Lecture Notes on Eigenvalues, Eigenvectors, and Operators

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March 28, 2025

Introduction

In quantum mechanics, the description of physical observables is fundamentally rooted in the concept of operators. A central postulate of quantum mechanics asserts that every observable quantity is represented by a Hermitian operator. While not every Hermitian operator necessarily corresponds to a directly observable quantity, this postulate establishes a crucial link between the mathematical formalism and physical reality.

This lecture aims to revisit and refine the concepts of eigenvalues and eigenvectors, with a particular focus on their significance for normal and Hermitian operators. These operators are not merely mathematical abstractions; they are essential tools for understanding and describing quantum systems. The eigenvalue problem, at first glance, might appear to be an abstract mathematical exercise. However, it provides an indispensable framework for analyzing operators and vectors in a way that is profoundly useful in quantum mechanics.

We will explore how the eigenvalue problem allows us to extract physically meaningful information from operators. Furthermore, we will extend our discussion to include normal and unitary operators, highlighting their unique properties and roles within the quantum mechanical framework. Finally, we will introduce the concept of the trace of an operator, an operation that becomes particularly important when we consider quantum subsystems and composite systems.

This lecture will lay the groundwork for understanding how these mathematical constructs are used to describe quantum phenomena, measurements, and the evolution of quantum states. By the end of this lecture, you will gain a deeper appreciation for the role of eigenvalues, eigenvectors, and specific classes of operators in the language of quantum mechanics.

Eigenvalues and Eigenvectors

Definition of Eigenvalues and Eigenvectors

When a linear operator acts on a vector, it generally transforms it into a different vector, altering both its magnitude and direction. However, a special class of vectors, known as eigenvectors, exhibit a simpler behavior under the action of a specific operator. When an operator acts on an eigenvector, the resulting vector is merely a scaled version of the original eigenvector, maintaining its direction. This scaling factor is termed the eigenvalue.

Definition 1 (Eigenvalue and Eigenvector). Let $A$ be a linear operator acting on a vector space $V$. A non-zero vector $\left| v \right\rangle \in V$ is called an eigenvector of $A$ if the application of $A$ to $\left| v \right\rangle$ results in a vector collinear with $\left| v \right\rangle$. Mathematically, this is expressed as: \[A\left| v \right\rangle = \lambda \left| v \right\rangle,\] where $\lambda \in \mathbb{C}$ is a complex number called the eigenvalue associated with the eigenvector $\left| v \right\rangle$. The set of all eigenvalues of an operator is referred to as its spectrum.

The equation $A\left| v \right\rangle = \lambda \left| v \right\rangle$ is the fundamental eigenvalue equation. It signifies that the operator $A$, when applied to the eigenvector $\left| v \right\rangle$, acts simply as a scalar multiplication by the eigenvalue $\lambda$. This concept is central to understanding the behavior of linear operators and has profound implications in quantum mechanics, where operators represent physical observables.

Matrix Formulation of the Eigenvalue Problem

To practically solve the eigenvalue equation, especially in finite-dimensional vector spaces, we transition to a matrix representation. Choosing a basis $\{\left| e_i \right\rangle\}$ for the vector space $V$, we can represent the operator $A$ as a matrix $[A]$ and the eigenvector $\left| v \right\rangle$ as a column vector $\mathbf{v}$. In this basis, the eigenvalue equation $A\left| v \right\rangle = \lambda \left| v \right\rangle$ is transformed into a matrix equation.

