qquad c_- = \langle-\vert\psi\rangle = \frac{1-i}{2}.$$

Normalization is preserved:

$$\lvert c_+\rvert^2+\lvert c_-\rvert^2 = \frac{2}{4}+\frac{2}{4} = 1.$$

2. Born Rule in Two Bases

Let

$$\lvert\psi\rangle = \frac{\sqrt3}{2}\lvert0\rangle + \frac{e^{i\phi}}{2}\lvert1\rangle.$$

Find the probabilities for measuring $\lvert0\rangle$ or $\lvert1\rangle$, and the probability for the $+$ outcome in the $\{\lvert+\rangle,\lvert-\rangle\}$ basis.

Solution

In the $\{\lvert0\rangle,\lvert1\rangle\}$ basis,

$$p(0) = \frac{3}{4}, \qquad p(1) = \frac{1}{4}.$$

For the $+$ outcome,

$$p(+) = \lvert\langle+\vert\psi\rangle\rvert^2 = \frac{1}{2} \left\lvert \frac{\sqrt3}{2} + \frac{e^{i\phi}}{2} \right\rvert^2.$$

Expanding the squared magnitude gives

$$p(+) = \frac{1}{2} + \frac{\sqrt3}{4}\cos\phi.$$

The relative phase does not affect the $\lvert0\rangle,\lvert1\rangle$ probabilities, but it does affect interference in the $\lvert+\rangle,\lvert-\rangle$ basis.

3. Expectation Value and Variance

For the same state as Problem 2, compute $\langle\sigma_z\rangle$ and $\operatorname{Var}(\sigma_z)$.

Solution

Since $\sigma_z\lvert0\rangle=\lvert0\rangle$ and $\sigma_z\lvert1\rangle=-\lvert1\rangle$,

$$\langle\sigma_z\rangle = p(0)-p(1) = \frac{3}{4}-\frac{1}{4} = \frac{1}{2}.$$

Also $\sigma_z^2=I$, so $\langle\sigma_z^2\rangle=1$. Therefore

$$\operatorname{Var}(\sigma_z) = \langle\sigma_z^2\rangle-\langle\sigma_z\rangle^2 = 1-\frac{1}{4} = \frac{3}{4}.$$

4. Commutator and Uncertainty Check

Use $[\sigma_x,\sigma_y]=2i\sigma_z$. In the state $\lvert0\rangle$, compute the Robertson lower bound for $\Delta\sigma_x\,\Delta\sigma_y$, and compare it with the actual product.

Solution

The Robertson bound gives

$$\Delta\sigma_x\,\Delta\sigma_y \ge \frac{1}{2} \left\lvert \langle[\sigma_x,\sigma_y]\rangle \right\rvert = \frac{1}{2} \left\lvert 2i\langle\sigma_z\rangle \right\rvert.$$

In $\lvert0\rangle$, $\langle\sigma_z\rangle=1$, so the lower bound is $1$.

Since $\langle\sigma_x\rangle=\langle\sigma_y\rangle=0$ and $\sigma_x^2=\sigma_y^2=I$,

$$\Delta\sigma_x = 1, \qquad \Delta\sigma_y = 1.$$

The product is $1$, so the bound is saturated.

5. Degenerate Projective Measurement

Let $\{\lvert1\rangle,\lvert2\rangle,\lvert3\rangle\}$ be an orthonormal basis. An ideal Lüders measurement has outcome $a$ on the subspace spanned by $\lvert1\rangle,\lvert2\rangle$ and outcome $b$ on the subspace spanned by $\lvert3\rangle$. For

$$\lvert\psi\rangle = \frac{\lvert1\rangle+\lvert2\rangle+\sqrt2\,\lvert3\rangle}{2},$$

find $p(a)$ and the selective post-measurement state after observing $a$.

Solution

The projector for outcome $a$ is

$$P_a = \lvert1\rangle\langle1\rvert + \lvert2\rangle\langle2\rvert.$$

Applying it gives

$$P_a\lvert\psi\rangle = \frac{\lvert1\rangle+\lvert2\rangle}{2}.$$

Thus

$$p(a) = \langle\psi\vert P_a\vert\psi\rangle = \frac{1}{2}.$$

The normalized selective state is

$$\lvert\psi_a\rangle = \frac{P_a\lvert\psi\rangle}{\sqrt{p(a)}} = \frac{\lvert1\rangle+\lvert2\rangle}{\sqrt2}.$$

The relative coherence inside the degenerate outcome subspace is preserved by the ideal Lüders update. If the apparatus instead resolves states $1$ and $2$ and then reports both as $a$, the conditional state is $(\lvert1\rangle\langle1\rvert+\lvert2\rangle\langle2\rvert)/2$. That different measurement has the same $p(a)$ but a different state change.

