Linear Operators & Matrices

Quantum gates, observables, Hamiltonians, and measurements are all linear operators. This lesson nails the operator toolkit: how an operator becomes a matrix once you fix a basis, how the adjoint is defined (it's where "Hermitian" and "unitary" will come from), and the basis-independent quantities — trace and determinant — we rely on constantly.

Learning Objectives

After this lesson you will be able to:

1. Represent a linear operator as a matrix in a chosen basis and compute its action on kets.
2. Define the adjoint $A^\dagger$ via the inner product and show it equals the conjugate transpose.
3. Perform a change of basis and recognize similarity transformations.
4. Compute and use the trace (cyclicity, basis-independence) and the determinant.
5. Evaluate commutators $[A,B]$ and explain why they matter physically.

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Intuition

A linear operator is a function on vectors that respects addition and scaling: it sends lines to lines and the origin to the origin. Fix a basis and the operator is completely captured by what it does to the basis vectors — a finite table of numbers, i.e. a matrix. Change the basis and the same operator gets a different matrix. The quantities that don't change under such relabeling — trace, determinant, eigenvalues — are the operator's physical fingerprints.

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Theory

Linear operators

A map $A:\mathcal{H}\to\mathcal{H}$ is a linear operator if for all $|\psi\rangle,|\phi\rangle$ and $a\in\mathbb{C}$,

$$A\big(a|\psi\rangle + |\phi\rangle\big) = a\,A|\psi\rangle + A|\phi\rangle .$$

Linearity means $A$ is determined by its action on a basis: if $A|j\rangle = \sum_i A_{ij}|i\rangle$, then for $|\psi\rangle = \sum_j c_j|j\rangle$,

$$A|\psi\rangle = \sum_j c_j A|j\rangle = \sum_{i} \Big(\sum_j A_{ij} c_j\Big)|i\rangle .$$

The coefficients $A_{ij} = \langle i|A|j\rangle$ (from 0.1.3) are the matrix of $A$; the formula above is matrix–vector multiplication. Composition is matrix multiplication: $(AB)_{ik} = \sum_j A_{ij}B_{jk}$, derived by inserting completeness $A B = A\,I\,B = \sum_j A|j\rangle\langle j|B$.

The adjoint

The adjoint (Hermitian conjugate) $A^\dagger$ is defined by the inner-product relation

$$\boxed{\ \langle\phi|A\psi\rangle = \langle A^\dagger\phi|\psi\rangle \quad\text{for all }|\phi\rangle,|\psi\rangle.\ }$$

Equivalently $\langle\phi|A|\psi\rangle = \overline{\langle\psi|A^\dagger|\phi\rangle}$. Claim: in any orthonormal basis, the matrix of $A^\dagger$ is the conjugate transpose of the matrix of $A$: $(A^\dagger)_{ij} = \overline{A_{ji}}$.

Proof. $(A^\dagger)_{ij} = \langle i|A^\dagger|j\rangle$. By the defining relation with $|\phi\rangle = |j\rangle$, $|\psi\rangle = |i\rangle$: $\langle j|A|i\rangle = \langle A^\dagger j|i\rangle = \overline{\langle i | A^\dagger | j\rangle} = \overline{(A^\dagger){ij}}$. Hence$(A^\dagger){ij} = \overline{\langle j|A|i\rangle} = \overline{A_{ji}}$. ∎

Properties (all from the definition): $(A^\dagger)^\dagger = A$, $(aA+B)^\dagger = \overline{a}A^\dagger + B^\dagger$, $(AB)^\dagger = B^\dagger A^\dagger$, and $\big(|\psi\rangle\langle\phi|\big)^\dagger = |\phi\rangle\langle\psi|$.

These three names — defined next lesson but anchored here — are adjoint conditions: Hermitian $A^\dagger = A$, unitary $U^\dagger = U^{-1}$, normal $A^\dagger A = A A^\dagger$.

Change of basis

Let $\{|k\rangle\}$ and $\{|k'\rangle\}$ be two orthonormal bases, related by a unitary $S$ with $|k'\rangle = S|k\rangle$ (columns of $S$ are the new basis vectors in old coordinates). A ket's coordinate column transforms as $c' = S^\dagger c$, and an operator's matrix transforms by a similarity transformation:

$$A' = S^\dagger A\, S .$$

Derivation. $A'_{ij} = \langle i'|A|j'\rangle = \langle S i|A|S j\rangle = \langle i|S^\dagger A S|j\rangle = (S^\dagger A S)_{ij}$. Same operator, new matrix. Quantities preserved by $A\mapsto S^\dagger A S$ are exactly the physical invariants below.

