1.1.3 Hermitian Operators#

Prompts

Why must observables be represented by Hermitian operators rather than arbitrary operators? What special property of Hermitian operators makes them suitable for describing measurements?
Find the eigenvalues and eigenvectors of \(\hat{X}\) and \(\hat{Z}\) by solving the characteristic equation. Why are all Pauli eigenvalues \(\pm 1\)?
For a general qubit state \(\vert \psi\rangle = \alpha\vert 0\rangle + \beta\vert 1\rangle\), write down the expectation values \(\langle\hat{X}\rangle\), \(\langle\hat{Y}\rangle\), \(\langle\hat{Z}\rangle\) in terms of \(\alpha\) and \(\beta\). What do these three numbers represent geometrically?
Prove that eigenvectors of a Hermitian operator with different eigenvalues are orthogonal. Does this result depend on the eigenvectors being normalized?
Interpret the Bloch vector: for a qubit state parameterized by angles \((\theta, \phi)\), explain how the expectation values of the Pauli operators encode position on the Bloch sphere.

Lecture Notes#

Overview#

In quantum mechanics, observables (measurable quantities) are represented by Hermitian operators. This is a fundamental postulate, not a mathematical curiosity. The connection between operator structure and measurement reality is profound: for measurement outcomes to be real numbers, the operators must have special properties. This section establishes the mathematical foundation—eigenvalues, spectral decomposition, and the Pauli matrices—that makes quantum predictions testable.

Why Hermitian?#

A quantum state is described by complex numbers (coefficients in a superposition). Yet every measurement outcome is a real number. How do we extract reality from complexity?

Analogy from complex numbers: For a complex number \(z = x + \mathrm{i}y\), the real part is \(x = \frac{z + z^*}{2}\). The key property: \(x = x^*\) (the real part equals its own complex conjugate).

Same idea for operators: For an observable to yield real measurement outcomes, we require:

(9)#\[\hat{O}^\dagger = \hat{O}\]

This is the definition of a Hermitian (or self-adjoint) operator, where \(\hat{O}^\dagger\) is the Hermitian conjugate.

Definition: Hermitian Conjugate

For an operator \(\hat{O}\), the Hermitian conjugate \(\hat{O}^\dagger\) is defined by:

\[\langle \psi \vert \hat{O} \phi \rangle = \langle \hat{O}^\dagger \psi \vert \phi \rangle\]

for all states \(\vert \psi\rangle, \vert \phi\rangle\).

Operators as Matrices#

Quantum operators act on the Hilbert space (spanned by basis states \(\vert 0\rangle, \vert 1\rangle, \ldots\)). We represent operators as matrices using the rule:

(10)#\[O_{ij} = \langle i \vert \hat{O} \vert j \rangle\]

How to build a matrix: If you know how \(\hat{O}\) acts on each basis state, fill in the matrix column by column.

Tip: Building a Matrix from Action Rules

If \(\hat{O}\vert j\rangle = \sum_i a_{ij}\vert i\rangle\), then the \(j\)-th column of the matrix is \([a_{0j}, a_{1j}, a_{2j}, \ldots]^T\).

Example

For \(\hat{X}\) on a qubit (basis \(\vert 0\rangle, \vert 1\rangle\)):

\[\begin{split} \hat{X}\vert 0\rangle = \vert 1\rangle = 0\vert 0\rangle + 1\vert 1\rangle \quad \Rightarrow \text{column 0} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{split}\]

\[\begin{split} \hat{X}\vert 1\rangle = \vert 0\rangle = 1\vert 0\rangle + 0\vert 1\rangle \quad \Rightarrow \text{column 1} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{split}\]

Therefore,

\[\begin{split} \quad \hat{X} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{split}\]

For a matrix, \(\hat{O}^\dagger = (\hat{O}^*)^T\) (conjugate transpose).

A matrix is Hermitian if all diagonal elements are real and off-diagonal elements are conjugate pairs: \(O_{ij} = O_{ji}^*\).

Eigenvalues and Eigenvectors#

Some special states \(\vert \psi\rangle\) are only rescaled by an operator:

(11)#\[\hat{O}\vert \psi\rangle = \lambda \vert \psi\rangle\]

Here \(\vert \psi\rangle\) is an eigenvector (eigenstate) and \(\lambda\) is its eigenvalue.

Theorem: Hermitian Operators Have Real Eigenvalues

If \(\hat{O}\) is Hermitian and \(\vert \psi\rangle\) is an eigenvector with eigenvalue \(\lambda\), then \(\lambda \in \mathbb{R}\) (is real).

Proof

Take the inner product of the eigenvalue equation with \(\langle \psi\vert \):

\[\langle \psi\vert \hat{O}\vert \psi\rangle = \lambda \langle\psi\vert \psi\rangle = \lambda\]

Take the complex conjugate of both sides:

\[\langle \psi\vert \hat{O}\vert \psi\rangle^* = \lambda^*\]

But using Hermiticity:

\[\langle \psi\vert \hat{O}\vert \psi\rangle^* = (\langle \psi\vert \hat{O}\vert \psi\rangle)^* = \langle \psi\vert \hat{O}^\dagger\vert \psi\rangle = \langle \psi\vert \hat{O}\vert \psi\rangle = \lambda\]

Thus \(\lambda = \lambda^*\), so \(\lambda\) is real.

Theorem: Orthogonality of Eigenvectors

If \(\hat{O}\) is Hermitian and \(\vert \psi_1\rangle, \vert \psi_2\rangle\) are eigenvectors with different eigenvalues \(\lambda_1 \neq \lambda_2\), then they are orthogonal:

\[\langle\psi_1\vert \psi_2\rangle = 0\]

Proof

Compute:

\[\langle \psi_1\vert \hat{O}\vert \psi_2\rangle = \lambda_2 \langle\psi_1\vert \psi_2\rangle\]

But also:

\[\langle \psi_1\vert \hat{O}\vert \psi_2\rangle = (\hat{O}\vert \psi_1\rangle)^\dagger\vert \psi_2\rangle = (\lambda_1\vert \psi_1\rangle)^\dagger\vert \psi_2\rangle = \lambda_1^* \langle\psi_1\vert \psi_2\rangle = \lambda_1 \langle\psi_1\vert \psi_2\rangle\]

(using Hermiticity: \(\lambda_1^* = \lambda_1\)). Equating:

\[\lambda_2 \langle\psi_1\vert \psi_2\rangle = \lambda_1 \langle\psi_1\vert \psi_2\rangle \quad \Rightarrow \quad (\lambda_2 - \lambda_1)\langle\psi_1\vert \psi_2\rangle = 0\]

Since \(\lambda_2 \neq \lambda_1\), we have \(\langle\psi_1\vert \psi_2\rangle = 0\).

Completeness for qubits: A 2×2 Hermitian matrix has exactly 2 linearly independent eigenvectors (counting multiplicity). These eigenvectors span the entire 2-dimensional Hilbert space and form an eigenbasis.

Spectral Decomposition#

Any Hermitian operator can be decomposed as:

(12)#\[\hat{O} = \sum_i \lambda_i \vert \psi_i\rangle\langle \psi_i\vert \]

where \(\lambda_i\) are eigenvalues and \(\vert \psi_i\rangle\langle \psi_i\vert \) is the projector onto eigenstate \(\vert \psi_i\rangle\).

Example: Spectral Decomposition of Pauli Operators

\(\hat{Z}\): Eigenstates \(\vert 0\rangle\) (eigenvalue \(+1\)) and \(\vert 1\rangle\) (eigenvalue \(-1\)).

\[\begin{split}\hat{Z} = (+1)\vert 0\rangle\langle 0\vert + (-1)\vert 1\rangle\langle 1\vert = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\end{split}\]

\(\hat{X}\): Eigenstates \(\vert +\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\) (eigenvalue \(+1\)) and \(\vert -\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle - \vert 1\rangle)\) (eigenvalue \(-1\)).

\[\hat{X} = (+1)\vert +\rangle\langle +\vert + (-1)\vert -\rangle\langle -\vert \]

As matrices:

\[\begin{split}|+\rangle\langle +| = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \quad |-\rangle\langle -| = \frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\end{split}\]

\[\begin{split}\hat{X} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} - \frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad \checkmark\end{split}\]

Pauli Operators#

How many independent Hermitian operators exist on a qubit? A general \(2\times 2\) Hermitian matrix has the form

\[\begin{split}\begin{pmatrix} a & b+\mathrm{i}c \\ b-\mathrm{i}c & d \end{pmatrix}, \quad a,b,c,d \in \mathbb{R}\end{split}\]

giving 4 real degrees of freedom. Thus there are exactly 4 linearly independent Hermitian \(2\times 2\) matrices. The standard choice is the identity \(\hat{I}\) and the three Pauli operators \(\hat{\sigma}^x, \hat{\sigma}^y, \hat{\sigma}^z\): every Hermitian \(2\times 2\) operator can be uniquely written as \(a_0\hat{I} + a_x\hat{\sigma}^x + a_y\hat{\sigma}^y + a_z\hat{\sigma}^z\) with \(a_0,a_x,a_y,a_z\in\mathbb{R}\).

Table: Pauli Operators

Spin Notation	QI Notation	Matrix	Action on \(\vert 0 \rangle\)	Action on \(\vert 1 \rangle\)
\(\hat{\sigma}^x\)	\(\hat{X}\)	\(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)	\(\vert 1 \rangle\)	\(\vert 0 \rangle\)
\(\hat{\sigma}^y\)	\(\hat{Y}\)	\(\begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix}\)	\(\mathrm{i}\vert 1 \rangle\)	\(-\mathrm{i}\vert 0 \rangle\)
\(\hat{\sigma}^z\)	\(\hat{Z}\)	\(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)	\(\vert 0 \rangle\)	\(-\vert 1 \rangle\)

Table: Spectral Decomposition of Pauli Operators

Operator \(\hat{O}\)	Eigenvalue	Eigenstate	Projector \(\hat{P}_{O=+1}\)
\(\hat{X}\)	\(+1\)	\(\vert + \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}\)	\(\vert + \rangle\langle + \vert = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \frac{1}{2}(\hat{I} + \hat{X})\)
\(\hat{Y}\)	\(+1\)	\(\vert \mathrm{i} \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ \mathrm{i} \end{pmatrix}\)	\(\vert \mathrm{i} \rangle\langle \mathrm{i} \vert = \frac{1}{2}\begin{pmatrix} 1 & -\mathrm{i} \\ \mathrm{i} & 1 \end{pmatrix} = \frac{1}{2}(\hat{I} + \hat{Y})\)
\(\hat{Z}\)	\(+1\)	\(\vert 0 \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\)	\(\vert 0 \rangle\langle 0 \vert = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \frac{1}{2}(\hat{I} + \hat{Z})\)

Operator \(\hat{O}\)	Eigenvalue	Eigenstate	Projector \(\hat{P}_{O=-1}\)
\(\hat{X}\)	\(-1\)	\(\vert - \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}\)	\(\vert - \rangle\langle - \vert = \frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = \frac{1}{2}(\hat{I} - \hat{X})\)
\(\hat{Y}\)	\(-1\)	\(\vert \bar{\mathrm{i}} \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -\mathrm{i} \end{pmatrix}\)	\(\vert \bar{\mathrm{i}} \rangle\langle \bar{\mathrm{i}} \vert = \frac{1}{2}\begin{pmatrix} 1 & \mathrm{i} \\ -\mathrm{i} & 1 \end{pmatrix} = \frac{1}{2}(\hat{I} - \hat{Y})\)
\(\hat{Z}\)	\(-1\)	\(\vert 1 \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\)	\(\vert 1 \rangle\langle 1 \vert = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \frac{1}{2}(\hat{I} - \hat{Z})\)

Pauli Algebra#

The Pauli matrices satisfy elegant algebraic relations:

(13)#\[\hat{\sigma}^i \hat{\sigma}^j = \delta_{ij}\hat{I} + \mathrm{i}\epsilon_{ijk}\hat{\sigma}^k\]

Explicitly:

\[\hat{\sigma}^x \hat{\sigma}^x = \hat{\sigma}^y \hat{\sigma}^y = \hat{\sigma}^z \hat{\sigma}^z = \hat{I}\]

\[\hat{\sigma}^x \hat{\sigma}^y = \mathrm{i}\hat{\sigma}^z, \quad \hat{\sigma}^y \hat{\sigma}^z = \mathrm{i}\hat{\sigma}^x, \quad \hat{\sigma}^z \hat{\sigma}^x = \mathrm{i}\hat{\sigma}^y\]

\[\hat{\sigma}^y \hat{\sigma}^x = -\mathrm{i}\hat{\sigma}^z, \quad \hat{\sigma}^z \hat{\sigma}^y = -\mathrm{i}\hat{\sigma}^x, \quad \hat{\sigma}^x \hat{\sigma}^z = -\mathrm{i}\hat{\sigma}^y\]

Commutation Relations

\[[\hat{\sigma}^i, \hat{\sigma}^j] = 2\mathrm{i}\epsilon_{ijk}\hat{\sigma}^k\]

Explicitly: \([\hat{\sigma}^x, \hat{\sigma}^y] = 2\mathrm{i}\hat{\sigma}^z\) and cyclic permutations.

Key insight: The Pauli operators do NOT commute. Measuring along one axis disturbs the result of measuring along another.

Anti-Commutation Relations

\[\{\hat{\sigma}^i, \hat{\sigma}^j\} = 2\delta_{ij}\hat{I}\]

where \(\{\hat{A}, \hat{B}\} = \hat{A}\hat{B} + \hat{B}\hat{A}\) is the anti-commutator.

Discussion

Are all physical observables necessarily Hermitian?

We argued that measurement outcomes must be real, so observables must be Hermitian. But consider:

The creation operator \(\hat{a}^\dagger\) is not Hermitian, yet it is physically meaningful (it adds a particle). Is it an “observable”?
In quantum optics, the electric field operator \(\hat{E}(t)\) is Hermitian, but its positive-frequency part \(\hat{E}^{(+)}(t)\) is not. Which one do we “measure”?
Recent work on non-Hermitian quantum mechanics (PT-symmetric Hamiltonians) shows that certain non-Hermitian operators can have entirely real spectra. Does this challenge the Hermiticity requirement, or is something subtler going on?

What is the minimal mathematical requirement for an operator to represent a physical observable?

Expectation Values#

Expectation Value

For a state \(\vert\psi\rangle\) and Hermitian observable \(\hat{O}\), the expectation value is:

(14)#\[\langle\hat{O}\rangle = \langle\psi\vert\hat{O}\vert\psi\rangle\]

This is the average outcome over many measurements on identically prepared copies of \(\vert\psi\rangle\).

The Pauli expectation values encode the qubit state as a 3D unit vector — the Bloch vector:

Pauli Expectation Values and Bloch Vector

For \(\vert\psi\rangle = \alpha\vert 0\rangle + \beta\vert 1\rangle\) (with \(|\alpha|^2 + |\beta|^2 = 1\)):

(15)#\[\langle\hat{X}\rangle = 2\operatorname{Re}(\alpha^*\beta), \quad \langle\hat{Y}\rangle = 2\operatorname{Im}(\alpha^*\beta), \quad \langle\hat{Z}\rangle = |\alpha|^2 - |\beta|^2\]

In the Bloch parametrization \(\vert\psi\rangle = \cos(\theta/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2)\vert 1\rangle\):

(16)#\[\langle\hat{X}\rangle = \sin\theta\cos\phi, \quad \langle\hat{Y}\rangle = \sin\theta\sin\phi, \quad \langle\hat{Z}\rangle = \cos\theta\]

The Bloch vector \(\boldsymbol{n} = (\sin\theta\cos\phi,\,\sin\theta\sin\phi,\,\cos\theta)\) lies on the unit sphere and completely specifies the qubit state.

Derivation: Pauli Expectation Values

In the \((\alpha,\beta)\) parametrization:

For \(\hat{X} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\):

\[\begin{split}\langle\hat{X}\rangle = (\alpha^*\langle 0\vert + \beta^*\langle 1\vert)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}(\alpha\vert 0\rangle + \beta\vert 1\rangle) = (\alpha^*\langle 0\vert + \beta^*\langle 1\vert)(\beta\vert 0\rangle + \alpha\vert 1\rangle) = \alpha^*\beta + \beta^*\alpha = 2\operatorname{Re}(\alpha^*\beta)\end{split}\]

Similarly \(\langle\hat{Y}\rangle = 2\operatorname{Im}(\alpha^*\beta)\) and \(\langle\hat{Z}\rangle = |\alpha|^2 - |\beta|^2\). ✓

From \((\alpha,\beta)\) to \((\theta,\phi)\):

Substitute \(\alpha = \cos(\theta/2)\), \(\beta = \mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2)\):

\[\alpha^*\beta = \cos(\theta/2)\,\mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2) = \tfrac{1}{2}\mathrm{e}^{\mathrm{i}\phi}\sin\theta\]

Therefore:

\[\langle\hat{X}\rangle = 2\operatorname{Re}\!\left(\tfrac{1}{2}\mathrm{e}^{\mathrm{i}\phi}\sin\theta\right) = \sin\theta\cos\phi \;\checkmark, \quad \langle\hat{Y}\rangle = 2\operatorname{Im}\!\left(\tfrac{1}{2}\mathrm{e}^{\mathrm{i}\phi}\sin\theta\right) = \sin\theta\sin\phi \;\checkmark\]

\[\langle\hat{Z}\rangle = \cos^2(\theta/2) - \sin^2(\theta/2) = \cos\theta \;\checkmark\]

Summary#

Hermitian operators have real eigenvalues and orthogonal eigenvectors—essential for describing physical observables.
The Pauli matrices \(\hat{\sigma}^x, \hat{\sigma}^y, \hat{\sigma}^z\) are the fundamental observables for a qubit, representing spin along three orthogonal directions.
Spectral decomposition expresses any observable as a sum of projectors onto its eigenstates, weighted by eigenvalues.
Pauli algebra (multiplication, commutation, anti-commutation relations) governs how spins interact and measure—non-commuting observables cannot be simultaneously diagonal.
Expectation values of Pauli operators (\(\hat{X}\), \(\hat{Y}\), \(\hat{Z}\)) encode the qubit state as a point on the Bloch sphere: a single state \(\vert \psi\rangle\) maps to one point \(\boldsymbol{n}\).

Homework#

1. Show that the Pauli matrix \(\hat{\sigma}^y = \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix}\) is Hermitian by explicitly computing \((\hat{\sigma}^y)^\dagger\) and verifying it equals \(\hat{\sigma}^y\). Then find its eigenvalues and normalized eigenvectors.

2. Consider the operator \(\hat{A} = \begin{pmatrix} 1 & \mathrm{i} \\ \mathrm{i} & 1 \end{pmatrix}\). Is \(\hat{A}\) Hermitian? If not, find \(\hat{A}^\dagger\) and construct a Hermitian operator from \(\hat{A}\) using the combination \(\frac{1}{2}(\hat{A} + \hat{A}^\dagger)\).

3. A qubit is in the state \(\vert \psi\rangle = \frac{1}{\sqrt{3}}\vert 0\rangle + \sqrt{\frac{2}{3}}\vert 1\rangle\). Compute the expectation values \(\langle\hat{X}\rangle\), \(\langle\hat{Y}\rangle\), and \(\langle\hat{Z}\rangle\). Verify that \(\langle\hat{X}\rangle^2 + \langle\hat{Y}\rangle^2 + \langle\hat{Z}\rangle^2 = 1\).

4. The spectral decomposition of \(\hat{X}\) is \(\hat{X} = (+1)\vert +\rangle\langle +\vert + (-1)\vert -\rangle\langle -\vert \), where \(\vert +\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\) and \(\vert -\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle - \vert 1\rangle)\). Verify this by explicitly computing the right-hand side as a \(2\times 2\) matrix.

5. Show that the eigenvalues of any projector \(\hat{P} = \vert \psi\rangle\langle \psi\vert \) are 0 and 1. What are the corresponding eigenstates? Is a projector Hermitian?

6. A spin-1/2 particle is in the state \(\vert \psi\rangle = \cos\alpha\,\vert \uparrow\rangle + \sin\alpha\,\vert \downarrow\rangle\) (with \(\alpha\) real). You measure \(\hat{\sigma}^z\). What are the possible outcomes and their probabilities? Then compute \(\langle \hat{\sigma}^z\rangle\) directly from the definition \(\langle\psi\vert \hat{\sigma}^z\vert \psi\rangle\) and verify it equals \(\sum_i \lambda_i P_i\).

7. The operator \(\hat{n}\cdot\hat{\boldsymbol{\sigma}} = n_x \hat{\sigma}^x + n_y \hat{\sigma}^y + n_z \hat{\sigma}^z\) represents spin along an arbitrary unit vector \(\hat{n} = (n_x, n_y, n_z)\) with \(|\hat{n}| = 1\). Show that \((\hat{n}\cdot\hat{\boldsymbol{\sigma}})^2 = \hat{I}\) (the identity matrix). What does this imply about the eigenvalues of \(\hat{n}\cdot\hat{\boldsymbol{\sigma}}\)?

8. Using the result of Problem 7, show that the eigenvalues of \(\hat{n}\cdot\hat{\boldsymbol{\sigma}}\) are \(\pm 1\) for any unit vector \(\hat{n}\). Find the eigenvector corresponding to eigenvalue \(+1\) for the case \(\hat{n} = (\sin\theta, 0, \cos\theta)\) (a vector in the \(xz\)-plane). Express your answer in terms of \(\theta\).

9. Prove that the expectation value of any Hermitian operator \(\hat{O}\) is real-valued for any state \(\vert \psi\rangle\). That is, show \(\langle \hat{O}\rangle^* = \langle\hat{O}\rangle\) using the definition \(\langle\hat{O}\rangle = \langle\psi\vert \hat{O}\vert \psi\rangle\) and the Hermiticity condition.