6. Two-Level Time Evolution

Let

$$H = \frac{\hbar\omega}{2}\sigma_z, \qquad \lvert\psi(0)\rangle = \lvert+\rangle.$$

Compute the probability of finding the system in $\lvert+\rangle$ at time $t$.

Solution

The evolution operator is

$$U(t) = e^{-iHt/\hbar} = e^{-i\omega t\sigma_z/2}.$$

Since $\lvert+\rangle=(\lvert0\rangle+\lvert1\rangle)/\sqrt2$,

$$\lvert\psi(t)\rangle = \frac{e^{-i\omega t/2}\lvert0\rangle + e^{i\omega t/2}\lvert1\rangle}{\sqrt2}.$$

The return amplitude is

$$\langle+\vert\psi(t)\rangle = \frac{e^{-i\omega t/2}+e^{i\omega t/2}}{2} = \cos\frac{\omega t}{2}.$$

Therefore

$$p_+(t) = \cos^2\frac{\omega t}{2}.$$

7. Product State or Entangled State

Classify the two states:

$$\lvert\psi\rangle = \frac{\lvert00\rangle+\lvert01\rangle}{\sqrt2}, \qquad \lvert\Phi^+\rangle = \frac{\lvert00\rangle+\lvert11\rangle}{\sqrt2}.$$

Which one factors into subsystem states? Find the reduced state of qubit $A$ for each. Then compare the Bell state with

$$\rho_{\rm corr}=\frac12\lvert00\rangle\langle00\rvert +\frac12\lvert11\rangle\langle11\rvert.$$

Do their reduced states distinguish them? Calculate their probabilities for both qubits to give $+$ in local $+,-$ measurements.

Solution

The first state factors:

$$\lvert\psi\rangle = \lvert0\rangle \otimes \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2} = \lvert0\rangle\otimes\lvert+\rangle.$$

Its reduced state on $A$ is $\lvert0\rangle\langle0\rvert$. For the Bell state, tracing over the orthogonal states $\lvert0\rangle_B,\lvert1\rangle_B$ gives

$$\rho_A=\frac12\lvert0\rangle\langle0\rvert +\frac12\lvert1\rangle\langle1\rvert=I_A/2.$$

A pure product state has a pure reduction, so this mixed reduction proves that the pure Bell state cannot factor. The separable mixture $\rho_{\rm corr}$ also has $I/2$ on each qubit; local states alone cannot distinguish the two joint states.

A joint measurement does distinguish them. Since $\langle++\vert\Phi^+\rangle=1/\sqrt2$, the Bell state gives $p(+,+)=1/2$. For $\rho_{\rm corr}$, add the probabilities of its two classical preparation alternatives:

$$p(+,+)=\frac12\left(\frac14\right) +\frac12\left(\frac14\right)=\frac14.$$

The distinction lies in joint coherences, which the reduced states do not retain.

8. Density-Matrix Positivity

For real $r$, consider

$$\rho(r) = \frac{1}{2} \begin{pmatrix} 1 & r \\ r & 1 \end{pmatrix}.$$

For which values of $r$ is this a valid density matrix? When is it pure?

Solution

The matrix is Hermitian and has trace one for every real $r$. Its eigenvectors are the $\lvert+\rangle$ and $\lvert-\rangle$ states, with eigenvalues

$$\lambda_\pm = \frac{1\pm r}{2}.$$

Positivity requires both eigenvalues to be nonnegative, so

$$-1 \le r \le 1.$$

The state is pure when one eigenvalue is $1$ and the other is $0$, which occurs at $r=1$ or $r=-1$. Equivalently, $\operatorname{Tr}\rho^2=1$ only for $\lvert r\rvert=1$.

9. Postulates and Scope

Someone writes: “The Born rule says that a quantum state assigns a pre-existing definite value to every observable.” Explain what the rule actually supplies. Does the probability rule alone settle whether a more detailed description exists?

Solution

The Born rule supplies outcome probabilities for a specified state and measurement. It does not postulate pre-existing definite values for all observables. For a projective measurement, support of the state within one eigenspace makes that outcome certain; a general preparation gives a distribution.

The probability rule alone does not settle every question about a more detailed description. Results restricting hidden-variable models need additional, explicit assumptions, such as locality or noncontextuality. They cannot be inferred merely from the presence of probabilities. See What the Postulates Do Not Say.

Checking your answers

Check normalization and nonnegative probabilities first, then try a simple limit: $\phi=0$ in Problem 2, $t=0$ in Problem 6, or $r=\pm1$ in Problem 8. Finally, describe what could be measured to test your result. Common Checks and Sanity Tests gives a broader set of checks, and Common Mistakes explains frequent conceptual errors.

References

These references develop the methods used here; the problems and solutions are written for this review set.

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018 — basic state and operator calculations.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020 — spin, measurement, uncertainty, and dynamics.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, Chapters 2 and 6 — density operators, measurements, and bipartite states.