Let $A_{ij} = \left\langle e_i \middle| A e_j \right\rangle$ be the matrix elements of $A$ and $v_j = \left\langle e_j \middle| v \right\rangle$ be the components of $\left| v \right\rangle$ in the chosen basis. The eigenvalue equation in component form becomes: \[\sum_{j} A_{ij} v_j = \lambda v_i.\] Rearranging terms to bring all terms to one side, we get: \[\sum_{j} A_{ij} v_j - \lambda v_i = 0.\] To express this as a matrix equation, we introduce the identity matrix $I$, whose matrix elements are given by the Kronecker delta $\delta_{ij}$ (i.e., $\delta_{ij} = 1$ if $i=j$ and $0$ if $i \neq j$). We can rewrite $\lambda v_i$ as $\lambda \sum_{j} \delta_{ij} v_j = \sum_{j} \lambda \delta_{ij} v_j$. Substituting this back into the equation: \[\sum_{j} A_{ij} v_j - \sum_{j} \lambda \delta_{ij} v_j = 0,\] \[\sum_{j} (A_{ij} - \lambda \delta_{ij}) v_j = 0.\] In matrix notation, this system of linear equations is compactly written as: \[(A - \lambda I) \mathbf{v} = \mathbf{0},\] where $I$ is the identity matrix of the same dimension as $A$, and $\mathbf{v}$ is the column vector representing the eigenvector $\left| v \right\rangle$. This equation is a system of homogeneous linear equations for the components of $\mathbf{v}$.

The Characteristic Equation and the Eigenvalues

The equation $(A - \lambda I) \mathbf{v} = \mathbf{0}$ has a non-trivial solution (i.e., $\mathbf{v} \neq \mathbf{0}$) if and only if the matrix $(A - \lambda I)$ is singular. A matrix is singular if and only if its determinant is zero. This condition leads to the characteristic equation, which is used to find the eigenvalues.

Proposition 1 (Characteristic Equation). The eigenvalues $\lambda$ of an operator $A$ are precisely the solutions to the characteristic equation: \[\det(A - \lambda I) = 0.\]

The determinant $\det(A - \lambda I)$ is a polynomial in $\lambda$ of degree $n$, where $n$ is the dimension of the matrix $A$. This polynomial, $p(\lambda) = \det(A - \lambda I)$, is called the characteristic polynomial of $A$. Setting the characteristic polynomial to zero, $p(\lambda) = 0$, yields the characteristic equation. According to the fundamental theorem of algebra, a polynomial equation of degree $n$ has exactly $n$ roots in the complex numbers, counting multiplicities. These roots are the eigenvalues of the operator $A$. Therefore, an $n \times n$ matrix (or an operator in an $n$-dimensional vector space) has exactly $n$ eigenvalues. These eigenvalues may be real or complex, and they may not all be distinct (degeneracy).

Procedure for Finding Eigenvalues and Eigenvectors

The process of finding eigenvalues and their corresponding eigenvectors is algorithmic and can be summarized in the following steps:

Form the Characteristic Matrix: Given an operator $A$, construct the characteristic matrix $(A - \lambda I)$, where $I$ is the identity matrix and $\lambda$ is an unknown scalar. Calculate the Characteristic Polynomial: Compute the determinant of the characteristic matrix: $p(\lambda) = \det(A - \lambda I)$. This will result in a polynomial in $\lambda$ of degree $n$, where $n$ is the dimension of $A$. Solve the Characteristic Equation for Eigenvalues: Solve the characteristic equation $p(\lambda) = 0$ for $\lambda$. The roots $\lambda_1, \lambda_2, \dots, \lambda_n$ of this equation are the eigenvalues of the operator $A$. Determine Eigenvectors for Each Eigenvalue: For each eigenvalue $\lambda_i$ ($i = 1, 2, \dots, n$), substitute $\lambda_i$ back into the homogeneous system of linear equations $(A - \lambda_i I) \mathbf{v} = \mathbf{0}$. Solve for Eigenvectors: Solve the system of linear equations $(A - \lambda_i I) \mathbf{v} = \mathbf{0}$ to find the eigenvector(s) $\mathbf{v}_i$ corresponding to the eigenvalue $\lambda_i$. Since $\det(A - \lambda_i I) = 0$, this system will have at least one non-trivial solution. In case of degeneracy (multiple linearly independent eigenvectors for the same eigenvalue), find a basis for the null space of $(A - \lambda_i I)$. Normalize Eigenvectors (Optional but Recommended): Normalize each eigenvector $\mathbf{v}_i$ so that it has unit length, i.e., $\left\langle v_i \middle| v_i \right\rangle = 1$. This is often convenient, especially in quantum mechanics.

Complexity Analysis:

Step 2 (Determinant Calculation): For an $n \times n$ matrix, determinant calculation using cofactor expansion is $O(n!)$, but using Gaussian elimination is $O(n^3)$.
Step 3 (Solving Polynomial Equation): Finding roots of a polynomial of degree $n$ can be computationally intensive and depends on the desired precision and method used. For small $n$, analytical solutions may exist; for larger $n$, numerical methods are typically employed.
Step 5 (Solving Linear System): Solving a system of linear equations using Gaussian elimination is $O(n^3)$. This step is repeated for each eigenvalue.

The overall complexity is dominated by determinant and root-finding, and solving linear systems, making it computationally feasible for matrices of moderate size.

Example 1. Let’s find the eigenvalues and eigenvectors of the Pauli-Z matrix $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.

Step 1 & 2: Characteristic Equation. Form the characteristic matrix $(\sigma_z - \lambda I)$ and compute its determinant: \[\det(\sigma_z - \lambda I) = \det \left( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) = \det \begin{pmatrix} 1-\lambda & 0 \\ 0 & -1-\lambda \end{pmatrix}.\] The determinant is $(1-\lambda)(-1-\lambda) - (0)(0) = (1-\lambda)(-1-\lambda)$.

Step 3: Solve for Eigenvalues. Set the determinant to zero to find the eigenvalues: \[(1-\lambda)(-1-\lambda) = 0.\] The solutions are $\lambda_1 = 1$ and $\lambda_2 = -1$. Thus, the eigenvalues of $\sigma_z$ are $1$ and $-1$.

Step 4 & 5: Solve for Eigenvectors. For $\lambda_1 = 1$, we solve $(\sigma_z - 1 I) \mathbf{v}_1 = \mathbf{0}$: \[\left( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - 1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.\] This system of equations is: \[\begin{aligned} 0\alpha + 0\beta &= 0 \\ 0\alpha - 2\beta &= 0 \end{aligned}\] From the second equation, we get $\beta = 0$, while $\alpha$ is undetermined and can be any non-zero complex number. Choosing $\alpha = 1$, we get an eigenvector $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.

For $\lambda_2 = -1$, we solve $(\sigma_z - (-1) I) \mathbf{v}_2 = \mathbf{0}$: \[\left( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - (-1) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.\] This system of equations is: \[\begin{aligned} 2\alpha + 0\beta &= 0 \\ 0\alpha + 0\beta &= 0 \end{aligned}\] From the first equation, we get $\alpha = 0$, while $\beta$ is undetermined. Choosing $\beta = 1$, we get an eigenvector $\mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.

Step 6: Normalize Eigenvectors. The eigenvectors $\left| v_1 \right\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\left| v_2 \right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are already normalized as $\sqrt{|1|^2 + |0|^2} = 1$ and $\sqrt{|0|^2 + |1|^2} = 1$.

Thus, the eigenvalues of $\sigma_z$ are $\lambda_1 = 1$ and $\lambda_2 = -1$, with corresponding normalized eigenvectors $\left| v_1 \right\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\left| v_2 \right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.

Normalization and Phase Freedom of Eigenvectors

Eigenvectors are determined by the eigenvalue equation up to a multiplicative constant. If $\left| v \right\rangle$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $c\left| v \right\rangle$ is also an eigenvector for any non-zero complex number $c$, since $A(c\left| v \right\rangle) = c(A\left| v \right\rangle) = c(\lambda\left| v \right\rangle) = \lambda(c\left| v \right\rangle)$. This scalar multiplication freedom allows us to normalize eigenvectors.

Normalization is the process of choosing a specific representative eigenvector by scaling it such that its norm is unity. For an eigenvector $\left| v \right\rangle$, the normalized eigenvector $\left| v \right\rangle_{\text{norm}}$ is obtained by dividing $\left| v \right\rangle$ by its norm $\|\left| v \right\rangle\| = \sqrt{\left\langle v \middle| v \right\rangle}$: \[\left| v \right\rangle_{\text{norm}} = \frac{\left| v \right\rangle}{\|\left| v \right\rangle\|} = \frac{\left| v \right\rangle}{\sqrt{\left\langle v \middle| v \right\rangle}}.\] Normalized eigenvectors are particularly useful in quantum mechanics, where probabilities are calculated using inner products and wave functions are often normalized.

Even after normalization, there remains a phase freedom. If $\left| v \right\rangle_{\text{norm}}$ is a normalized eigenvector, then $e^{i\phi} \left| v \right\rangle_{\text{norm}}$ is also a normalized eigenvector for any real number $\phi$, because $|e^{i\phi}| = 1$ and the eigenvalue equation remains satisfied: $A(e^{i\phi} \left| v \right\rangle_{\text{norm}}) = e^{i\phi} (A\left| v \right\rangle_{\text{norm}}) = e^{i\phi} (\lambda \left| v \right\rangle_{\text{norm}}) = \lambda (e^{i\phi} \left| v \right\rangle_{\text{norm}})$. This phase factor $e^{i\phi}$ does not change the physical state represented by the eigenvector in quantum mechanics, as physical observables are typically independent of the global phase.

Normal Operators

Definition and Significance of Normal Operators

Normal operators constitute a particularly important class of linear operators in linear algebra and quantum mechanics. Their significance stems from their well-behaved spectral properties, which generalize those of both Hermitian and unitary operators.

Definition 2 (Normal Operator). A linear operator $A$ on a vector space is defined as normal if it commutes with its adjoint operator $A^\dagger$. Mathematically, the condition for normality is: \[A A^\dagger = A^\dagger A.\]

This commutativity condition is the defining characteristic of normal operators. It ensures that $A$ and $A^\dagger$ share a common set of eigenvectors, a property that greatly simplifies their analysis and application.

It is crucial to recognize that both Hermitian and unitary operators are special cases of normal operators.

Hermitian Operators: By definition, a Hermitian operator $H$ is self-adjoint, meaning $H^\dagger = H$. Consequently, $H H^\dagger = H H = H^\dagger H$, satisfying the normality condition. Hermitian operators are fundamental in quantum mechanics as they represent physical observables.
Unitary Operators: A unitary operator $U$ is defined by the property $U^\dagger U = U U^\dagger = I$, where $I$ is the identity operator. From this definition, it is evident that $U U^\dagger = U^\dagger U$, thus unitary operators are also normal. Unitary operators are essential for describing time evolution and transformations in quantum systems.

The class of normal operators is broader than just Hermitian and unitary operators, encompassing any operator that satisfies the commutativity condition $A A^\dagger = A^\dagger A$.

Eigenvalues and Eigenvectors of Normal Operators

Normal operators exhibit several key properties concerning their eigenvalues and eigenvectors that make them particularly tractable and important in applications.

Eigenvalues of the Adjoint Operator

A fundamental property of normal operators is the relationship between the eigenvalues of a normal operator and its adjoint.

Proposition 2 (Eigenvalues of the Adjoint of a Normal Operator). If $A$ is a normal operator, and $\left| v \right\rangle$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, i.e., $A\left| v \right\rangle = \lambda \left| v \right\rangle$, then $\left| v \right\rangle$ is also an eigenvector of the adjoint operator $A^\dagger$, but with eigenvalue $\lambda^*$, the complex conjugate of $\lambda$. That is, $A^\dagger\left| v \right\rangle = \lambda^* \left| v \right\rangle$.

Proof. Proof. To prove this, we consider the operator $B = A - \lambda I$, where $I$ is the identity operator. If $A$ is normal, then $B$ is also normal. To see this, we compute $BB^\dagger$ and $B^\dagger B$: \[\begin{aligned} BB^\dagger &= (A - \lambda I)(A - \lambda I)^\dagger = (A - \lambda I)(A^\dagger - \lambda^* I) \\ &= AA^\dagger - \lambda^* A - \lambda A^\dagger + \lambda \lambda^* I \\ B^\dagger B &= (A - \lambda I)^\dagger (A - \lambda I) = (A^\dagger - \lambda^* I)(A - \lambda I) \\ &= A^\dagger A - \lambda A^\dagger - \lambda^* A + \lambda^* \lambda I \end{aligned}\] Since $A$ is normal, $AA^\dagger = A^\dagger A$, and thus $BB^\dagger = B^\dagger B$, meaning $B$ is also normal.

Now, if $\left| v \right\rangle$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A\left| v \right\rangle = \lambda \left| v \right\rangle$, which implies $B\left| v \right\rangle = (A - \lambda I)\left| v \right\rangle = A\left| v \right\rangle - \lambda \left| v \right\rangle = \mathbf{0}$. For any vector $\left| u \right\rangle$, we have $\|B\left| u \right\rangle\|^2 = \left\langle B\left| u \right\rangle \middle| B\left| u \right\rangle \right\rangle = \left\langle u \middle| B^\dagger B \left| u \right\rangle \right\rangle$. If $B\left| v \right\rangle = \mathbf{0}$, then $\|B\left| v \right\rangle\|^2 = \left\langle v \middle| B^\dagger B \left| v \right\rangle \right\rangle = 0$. This implies $\left\langle v \middle| B^\dagger B \left| v \right\rangle \right\rangle = \|B^\dagger \left| v \right\rangle\|^2 = 0$ because $B^\dagger B$ is a positive semi-definite operator (for normal $B$, $BB^\dagger = B^\dagger B$, and both are positive semi-definite). Therefore, $\|B^\dagger \left| v \right\rangle\|^2 = 0$ implies $B^\dagger \left| v \right\rangle = \mathbf{0}$.

Substituting back $B^\dagger = (A - \lambda I)^\dagger = A^\dagger - \lambda^* I$, we have $(A^\dagger - \lambda^* I)\left| v \right\rangle = \mathbf{0}$, which gives $A^\dagger \left| v \right\rangle = \lambda^* \left| v \right\rangle$. Thus, if $\left| v \right\rangle$ is an eigenvector of a normal operator $A$ with eigenvalue $\lambda$, it is also an eigenvector of $A^\dagger$ with eigenvalue $\lambda^*$. ◻

Orthogonality of Eigenvectors

Another crucial property of normal operators is the orthogonality of their eigenvectors corresponding to distinct eigenvalues.

Theorem 3 (Orthogonality of Eigenvectors of Normal Operators). Eigenvectors of a normal operator corresponding to distinct eigenvalues are mutually orthogonal.

Proof. Proof. Let $A$ be a normal operator, and let $\left| v_1 \right\rangle$ and $\left| v_2 \right\rangle$ be eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_1$ and $\lambda_2$, respectively, so $A\left| v_1 \right\rangle = \lambda_1 \left| v_1 \right\rangle$ and $A\left| v_2 \right\rangle = \lambda_2 \left| v_2 \right\rangle$, with $\lambda_1 \neq \lambda_2$. We want to show that $\left\langle v_1 \middle| v_2 \right\rangle = 0$. Consider the inner product $\left\langle v_1 \middle| A v_2 \right\rangle$. Using the eigenvalue equation for $\left| v_2 \right\rangle$, we have: \[\left\langle v_1 \middle| A v_2 \right\rangle = \left\langle v_1 \middle| \lambda_2 v_2 \right\rangle = \lambda_2 \left\langle v_1 \middle| v_2 \right\rangle.\] Alternatively, we can use the adjoint operator $A^\dagger$ and the property of the adjoint of an operator acting on the bra vector: \[\left\langle v_1 \middle| A v_2 \right\rangle = \left\langle A^\dagger v_1 \middle| v_2 \right\rangle.\] Since $A$ is normal and $A\left| v_1 \right\rangle = \lambda_1 \left| v_1 \right\rangle$, from Proposition [proposition:Eigenvalues of Adjoint of Normal Operator], we know that $A^\dagger \left| v_1 \right\rangle = \lambda_1^* \left| v_1 \right\rangle$. Thus, \[\left\langle A^\dagger v_1 \middle| v_2 \right\rangle = \left\langle \lambda_1^* v_1 \middle| v_2 \right\rangle = \lambda_1 \left\langle v_1 \middle| v_2 \right\rangle.\] Equating the two expressions for $\left\langle v_1 \middle| A v_2 \right\rangle$, we get: \[\lambda_2 \left\langle v_1 \middle| v_2 \right\rangle = \lambda_1 \left\langle v_1 \middle| v_2 \right\rangle.\] