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Lecture 03 · Operators

Operators, Hermitian Matrices & the Four Postulates

Dirac notation and completeness, linear operators in quantum mechanics, Hermitian operators and their key theorems, the four foundational postulates, systematic derivation of the Pauli matrices, and the geometry of the Bloch sphere.

Theory Exercises

Part I — Dirac Notation & the Completeness Relation

A quantum state lives in a complex vector space — a Hilbert space. Any orthonormal basis $\{|i\rangle\}$ of this space satisfies:

$$\langle i|j\rangle = \delta_{ij}$$

Any state $|a\rangle$ can be expanded in this basis. The expansion coefficient onto basis vector $|i\rangle$ is the inner product:

$$|a\rangle = \sum_i \alpha_i|i\rangle, \qquad \alpha_i = \langle i|a\rangle$$

Substituting the second into the first:

$$|a\rangle = \sum_i \langle i|a\rangle\,|i\rangle = \left(\sum_i |i\rangle\langle i|\right)|a\rangle$$

Since this holds for every $|a\rangle$, the operator in parentheses must be the identity:

Completeness Relation
$$\sum_i |i\rangle\langle i| = \mathbf{1}$$

This is one of the most powerful tools in quantum mechanics. Inserting a complete set of states between any two expressions allows you to change bases, evaluate matrix elements, and derive results without ever writing out components explicitly.

The bra vector $\langle a|$ is the linear functional corresponding to the ket $|a\rangle$. If $|a\rangle = \sum_i \alpha_i|i\rangle$, then $\langle a| = \sum_i \alpha_i^*\langle i|$. The bra is the "dual" of the ket — a row vector of complex conjugates.

Part II — Linear Operators

A linear operator $M$ is a map from the Hilbert space to itself satisfying:

$$M(z|a\rangle) = z\cdot M|a\rangle \qquad \text{(scalar multiplication)}$$ $$M(|a\rangle + |b\rangle) = M|a\rangle + M|b\rangle \qquad \text{(additivity)}$$

In any orthonormal basis, a linear operator is represented by a matrix. The matrix element $M_{ij}$ is:

Matrix Element Formula
$$M_{ij} = \langle i|M|j\rangle$$

This formula is central: it tells us how to construct the matrix representation of any operator in any basis. The row index $i$ and column index $j$ correspond to the "output" and "input" basis states respectively.

The Hermitian Conjugate

The Hermitian conjugate (or adjoint) $M^\dagger$ of an operator $M$ is defined by:

$$\langle b|M|a\rangle^* = \langle a|M^\dagger|b\rangle \qquad \text{for all }|a\rangle,|b\rangle$$

In matrix form: take the transpose and then complex-conjugate every element (or equivalently, complex-conjugate then transpose — the order doesn't matter):

$$M^\dagger = (M^T)^* = (M^*)^T$$

Part III — Eigenvectors and Eigenvalues

A vector $|\lambda\rangle$ is an eigenvector of operator $M$ with eigenvalue $\lambda$ if:

$$M|\lambda\rangle = \lambda|\lambda\rangle$$

Geometrically: the operator does not rotate this vector — it only scales it by $\lambda$. In quantum mechanics, the eigenvectors of an observable are the states in which that observable has a definite value, and the eigenvalues are the possible measurement outcomes.

Part IV — Hermitian Operators

An operator is Hermitian (or self-adjoint) if it equals its own conjugate:

$$M = M^\dagger$$

Hermitian operators are the mathematical objects that represent physical observables in quantum mechanics. There are two fundamental theorems that explain why.