Trace

The trace is $\operatorname{Tr}(A) = \sum_k A_{kk} = \sum_k\langle k|A|k\rangle$. Key properties:

• Linear: $\operatorname{Tr}(aA+B) = a\operatorname{Tr}(A) + \operatorname{Tr}(B)$.
• Cyclic: $\operatorname{Tr}(ABC) = \operatorname{Tr}(CAB) = \operatorname{Tr}(BCA)$. (Proof: expand indices; $\operatorname{Tr}(AB) = \sum_{i,j}A_{ij}B_{ji} = \operatorname{Tr}(BA)$, then iterate.)
• Basis-independent: $\operatorname{Tr}(S^\dagger A S) = \operatorname{Tr}(A S S^\dagger) = \operatorname{Tr}(A)$by cyclicity and$SS^\dagger = I$. (We proved basis-independence directly in 0.1.3 E3 too.)
• Equals the sum of eigenvalues (next lesson).

The trace is how expectation values and probabilities are computed in the density-matrix formalism: $\langle A\rangle = \operatorname{Tr}(\rho A)$ (Term 1.5).

Determinant and commutator

The determinant $\det A$ is the product of eigenvalues; it is multiplicative ($\det(AB) = \det A\,\det B$), basis-independent ($\det(S^{-1}AS) = \det A$), and $A$ is invertible iff $\det A \neq 0$. For unitaries $|\det U| = 1$.

The commutator $[A,B] = AB - BA$ measures failure to commute; the anticommutator is $\{A,B\} = AB+BA$. Commutators are central in quantum mechanics: observables commute iff they are simultaneously diagonalizable (Term 0.1.5) and can be measured together without disturbance; the canonical $[\hat x,\hat p] = i\hbar$ gives the uncertainty principle (Term 1.3). For Pauli matrices, $[X,Y] = 2iZ$ (see Appendix E).

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Worked Examples

Example 1 — Building a matrix from its action

Define $A$ on a qubit by $A|0\rangle = |1\rangle$, $A|1\rangle = |0\rangle$ (it swaps basis states). The columns of the matrix are the images of the basis kets in coordinates:

$$A = \big(\,A|0\rangle \ \big|\ A|1\rangle\,\big) = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} = X .$$

This is the Pauli-$X$ / NOT gate. Its adjoint is $X^\dagger = \overline{X}^T = X$ (Hermitian), and since $X^2 = I$ also $X^\dagger = X^{-1}$ (unitary): $X$ is both Hermitian and unitary — a fact we'll see is special to involutions.

Example 2 — Change of basis: $Z$ seen in the Hadamard basis

The Pauli-$Z = \operatorname{diag}(1,-1)$ is diagonal in $\{|0\rangle,|1\rangle\}$. Change to the Hadamard basis via S = H = \tfrac1{\sqrt2}\begin{psmallmatrix}1 & 1\\ 1 & -1\end{psmallmatrix} (which is unitary and Hermitian, so $S^\dagger = S$):

$$Z' = S^\dagger Z S = H Z H = \tfrac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix} = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix} = X .$$

So $Z$ in the $\{|+\rangle,|-\rangle\}$ basis is $X$ — the famous identity $HZH = X$. Same operator, different matrix; note $\operatorname{Tr}$ and $\det$ are unchanged ($0$ and $-1$ respectively).

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Hands-on (Python)

import numpy as np

def dag(A):
    return A.conj().T                       # adjoint = conjugate transpose

X = np.array([[0, 1], [1, 0]], dtype=complex)
Y = np.array([[0, -1j], [1j, 0]], dtype=complex)
Z = np.array([[1, 0], [0, -1]], dtype=complex)
H = np.array([[1, 1], [1, -1]], dtype=complex) / np.sqrt(2)

# Build an operator column-by-column from its action on the basis:
ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)
A = np.column_stack([X @ ket0, X @ ket1])   # columns are A|0>, A|1>
print(np.allclose(A, X))                    # True

# Adjoint, trace, determinant:
print(np.allclose(dag(X), X))               # X is Hermitian
print(np.trace(Z), np.linalg.det(Z))        # 0,  (-1+0j)

# Change of basis HZH = X (similarity transform with S = H, S† = H):
print(np.allclose(dag(H) @ Z @ H, X))       # True
print(np.isclose(np.trace(dag(H) @ Z @ H), np.trace(Z)))  # trace invariant: True

# Commutators / anticommutators of Paulis:
def comm(A, B):  return A @ B - B @ A
def acomm(A, B): return A @ B + B @ A

print(np.allclose(comm(X, Y), 2j * Z))      # [X,Y] = 2iZ
print(np.allclose(acomm(X, Y), np.zeros((2, 2))))  # Paulis anticommute: {X,Y}=0

Braket preview. When you build Circuit().x(0) in Term 2, the SDK is composing exactly these operator matrices (as a unitary) and applying them to the state vector. The math here is what the simulator does under the hood.

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Exercises

E1 (easy). Find the matrix of the operator $A = |0\rangle\langle 1| + |1\rangle\langle 0|$ and identify it.

Solution

|0\rangle\langle1| = \begin{psmallmatrix}0&1\\0&0\end{psmallmatrix}, |1\rangle\langle0| = \begin{psmallmatrix}0&0\\1&0\end{psmallmatrix}; their sum is \begin{psmallmatrix}0&1\\1&0\end{psmallmatrix} = X, the NOT gate.

E2 (easy). Verify $(AB)^\dagger = B^\dagger A^\dagger$ for $A = X$, $B = S = \operatorname{diag}(1,i)$.

Solution

AB = X S = \begin{psmallmatrix}0&i\\1&0\end{psmallmatrix}, so (AB)^\dagger = \begin{psmallmatrix}0&1\\-i&0\end{psmallmatrix}. $B^\dagger = \operatorname{diag}(1,-i)$, $A^\dagger = X$, so $B^\dagger A^\dagger = \operatorname{diag}(1,-i)X = \begin{psmallmatrix}0&1\-i&0\end{psmallmatrix}$. They match. ∎

E3 (medium). Prove $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$ from the index definition, and give a counterexample to $\operatorname{Tr}(ABC) = \operatorname{Tr}(ACB)$.

Solution

$\operatorname{Tr}(AB) = \sum_i (AB){ii} = \sum{i,j}A_{ij}B_{ji} = \sum_{j,i}B_{ji}A_{ij} = \sum_j (BA)_{jj} = \operatorname{Tr}(BA)$. But cyclicity allows only rotations, not arbitrary swaps. Counterexample with Paulis: $\operatorname{Tr}(XYZ)$ vs $\operatorname{Tr}(XZY)$. Since $XYZ = X(iX) = iX^2 = iI$... let's use $\operatorname{Tr}(XYZ)$: $YZ = iX$ so $XYZ = iX^2 = iI$, trace $2i$. And $XZY$: $ZY = -iX$, so $XZY = -iX^2 = -iI$, trace $-2i \neq 2i$. ∎

E4 (medium). Show that the trace of a commutator is always zero, $\operatorname{Tr}([A,B]) = 0$. What does this imply about whether $[\hat x,\hat p] = i\hbar I$ can hold for finite matrices?

Solution

$\operatorname{Tr}(AB - BA) = \operatorname{Tr}(AB) - \operatorname{Tr}(BA) = 0$ by cyclicity. But $\operatorname{Tr}(i\hbar I) = i\hbar\,n \neq 0$ in dimension $n$. So the canonical commutation relation $[\hat x,\hat p] = i\hbar I$ has no finite-dimensional matrix realization — position and momentum require infinite-dimensional Hilbert spaces. (Qubits, being finite, only ever use bounded observables like the Paulis.)

E5 (hard). Prove that the trace is the unique (up to scale) linear functional $f$ on operators satisfying $f(AB) = f(BA)$ for all $A,B$ — i.e. cyclicity essentially characterizes the trace.

Solution sketch

Using $f(AB)=f(BA)$ with $A = |i\rangle\langle j|$, $B = |k\rangle\langle l|$: $AB = \delta_{jk}|i\rangle\langle l|$ and $BA = \delta_{li}|k\rangle\langle j|$, so $f(|i\rangle\langle l|)\delta_{jk} = f(|k\rangle\langle j|)\delta_{li}$. Take $j=k$, $i\neq l$: forces $f(|i\rangle\langle l|) = 0$ for $i\neq l$. Take $i=l$, $j=k$: forces all diagonal values $f(|i\rangle\langle i|)$ equal, say to $c$. Then by linearity $f(A) = \sum_i A_{ii},c = c\operatorname{Tr}(A)$. Hence$f$ is a scalar multiple of the trace. ∎

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Checkpoint

1. How is an operator's matrix determined by its action on a basis?
2. Give the defining inner-product relation for the adjoint and the resulting matrix rule.
3. How does an operator's matrix transform under a change of orthonormal basis?
4. List three properties of the trace and explain why basis-independence matters physically.
5. What is the physical significance of two observables commuting?

Answers

1. The $j$-th column is $A|j\rangle$ in coordinates; equivalently $A_{ij} = \langle i|A|j\rangle$.
2. $\langle\phi|A\psi\rangle = \langle A^\dagger\phi|\psi\rangle$; in an ONB, $(A^\dagger)_{ij} = \overline{A_{ji}}$ (conjugate transpose).
3. $A' = S^\dagger A S$ with $S$ unitary (similarity transform).
4. Linear, cyclic ($\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$), basis-independent, equals sum of eigenvalues. Basis-independence means physical predictions (e.g. $\operatorname{Tr}(\rho A)$) don't depend on the arbitrary choice of coordinates.
5. Commuting observables share an eigenbasis, so they can be measured simultaneously with definite joint values and without mutual disturbance.

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Further Reading

• [Axl] Axler, Linear Algebra Done Right, Ch. 3, 5, 7, 10 — operators, adjoints, trace, determinant.
• [NC] Nielsen & Chuang, §2.1.5–2.1.7 — operators, adjoints, the outer-product representation, trace.
• [HJ] Horn & Johnson, Matrix Analysis — the definitive matrix reference.

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