Lecture Notes on Matrix Representation of Linear Operators and Basis Transformations

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February 5, 2025

Introduction

In this lecture, we will explore the matrix representation of linear operators, a fundamental concept in linear algebra and quantum mechanics. We will begin by demonstrating how the action of a linear operator on an orthonormal basis fully defines its operation on any vector within the vector space. This approach allows us to represent linear operators as matrices, facilitating calculations and providing a powerful framework for analysis.

We will then discuss the product of linear operators and its representation through matrix multiplication. A key concept we will address is the non-commutativity of operator products, a feature that is particularly significant in quantum mechanics. The commutator will be introduced as a measure to quantify this non-commutativity.

To illustrate these principles, we will introduce the Pauli sigma matrices ($\sigma_x, \sigma_y, \sigma_z$) as concrete examples of linear operators and examine their matrix representations and actions on state vectors. We will perform explicit calculations of matrix products and commutators involving Pauli matrices to solidify understanding.

Furthermore, we will cover the practical procedures for constructing orthonormal bases, including normalization and orthogonalization. A detailed example will demonstrate how to find an orthonormal basis given an initial vector and how to represent a quantum state in this new basis using projection operators.

Finally, we will discuss the concept of basis transformations in a more general context, introducing transformation matrices and highlighting the crucial property that vector norms remain invariant under orthonormal basis changes. This lecture aims to provide a comprehensive introduction to matrix representations of operators and basis transformations, laying the groundwork for more advanced topics in quantum mechanics and related fields.

Matrix Representation of Linear Operators

Linear Operators and Orthonormal Bases

Consider a linear operator $A$ acting within a vector space $\mathcal{V}$. In quantum mechanics, $\mathcal{V}$ is often a Hilbert space. Any vector $\left| \psi \right\rangle \in \mathcal{V}$ can be expressed as a linear combination of orthonormal basis vectors $\{\left| v_j \right\rangle\}$: \[\left| \psi \right\rangle = \sum_j c_j \left| v_j \right\rangle,\] where $c_j$ are complex coefficients. Due to the linearity of $A$, its action on $\left| \psi \right\rangle$ is determined by its action on each basis vector $\left| v_j \right\rangle$: \[A\left| \psi \right\rangle = A\left(\sum_j c_j \left| v_j \right\rangle\right) = \sum_j c_j A\left| v_j \right\rangle.\label{eq:linear_action_basis}\] This fundamental property implies that to fully understand the behavior of a linear operator $A$ on any vector $\left| \psi \right\rangle$, it suffices to know its action on all vectors of an orthonormal basis $\{\left| v_j \right\rangle\}$.

Action of an Operator on Basis Vectors and Matrix Elements

Definition 1 (Matrix Elements of an Operator). When a linear operator $A$ acts on a basis vector $\left| v_j \right\rangle$, the result is generally another vector in the same vector space $\mathcal{V}$. This resulting vector can be expressed as a linear combination of the same orthonormal basis vectors $\{\left| v_i \right\rangle\}$: \[A\left| v_j \right\rangle = \sum_i A_{ij} \left| v_i \right\rangle.\label{eq:action_on_basis_vector}\] The coefficients $A_{ij}$ are known as the matrix elements of the operator $A$ in the basis $\{\left| v_i \right\rangle\}$. They quantify the component of $A\left| v_j \right\rangle$ along each basis vector $\left| v_i \right\rangle$.

To determine these matrix elements $A_{ij}$, we project $A\left| v_j \right\rangle$ onto the basis vector $\left| v_i \right\rangle$ by taking the inner product: \[\left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle = \left\langle v_i \middle| \sum_k A_{kj} \left| v_k \right\rangle \right\rangle = \sum_k A_{kj} \left\langle v_i \middle| v_k \right\rangle.\label{eq:inner_product_projection}\] Since $\{\left| v_i \right\rangle\}$ is an orthonormal basis, the inner product $\left\langle v_i \middle| v_k \right\rangle = \delta_{ik}$, where $\delta_{ik}$ is the Kronecker delta (equal to 1 if $i=k$ and 0 if $i \neq k$). Therefore, Equation [eq:inner_product_projection] simplifies to: \[\left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle = \sum_k A_{kj} \delta_{ik} = A_{ij}.\label{eq:matrix_elements_inner_product}\] Thus, the matrix element $A_{ij}$ is precisely the inner product of $\left| v_i \right\rangle$ with $A\left| v_j \right\rangle$.

Theorem 1 (Matrix Elements via Inner Product). The matrix element $A_{ij}$ of a linear operator $A$ in an orthonormal basis $\{\left| v_i \right\rangle\}$ is given by the inner product of the $i$-th basis vector with the action of $A$ on the $j$-th basis vector: \[A_{ij} = \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle.\]

Matrix Representation of a Linear Operator

Definition 2 (Matrix Representation of a Linear Operator). Using the matrix elements $A_{ij}$, we can construct a representation of the linear operator $A$ in the orthonormal basis $\{\left| v_i \right\rangle\}$. The operator $A$ can be expressed as: \[A = \sum_{i,j} A_{ij} \left| v_i \right\rangle\left\langle v_j \right| = \sum_{i,j} \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle \left| v_i \right\rangle\left\langle v_j \right|.\label{eq:operator_representation}\] This representation explicitly shows how the operator $A$ acts in the chosen orthonormal basis.

Applying $A$ to a basis vector $\left| v_k \right\rangle$ using this representation yields: \[\begin{aligned}A\left| v_k \right\rangle &= \left(\sum_{i,j} A_{ij} \left| v_i \right\rangle\left\langle v_j \right|\right)\left| v_k \right\rangle \\&= \sum_{i,j} A_{ij} \left| v_i \right\rangle \left\langle v_j \middle| v_k \right\rangle \\&= \sum_{i,j} A_{ij} \left| v_i \right\rangle \delta_{jk} \\&= \sum_{i} A_{ik} \left| v_i \right\rangle,\end{aligned}\] which is consistent with the definition of matrix elements $A_{ij}$ in Equation [eq:action_on_basis_vector].

Definition 3 (Matrix Form of a Linear Operator). In an $n$-dimensional vector space, the matrix representation of $A$, denoted as $[A]$, is an $n \times n$ matrix where the element in the $i$-th row and $j$-th column is $A_{ij}$: \[[A] = \begin{pmatrix}A_{11} & A_{12} & \cdots & A_{1n} \\A_{21} & A_{22} & \cdots & A_{2n} \\\vdots & \vdots & \ddots & \vdots \\A_{n1} & A_{n2} & \cdots & A_{nn}\end{pmatrix}.\label{eq:matrix_representation_explicit}\] Each entry $A_{ij}$ in this matrix quantifies the transition from the $j$-th basis state to the $i$-th basis state under the action of the operator $A$.

Products and Non-Commutativity of Operators

Operator Product and Matrix Multiplication

Consider two linear operators $A$ and $B$. We want to determine the matrix representation of their product $C = AB$ in a given orthonormal basis $\{\left| v_i \right\rangle\}$. Let $[A]$ and $[B]$ be the matrix representations of $A$ and $B$ respectively, with matrix elements $A_{ij} = \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle$ and $B_{lm} = \left\langle v_l \middle| B\left| v_m \right\rangle \right\rangle$.

Using the representation of operators in terms of orthonormal basis vectors (Equation [eq:operator_representation]), we can express the product $AB$ as: \[\begin{aligned}AB &= \left(\sum_{i,j} A_{ij} \left| v_i \right\rangle\left\langle v_j \right|\right) \left(\sum_{l,m} B_{lm} \left| v_l \right\rangle\left\langle v_m \right|\right) \\&= \sum_{i,j,l,m} A_{ij} B_{lm} \left| v_i \right\rangle\left\langle v_j \right| \left| v_l \right\rangle\left\langle v_m \right| \\&= \sum_{i,j,l,m} A_{ij} B_{lm} \left| v_i \right\rangle \underbrace{\left\langle v_j \middle| v_l \right\rangle}_{\delta_{jl}} \left\langle v_m \right| \\&= \sum_{i,j,m} A_{ij} B_{jm} \left| v_i \right\rangle\left\langle v_m \right| \\&= \sum_{i,m} \left(\sum_j A_{ij} B_{jm}\right) \left| v_i \right\rangle\left\langle v_m \right|.\end{aligned}\] Let $C_{im} = \sum_j A_{ij} B_{jm}$. Then the product operator $C = AB$ can be written as $C = \sum_{i,m} C_{im} \left| v_i \right\rangle\left\langle v_m \right|$. The coefficients $C_{im}$ are precisely the elements of the matrix product of $[A]$ and $[B]$.

Theorem 2 (Matrix Multiplication for Operator Product). The matrix representation of the product of two linear operators is obtained by the matrix multiplication of their respective matrix representations. If $[A]$ is an $n \times n$ matrix and $[B]$ is an $n \times n$ matrix, then their matrix product $[C] = [A][B]$ is also an $n \times n$ matrix, with elements given by $C_{im} = \sum_j A_{ij} B_{jm}$.

Non-Commutativity of Operator Products

Remark. Remark 1 (Non-Commutativity of Operators). In general, matrix multiplication is not commutative, meaning that for two matrices $[A]$ and $[B]$, it is not always true that $[A][B] = [B][A]$. Consequently, for linear operators $A$ and $B$, the order of application matters, and in general $AB \neq BA$.

When $AB = BA$, we say that the operators $A$ and $B$ commute. If $AB \neq BA$, they are said to be non-commuting. This property of non-commutativity is a cornerstone of quantum mechanics and distinguishes it fundamentally from classical mechanics, where observables are typically represented by commuting quantities.

A simple example from classical mechanics involves rotations in three dimensions. Rotations about different axes generally do not commute; the final orientation depends on the order in which rotations are performed. Similarly, in quantum mechanics, many important operators, such as those representing position and momentum, or different components of spin, do not commute.

The Commutator

Definition 4 (Commutator of Two Operators). To quantify the non-commutativity of two operators $A$ and $B$, we define the commutator* as: \[[A, B] = AB - BA.\label{eq:commutator_definition}\] The commutator $[A, B]$ itself is a linear operator. If $A$ and $B$ commute, then $AB - BA = 0$, so $[A, B] = 0$. Conversely, if $[A, B] \neq 0$, then $A$ and $B$ do not commute. The commutator thus provides a measure of the degree to which two operators fail to commute.*

Remark. Remark 2 (Importance of Commutator in Quantum Mechanics). In quantum mechanics, commutation relations between operators are fundamental. For instance, the canonical commutation relation between position and momentum operators is a basic postulate of quantum theory and has profound implications for the uncertainty principle. We will explore examples of commutators, particularly using Pauli matrices, in the next section.

Example: Pauli Sigma Matrices

Introduction to Pauli Matrices and Spin Operators

Definition 5 (Pauli Sigma Matrices). The Pauli sigma matrices are a set of three fundamental $2 \times 2$ complex matrices in quantum mechanics. They are particularly crucial in describing spin-$1/2$ particles, such as electrons, and are defined as: \[\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.\label{eq:pauli_matrices_definitions}\] These matrices serve as representations of the spin operators along the $x$, $y$, and $z$ axes, respectively, in a dimensionless form. When considering spin angular momentum, they are often multiplied by $\hbar/2$.

Action of Pauli Matrices on Computational Basis States

Let’s examine how Pauli matrices act on a general two-component state vector $\left| \psi \right\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$ expressed in the computational basis $\{\left| 0 \right\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \left| 1 \right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\}$. Here, $\alpha$ and $\beta$ are complex coefficients representing the amplitudes of the states $\left| 0 \right\rangle$ and $\left| 1 \right\rangle$ respectively.

Example 1 (Action of $\sigma_z$ on a state vector). Action of $\sigma_z$: \[\sigma_z \left| \psi \right\rangle = \sigma_z \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \alpha \\ -\beta \end{pmatrix} = \alpha \left| 0 \right\rangle - \beta \left| 1 \right\rangle. \label{eq:sigma_z_action}\] The operator $\sigma_z$ leaves the amplitude of the $\left| 0 \right\rangle$ state unchanged while flipping the sign of the amplitude of the $\left| 1 \right\rangle$ state.

Example 2 (Action of $\sigma_x$ on a state vector). Action of $\sigma_x$: \[\sigma_x \left| \psi \right\rangle = \sigma_x \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \beta \\ \alpha \end{pmatrix} = \beta \left| 0 \right\rangle + \alpha \left| 1 \right\rangle. \label{eq:sigma_x_action}\] The operator $\sigma_x$ swaps the amplitudes of the $\left| 0 \right\rangle$ and $\left| 1 \right\rangle$ states.