Theorem 1: Eigenvalues of Hermitian Operators Are Real

Proof. Let $M|\lambda\rangle = \lambda|\lambda\rangle$ with $|\lambda\rangle$ normalized ($\langle\lambda|\lambda\rangle = 1$). Compute $\langle\lambda|M|\lambda\rangle$ two ways:

$$\text{Using eigenvalue: } \langle\lambda|M|\lambda\rangle = \langle\lambda|\lambda\lambda\rangle = \lambda\langle\lambda|\lambda\rangle = \lambda$$ $$\text{Using }M=M^\dagger\text{: } \langle\lambda|M|\lambda\rangle = \langle\lambda|M^\dagger|\lambda\rangle = (\langle\lambda|M|\lambda\rangle)^* = \lambda^*$$

Therefore $\lambda = \lambda^*$, which means $\lambda \in \mathbb{R}$. $\square$

This is why observables must be Hermitian: measurement outcomes are real numbers, and Hermitian operators guarantee real eigenvalues.

Theorem 2: Eigenvectors with Different Eigenvalues Are Orthogonal

Proof. Let $M|a\rangle = \lambda_a|a\rangle$ and $M|b\rangle = \lambda_b|b\rangle$ with $\lambda_a \neq \lambda_b$. Compute $\langle b|M|a\rangle$ two ways:

$$\text{Acting right: } \langle b|M|a\rangle = \lambda_a\langle b|a\rangle$$ $$\text{Acting left (since }M\text{ Hermitian, }\langle b|M = \lambda_b\langle b|\text{): } \langle b|M|a\rangle = \lambda_b\langle b|a\rangle$$

Subtracting: $(\lambda_a - \lambda_b)\langle b|a\rangle = 0$. Since $\lambda_a \neq \lambda_b$, we have $\langle b|a\rangle = 0$. $\square$

Hermitian operators always have enough eigenvectors to form a complete basis (a fact requiring more work to prove in general). In their own eigenbasis, a Hermitian operator is diagonal:

$$M = \begin{pmatrix}\lambda_1 & 0 & \cdots \\ 0 & \lambda_2 & \cdots \\ \vdots & \vdots & \ddots\end{pmatrix}$$

Part V — The Four Postulates

All of quantum mechanics for isolated systems rests on four postulates. Together they form a complete logical framework — a precise mathematical language in which to express all quantum phenomena.

Postulate 1 — Observables
Observables are represented by Hermitian operators acting on the state space.
Postulate 2 — Measurement Outcomes
The only possible outcomes of a measurement of observable $M$ are its eigenvalues $\lambda_i$. These are real numbers (by the Hermiticity theorem).
Postulate 3 — Definite States
If the system is in an eigenstate $|i\rangle$ of $M$, then measuring $M$ yields the corresponding eigenvalue $\lambda_i$ with certainty: $P(\lambda_i\,|\,|i\rangle) = 1$.
Postulate 4 — The Born Rule
$$P(\lambda_i\,|\,|a\rangle) = |\langle i|a\rangle|^2$$

Consistency: Normalization

The Born rule is consistent with normalization. Using the completeness relation:

$$\sum_i P(\lambda_i) = \sum_i|\langle i|a\rangle|^2 = \langle a|\left(\sum_i|i\rangle\langle i|\right)|a\rangle = \langle a|\mathbf{1}|a\rangle = \langle a|a\rangle = 1$$

Global Phase

Multiplying a state by an overall phase $e^{i\theta}$ changes nothing observable. For any $|\phi\rangle$:

$$|\langle\phi|\,e^{i\theta}|\psi\rangle\,|^2 = |e^{i\theta}|^2|\langle\phi|\psi\rangle|^2 = |\langle\phi|\psi\rangle|^2$$

Therefore $e^{i\theta}|\psi\rangle \equiv |\psi\rangle$ as physical states. Quantum states are not individual vectors — they are rays (equivalence classes of vectors differing only by a global phase) in Hilbert space.

Part VI — Deriving the Pauli Matrices

Rather than postulating the Pauli matrices, we can derive them systematically from their eigenstates using the formula $M_{ij} = \langle i|M|j\rangle$.

Deriving $\sigma_z$

The eigenstates are $|u\rangle = (1,0)^T$ with eigenvalue $+1$ and $|d\rangle = (0,1)^T$ with eigenvalue $-1$. Using $\{|u\rangle, |d\rangle\}$ as the basis, compute each matrix element:

$$(\sigma_z)_{uu} = \langle u|\sigma_z|u\rangle = +1\langle u|u\rangle = +1$$ $$(\sigma_z)_{dd} = \langle d|\sigma_z|d\rangle = -1\langle d|d\rangle = -1$$ $$(\sigma_z)_{ud} = \langle u|\sigma_z|d\rangle = -1\langle u|d\rangle = 0$$ $$(\sigma_z)_{du} = \langle d|\sigma_z|u\rangle = +1\langle d|u\rangle = 0$$
$$\sigma_z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

Deriving $\sigma_x$

The eigenstates are $|r\rangle = \frac{1}{\sqrt{2}}(1,1)^T$ ($+1$) and $|l\rangle = \frac{1}{\sqrt{2}}(1,-1)^T$ ($-1$). Using the completeness relation $|r\rangle\langle r| + |l\rangle\langle l| = \mathbf{1}$, each matrix element is:

$$(\sigma_x)_{uu} = \langle u|\sigma_x|u\rangle = +1\,|\langle r|u\rangle|^2\cdot\text{sign}(\ldots) - \ldots$$

More directly, $\sigma_x = (+1)|r\rangle\langle r| + (-1)|l\rangle\langle l|$. Computing in the $\{|u\rangle, |d\rangle\}$ basis:

$$(\sigma_x)_{uu} = \langle u|\sigma_x|u\rangle = \langle u|(|r\rangle\langle r| - |l\rangle\langle l|)|u\rangle = |\langle r|u\rangle|^2 - |\langle l|u\rangle|^2 = \frac{1}{2} - \frac{1}{2} = 0$$ $$(\sigma_x)_{ud} = \langle u|\sigma_x|d\rangle = \langle u|r\rangle\langle r|d\rangle - \langle u|l\rangle\langle l|d\rangle = \frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\cdot\frac{-1}{\sqrt{2}} = \frac{1}{2}+\frac{1}{2} = 1$$ $$(\sigma_x)_{du} = \langle d|\sigma_x|u\rangle = \langle d|r\rangle\langle r|u\rangle - \langle d|l\rangle\langle l|u\rangle = \frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} - \frac{-1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} = \frac{1}{2}+\frac{1}{2} = 1$$ $$(\sigma_x)_{dd} = 0 \quad\text{(same argument as }(\sigma_x)_{uu}\text{)}$$
$$\sigma_x = \begin{pmatrix}0&1\\1&0\end{pmatrix}$$

Deriving $\sigma_y$

The eigenstates are $|i_s\rangle = \frac{1}{\sqrt{2}}(1,i)^T$ ($+1$) and $|o\rangle = \frac{1}{\sqrt{2}}(1,-i)^T$ ($-1$). Using $\sigma_y = |i_s\rangle\langle i_s| - |o\rangle\langle o|$:

$$(\sigma_y)_{ud} = \langle u|i_s\rangle\langle i_s|d\rangle - \langle u|o\rangle\langle o|d\rangle = \frac{1}{\sqrt{2}}\cdot\frac{-i}{\sqrt{2}} - \frac{1}{\sqrt{2}}\cdot\frac{i}{\sqrt{2}} = \frac{-i}{2} - \frac{i}{2} = -i$$ $$(\sigma_y)_{du} = (\sigma_y)_{ud}^* = (-i)^* = i, \qquad (\sigma_y)_{uu} = (\sigma_y)_{dd} = 0$$
$$\sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$$

Verification: $\sigma_y|i_s\rangle = \begin{pmatrix}0&-i\\i&0\end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}-i\cdot i\\i\cdot 1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix} = +1\cdot|i_s\rangle$ ✓

Properties of All Three Pauli Matrices

PropertyMeaning
Hermitian: $\sigma_i^\dagger = \sigma_i$Represent valid observables (real eigenvalues)
Unitary: $\sigma_i\sigma_i^\dagger = \mathbf{1}$Eigenvalues have unit magnitude
Traceless: $\text{tr}(\sigma_i) = 0$Eigenvalues sum to zero: $+1 + (-1) = 0$
$\sigma_i^2 = \mathbf{1}$Idempotent in a sense; measurement twice = once

The full Pauli algebra is encoded in one compact formula:

Pauli Algebra
$$\sigma_i\sigma_j = \delta_{ij}\,\mathbf{1} + i\,\varepsilon_{ijk}\,\sigma_k$$

Part VII — Physical Meaning of Matrix Elements

The matrix element $M_{ij} = \langle i|M|j\rangle$ has a direct physical interpretation: it is the amplitude for a system in state $|j\rangle$ to contribute to the observable value when the output is projected onto $|i\rangle$. The squared modulus $|M_{ij}|^2$ gives the corresponding probability (in a suitable sense).

The expectation value of $M$ in state $|a\rangle$ is:

$$\langle M\rangle = \langle a|M|a\rangle = \sum_n \lambda_n |\langle n|a\rangle|^2$$

The second form (expanding in the eigenbasis $\{|n\rangle\}$ of $M$) shows that $\langle M\rangle$ is literally the weighted average of eigenvalues, with weights given by the Born rule probabilities. This is the quantum generalization of a classical expectation value.

Part VIII — Degrees of Freedom and the Bloch Sphere

How many real parameters specify a quantum state? Naively, a vector in $\mathbb{C}^N$ has $2N$ real numbers (real and imaginary parts of each component). But physics reduces this count:

ConstraintRemovesRemaining parameters
Start with $|a\rangle\in\mathbb{C}^N$—$2N$
Normalization: $\langle a|a\rangle = 1$1 real constraint$2N - 1$
Global phase: $e^{i\theta}|a\rangle \equiv |a\rangle$1 more$2N - 2$

For a spin-½ system ($N = 2$): $2(2) - 2 = 2$ real parameters. Two real parameters label a point on a sphere — the Bloch sphere. The general qubit state (up to global phase) is:

$$|\psi\rangle = \cos\frac{\theta}{2}\,|u\rangle + e^{i\varphi}\sin\frac{\theta}{2}\,|d\rangle, \qquad \theta\in[0,\pi],\;\varphi\in[0,2\pi)$$

The expectation values of the three spin components are:

$$\langle\sigma_z\rangle = \cos\theta, \qquad \langle\sigma_x\rangle = \sin\theta\cos\varphi, \qquad \langle\sigma_y\rangle = \sin\theta\sin\varphi$$

These are exactly the Cartesian components of a unit vector in spherical coordinates $(\theta, \varphi)$. The "Bloch vector" $(\langle\sigma_x\rangle, \langle\sigma_y\rangle, \langle\sigma_z\rangle)$ lies exactly on the unit sphere. Every pure qubit state corresponds to a unique point on the Bloch sphere.

Exercises — Lecture 3

Nine exercises on operators, Hermitian matrices, the four postulates, and the Bloch sphere. Click any exercise to reveal the full worked solution.

1
Completeness Relation Verification
EasyDirac Notation

In the $\{|u\rangle, |d\rangle\}$ basis, verify the completeness relation by computing $|u\rangle\langle u| + |d\rangle\langle d|$ as explicit $2\times 2$ matrices and confirming the sum is the identity.

Complete Solution
1

Recall $|u\rangle = (1,0)^T$ so $\langle u| = (1,0)$. The outer product is a matrix:

$$|u\rangle\langle u| = \begin{pmatrix}1\\0\end{pmatrix}(1,0) = \begin{pmatrix}1\cdot1 & 1\cdot0\\0\cdot1 & 0\cdot0\end{pmatrix} = \begin{pmatrix}1&0\\0&0\end{pmatrix}$$
2

Similarly, $|d\rangle = (0,1)^T$ so $\langle d| = (0,1)$:

$$|d\rangle\langle d| = \begin{pmatrix}0\\1\end{pmatrix}(0,1) = \begin{pmatrix}0&0\\0&1\end{pmatrix}$$
3
$$|u\rangle\langle u| + |d\rangle\langle d| = \begin{pmatrix}1&0\\0&0\end{pmatrix} + \begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = \mathbf{1} \checkmark$$
Note
The completeness relation holds in any orthonormal basis. The outer products $|i\rangle\langle i|$ are projection operators — they project any vector onto the $|i\rangle$ direction. Their sum reconstructs the full identity, meaning no direction is missed.
2
Eigenvalue Equation Check
EasyOperators

Verify that $|r\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$ is an eigenvector of $\sigma_x = \begin{pmatrix}0&1\\1&0\end{pmatrix}$ with eigenvalue $+1$.

Complete Solution
1

Compute $\sigma_x|r\rangle$ directly:

$$\sigma_x|r\rangle = \begin{pmatrix}0&1\\1&0\end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}0\cdot1+1\cdot1\\1\cdot1+0\cdot1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$$
2
$$\sigma_x|r\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = +1 \cdot |r\rangle \checkmark$$
Conclusion
$|r\rangle$ is an eigenvector of $\sigma_x$ with eigenvalue $+1$. Applying $\sigma_x$ merely returns the same vector — no rotation, only scaling by $+1$.
3
Checking Hermiticity of $\sigma_y$
EasyHermitian

Show that $\sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$ is Hermitian by computing $(\sigma_y)^\dagger$ and verifying $(\sigma_y)^\dagger = \sigma_y$.

Complete Solution
Transpose

Swap rows and columns of $\sigma_y$:

$$\sigma_y^T = \begin{pmatrix}0&i\\-i&0\end{pmatrix}$$
Complex conjugate

Complex-conjugate each entry:

$$(\sigma_y^T)^* = \begin{pmatrix}0^*&i^*\\(-i)^*&0^*\end{pmatrix} = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$$
Conclusion
$$(\sigma_y)^\dagger = (σ_y^T)^* = \begin{pmatrix}0&-i\\i&0\end{pmatrix} = \sigma_y \checkmark$$
Verified
$\sigma_y$ is Hermitian: $\sigma_y^\dagger = \sigma_y$. Therefore its eigenvalues are guaranteed to be real, and its eigenstates form a complete orthonormal basis.
4
$\sigma_z$ in the $x$-Eigenbasis
MediumMatrix Elements

Using $M_{ij} = \langle i|M|j\rangle$, compute all four matrix elements of $\sigma_z$ in the $\{|r\rangle, |l\rangle\}$ basis. What does the result tell you about $|r\rangle$ and $|l\rangle$ as states of $\sigma_z$? What does the off-diagonal element signify?

Complete Solution
Setup

First find how $\sigma_z$ acts on $|r\rangle$ and $|l\rangle$:

$$\sigma_z|r\rangle = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix} = |l\rangle$$
$$\sigma_z|l\rangle = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = |r\rangle$$
Matrix elements
$$(\sigma_z)_{rr} = \langle r|\sigma_z|r\rangle = \langle r|l\rangle = 0$$
$$(\sigma_z)_{rl} = \langle r|\sigma_z|l\rangle = \langle r|r\rangle = 1$$
$$(\sigma_z)_{lr} = \langle l|\sigma_z|r\rangle = \langle l|l\rangle = 1$$
$$(\sigma_z)_{ll} = \langle l|\sigma_z|l\rangle = \langle l|r\rangle = 0$$
Result

In the $\{|r\rangle, |l\rangle\}$ basis:

$$\sigma_z \overset{\{|r\rangle,|l\rangle\}}{=} \begin{pmatrix}0&1\\1&0\end{pmatrix} = \sigma_x \text{ (in the standard basis!)}$$
Interpretation
$\sigma_z$ looks like $\sigma_x$ in the $x$-eigenbasis — because we've just relabeled our basis vectors. The zero diagonal means $|r\rangle$ and $|l\rangle$ are not eigenstates of $\sigma_z$. The off-diagonal entry $= 1$ means $\sigma_z$ mixes $|r\rangle$ and $|l\rangle$ — it transitions between them. Non-zero off-diagonal elements signal that the operator does not share an eigenbasis with the current basis.
5
Born Rule with a Superposition State
MediumBorn Rule

A spin is in state $|\psi\rangle = \dfrac{1}{2}|u\rangle + \dfrac{\sqrt{3}}{2}|d\rangle$.

(a) Verify normalization.

(b) Find $P(\sigma_z = +1)$ and $P(\sigma_z = -1)$.

(c) Compute $\langle\sigma_z\rangle$ two ways: using the Born rule, and using the matrix sandwich $\langle\psi|\sigma_z|\psi\rangle$.

Complete Solution
a
$$\langle\psi|\psi\rangle = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1 \checkmark$$
b

By the Born rule (squaring the amplitudes):

$$P(\sigma_z=+1) = |\langle u|\psi\rangle|^2 = \left|\frac{1}{2}\right|^2 = \frac{1}{4}$$
$$P(\sigma_z=-1) = |\langle d|\psi\rangle|^2 = \left|\frac{\sqrt{3}}{2}\right|^2 = \frac{3}{4}$$
c

Via Born rule:

$$\langle\sigma_z\rangle = (+1)\cdot\frac{1}{4} + (-1)\cdot\frac{3}{4} = \frac{1}{4} - \frac{3}{4} = -\frac{1}{2}$$

Via matrix sandwich (with $|\psi\rangle = (1/2,\,\sqrt{3}/2)^T$):

$$\langle\psi|\sigma_z|\psi\rangle = \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1/2\\\sqrt{3}/2\end{pmatrix} = \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\begin{pmatrix}1/2\\-\sqrt{3}/2\end{pmatrix}$$
$$= \frac{1}{2}\cdot\frac{1}{2} + \frac{\sqrt{3}}{2}\cdot\left(-\frac{\sqrt{3}}{2}\right) = \frac{1}{4} - \frac{3}{4} = -\frac{1}{2} \checkmark$$
Both methods agree
$\langle\sigma_z\rangle = -\frac{1}{2}$. The two calculations are always equivalent — they are different expressions of the same underlying formula.
6
Proving Eigenvalue Reality
MediumProof

Let $M$ be a Hermitian $2\times 2$ matrix with eigenvalue $\lambda$ and normalized eigenvector $|v\rangle$. Prove that $\lambda$ is real by computing $\langle v|M|v\rangle$ in two different ways and comparing.

Complete Solution
Way 1

Use the eigenvalue equation $M|v\rangle = \lambda|v\rangle$:

$$\langle v|M|v\rangle = \langle v|(\lambda|v\rangle) = \lambda\langle v|v\rangle = \lambda \cdot 1 = \lambda$$
Way 2

Use $M = M^\dagger$ and the definition of the Hermitian conjugate. Since $M = M^\dagger$:

$$\langle v|M|v\rangle = \langle v|M^\dagger|v\rangle = (\langle v|M|v\rangle)^* = \lambda^*$$

(The last step uses the general identity $\langle b|M^\dagger|a\rangle = \langle a|M|b\rangle^*$ with $|a\rangle = |b\rangle = |v\rangle$.)

Conclusion

Both computations give $\langle v|M|v\rangle$, so their results must be equal:

$$\lambda = \lambda^*$$
Proven
A complex number equals its own conjugate if and only if it is real. Therefore $\lambda \in \mathbb{R}$. $\square$ This is why Hermitian operators represent observables — their eigenvalues (possible measurement outcomes) are guaranteed to be real numbers.
7
Orthogonality of Eigenvectors from Algebra
MediumProof

Given $\sigma_z|u\rangle = +1\cdot|u\rangle$ and $\sigma_z|d\rangle = -1\cdot|d\rangle$. Use Theorem 2 to prove $\langle d|u\rangle = 0$ without computing the inner product directly. Show each algebraic step.

Complete Solution
Setup

We will compute $\langle d|\sigma_z|u\rangle$ in two ways by acting on the right and on the left.

Acting right

Use the eigenvalue equation $\sigma_z|u\rangle = +1\cdot|u\rangle$:

$$\langle d|\sigma_z|u\rangle = \langle d|(+1)|u\rangle = +1\cdot\langle d|u\rangle$$
Acting left

Since $\sigma_z$ is Hermitian, it acts to the left on $\langle d|$. From $\sigma_z|d\rangle = -|d\rangle$, taking the Hermitian conjugate gives $\langle d|\sigma_z = -1\cdot\langle d|$:

$$\langle d|\sigma_z|u\rangle = (-1\cdot\langle d|)|u\rangle = -1\cdot\langle d|u\rangle$$
Equate

Both expressions equal $\langle d|\sigma_z|u\rangle$:

$$+1\cdot\langle d|u\rangle = -1\cdot\langle d|u\rangle \implies 2\langle d|u\rangle = 0 \implies \langle d|u\rangle = 0 \checkmark$$
Key Step
The eigenvalues $+1 \neq -1$ — this is what forced $\langle d|u\rangle = 0$. If the eigenvalues were equal, the argument would give $0 = 0$, which is uninformative (degenerate case).
8
Deriving $\sigma_y$ Matrix Elements
HardDerivation

Using the eigenstates $|i_s\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}$ (eigenvalue $+1$) and $|o\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}$ (eigenvalue $-1$), compute all four matrix elements of $\sigma_y$ in the $\{|u\rangle, |d\rangle\}$ basis using $M_{ij} = \langle i|M|j\rangle$, expanding via $\sigma_y = |i_s\rangle\langle i_s| - |o\rangle\langle o|$.

Complete Solution
Spectral decomposition

Any Hermitian operator with eigenstates $|n\rangle$ and eigenvalues $\lambda_n$ can be written as $M = \sum_n \lambda_n|n\rangle\langle n|$. For $\sigma_y$:

$$\sigma_y = (+1)|i_s\rangle\langle i_s| + (-1)|o\rangle\langle o| = |i_s\rangle\langle i_s| - |o\rangle\langle o|$$

So $\langle i|\sigma_y|j\rangle = \langle i|i_s\rangle\langle i_s|j\rangle - \langle i|o\rangle\langle o|j\rangle$ where $i,j \in \{u,d\}$.

Needed amplitudes

Compute the overlaps (bra of $|i_s\rangle$ is $\frac{1}{\sqrt{2}}(1,-i)$, bra of $|o\rangle$ is $\frac{1}{\sqrt{2}}(1,i)$):

$$\langle i_s|u\rangle = \frac{1}{\sqrt{2}}, \quad \langle i_s|d\rangle = \frac{-i}{\sqrt{2}}, \quad \langle o|u\rangle = \frac{1}{\sqrt{2}}, \quad \langle o|d\rangle = \frac{i}{\sqrt{2}}$$
$$\langle u|i_s\rangle = \frac{1}{\sqrt{2}}, \quad \langle d|i_s\rangle = \frac{i}{\sqrt{2}}, \quad \langle u|o\rangle = \frac{1}{\sqrt{2}}, \quad \langle d|o\rangle = \frac{-i}{\sqrt{2}}$$
Diagonal elements
$$(\sigma_y)_{uu} = \langle u|i_s\rangle\langle i_s|u\rangle - \langle u|o\rangle\langle o|u\rangle = \frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} = \frac{1}{2} - \frac{1}{2} = 0$$
$$(\sigma_y)_{dd} = \frac{i}{\sqrt{2}}\cdot\frac{-i}{\sqrt{2}} - \frac{-i}{\sqrt{2}}\cdot\frac{i}{\sqrt{2}} = \frac{-i^2}{2} - \frac{-i^2}{2} = \frac{1}{2} - \frac{1}{2} = 0$$
Off-diagonal elements
$$(\sigma_y)_{ud} = \langle u|i_s\rangle\langle i_s|d\rangle - \langle u|o\rangle\langle o|d\rangle = \frac{1}{\sqrt{2}}\cdot\frac{-i}{\sqrt{2}} - \frac{1}{\sqrt{2}}\cdot\frac{i}{\sqrt{2}} = \frac{-i}{2} - \frac{i}{2} = -i$$
$$(\sigma_y)_{du} = \langle d|i_s\rangle\langle i_s|u\rangle - \langle d|o\rangle\langle o|u\rangle = \frac{i}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} - \frac{-i}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} = \frac{i}{2} + \frac{i}{2} = i$$
Result
$$\sigma_y = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} \checkmark$$ The derivation from eigenstates reproduces the correct Pauli matrix — no guesswork required.
9
Global Phase and the Bloch Sphere
HardBloch Sphere

The general normalized qubit state (up to global phase) is:

$$|\psi\rangle = \cos\frac{\theta}{2}\,|u\rangle + e^{i\varphi}\sin\frac{\theta}{2}\,|d\rangle, \qquad \theta\in[0,\pi],\;\varphi\in[0,2\pi)$$

(a) How many real parameters specify this state? Why is this the Bloch sphere?

(b) Compute $\langle\sigma_z\rangle$, $\langle\sigma_x\rangle$, and $\langle\sigma_y\rangle$ for this general state.

(c) Show that $\langle\sigma_x\rangle^2 + \langle\sigma_y\rangle^2 + \langle\sigma_z\rangle^2 = 1$. What does this mean geometrically?

Complete Solution
a

The state has two real parameters: $\theta \in [0,\pi]$ and $\varphi \in [0,2\pi)$. These are exactly the polar and azimuthal angles of a point on a unit sphere $S^2$. Hence the space of physically distinct qubit states is the Bloch sphere — a 2-sphere. The general $\mathbb{C}^2$ vector has 4 real degrees of freedom, reduced by normalization (−1) and global phase (−1) to exactly 2.

b

Write $\alpha_1 = \cos(\theta/2)$ and $\alpha_2 = e^{i\varphi}\sin(\theta/2)$. Compute each expectation value:

$$\langle\sigma_z\rangle = |\alpha_1|^2(+1) + |\alpha_2|^2(-1) = \cos^2\!\frac{\theta}{2} - \sin^2\!\frac{\theta}{2} = \cos\theta$$

For $\langle\sigma_x\rangle = 2\,\text{Re}(\alpha_1^*\alpha_2)$:

$$\alpha_1^*\alpha_2 = \cos\frac{\theta}{2}\cdot e^{i\varphi}\sin\frac{\theta}{2} \implies \text{Re}(\alpha_1^*\alpha_2) = \cos\frac{\theta}{2}\sin\frac{\theta}{2}\cos\varphi$$
$$\langle\sigma_x\rangle = 2\cos\frac{\theta}{2}\sin\frac{\theta}{2}\cos\varphi = \sin\theta\cos\varphi$$

For $\langle\sigma_y\rangle = 2\,\text{Im}(\alpha_1^*\alpha_2)$:

$$\langle\sigma_y\rangle = 2\cos\frac{\theta}{2}\sin\frac{\theta}{2}\sin\varphi = \sin\theta\sin\varphi$$
c
$$\langle\sigma_x\rangle^2 + \langle\sigma_y\rangle^2 + \langle\sigma_z\rangle^2 = \sin^2\!\theta\cos^2\!\varphi + \sin^2\!\theta\sin^2\!\varphi + \cos^2\!\theta$$ $$= \sin^2\!\theta(\cos^2\!\varphi + \sin^2\!\varphi) + \cos^2\!\theta = \sin^2\!\theta + \cos^2\!\theta = 1$$
Geometric Meaning
The expectation value vector $(\langle\sigma_x\rangle, \langle\sigma_y\rangle, \langle\sigma_z\rangle) = (\sin\theta\cos\varphi,\,\sin\theta\sin\varphi,\,\cos\theta)$ has magnitude 1. It is a unit vector — precisely the Bloch vector pointing to the corresponding point on the Bloch sphere. Every pure qubit state corresponds to a unique unit vector in $\mathbb{R}^3$, and the measurement statistics are fully encoded in that single Bloch vector.