Example 3 (Action of $\sigma_y$ on a state vector). Action of $\sigma_y$: \[\sigma_y \left| \psi \right\rangle = \sigma_y \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} -i\beta \\ i\alpha \end{pmatrix} = -i\beta \left| 0 \right\rangle + i\alpha \left| 1 \right\rangle. \label{eq:sigma_y_action}\] The operator $\sigma_y$ swaps the amplitudes and introduces phase factors of $\pm i$.

Matrix Product Examples: $\sigma_z \sigma_x$ and $\sigma_x \sigma_z$

Let’s compute the matrix product $\sigma_z \sigma_x$: \[\begin{aligned}\sigma_z \sigma_x &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (1\cdot 0 + 0\cdot 1) & (1\cdot 1 + 0\cdot 0) \\ (0\cdot 0 + (-1)\cdot 1) & (0\cdot 1 + (-1)\cdot 0) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.\label{eq:sigma_z_sigma_x_product}\end{aligned}\] Now, let’s compute the product $\sigma_x \sigma_z$: \[\begin{aligned}\sigma_x \sigma_z &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (0\cdot 1 + 1\cdot 0) & (0\cdot 0 + 1\cdot (-1)) \\ (1\cdot 1 + 0\cdot 0) & (1\cdot 0 + 0\cdot (-1)) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.\label{eq:sigma_x_sigma_z_product}\end{aligned}\]

Remark. Remark 3 (Non-Commutativity of $\sigma_x$ and $\sigma_z$). Comparing Equations [eq:sigma_z_sigma_x_product] and [eq:sigma_x_sigma_z_product], we clearly see that $\sigma_z \sigma_x \neq \sigma_x \sigma_z$, demonstrating the non-commutativity of Pauli matrices.

Commutator of $\sigma_z$ and $\sigma_x$

Example 4 ([_z, _x). ] The commutator of $\sigma_z$ and $\sigma_x$ is given by $[\sigma_z, \sigma_x] = \sigma_z \sigma_x - \sigma_x \sigma_z$: \[\begin{aligned}[\sigma_z, \sigma_x] &= \sigma_z \sigma_x - \sigma_x \sigma_z \nonumber \\&= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (0-0) & (1-(-1)) \\ (-1-1) & (0-0) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}.\label{eq:sigma_zx_commutator_intermediate}\end{aligned}\] We want to express this commutator in terms of another Pauli matrix. Let’s compare Equation [eq:sigma_zx_commutator_intermediate] with $\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$. We observe that: \[2i \sigma_y = 2i \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2i^2 \\ 2i^2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}.\] Thus, we find the commutation relation: \[[\sigma_z, \sigma_x] = 2i \sigma_y.\label{eq:sigma_zx_commutator_result}\] This result is a fundamental commutation relation in quantum mechanics for spin operators. Cyclic permutations of indices yield similar relations: $[\sigma_x, \sigma_y] = 2i \sigma_z$ and $[\sigma_y, \sigma_z] = 2i \sigma_x$.

Example: Triple Product $\sigma_z \sigma_x \sigma_z$

Example 5 (Triple Product $\sigma_z \sigma_x \sigma_z$). Let’s calculate the triple product $\sigma_z \sigma_x \sigma_z = (\sigma_z \sigma_x) \sigma_z$. Using the result from Equation [eq:sigma_z_sigma_x_product] for $\sigma_z \sigma_x$: \[\begin{aligned}\sigma_z \sigma_x \sigma_z &= \left(\sigma_z \sigma_x\right) \sigma_z = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (0\cdot 1 + 1\cdot 0) & (0\cdot 0 + 1\cdot (-1)) \\ ((-1)\cdot 1 + 0\cdot 0) & ((-1)\cdot 0 + 0\cdot (-1)) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}.\label{eq:sigma_z_sigma_x_sigma_z_intermediate}\end{aligned}\] Comparing Equation [eq:sigma_z_sigma_x_sigma_z_intermediate] with the definition of $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, we observe that: \[\sigma_z \sigma_x \sigma_z = - \sigma_x.\label{eq:sigma_z_sigma_x_sigma_z_result}\] This type of manipulation of Pauli matrices is frequently encountered in quantum mechanical calculations, particularly in spin systems and quantum information theory.

Constructing Orthonormal Bases: A Practical Example

Problem Statement: Constructing an Orthonormal Basis

Example 6 (Constructing an Orthonormal Basis). Given a vector $\left| \psi \right\rangle = \frac{1}{2}\left| 0 \right\rangle + \frac{3}{2}i\left| 1 \right\rangle = \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix}$, our objective is to construct an orthonormal basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ such that the first basis vector $\left| v_1 \right\rangle$ is aligned with the direction of $\left| \psi \right\rangle$. This process involves normalization and orthogonalization to ensure that both basis vectors are mutually orthogonal and normalized to unity.

Normalization of $\left| \psi \right\rangle$ to Obtain $\left| v_1 \right\rangle$

First, we normalize the vector $\left| \psi \right\rangle$ to obtain the first orthonormal basis vector $\left| v_1 \right\rangle$. To do this, we calculate the squared norm of $\left| \psi \right\rangle$: \[\begin{aligned}\left\langle \psi \middle| \psi \right\rangle &= \left\langle \psi \right|\left| \psi \right\rangle = \begin{pmatrix} 1/2 & -\frac{3}{2}i \end{pmatrix} \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix} \\&= \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(-\frac{3}{2}i\right)\left(\frac{3}{2}i\right) \\&= \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2}.\end{aligned}\] The norm of $\left| \psi \right\rangle$ is $\|\left| \psi \right\rangle\| = \sqrt{\left\langle \psi \middle| \psi \right\rangle} = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}$. We normalize $\left| \psi \right\rangle$ by dividing it by its norm to obtain $\left| v_1 \right\rangle$: \[\left| v_1 \right\rangle = \frac{\left| \psi \right\rangle}{\|\left| \psi \right\rangle\|} = \frac{1}{\sqrt{5/2}} \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix} = \sqrt{\frac{2}{5}} \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix} = \begin{pmatrix} 1/\sqrt{10} \\ 3i/\sqrt{10} \end{pmatrix} = \frac{1}{\sqrt{10}} \left| 0 \right\rangle + \frac{3i}{\sqrt{10}} \left| 1 \right\rangle.\label{eq:normalized_v1}\] To verify normalization, we compute $\left\langle v_1 \middle| v_1 \right\rangle$: \[\begin{aligned}\left\langle v_1 \middle| v_1 \right\rangle &= \left\langle v_1 \right|\left| v_1 \right\rangle = \begin{pmatrix} 1/\sqrt{10} & -\frac{3}{\sqrt{10}}i \end{pmatrix} \begin{pmatrix} 1/\sqrt{10} \\ 3i/\sqrt{10} \end{pmatrix} \\&= \left| \frac{1}{\sqrt{10}} \right|^2 + \left| \frac{3i}{\sqrt{10}} \right|^2 = \frac{1}{10} + \frac{9}{10} = 1.\end{aligned}\] Thus, $\left| v_1 \right\rangle$ is indeed normalized.

Finding a Vector $\left| v_2 \right\rangle$ Orthogonal to $\left| v_1 \right\rangle$

Next, we need to find a vector $\left| v_2 \right\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle$ that is orthogonal to $\left| v_1 \right\rangle$. The orthogonality condition $\left\langle v_1 \middle| v_2 \right\rangle = 0$ is expressed as: \[\left\langle v_1 \middle| v_2 \right\rangle = \left\langle v_1 \right|\left| v_2 \right\rangle = \begin{pmatrix} 1/\sqrt{10} & -\frac{3i}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \frac{1}{\sqrt{10}} \alpha - \frac{3i}{\sqrt{10}} \beta = 0.\label{eq:orthogonality_condition}\] Multiplying by $\sqrt{10}$, we get $\alpha - 3i\beta = 0$, which implies $\alpha = 3i\beta$. Let us choose $\beta = c$, where $c$ is a constant to be determined by normalization. Then $\alpha = 3ic$, and $\left| v_2 \right\rangle = \begin{pmatrix} 3ic \\ c \end{pmatrix} = 3ic \left| 0 \right\rangle + c \left| 1 \right\rangle$.

Normalization of $\left| v_2 \right\rangle$

Now we normalize $\left| v_2 \right\rangle$. We calculate the squared norm of $\left| v_2 \right\rangle$: \[\begin{aligned}\left\langle v_2 \middle| v_2 \right\rangle &= \left\langle v_2 \right|\left| v_2 \right\rangle = \begin{pmatrix} -3ic^* & c^* \end{pmatrix} \begin{pmatrix} 3ic \\ c \end{pmatrix} \\&= (-3ic^*)(3ic) + (c^*)(c) \\&= 9\left| c \right|^2 + \left| c \right|^2 = 10\left| c \right|^2.\end{aligned}\] For $\left| v_2 \right\rangle$ to be normalized, we require $\left\langle v_2 \middle| v_2 \right\rangle = 1$, so $10\left| c \right|^2 = 1$, which gives $\left| c \right| = \frac{1}{\sqrt{10}}$. We can choose $c = \frac{1}{\sqrt{10}}$ for simplicity. Then $\beta = \frac{1}{\sqrt{10}}$ and $\alpha = \frac{3i}{\sqrt{10}}$. Thus, \[\left| v_2 \right\rangle = \begin{pmatrix} 3i/\sqrt{10} \\ 1/\sqrt{10} \end{pmatrix} = \frac{3i}{\sqrt{10}} \left| 0 \right\rangle + \frac{1}{\sqrt{10}} \left| 1 \right\rangle.\label{eq:normalized_v2}\] We verify the normalization of $\left| v_2 \right\rangle$: \[\begin{aligned}\braket{v_2 | v_2} &= \left\langle v_2 \right| \left| v_2 \right\rangle = \begin{pmatrix} -\frac{3i}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} \frac{3i}{\sqrt{10}} \\ \frac{1}{\sqrt{10}} \end{pmatrix} \\&= \left| \frac{3i}{\sqrt{10}} \right|^2 + \left| \frac{1}{\sqrt{10}} \right|^2 = \frac{9}{10} + \frac{1}{10} = 1.\end{aligned}\]

Verification of Orthonormality of $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$

We have constructed the set of vectors $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ with \[\left| v_1 \right\rangle = \frac{1}{\sqrt{10}} \left| 0 \right\rangle + \frac{3i}{\sqrt{10}} \left| 1 \right\rangle, \quad\left| v_2 \right\rangle = \frac{3i}{\sqrt{10}} \left| 0 \right\rangle + \frac{1}{\sqrt{10}} \left| 1 \right\rangle.\label{eq:basis_vectors_v1_v2}\] We have already verified that $\left\langle v_1 \middle| v_1 \right\rangle = 1$ and $\left\langle v_2 \middle| v_2 \right\rangle = 1$. Now we verify orthogonality $\left\langle v_1 \middle| v_2 \right\rangle = 0$: \[\begin{aligned}\left\langle v_1 \middle| v_2 \right\rangle &= \left\langle v_1 \right|\left| v_2 \right\rangle = \begin{pmatrix} 1/\sqrt{10} & -\frac{3i}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} 3i/\sqrt{10} \\ 1/\sqrt{10} \end{pmatrix} \\&= \left(\frac{1}{\sqrt{10}}\right)\left(\frac{3i}{\sqrt{10}}\right) + \left(-\frac{3i}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right) \\&= \frac{3i}{10}- \frac{3i}{10} = 0.\end{aligned}\] Since $\left\langle v_1 \middle| v_1 \right\rangle = 1$, $\left\langle v_2 \middle| v_2 \right\rangle = 1$, and $\left\langle v_1 \middle| v_2 \right\rangle = 0$, the set $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ forms an orthonormal basis.

Representing the State $\left| + \right\rangle$ in the Basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$

Using Projectors for Basis Transformation

To express the state $\left| + \right\rangle = \frac{1}{\sqrt{2}}(\left| 0 \right\rangle + \left| 1 \right\rangle)$ in the new orthonormal basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$, we use projection operators. The projector onto $\left| v_1 \right\rangle$ is $P_{v_1} = \left| v_1 \right\rangle\left\langle v_1 \right|$, and onto $\left| v_2 \right\rangle$ is $P_{v_2} = \left| v_2 \right\rangle\left\langle v_2 \right|$. The components of $\left| + \right\rangle$ in the new basis are given by the projection coefficients $c_1 = \left\langle v_1 \middle| + \right\rangle$ and $c_2 = \left\langle v_2 \middle| + \right\rangle$.

Calculating Coefficients $\left\langle v_1 \middle| + \right\rangle$ and $\left\langle v_2 \middle| + \right\rangle$

Let $\left| + \right\rangle = \frac{1}{\sqrt{2}}(\left| 0 \right\rangle + \left| 1 \right\rangle) = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}$. We calculate the coefficients: \[\begin{aligned}c_1 = \left\langle v_1 \middle| + \right\rangle &= \left\langle v_1 \right|\left| + \right\rangle = \begin{pmatrix} 1/\sqrt{10} & -\frac{3i}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} \\&= \left(\frac{1}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(-\frac{3i}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{20}} - \frac{3i}{\sqrt{20}} = \frac{1-3i}{\sqrt{20}}. \\c_2 = \left\langle v_2 \middle| + \right\rangle &= \left\langle v_2 \right|\left| + \right\rangle = \begin{pmatrix} -\frac{3i}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} \\&= \left(-\frac{3i}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{2}}\right) = -\frac{3i}{\sqrt{20}} + \frac{1}{\sqrt{20}} = \frac{1+3i}{\sqrt{20}}.\end{aligned}\] Thus, in the new basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$, the state $\left| + \right\rangle$ is represented as: \[\left| + \right\rangle = c_1 \left| v_1 \right\rangle + c_2 \left| v_2 \right\rangle = \frac{1-3i}{\sqrt{20}} \left| v_1 \right\rangle + \frac{1+3i}{\sqrt{20}} \left| v_2 \right\rangle.\label{eq:plus_state_in_v1v2}\]

Verification of Normalization of $\left| + \right\rangle$ in the $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ Basis

Finally, we verify that $\left| + \right\rangle$ remains normalized when expressed in the new basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$: \[\begin{aligned}\text{Norm}^2 &= \left| c_1 \right|^2 + \left| c_2 \right|^2 = \left| \frac{1-3i}{\sqrt{20}} \right|^2 + \left| \frac{1+3i}{\sqrt{20}} \right|^2 \\&= \frac{\left| 1-3i \right|^2}{20} + \frac{\left| 1+3i \right|^2}{20} \\&= \frac{(1)^2 + (-3)^2}{20} + \frac{(1)^2 + 3^2}{20} \\&= \frac{10}{20} + \frac{10}{20} = 1.\end{aligned}\] The norm squared is indeed 1, confirming that the state $\left| + \right\rangle$ remains normalized in the new orthonormal basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$, as expected for a change of orthonormal basis.

Remark. Remark 4 (Significance of Orthonormal Basis Construction). The process of constructing orthonormal bases and changing basis representations is fundamental in quantum mechanics. It allows us to analyze quantum systems in different bases that may be more convenient or physically relevant for specific problems.

Change of Basis and Transformation Matrices

General Basis Transformations Between Orthonormal Bases

Definition 6 (Basis Transformation). In linear algebra and quantum mechanics, it is often necessary to change the basis in which vectors and operators are represented. Consider two orthonormal bases for a vector space $\mathcal{V}$: $V = \{\left| v_1 \right\rangle, \left| v_2 \right\rangle, \ldots, \left| v_n \right\rangle\}$ and $U = \{\left| u_1 \right\rangle, \left| u_2 \right\rangle, \ldots, \left| u_n \right\rangle\}$. Since both $V$ and $U$ are bases, any vector in $\mathcal{V}$, and in particular each vector in basis $V$, can be expressed as a linear combination of vectors in basis $U$, and vice versa.

Transformation Matrix from Basis $V$ to Basis $U$

Definition 7 (Transformation Matrix). We can express each vector $\left| v_j \right\rangle$ of basis $V$ as a linear combination of the vectors of basis $U$: \[\left| v_j \right\rangle = \sum_{i=1}^{n} R_{ij} \left| u_i \right\rangle, \quad \text{for } j = 1, 2, \ldots, n,\label{eq:v_in_terms_of_u}\] where $R_{ij}$ are the transformation coefficients. The transformation matrix $R$ from basis $V$ to basis $U$ has elements $R_{ij} = \left\langle u_i \middle| v_j \right\rangle$. $R_{ij}$ is the component of $\left| v_j \right\rangle$ along $\left| u_i \right\rangle$.

Similarly, we can express each vector $\left| u_i \right\rangle$ of basis $U$ in terms of the vectors of basis $V$: \[\left| u_i \right\rangle = \sum_{j=1}^{n} (R')_{ji} \left| v_j \right\rangle, \quad \text{for } i = 1, 2, \ldots, n.\label{eq:u_in_terms_of_v}\] By analogy, the coefficients $(R')_{ji} = \left\langle v_j \middle| u_i \right\rangle$. Let $R'$ be the transformation matrix from basis $U$ to basis $V$ with elements $(R')_{ij} = \left\langle v_i \middle| u_j \right\rangle$. Then $(R')_{ji} = R_{ij}^*$. Thus, $R' = R^\dagger$, the adjoint (conjugate transpose) of $R$.

Relationship Between Transformation Matrices: Unitary Transformation

Theorem 3 (Unitary Transformation Matrix). The transformation matrix between two orthonormal bases is a unitary matrix, meaning it satisfies $R^\dagger R = R R^\dagger = I$, where $I$ is the identity matrix and $R^\dagger$ is the adjoint of $R$.

Description: This theorem states that when we change from one orthonormal basis to another, the matrix that performs this transformation is unitary. Unitary matrices are crucial in quantum mechanics because they preserve probabilities and norms of vectors.

Substituting Equation [eq:u_in_terms_of_v] into Equation [eq:v_in_terms_of_u], we get: \[\begin{aligned}\left| v_j \right\rangle &= \sum_{i=1}^{n} R_{ij} \left| u_i \right\rangle = \sum_{i=1}^{n} R_{ij} \left(\sum_{k=1}^{n} (R')_{ki} \left| v_k \right\rangle\right) \\&= \sum_{k=1}^{n} \left(\sum_{i=1}^{n} (R')_{ki} R_{ij}\right) \left| v_k \right\rangle.\end{aligned}\] For this to hold for all $\left| v_j \right\rangle$ in the basis $V$, we must have: \[\sum_{i=1}^{n} (R')_{ki} R_{ij} = \delta_{kj}.\label{eq:identity_relation_matrix}\] In matrix notation, this is $(R'R)^T = I$ or more directly $R'R = I$, where $I$ is the identity matrix. Since $R' = R^\dagger$, we have $R^\dagger R = I$. Similarly, substituting Equation [eq:v_in_terms_of_u] into Equation [eq:u_in_terms_of_v] leads to $R R^\dagger = I$. Matrices that satisfy $R^\dagger R = R R^\dagger = I$ are called unitary matrices. Therefore, the transformation between two orthonormal bases is a unitary transformation, and the transformation matrix is a unitary matrix.

Invariance of Vector Norms Under Basis Transformation

Theorem 4 (Invariance of Vector Norms). Vector norms remain invariant under orthonormal basis changes. If $\left| \psi \right\rangle$ is a vector, its norm calculated in any orthonormal basis will be the same.

Description: This theorem highlights a fundamental property of orthonormal basis transformations: they preserve the length of vectors. This is essential in physics, as physical quantities like probabilities (derived from vector norms) must be independent of the choice of basis.

Let $\left| \psi \right\rangle$ be a vector expressed in basis $V$ as $\left| \psi \right\rangle = \sum_{j} c_j \left| v_j \right\rangle$ and in basis $U$ as $\left| \psi \right\rangle = \sum_{i} c'_i \left| u_i \right\rangle$. The norm squared of $\left| \psi \right\rangle$ in basis $V$ is $\|\left| \psi \right\rangle\|^2 = \left\langle \psi \middle| \psi \right\rangle = \sum_{j} |c_j|^2$ due to orthonormality of $\{\left| v_j \right\rangle\}$. Similarly, in basis $U$, $\|\left| \psi \right\rangle\|^2 = \sum_{i} |c'_i|^2$.

The transformation of coefficients from basis $V$ to basis $U$ is given by: \[c'_i = \left\langle u_i \middle| \left| \psi \right\rangle \right\rangle = \left\langle u_i \middle| \sum_{j} c_j \left| v_j \right\rangle \right\rangle = \sum_{j} c_j \left\langle u_i \middle| v_j \right\rangle = \sum_{j} R_{ij} c_j.\] In matrix notation, $\mathbf{c'} = R \mathbf{c}$, where $\mathbf{c}$ and $\mathbf{c'}$ are column vectors of coefficients in bases $V$ and $U$ respectively.

The invariance of the norm under unitary transformations can be shown as follows: \[\begin{aligned}\|\left| \psi \right\rangle\|^2_{U} = \sum_{i} |c'_i|^2 = (\mathbf{c'})^\dagger \mathbf{c'} = (R\mathbf{c})^\dagger (R\mathbf{c}) = \mathbf{c}^\dagger R^\dagger R \mathbf{c} = \mathbf{c}^\dagger I \mathbf{c} = \mathbf{c}^\dagger \mathbf{c} = \sum_{j} |c_j|^2 = \|\left| \psi \right\rangle\|^2_{V}.\end{aligned}\] Thus, the norm of the vector $\left| \psi \right\rangle$ is invariant under a change of orthonormal basis, which is a consequence of the unitary nature of the transformation matrix $R$. This reflects the physical intuition that the length of a vector is an intrinsic property, independent of the coordinate system used to describe it.

Remark. Remark 5 (Importance of Basis Transformation). The concept of basis transformation is essential for understanding how physical quantities are represented in different coordinate systems. In quantum mechanics, choosing an appropriate basis can significantly simplify calculations and provide deeper insights into the system’s properties. The unitary nature of basis transformations ensures that fundamental physical quantities like probabilities (related to squared norms) are preserved, regardless of the chosen basis.

Conclusion

In 6 we summarize the lecture.

Remark. Remark 6 (Summary of the Lecture). This lecture has established the fundamental concepts of matrix representations for linear operators and basis transformations, which are crucial for quantum mechanics. We have covered the following key points:

Matrix Representation of Operators: Linear operators can be represented as matrices once an orthonormal basis is chosen. The matrix elements are given by the inner product $\left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle$.
Operator Products and Matrix Multiplication: The matrix representation of the product of two operators is equivalent to the matrix product of their individual matrix representations.
Non-Commutativity and the Commutator: Operator products are generally non-commutative, a key feature in quantum mechanics. The commutator $[A, B] = AB - BA$ quantifies this non-commutativity.
Pauli Matrices Example: Pauli matrices ($\sigma_x, \sigma_y, \sigma_z$) were introduced as examples of non-commuting operators, demonstrating their actions on state vectors and their commutation relations.
Constructing Orthonormal Bases: We demonstrated a practical method for constructing orthonormal bases, starting from an arbitrary vector and using normalization and orthogonalization procedures.
Basis Transformations and Unitary Matrices: Transformations between orthonormal bases are unitary, described by unitary transformation matrices. These transformations preserve vector norms and inner products, ensuring physical consistency.

It is crucial to remember that while matrix representations are basis-dependent, the underlying vectors and operators are basis-independent entities. The choice of basis is a matter of convenience and can be adapted to simplify specific problems.

In the upcoming lectures, we will build upon these concepts by exploring the properties and applications of transformation matrices in more detail. We will investigate how basis transformations are used to simplify quantum problems and potentially delve into the concept of time evolution as a change of basis in quantum mechanics.

When a linear operator $A$ acts on a basis vector $\left| v_j \right\rangle$, the result is generally another vector in the same vector space $\mathcal{V}$. This resulting vector can be expressed as a linear combination of the same orthonormal basis vectors $\{\left| v_i \right\rangle\}$: \[A\left| v_j \right\rangle = \sum_i A_{ij} \left| v_i \right\rangle.\label{eq:action_on_basis_vector-box}\] The coefficients $A_{ij}$ are known as the matrix elements of the operator $A$ in the basis $\{\left| v_i \right\rangle\}$. They quantify the component of $A\left| v_j \right\rangle$ along each basis vector $\left| v_i \right\rangle$.

The matrix element $A_{ij}$ of a linear operator $A$ in an orthonormal basis $\{\left| v_i \right\rangle\}$ is given by the inner product of the $i$-th basis vector with the action of $A$ on the $j$-th basis vector: \[A_{ij} = \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle.\]

Using the matrix elements $A_{ij}$, we can construct a representation of the linear operator $A$ in the orthonormal basis $\{\left| v_i \right\rangle\}$. The operator $A$ can be expressed as: \[A = \sum_{i,j} A_{ij} \left| v_i \right\rangle\left\langle v_j \right| = \sum_{i,j} \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle \left| v_i \right\rangle\left\langle v_j \right|.\label{eq:operator_representation-box}\] This representation explicitly shows how the operator $A$ acts in the chosen orthonormal basis.

In an $n$-dimensional vector space, the matrix representation of $A$, denoted as $[A]$, is an $n \times n$ matrix where the element in the $i$-th row and $j$-th column is $A_{ij}$: \[[A] = \begin{pmatrix}A_{11} & A_{12} & \cdots & A_{1n} \\A_{21} & A_{22} & \cdots & A_{2n} \\\vdots & \vdots & \ddots & \vdots \\A_{n1} & A_{n2} & \cdots & A_{nn}\end{pmatrix}.\label{eq:matrix_representation_explicit-box}\] Each entry $A_{ij}$ in this matrix quantifies the transition from the $j$-th basis state to the $i$-th basis state under the action of the operator $A$.

The matrix representation of the product of two linear operators is obtained by the matrix multiplication of their respective matrix representations. If $[A]$ is an $n \times n$ matrix and $[B]$ is an $n \times n$ matrix, then their matrix product $[C] = [A][B]$ is also an $n \times n$ matrix, with elements given by $C_{im} = \sum_j A_{ij} B_{jm}$.

In general, matrix multiplication is not commutative, meaning that for two matrices $[A]$ and $[B]$, it is not always true that $[A][B] = [B][A]$. Consequently, for linear operators $A$ and $B$, the order of application matters, and in general $AB \neq BA$.

When $AB = BA$, we say that the operators $A$ and $B$ commute. If $AB \neq BA$, they are said to be non-commuting. This property of non-commutativity is a cornerstone of quantum mechanics and distinguishes it fundamentally from classical mechanics, where observables are typically represented by commuting quantities.

To quantify the non-commutativity of two operators $A$ and $B$, we define the commutator as: \[[A, B] = AB - BA.\label{eq:commutator_definition-box}\] The commutator $[A, B]$ itself is a linear operator. If $A$ and $B$ commute, then $AB - BA = 0$, so $[A, B] = 0$. Conversely, if $[A, B] \neq 0$, then $A$ and $B$ do not commute. The commutator thus provides a measure of the degree to which two operators fail to commute.

In quantum mechanics, commutation relations between operators are fundamental. For instance, the canonical commutation relation between position and momentum operators is a basic postulate of quantum theory and has profound implications for the uncertainty principle. We will explore examples of commutators, particularly using Pauli matrices, in the next section.

The Pauli sigma matrices are a set of three fundamental $2 \times 2$ complex matrices in quantum mechanics. They are particularly crucial in describing spin-$1/2$ particles, such as electrons, and are defined as: \[\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.\label{eq:pauli_matrices_definitions-box}\] These matrices serve as representations of the spin operators along the $x$, $y$, and $z$ axes, respectively, in a dimensionless form. When considering spin angular momentum, they are often multiplied by $\hbar/2$.

Action of $\sigma_z$: \[\sigma_z \left| \psi \right\rangle = \sigma_z \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \alpha \\ -\beta \end{pmatrix} = \alpha \left| 0 \right\rangle - \beta \left| 1 \right\rangle. \label{eq:sigma_z_action-box}\] The operator $\sigma_z$ leaves the amplitude of the $\left| 0 \right\rangle$ state unchanged while flipping the sign of the amplitude of the $\left| 1 \right\rangle$ state.

Action of $\sigma_x$: \[\sigma_x \left| \psi \right\rangle = \sigma_x \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \beta \\ \alpha \end{pmatrix} = \beta \left| 0 \right\rangle + \alpha \left| 1 \right\rangle. \label{eq:sigma_x_action-box}\] The operator $\sigma_x$ swaps the amplitudes of the $\left| 0 \right\rangle$ and $\left| 1 \right\rangle$ states.

Action of $\sigma_y$: \[\sigma_y \left| \psi \right\rangle = \sigma_y \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} -i\beta \\ i\alpha \end{pmatrix} = -i\beta \left| 0 \right\rangle + i\alpha \left| 1 \right\rangle. \label{eq:sigma_y_action-box}\] The operator $\sigma_y$ swaps the amplitudes and introduces phase factors of $\pm i$.

Comparing Equations [eq:sigma_z_sigma_x_product] and [eq:sigma_x_sigma_z_product], we clearly see that $\sigma_z \sigma_x \neq \sigma_x \sigma_z$, demonstrating the non-commutativity of Pauli matrices.

$] The commutator of$_z$and$_x$is given by$[_z, _x] = _z _x - _x _z$: \begin{align*} [\sigma_z, \sigma_x] &= \sigma_z \sigma_x - \sigma_x \sigma_z \nonumber \\ &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \nonumber \\ &= \begin{pmatrix} (0-0) & (1-(-1)) \\ (-1-1) & (0-0) \end{pmatrix} \nonumber \\ &= \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} = 2i \sigma_y. \end{align*} This result is a fundamental commutation relation in quantum mechanics for spin operators. Cyclic permutations of indices yield similar relations:$[_x, _y] = 2i _z$and$[_y, $\sigma_z] = 2i \sigma_x$.

Let’s calculate the triple product $\sigma_z \sigma_x \sigma_z = (\sigma_z \sigma_x) \sigma_z$. Using the result from Equation [eq:sigma_z_sigma_x_product] for $\sigma_z \sigma_x$: \[\begin{aligned}\sigma_z \sigma_x \sigma_z &= \left(\sigma_z \sigma_x\right) \sigma_z = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (0\cdot 1 + 1\cdot 0) & (0\cdot 0 + 1\cdot (-1)) \\ ((-1)\cdot 1 + 0\cdot 0) & ((-1)\cdot 0 + 0\cdot (-1)) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = - \sigma_x.\end{aligned}\] This type of manipulation of Pauli matrices is frequently encountered in quantum mechanical calculations, particularly in spin systems and quantum information theory.

Given a vector $\left| \psi \right\rangle = \frac{1}{2}\left| 0 \right\rangle + \frac{3}{2}i\left| 1 \right\rangle = \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix}$, our objective is to construct an orthonormal basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ such that the first basis vector $\left| v_1 \right\rangle$ is aligned with the direction of $\left| \psi \right\rangle$. This process involves normalization and orthogonalization to ensure that both basis vectors are mutually orthogonal and normalized to unity.

The process of constructing orthonormal bases and changing basis representations is fundamental in quantum mechanics. It allows us to analyze quantum systems in different bases that may be more convenient or physically relevant for specific problems.

In linear algebra and quantum mechanics, it is often necessary to change the basis in which vectors and operators are represented. Consider two orthonormal bases for a vector space $\mathcal{V}$: $V = \{\left| v_1 \right\rangle, \left| v_2 \right\rangle, \ldots, \left| v_n \right\rangle\}$ and $U = \{\left| u_1 \right\rangle, \left| u_2 \right\rangle, \ldots, \left| u_n \right\rangle\}$. Since both $V$ and $U$ are bases, any vector in $\mathcal{V}$, and in particular each vector in basis $V$, can be expressed as a linear combination of vectors in basis $U$, and vice versa.

We can express each vector $\left| v_j \right\rangle$ of basis $V$ as a linear combination of the vectors of basis $U$: \[\left| v_j \right\rangle = \sum_{i=1}^{n} R_{ij} \left| u_i \right\rangle, \quad \text{for } j = 1, 2, \ldots, n,\label{eq:v_in_terms_of_u-box}\] where $R_{ij}$ are the transformation coefficients. The transformation matrix $R$ from basis $V$ to basis $U$ has elements $R_{ij} = \left\langle u_i \middle| v_j \right\rangle$. $R_{ij}$ is the component of $\left| v_j \right\rangle$ along $\left| u_i \right\rangle$.

The transformation matrix between two orthonormal bases is a unitary matrix, meaning it satisfies $R^\dagger R = R R^\dagger = I$, where $I$ is the identity matrix and $R^\dagger$ is the adjoint of $R$. Description: This theorem states that when we change from one orthonormal basis to another, the matrix that performs this transformation is unitary. Unitary matrices are crucial in quantum mechanics because they preserve probabilities and norms of vectors.

Vector norms remain invariant under orthonormal basis changes. If $\left| \psi \right\rangle$ is a vector, its norm calculated in any orthonormal basis will be the same. Description: This theorem highlights a fundamental property of orthonormal basis transformations: they preserve the length of vectors. This is essential in physics, as physical quantities like probabilities (derived from vector norms) must be independent of the choice of basis.

The concept of basis transformation is essential for understanding how physical quantities arerepresented in different coordinate systems. In quantum mechanics, choosing an appropriate basis can significantly simplify calculations and provide deeper insights into the system’s properties. The unitary nature of basis transformations ensures that fundamental physical quantities like probabilities (related to squared norms) are preserved, regardless of the chosen basis.

--- title: "Lecture Notes on Matrix Representation of Linear Operators and Basis Transformations" author: "Your Name" date: "2025-02-05" format: html: toc: true # Table of Contents toc-depth: 2 code-tools: true theme: cosmo # Or "journal" for Distill-like minimalism --- # Introduction In this lecture, we will explore the matrix representation of linear operators, a fundamental concept in linear algebra and quantum mechanics. We will begin by demonstrating how the action of a linear operator on an orthonormal basis fully defines its operation on any vector within the vector space. This approach allows us to represent linear operators as matrices, facilitating calculations and providing a powerful framework for analysis. We will then discuss the product of linear operators and its representation through matrix multiplication. A key concept we will address is the non-commutativity of operator products, a feature that is particularly significant in quantum mechanics. The commutator will be introduced as a measure to quantify this non-commutativity. To illustrate these principles, we will introduce the Pauli sigma matrices ($\sigma_x, \sigma_y, \sigma_z$) as concrete examples of linear operators and examine their matrix representations and actions on state vectors. We will perform explicit calculations of matrix products and commutators involving Pauli matrices to solidify understanding. Furthermore, we will cover the practical procedures for constructing orthonormal bases, including normalization and orthogonalization. A detailed example will demonstrate how to find an orthonormal basis given an initial vector and how to represent a quantum state in this new basis using projection operators. Finally, we will discuss the concept of basis transformations in a more general context, introducing transformation matrices and highlighting the crucial property that vector norms remain invariant under orthonormal basis changes. This lecture aims to provide a comprehensive introduction to matrix representations of operators and basis transformations, laying the groundwork for more advanced topics in quantum mechanics and related fields. # Matrix Representation of Linear Operators ## Linear Operators and Orthonormal Bases {#subsec:linear_operators_orthonormal_bases} Consider a linear operator $A$ acting within a vector space $\mathcal{V}$. In quantum mechanics, $\mathcal{V}$ is often a Hilbert space. Any vector $\left| \psi \right\rangle \in \mathcal{V}$ can be expressed as a linear combination of orthonormal basis vectors $\{\left| v_j \right\rangle\}$: $$\left| \psi \right\rangle = \sum_j c_j \left| v_j \right\rangle,$$ where $c_j$ are complex coefficients. Due to the linearity of $A$, its action on $\left| \psi \right\rangle$ is determined by its action on each basis vector $\left| v_j \right\rangle$: $$A\left| \psi \right\rangle = A\left(\sum_j c_j \left| v_j \right\rangle\right) = \sum_j c_j A\left| v_j \right\rangle.\label{eq:linear_action_basis}$$ This fundamental property implies that to fully understand the behavior of a linear operator $A$ on any vector $\left| \psi \right\rangle$, it suffices to know its action on all vectors of an orthonormal basis $\{\left| v_j \right\rangle\}$. ## Action of an Operator on Basis Vectors and Matrix Elements {#subsec:action_operator_basis_vectors} ::: definition **Definition 1** (Matrix Elements of an Operator). *When a linear operator $A$ acts on a basis vector $\left| v_j \right\rangle$, the result is generally another vector in the same vector space $\mathcal{V}$. This resulting vector can be expressed as a linear combination of the same orthonormal basis vectors $\{\left| v_i \right\rangle\}$: $$A\left| v_j \right\rangle = \sum_i A_{ij} \left| v_i \right\rangle.\label{eq:action_on_basis_vector}$$ The coefficients $A_{ij}$ are known as the matrix elements of the operator $A$ in the basis $\{\left| v_i \right\rangle\}$. They quantify the component of $A\left| v_j \right\rangle$ along each basis vector $\left| v_i \right\rangle$.* ::: To determine these matrix elements $A_{ij}$, we project $A\left| v_j \right\rangle$ onto the basis vector $\left| v_i \right\rangle$ by taking the inner product: $$\left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle = \left\langle v_i \middle| \sum_k A_{kj} \left| v_k \right\rangle \right\rangle = \sum_k A_{kj} \left\langle v_i \middle| v_k \right\rangle.\label{eq:inner_product_projection}$$ Since $\{\left| v_i \right\rangle\}$ is an orthonormal basis, the inner product $\left\langle v_i \middle| v_k \right\rangle = \delta_{ik}$, where $\delta_{ik}$ is the Kronecker delta (equal to 1 if $i=k$ and 0 if $i \neq k$). Therefore, Equation [\[eq:inner_product_projection\]](#eq:inner_product_projection){reference-type="eqref" reference="eq:inner_product_projection"} simplifies to: $$\left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle = \sum_k A_{kj} \delta_{ik} = A_{ij}.\label{eq:matrix_elements_inner_product}$$ Thus, the matrix element $A_{ij}$ is precisely the inner product of $\left| v_i \right\rangle$ with $A\left| v_j \right\rangle$. ::: theorem **Theorem 1** (Matrix Elements via Inner Product). *The matrix element $A_{ij}$ of a linear operator $A$ in an orthonormal basis $\{\left| v_i \right\rangle\}$ is given by the inner product of the $i$-th basis vector with the action of $A$ on the $j$-th basis vector: $$A_{ij} = \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle.$$* ::: ## Matrix Representation of a Linear Operator {#subsec:matrix_representation} ::: definition **Definition 2** (Matrix Representation of a Linear Operator). *Using the matrix elements $A_{ij}$, we can construct a representation of the linear operator $A$ in the orthonormal basis $\{\left| v_i \right\rangle\}$. The operator $A$ can be expressed as: $$A = \sum_{i,j} A_{ij} \left| v_i \right\rangle\left\langle v_j \right| = \sum_{i,j} \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle \left| v_i \right\rangle\left\langle v_j \right|.\label{eq:operator_representation}$$ This representation explicitly shows how the operator $A$ acts in the chosen orthonormal basis.* ::: Applying $A$ to a basis vector $\left| v_k \right\rangle$ using this representation yields: $$\begin{aligned}A\left| v_k \right\rangle &= \left(\sum_{i,j} A_{ij} \left| v_i \right\rangle\left\langle v_j \right|\right)\left| v_k \right\rangle \\&= \sum_{i,j} A_{ij} \left| v_i \right\rangle \left\langle v_j \middle| v_k \right\rangle \\&= \sum_{i,j} A_{ij} \left| v_i \right\rangle \delta_{jk} \\&= \sum_{i} A_{ik} \left| v_i \right\rangle,\end{aligned}$$ which is consistent with the definition of matrix elements $A_{ij}$ in Equation [\[eq:action_on_basis_vector\]](#eq:action_on_basis_vector){reference-type="eqref" reference="eq:action_on_basis_vector"}. ::: definition **Definition 3** (Matrix Form of a Linear Operator). *In an $n$-dimensional vector space, the matrix representation of $A$, denoted as $[A]$, is an $n \times n$ matrix where the element in the $i$-th row and $j$-th column is $A_{ij}$: $$[A] = \begin{pmatrix}A_{11} & A_{12} & \cdots & A_{1n} \\A_{21} & A_{22} & \cdots & A_{2n} \\\vdots & \vdots & \ddots & \vdots \\A_{n1} & A_{n2} & \cdots & A_{nn}\end{pmatrix}.\label{eq:matrix_representation_explicit}$$ Each entry $A_{ij}$ in this matrix quantifies the transition from the $j$-th basis state to the $i$-th basis state under the action of the operator $A$.* ::: # Products and Non-Commutativity of Operators ## Operator Product and Matrix Multiplication {#subsec:operator_product_matrix_multiplication} Consider two linear operators $A$ and $B$. We want to determine the matrix representation of their product $C = AB$ in a given orthonormal basis $\{\left| v_i \right\rangle\}$. Let $[A]$ and $[B]$ be the matrix representations of $A$ and $B$ respectively, with matrix elements $A_{ij} = \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle$ and $B_{lm} = \left\langle v_l \middle| B\left| v_m \right\rangle \right\rangle$. Using the representation of operators in terms of orthonormal basis vectors (Equation [\[eq:operator_representation\]](#eq:operator_representation){reference-type="eqref" reference="eq:operator_representation"}), we can express the product $AB$ as: $$\begin{aligned}AB &= \left(\sum_{i,j} A_{ij} \left| v_i \right\rangle\left\langle v_j \right|\right) \left(\sum_{l,m} B_{lm} \left| v_l \right\rangle\left\langle v_m \right|\right) \\&= \sum_{i,j,l,m} A_{ij} B_{lm} \left| v_i \right\rangle\left\langle v_j \right| \left| v_l \right\rangle\left\langle v_m \right| \\&= \sum_{i,j,l,m} A_{ij} B_{lm} \left| v_i \right\rangle \underbrace{\left\langle v_j \middle| v_l \right\rangle}_{\delta_{jl}} \left\langle v_m \right| \\&= \sum_{i,j,m} A_{ij} B_{jm} \left| v_i \right\rangle\left\langle v_m \right| \\&= \sum_{i,m} \left(\sum_j A_{ij} B_{jm}\right) \left| v_i \right\rangle\left\langle v_m \right|.\end{aligned}$$ Let $C_{im} = \sum_j A_{ij} B_{jm}$. Then the product operator $C = AB$ can be written as $C = \sum_{i,m} C_{im} \left| v_i \right\rangle\left\langle v_m \right|$. The coefficients $C_{im}$ are precisely the elements of the matrix product of $[A]$ and $[B]$. ::: theorem **Theorem 2** (Matrix Multiplication for Operator Product). *The matrix representation of the product of two linear operators is obtained by the matrix multiplication of their respective matrix representations. If $[A]$ is an $n \times n$ matrix and $[B]$ is an $n \times n$ matrix, then their matrix product $[C] = [A][B]$ is also an $n \times n$ matrix, with elements given by $C_{im} = \sum_j A_{ij} B_{jm}$.* ::: ## Non-Commutativity of Operator Products {#subsec:non_commutativity} ::: remark **Remark 1** (Non-Commutativity of Operators). *In general, matrix multiplication is not commutative, meaning that for two matrices $[A]$ and $[B]$, it is not always true that $[A][B] = [B][A]$. Consequently, for linear operators $A$ and $B$, the order of application matters, and in general $AB \neq BA$.* *When $AB = BA$, we say that the operators $A$ and $B$ *commute*. If $AB \neq BA$, they are said to be *non-commuting*. This property of non-commutativity is a cornerstone of quantum mechanics and distinguishes it fundamentally from classical mechanics, where observables are typically represented by commuting quantities.* *A simple example from classical mechanics involves rotations in three dimensions. Rotations about different axes generally do not commute; the final orientation depends on the order in which rotations are performed. Similarly, in quantum mechanics, many important operators, such as those representing position and momentum, or different components of spin, do not commute.* ::: ## The Commutator {#subsec:commutator} ::: definition **Definition 4** (Commutator of Two Operators). *To quantify the non-commutativity of two operators $A$ and $B$, we define the *commutator* as: $$[A, B] = AB - BA.\label{eq:commutator_definition}$$ The commutator $[A, B]$ itself is a linear operator. If $A$ and $B$ commute, then $AB - BA = 0$, so $[A, B] = 0$. Conversely, if $[A, B] \neq 0$, then $A$ and $B$ do not commute. The commutator thus provides a measure of the degree to which two operators fail to commute.* ::: ::: remark **Remark 2** (Importance of Commutator in Quantum Mechanics). *In quantum mechanics, commutation relations between operators are fundamental. For instance, the canonical commutation relation between position and momentum operators is a basic postulate of quantum theory and has profound implications for the uncertainty principle. We will explore examples of commutators, particularly using Pauli matrices, in the next section.* ::: # Example: Pauli Sigma Matrices ## Introduction to Pauli Matrices and Spin Operators {#subsec:pauli_matrices_introduction} ::: definition **Definition 5** (Pauli Sigma Matrices). *The Pauli sigma matrices are a set of three fundamental $2 \times 2$ complex matrices in quantum mechanics. They are particularly crucial in describing spin-$1/2$ particles, such as electrons, and are defined as: $$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.\label{eq:pauli_matrices_definitions}$$ These matrices serve as representations of the spin operators along the $x$, $y$, and $z$ axes, respectively, in a dimensionless form. When considering spin angular momentum, they are often multiplied by $\hbar/2$.* ::: ## Action of Pauli Matrices on Computational Basis States {#subsec:pauli_actions_basis_states} Let's examine how Pauli matrices act on a general two-component state vector $\left| \psi \right\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$ expressed in the computational basis $\{\left| 0 \right\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \left| 1 \right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\}$. Here, $\alpha$ and $\beta$ are complex coefficients representing the amplitudes of the states $\left| 0 \right\rangle$ and $\left| 1 \right\rangle$ respectively. ::: example **Example 1** (Action of $\sigma_z$ on a state vector). ***Action of $\sigma_z$:** $$\sigma_z \left| \psi \right\rangle = \sigma_z \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \alpha \\ -\beta \end{pmatrix} = \alpha \left| 0 \right\rangle - \beta \left| 1 \right\rangle. \label{eq:sigma_z_action}$$ The operator $\sigma_z$ leaves the amplitude of the $\left| 0 \right\rangle$ state unchanged while flipping the sign of the amplitude of the $\left| 1 \right\rangle$ state.* ::: ::: example **Example 2** (Action of $\sigma_x$ on a state vector). ***Action of $\sigma_x$:** $$\sigma_x \left| \psi \right\rangle = \sigma_x \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \beta \\ \alpha \end{pmatrix} = \beta \left| 0 \right\rangle + \alpha \left| 1 \right\rangle. \label{eq:sigma_x_action}$$ The operator $\sigma_x$ swaps the amplitudes of the $\left| 0 \right\rangle$ and $\left| 1 \right\rangle$ states.* ::: ::: example **Example 3** (Action of $\sigma_y$ on a state vector). ***Action of $\sigma_y$:** $$\sigma_y \left| \psi \right\rangle = \sigma_y \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} -i\beta \\ i\alpha \end{pmatrix} = -i\beta \left| 0 \right\rangle + i\alpha \left| 1 \right\rangle. \label{eq:sigma_y_action}$$ The operator $\sigma_y$ swaps the amplitudes and introduces phase factors of $\pm i$.* ::: ## Matrix Product Examples: $\sigma_z \sigma_x$ and $\sigma_x \sigma_z$ {#subsec:pauli_matrix_products} Let's compute the matrix product $\sigma_z \sigma_x$: $$\begin{aligned}\sigma_z \sigma_x &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (1\cdot 0 + 0\cdot 1) & (1\cdot 1 + 0\cdot 0) \\ (0\cdot 0 + (-1)\cdot 1) & (0\cdot 1 + (-1)\cdot 0) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.\label{eq:sigma_z_sigma_x_product}\end{aligned}$$ Now, let's compute the product $\sigma_x \sigma_z$: $$\begin{aligned}\sigma_x \sigma_z &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (0\cdot 1 + 1\cdot 0) & (0\cdot 0 + 1\cdot (-1)) \\ (1\cdot 1 + 0\cdot 0) & (1\cdot 0 + 0\cdot (-1)) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.\label{eq:sigma_x_sigma_z_product}\end{aligned}$$ ::: remark **Remark 3** (Non-Commutativity of $\sigma_x$ and $\sigma_z$). *Comparing Equations [\[eq:sigma_z_sigma_x_product\]](#eq:sigma_z_sigma_x_product){reference-type="eqref" reference="eq:sigma_z_sigma_x_product"} and [\[eq:sigma_x_sigma_z_product\]](#eq:sigma_x_sigma_z_product){reference-type="eqref" reference="eq:sigma_x_sigma_z_product"}, we clearly see that $\sigma_z \sigma_x \neq \sigma_x \sigma_z$, demonstrating the non-commutativity of Pauli matrices.* ::: ## Commutator of $\sigma_z$ and $\sigma_x$ {#subsec:commutator_sigma_zx} ::: example **Example 4** (\[\_z, \_x). *\] The commutator of $\sigma_z$ and $\sigma_x$ is given by $[\sigma_z, \sigma_x] = \sigma_z \sigma_x - \sigma_x \sigma_z$: $$\begin{aligned}[\sigma_z, \sigma_x] &= \sigma_z \sigma_x - \sigma_x \sigma_z \nonumber \\&= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (0-0) & (1-(-1)) \\ (-1-1) & (0-0) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}.\label{eq:sigma_zx_commutator_intermediate}\end{aligned}$$ We want to express this commutator in terms of another Pauli matrix. Let's compare Equation [\[eq:sigma_zx_commutator_intermediate\]](#eq:sigma_zx_commutator_intermediate){reference-type="eqref" reference="eq:sigma_zx_commutator_intermediate"} with $\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$. We observe that: $$2i \sigma_y = 2i \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2i^2 \\ 2i^2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}.$$ Thus, we find the commutation relation: $$[\sigma_z, \sigma_x] = 2i \sigma_y.\label{eq:sigma_zx_commutator_result}$$ This result is a fundamental commutation relation in quantum mechanics for spin operators. Cyclic permutations of indices yield similar relations: $[\sigma_x, \sigma_y] = 2i \sigma_z$ and $[\sigma_y, \sigma_z] = 2i \sigma_x$.* ::: ## Example: Triple Product $\sigma_z \sigma_x \sigma_z$ {#subsec:pauli_triple_product} ::: example **Example 5** (Triple Product $\sigma_z \sigma_x \sigma_z$). *Let's calculate the triple product $\sigma_z \sigma_x \sigma_z = (\sigma_z \sigma_x) \sigma_z$. Using the result from Equation [\[eq:sigma_z_sigma_x_product\]](#eq:sigma_z_sigma_x_product){reference-type="eqref" reference="eq:sigma_z_sigma_x_product"} for $\sigma_z \sigma_x$: $$\begin{aligned}\sigma_z \sigma_x \sigma_z &= \left(\sigma_z \sigma_x\right) \sigma_z = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (0\cdot 1 + 1\cdot 0) & (0\cdot 0 + 1\cdot (-1)) \\ ((-1)\cdot 1 + 0\cdot 0) & ((-1)\cdot 0 + 0\cdot (-1)) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}.\label{eq:sigma_z_sigma_x_sigma_z_intermediate}\end{aligned}$$ Comparing Equation [\[eq:sigma_z_sigma_x_sigma_z_intermediate\]](#eq:sigma_z_sigma_x_sigma_z_intermediate){reference-type="eqref" reference="eq:sigma_z_sigma_x_sigma_z_intermediate"} with the definition of $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, we observe that: $$\sigma_z \sigma_x \sigma_z = - \sigma_x.\label{eq:sigma_z_sigma_x_sigma_z_result}$$ This type of manipulation of Pauli matrices is frequently encountered in quantum mechanical calculations, particularly in spin systems and quantum information theory.* ::: # Constructing Orthonormal Bases: A Practical Example ## Problem Statement: Constructing an Orthonormal Basis {#subsec:problem_statement_orthonormal_basis} ::: example **Example 6** (Constructing an Orthonormal Basis). *Given a vector $\left| \psi \right\rangle = \frac{1}{2}\left| 0 \right\rangle + \frac{3}{2}i\left| 1 \right\rangle = \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix}$, our objective is to construct an orthonormal basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ such that the first basis vector $\left| v_1 \right\rangle$ is aligned with the direction of $\left| \psi \right\rangle$. This process involves normalization and orthogonalization to ensure that both basis vectors are mutually orthogonal and normalized to unity.* ::: ## Normalization of $\left| \psi \right\rangle$ to Obtain $\left| v_1 \right\rangle$ {#subsec:normalization_psi_v1} First, we normalize the vector $\left| \psi \right\rangle$ to obtain the first orthonormal basis vector $\left| v_1 \right\rangle$. To do this, we calculate the squared norm of $\left| \psi \right\rangle$: $$\begin{aligned}\left\langle \psi \middle| \psi \right\rangle &= \left\langle \psi \right|\left| \psi \right\rangle = \begin{pmatrix} 1/2 & -\frac{3}{2}i \end{pmatrix} \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix} \\&= \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(-\frac{3}{2}i\right)\left(\frac{3}{2}i\right) \\&= \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2}.\end{aligned}$$ The norm of $\left| \psi \right\rangle$ is $\|\left| \psi \right\rangle\| = \sqrt{\left\langle \psi \middle| \psi \right\rangle} = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}$. We normalize $\left| \psi \right\rangle$ by dividing it by its norm to obtain $\left| v_1 \right\rangle$: $$\left| v_1 \right\rangle = \frac{\left| \psi \right\rangle}{\|\left| \psi \right\rangle\|} = \frac{1}{\sqrt{5/2}} \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix} = \sqrt{\frac{2}{5}} \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix} = \begin{pmatrix} 1/\sqrt{10} \\ 3i/\sqrt{10} \end{pmatrix} = \frac{1}{\sqrt{10}} \left| 0 \right\rangle + \frac{3i}{\sqrt{10}} \left| 1 \right\rangle.\label{eq:normalized_v1}$$ To verify normalization, we compute $\left\langle v_1 \middle| v_1 \right\rangle$: $$\begin{aligned}\left\langle v_1 \middle| v_1 \right\rangle &= \left\langle v_1 \right|\left| v_1 \right\rangle = \begin{pmatrix} 1/\sqrt{10} & -\frac{3}{\sqrt{10}}i \end{pmatrix} \begin{pmatrix} 1/\sqrt{10} \\ 3i/\sqrt{10} \end{pmatrix} \\&= \left| \frac{1}{\sqrt{10}} \right|^2 + \left| \frac{3i}{\sqrt{10}} \right|^2 = \frac{1}{10} + \frac{9}{10} = 1.\end{aligned}$$ Thus, $\left| v_1 \right\rangle$ is indeed normalized. ## Finding a Vector $\left| v_2 \right\rangle$ Orthogonal to $\left| v_1 \right\rangle$ {#subsec:orthogonal_vector_v2} Next, we need to find a vector $\left| v_2 \right\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle$ that is orthogonal to $\left| v_1 \right\rangle$. The orthogonality condition $\left\langle v_1 \middle| v_2 \right\rangle = 0$ is expressed as: $$\left\langle v_1 \middle| v_2 \right\rangle = \left\langle v_1 \right|\left| v_2 \right\rangle = \begin{pmatrix} 1/\sqrt{10} & -\frac{3i}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \frac{1}{\sqrt{10}} \alpha - \frac{3i}{\sqrt{10}} \beta = 0.\label{eq:orthogonality_condition}$$ Multiplying by $\sqrt{10}$, we get $\alpha - 3i\beta = 0$, which implies $\alpha = 3i\beta$. Let us choose $\beta = c$, where $c$ is a constant to be determined by normalization. Then $\alpha = 3ic$, and $\left| v_2 \right\rangle = \begin{pmatrix} 3ic \\ c \end{pmatrix} = 3ic \left| 0 \right\rangle + c \left| 1 \right\rangle$. ## Normalization of $\left| v_2 \right\rangle$ {#subsec:normalization_v2} Now we normalize $\left| v_2 \right\rangle$. We calculate the squared norm of $\left| v_2 \right\rangle$: $$\begin{aligned}\left\langle v_2 \middle| v_2 \right\rangle &= \left\langle v_2 \right|\left| v_2 \right\rangle = \begin{pmatrix} -3ic^* & c^* \end{pmatrix} \begin{pmatrix} 3ic \\ c \end{pmatrix} \\&= (-3ic^*)(3ic) + (c^*)(c) \\&= 9\left| c \right|^2 + \left| c \right|^2 = 10\left| c \right|^2.\end{aligned}$$ For $\left| v_2 \right\rangle$ to be normalized, we require $\left\langle v_2 \middle| v_2 \right\rangle = 1$, so $10\left| c \right|^2 = 1$, which gives $\left| c \right| = \frac{1}{\sqrt{10}}$. We can choose $c = \frac{1}{\sqrt{10}}$ for simplicity. Then $\beta = \frac{1}{\sqrt{10}}$ and $\alpha = \frac{3i}{\sqrt{10}}$. Thus, $$\left| v_2 \right\rangle = \begin{pmatrix} 3i/\sqrt{10} \\ 1/\sqrt{10} \end{pmatrix} = \frac{3i}{\sqrt{10}} \left| 0 \right\rangle + \frac{1}{\sqrt{10}} \left| 1 \right\rangle.\label{eq:normalized_v2}$$ We verify the normalization of $\left| v_2 \right\rangle$: $$\begin{aligned}\braket{v_2 | v_2} &= \left\langle v_2 \right| \left| v_2 \right\rangle = \begin{pmatrix} -\frac{3i}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} \frac{3i}{\sqrt{10}} \\ \frac{1}{\sqrt{10}} \end{pmatrix} \\&= \left| \frac{3i}{\sqrt{10}} \right|^2 + \left| \frac{1}{\sqrt{10}} \right|^2 = \frac{9}{10} + \frac{1}{10} = 1.\end{aligned}$$ ## Verification of Orthonormality of $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ {#subsec:verification_orthonormality} We have constructed the set of vectors $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ with $$\left| v_1 \right\rangle = \frac{1}{\sqrt{10}} \left| 0 \right\rangle + \frac{3i}{\sqrt{10}} \left| 1 \right\rangle, \quad\left| v_2 \right\rangle = \frac{3i}{\sqrt{10}} \left| 0 \right\rangle + \frac{1}{\sqrt{10}} \left| 1 \right\rangle.\label{eq:basis_vectors_v1_v2}$$ We have already verified that $\left\langle v_1 \middle| v_1 \right\rangle = 1$ and $\left\langle v_2 \middle| v_2 \right\rangle = 1$. Now we verify orthogonality $\left\langle v_1 \middle| v_2 \right\rangle = 0$: $$\begin{aligned}\left\langle v_1 \middle| v_2 \right\rangle &= \left\langle v_1 \right|\left| v_2 \right\rangle = \begin{pmatrix} 1/\sqrt{10} & -\frac{3i}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} 3i/\sqrt{10} \\ 1/\sqrt{10} \end{pmatrix} \\&= \left(\frac{1}{\sqrt{10}}\right)\left(\frac{3i}{\sqrt{10}}\right) + \left(-\frac{3i}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right) \\&= \frac{3i}{10}- \frac{3i}{10} = 0.\end{aligned}$$ Since $\left\langle v_1 \middle| v_1 \right\rangle = 1$, $\left\langle v_2 \middle| v_2 \right\rangle = 1$, and $\left\langle v_1 \middle| v_2 \right\rangle = 0$, the set $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ forms an orthonormal basis. ## Representing the State $\left| + \right\rangle$ in the Basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ {#subsec:representing_plus_in_v1v2} ### Using Projectors for Basis Transformation {#subsubsec:projectors_basis_transformation} To express the state $\left| + \right\rangle = \frac{1}{\sqrt{2}}(\left| 0 \right\rangle + \left| 1 \right\rangle)$ in the new orthonormal basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$, we use projection operators. The projector onto $\left| v_1 \right\rangle$ is $P_{v_1} = \left| v_1 \right\rangle\left\langle v_1 \right|$, and onto $\left| v_2 \right\rangle$ is $P_{v_2} = \left| v_2 \right\rangle\left\langle v_2 \right|$. The components of $\left| + \right\rangle$ in the new basis are given by the projection coefficients $c_1 = \left\langle v_1 \middle| + \right\rangle$ and $c_2 = \left\langle v_2 \middle| + \right\rangle$. ### Calculating Coefficients $\left\langle v_1 \middle| + \right\rangle$ and $\left\langle v_2 \middle| + \right\rangle$ {#subsubsec:calculating_coefficients} Let $\left| + \right\rangle = \frac{1}{\sqrt{2}}(\left| 0 \right\rangle + \left| 1 \right\rangle) = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}$. We calculate the coefficients: $$\begin{aligned}c_1 = \left\langle v_1 \middle| + \right\rangle &= \left\langle v_1 \right|\left| + \right\rangle = \begin{pmatrix} 1/\sqrt{10} & -\frac{3i}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} \\&= \left(\frac{1}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(-\frac{3i}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{20}} - \frac{3i}{\sqrt{20}} = \frac{1-3i}{\sqrt{20}}. \\c_2 = \left\langle v_2 \middle| + \right\rangle &= \left\langle v_2 \right|\left| + \right\rangle = \begin{pmatrix} -\frac{3i}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} \\&= \left(-\frac{3i}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{2}}\right) = -\frac{3i}{\sqrt{20}} + \frac{1}{\sqrt{20}} = \frac{1+3i}{\sqrt{20}}.\end{aligned}$$ Thus, in the new basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$, the state $\left| + \right\rangle$ is represented as: $$\left| + \right\rangle = c_1 \left| v_1 \right\rangle + c_2 \left| v_2 \right\rangle = \frac{1-3i}{\sqrt{20}} \left| v_1 \right\rangle + \frac{1+3i}{\sqrt{20}} \left| v_2 \right\rangle.\label{eq:plus_state_in_v1v2}$$ ### Verification of Normalization of $\left| + \right\rangle$ in the $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ Basis {#subsubsec:verification_normalization_plus_v1v2} Finally, we verify that $\left| + \right\rangle$ remains normalized when expressed in the new basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$: $$\begin{aligned}\text{Norm}^2 &= \left| c_1 \right|^2 + \left| c_2 \right|^2 = \left| \frac{1-3i}{\sqrt{20}} \right|^2 + \left| \frac{1+3i}{\sqrt{20}} \right|^2 \\&= \frac{\left| 1-3i \right|^2}{20} + \frac{\left| 1+3i \right|^2}{20} \\&= \frac{(1)^2 + (-3)^2}{20} + \frac{(1)^2 + 3^2}{20} \\&= \frac{10}{20} + \frac{10}{20} = 1.\end{aligned}$$ The norm squared is indeed 1, confirming that the state $\left| + \right\rangle$ remains normalized in the new orthonormal basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$, as expected for a change of orthonormal basis. ::: remark **Remark 4** (Significance of Orthonormal Basis Construction). *The process of constructing orthonormal bases and changing basis representations is fundamental in quantum mechanics. It allows us to analyze quantum systems in different bases that may be more convenient or physically relevant for specific problems.* ::: # Change of Basis and Transformation Matrices ## General Basis Transformations Between Orthonormal Bases {#subsec:general_basis_transformations} ::: definition **Definition 6** (Basis Transformation). *In linear algebra and quantum mechanics, it is often necessary to change the basis in which vectors and operators are represented. Consider two orthonormal bases for a vector space $\mathcal{V}$: $V = \{\left| v_1 \right\rangle, \left| v_2 \right\rangle, \ldots, \left| v_n \right\rangle\}$ and $U = \{\left| u_1 \right\rangle, \left| u_2 \right\rangle, \ldots, \left| u_n \right\rangle\}$. Since both $V$ and $U$ are bases, any vector in $\mathcal{V}$, and in particular each vector in basis $V$, can be expressed as a linear combination of vectors in basis $U$, and vice versa.* ::: ## Transformation Matrix from Basis $V$ to Basis $U$ {#subsec:transformation_matrix_v_to_u} ::: definition **Definition 7** (Transformation Matrix). *We can express each vector $\left| v_j \right\rangle$ of basis $V$ as a linear combination of the vectors of basis $U$: $$\left| v_j \right\rangle = \sum_{i=1}^{n} R_{ij} \left| u_i \right\rangle, \quad \text{for } j = 1, 2, \ldots, n,\label{eq:v_in_terms_of_u}$$ where $R_{ij}$ are the transformation coefficients. The transformation matrix $R$ from basis $V$ to basis $U$ has elements $R_{ij} = \left\langle u_i \middle| v_j \right\rangle$. $R_{ij}$ is the component of $\left| v_j \right\rangle$ along $\left| u_i \right\rangle$.* ::: Similarly, we can express each vector $\left| u_i \right\rangle$ of basis $U$ in terms of the vectors of basis $V$: $$\left| u_i \right\rangle = \sum_{j=1}^{n} (R')_{ji} \left| v_j \right\rangle, \quad \text{for } i = 1, 2, \ldots, n.\label{eq:u_in_terms_of_v}$$ By analogy, the coefficients $(R')_{ji} = \left\langle v_j \middle| u_i \right\rangle$. Let $R'$ be the transformation matrix from basis $U$ to basis $V$ with elements $(R')_{ij} = \left\langle v_i \middle| u_j \right\rangle$. Then $(R')_{ji} = R_{ij}^*$. Thus, $R' = R^\dagger$, the adjoint (conjugate transpose) of $R$. ## Relationship Between Transformation Matrices: Unitary Transformation {#subsec:transformation_matrices_relationship} ::: theorem **Theorem 3** (Unitary Transformation Matrix). *The transformation matrix between two orthonormal bases is a unitary matrix, meaning it satisfies $R^\dagger R = R R^\dagger = I$, where $I$ is the identity matrix and $R^\dagger$ is the adjoint of $R$.* ::: *Description:* This theorem states that when we change from one orthonormal basis to another, the matrix that performs this transformation is unitary. Unitary matrices are crucial in quantum mechanics because they preserve probabilities and norms of vectors. Substituting Equation [\[eq:u_in_terms_of_v\]](#eq:u_in_terms_of_v){reference-type="eqref" reference="eq:u_in_terms_of_v"} into Equation [\[eq:v_in_terms_of_u\]](#eq:v_in_terms_of_u){reference-type="eqref" reference="eq:v_in_terms_of_u"}, we get: $$\begin{aligned}\left| v_j \right\rangle &= \sum_{i=1}^{n} R_{ij} \left| u_i \right\rangle = \sum_{i=1}^{n} R_{ij} \left(\sum_{k=1}^{n} (R')_{ki} \left| v_k \right\rangle\right) \\&= \sum_{k=1}^{n} \left(\sum_{i=1}^{n} (R')_{ki} R_{ij}\right) \left| v_k \right\rangle.\end{aligned}$$ For this to hold for all $\left| v_j \right\rangle$ in the basis $V$, we must have: $$\sum_{i=1}^{n} (R')_{ki} R_{ij} = \delta_{kj}.\label{eq:identity_relation_matrix}$$ In matrix notation, this is $(R'R)^T = I$ or more directly $R'R = I$, where $I$ is the identity matrix. Since $R' = R^\dagger$, we have $R^\dagger R = I$. Similarly, substituting Equation [\[eq:v_in_terms_of_u\]](#eq:v_in_terms_of_u){reference-type="eqref" reference="eq:v_in_terms_of_u"} into Equation [\[eq:u_in_terms_of_v\]](#eq:u_in_terms_of_v){reference-type="eqref" reference="eq:u_in_terms_of_v"} leads to $R R^\dagger = I$. Matrices that satisfy $R^\dagger R = R R^\dagger = I$ are called *unitary matrices*. Therefore, the transformation between two orthonormal bases is a unitary transformation, and the transformation matrix is a unitary matrix. ## Invariance of Vector Norms Under Basis Transformation {#subsec:invariance_vector_norms} ::: theorem **Theorem 4** (Invariance of Vector Norms). *Vector norms remain invariant under orthonormal basis changes. If $\left| \psi \right\rangle$ is a vector, its norm calculated in any orthonormal basis will be the same.* ::: *Description:* This theorem highlights a fundamental property of orthonormal basis transformations: they preserve the length of vectors. This is essential in physics, as physical quantities like probabilities (derived from vector norms) must be independent of the choice of basis. Let $\left| \psi \right\rangle$ be a vector expressed in basis $V$ as $\left| \psi \right\rangle = \sum_{j} c_j \left| v_j \right\rangle$ and in basis $U$ as $\left| \psi \right\rangle = \sum_{i} c'_i \left| u_i \right\rangle$. The norm squared of $\left| \psi \right\rangle$ in basis $V$ is $\|\left| \psi \right\rangle\|^2 = \left\langle \psi \middle| \psi \right\rangle = \sum_{j} |c_j|^2$ due to orthonormality of $\{\left| v_j \right\rangle\}$. Similarly, in basis $U$, $\|\left| \psi \right\rangle\|^2 = \sum_{i} |c'_i|^2$. The transformation of coefficients from basis $V$ to basis $U$ is given by: $$c'_i = \left\langle u_i \middle| \left| \psi \right\rangle \right\rangle = \left\langle u_i \middle| \sum_{j} c_j \left| v_j \right\rangle \right\rangle = \sum_{j} c_j \left\langle u_i \middle| v_j \right\rangle = \sum_{j} R_{ij} c_j.$$ In matrix notation, $\mathbf{c'} = R \mathbf{c}$, where $\mathbf{c}$ and $\mathbf{c'}$ are column vectors of coefficients in bases $V$ and $U$ respectively. The invariance of the norm under unitary transformations can be shown as follows: $$\begin{aligned}\|\left| \psi \right\rangle\|^2_{U} = \sum_{i} |c'_i|^2 = (\mathbf{c'})^\dagger \mathbf{c'} = (R\mathbf{c})^\dagger (R\mathbf{c}) = \mathbf{c}^\dagger R^\dagger R \mathbf{c} = \mathbf{c}^\dagger I \mathbf{c} = \mathbf{c}^\dagger \mathbf{c} = \sum_{j} |c_j|^2 = \|\left| \psi \right\rangle\|^2_{V}.\end{aligned}$$ Thus, the norm of the vector $\left| \psi \right\rangle$ is invariant under a change of orthonormal basis, which is a consequence of the unitary nature of the transformation matrix $R$. This reflects the physical intuition that the length of a vector is an intrinsic property, independent of the coordinate system used to describe it. ::: remark **Remark 5** (Importance of Basis Transformation). *The concept of basis transformation is essential for understanding how physical quantities are represented in different coordinate systems. In quantum mechanics, choosing an appropriate basis can significantly simplify calculations and provide deeper insights into the system's properties. The unitary nature of basis transformations ensures that fundamental physical quantities like probabilities (related to squared norms) are preserved, regardless of the chosen basis.* ::: # Conclusion In [6](#remark-conclusion){reference-type="ref+label" reference="remark-conclusion"} we summarize the lecture. ::: {.remark-conclusion .remark} **Remark 6** (Summary of the Lecture). *This lecture has established the fundamental concepts of matrix representations for linear operators and basis transformations, which are crucial for quantum mechanics. We have covered the following key points:* - ***Matrix Representation of Operators:** Linear operators can be represented as matrices once an orthonormal basis is chosen. The matrix elements are given by the inner product $\left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle$.* - ***Operator Products and Matrix Multiplication:** The matrix representation of the product of two operators is equivalent to the matrix product of their individual matrix representations.* - ***Non-Commutativity and the Commutator:** Operator products are generally non-commutative, a key feature in quantum mechanics. The commutator $[A, B] = AB - BA$ quantifies this non-commutativity.* - ***Pauli Matrices Example:** Pauli matrices ($\sigma_x, \sigma_y, \sigma_z$) were introduced as examples of non-commuting operators, demonstrating their actions on state vectors and their commutation relations.* - ***Constructing Orthonormal Bases:** We demonstrated a practical method for constructing orthonormal bases, starting from an arbitrary vector and using normalization and orthogonalization procedures.* - ***Basis Transformations and Unitary Matrices:** Transformations between orthonormal bases are unitary, described by unitary transformation matrices. These transformations preserve vector norms and inner products, ensuring physical consistency.* *It is crucial to remember that while matrix representations are basis-dependent, the underlying vectors and operators are basis-independent entities. The choice of basis is a matter of convenience and can be adapted to simplify specific problems.* *In the upcoming lectures, we will build upon these concepts by exploring the properties and applications of transformation matrices in more detail. We will investigate how basis transformations are used to simplify quantum problems and potentially delve into the concept of time evolution as a change of basis in quantum mechanics.* ::: ::: tcolorbox When a linear operator $A$ acts on a basis vector $\left| v_j \right\rangle$, the result is generally another vector in the same vector space $\mathcal{V}$. This resulting vector can be expressed as a linear combination of the same orthonormal basis vectors $\{\left| v_i \right\rangle\}$: $$A\left| v_j \right\rangle = \sum_i A_{ij} \left| v_i \right\rangle.\label{eq:action_on_basis_vector-box}$$ The coefficients $A_{ij}$ are known as the matrix elements of the operator $A$ in the basis $\{\left| v_i \right\rangle\}$. They quantify the component of $A\left| v_j \right\rangle$ along each basis vector $\left| v_i \right\rangle$. ::: ::: tcolorbox The matrix element $A_{ij}$ of a linear operator $A$ in an orthonormal basis $\{\left| v_i \right\rangle\}$ is given by the inner product of the $i$-th basis vector with the action of $A$ on the $j$-th basis vector: $$A_{ij} = \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle.$$ ::: ::: tcolorbox Using the matrix elements $A_{ij}$, we can construct a representation of the linear operator $A$ in the orthonormal basis $\{\left| v_i \right\rangle\}$. The operator $A$ can be expressed as: $$A = \sum_{i,j} A_{ij} \left| v_i \right\rangle\left\langle v_j \right| = \sum_{i,j} \left\langle v_i \middle| A\left| v_j \right\rangle \right\rangle \left| v_i \right\rangle\left\langle v_j \right|.\label{eq:operator_representation-box}$$ This representation explicitly shows how the operator $A$ acts in the chosen orthonormal basis. ::: ::: tcolorbox In an $n$-dimensional vector space, the matrix representation of $A$, denoted as $[A]$, is an $n \times n$ matrix where the element in the $i$-th row and $j$-th column is $A_{ij}$: $$[A] = \begin{pmatrix}A_{11} & A_{12} & \cdots & A_{1n} \\A_{21} & A_{22} & \cdots & A_{2n} \\\vdots & \vdots & \ddots & \vdots \\A_{n1} & A_{n2} & \cdots & A_{nn}\end{pmatrix}.\label{eq:matrix_representation_explicit-box}$$ Each entry $A_{ij}$ in this matrix quantifies the transition from the $j$-th basis state to the $i$-th basis state under the action of the operator $A$. ::: ::: tcolorbox The matrix representation of the product of two linear operators is obtained by the matrix multiplication of their respective matrix representations. If $[A]$ is an $n \times n$ matrix and $[B]$ is an $n \times n$ matrix, then their matrix product $[C] = [A][B]$ is also an $n \times n$ matrix, with elements given by $C_{im} = \sum_j A_{ij} B_{jm}$. ::: ::: tcolorbox In general, matrix multiplication is not commutative, meaning that for two matrices $[A]$ and $[B]$, it is not always true that $[A][B] = [B][A]$. Consequently, for linear operators $A$ and $B$, the order of application matters, and in general $AB \neq BA$. When $AB = BA$, we say that the operators $A$ and $B$ *commute*. If $AB \neq BA$, they are said to be *non-commuting*. This property of non-commutativity is a cornerstone of quantum mechanics and distinguishes it fundamentally from classical mechanics, where observables are typically represented by commuting quantities. A simple example from classical mechanics involves rotations in three dimensions. Rotations about different axes generally do not commute; the final orientation depends on the order in which rotations are performed. Similarly, in quantum mechanics, many important operators, such as those representing position and momentum, or different components of spin, do not commute. ::: ::: tcolorbox To quantify the non-commutativity of two operators $A$ and $B$, we define the *commutator* as: $$[A, B] = AB - BA.\label{eq:commutator_definition-box}$$ The commutator $[A, B]$ itself is a linear operator. If $A$ and $B$ commute, then $AB - BA = 0$, so $[A, B] = 0$. Conversely, if $[A, B] \neq 0$, then $A$ and $B$ do not commute. The commutator thus provides a measure of the degree to which two operators fail to commute. ::: ::: tcolorbox In quantum mechanics, commutation relations between operators are fundamental. For instance, the canonical commutation relation between position and momentum operators is a basic postulate of quantum theory and has profound implications for the uncertainty principle. We will explore examples of commutators, particularly using Pauli matrices, in the next section. ::: ::: tcolorbox The Pauli sigma matrices are a set of three fundamental $2 \times 2$ complex matrices in quantum mechanics. They are particularly crucial in describing spin-$1/2$ particles, such as electrons, and are defined as: $$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.\label{eq:pauli_matrices_definitions-box}$$ These matrices serve as representations of the spin operators along the $x$, $y$, and $z$ axes, respectively, in a dimensionless form. When considering spin angular momentum, they are often multiplied by $\hbar/2$. ::: ::: tcolorbox **Action of $\sigma_z$:** $$\sigma_z \left| \psi \right\rangle = \sigma_z \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \alpha \\ -\beta \end{pmatrix} = \alpha \left| 0 \right\rangle - \beta \left| 1 \right\rangle. \label{eq:sigma_z_action-box}$$ The operator $\sigma_z$ leaves the amplitude of the $\left| 0 \right\rangle$ state unchanged while flipping the sign of the amplitude of the $\left| 1 \right\rangle$ state. ::: ::: tcolorbox **Action of $\sigma_x$:** $$\sigma_x \left| \psi \right\rangle = \sigma_x \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \beta \\ \alpha \end{pmatrix} = \beta \left| 0 \right\rangle + \alpha \left| 1 \right\rangle. \label{eq:sigma_x_action-box}$$ The operator $\sigma_x$ swaps the amplitudes of the $\left| 0 \right\rangle$ and $\left| 1 \right\rangle$ states. ::: ::: tcolorbox **Action of $\sigma_y$:** $$\sigma_y \left| \psi \right\rangle = \sigma_y \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} -i\beta \\ i\alpha \end{pmatrix} = -i\beta \left| 0 \right\rangle + i\alpha \left| 1 \right\rangle. \label{eq:sigma_y_action-box}$$ The operator $\sigma_y$ swaps the amplitudes and introduces phase factors of $\pm i$. ::: ::: tcolorbox Comparing Equations [\[eq:sigma_z_sigma_x_product\]](#eq:sigma_z_sigma_x_product){reference-type="eqref" reference="eq:sigma_z_sigma_x_product"} and [\[eq:sigma_x_sigma_z_product\]](#eq:sigma_x_sigma_z_product){reference-type="eqref" reference="eq:sigma_x_sigma_z_product"}, we clearly see that $\sigma_z \sigma_x \neq \sigma_x \sigma_z$, demonstrating the non-commutativity of Pauli matrices. ::: ::: tcolorbox $] The commutator of$\_z$and$\_x$is given by$\[\_z, \_x\] = \_z \_x - \_x \_z$: \begin{align*} [\sigma_z, \sigma_x] &= \sigma_z \sigma_x - \sigma_x \sigma_z \nonumber \\ &= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \nonumber \\ &= \begin{pmatrix} (0-0) & (1-(-1)) \\ (-1-1) & (0-0) \end{pmatrix} \nonumber \\ &= \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} = 2i \sigma_y. \end{align*} This result is a fundamental commutation relation in quantum mechanics for spin operators. Cyclic permutations of indices yield similar relations:$\[\_x, \_y\] = 2i \_z$and$\[\_y, $\sigma_z] = 2i \sigma_x$. ::: ::: tcolorbox Let's calculate the triple product $\sigma_z \sigma_x \sigma_z = (\sigma_z \sigma_x) \sigma_z$. Using the result from Equation [\[eq:sigma_z_sigma_x_product\]](#eq:sigma_z_sigma_x_product){reference-type="eqref" reference="eq:sigma_z_sigma_x_product"} for $\sigma_z \sigma_x$: $$\begin{aligned}\sigma_z \sigma_x \sigma_z &= \left(\sigma_z \sigma_x\right) \sigma_z = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \nonumber \\&= \begin{pmatrix} (0\cdot 1 + 1\cdot 0) & (0\cdot 0 + 1\cdot (-1)) \\ ((-1)\cdot 1 + 0\cdot 0) & ((-1)\cdot 0 + 0\cdot (-1)) \end{pmatrix} \nonumber \\&= \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} = - \sigma_x.\end{aligned}$$ This type of manipulation of Pauli matrices is frequently encountered in quantum mechanical calculations, particularly in spin systems and quantum information theory. ::: ::: tcolorbox Given a vector $\left| \psi \right\rangle = \frac{1}{2}\left| 0 \right\rangle + \frac{3}{2}i\left| 1 \right\rangle = \begin{pmatrix} 1/2 \\ 3i/2 \end{pmatrix}$, our objective is to construct an orthonormal basis $\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}$ such that the first basis vector $\left| v_1 \right\rangle$ is aligned with the direction of $\left| \psi \right\rangle$. This process involves normalization and orthogonalization to ensure that both basis vectors are mutually orthogonal and normalized to unity. ::: ::: tcolorbox The process of constructing orthonormal bases and changing basis representations is fundamental in quantum mechanics. It allows us to analyze quantum systems in different bases that may be more convenient or physically relevant for specific problems. ::: ::: tcolorbox In linear algebra and quantum mechanics, it is often necessary to change the basis in which vectors and operators are represented. Consider two orthonormal bases for a vector space $\mathcal{V}$: $V = \{\left| v_1 \right\rangle, \left| v_2 \right\rangle, \ldots, \left| v_n \right\rangle\}$ and $U = \{\left| u_1 \right\rangle, \left| u_2 \right\rangle, \ldots, \left| u_n \right\rangle\}$. Since both $V$ and $U$ are bases, any vector in $\mathcal{V}$, and in particular each vector in basis $V$, can be expressed as a linear combination of vectors in basis $U$, and vice versa. ::: ::: tcolorbox We can express each vector $\left| v_j \right\rangle$ of basis $V$ as a linear combination of the vectors of basis $U$: $$\left| v_j \right\rangle = \sum_{i=1}^{n} R_{ij} \left| u_i \right\rangle, \quad \text{for } j = 1, 2, \ldots, n,\label{eq:v_in_terms_of_u-box}$$ where $R_{ij}$ are the transformation coefficients. The transformation matrix $R$ from basis $V$ to basis $U$ has elements $R_{ij} = \left\langle u_i \middle| v_j \right\rangle$. $R_{ij}$ is the component of $\left| v_j \right\rangle$ along $\left| u_i \right\rangle$. ::: ::: tcolorbox The transformation matrix between two orthonormal bases is a unitary matrix, meaning it satisfies $R^\dagger R = R R^\dagger = I$, where $I$ is the identity matrix and $R^\dagger$ is the adjoint of $R$. *Description:* This theorem states that when we change from one orthonormal basis to another, the matrix that performs this transformation is unitary. Unitary matrices are crucial in quantum mechanics because they preserve probabilities and norms of vectors. ::: ::: tcolorbox Vector norms remain invariant under orthonormal basis changes. If $\left| \psi \right\rangle$ is a vector, its norm calculated in any orthonormal basis will be the same. *Description:* This theorem highlights a fundamental property of orthonormal basis transformations: they preserve the length of vectors. This is essential in physics, as physical quantities like probabilities (derived from vector norms) must be independent of the choice of basis. ::: ::: tcolorbox The concept of basis transformation is essential for understanding how physical quantities arerepresented in different coordinate systems. In quantum mechanics, choosing an appropriate basis can significantly simplify calculations and provide deeper insights into the system's properties. The unitary nature of basis transformations ensures that fundamental physical quantities like probabilities (related to squared norms) are preserved, regardless of the chosen basis. :::

Lecture Notes on Matrix Representation of Linear Operators and Basis Transformations

Introduction

Matrix Representation of Linear Operators

Linear Operators and Orthonormal Bases

Action of an Operator on Basis Vectors and Matrix Elements

Matrix Representation of a Linear Operator

Products and Non-Commutativity of Operators

Operator Product and Matrix Multiplication

Non-Commutativity of Operator Products

The Commutator

Example: Pauli Sigma Matrices

Introduction to Pauli Matrices and Spin Operators

Action of Pauli Matrices on Computational Basis States

Matrix Product Examples: \(\sigma_z \sigma_x\) and \(\sigma_x \sigma_z\)

Commutator of \(\sigma_z\) and \(\sigma_x\)

Example: Triple Product \(\sigma_z \sigma_x \sigma_z\)

Constructing Orthonormal Bases: A Practical Example

Problem Statement: Constructing an Orthonormal Basis

Normalization of \(\left| \psi \right\rangle\) to Obtain \(\left| v_1 \right\rangle\)

Finding a Vector \(\left| v_2 \right\rangle\) Orthogonal to \(\left| v_1 \right\rangle\)

Normalization of \(\left| v_2 \right\rangle\)

Verification of Orthonormality of \(\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}\)

Representing the State \(\left| + \right\rangle\) in the Basis \(\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}\)

Using Projectors for Basis Transformation

Calculating Coefficients \(\left\langle v_1 \middle| + \right\rangle\) and \(\left\langle v_2 \middle| + \right\rangle\)

Verification of Normalization of \(\left| + \right\rangle\) in the \(\{\left| v_1 \right\rangle, \left| v_2 \right\rangle\}\) Basis

Change of Basis and Transformation Matrices

General Basis Transformations Between Orthonormal Bases

Transformation Matrix from Basis \(V\) to Basis \(U\)

Relationship Between Transformation Matrices: Unitary Transformation

Invariance of Vector Norms Under Basis Transformation

Conclusion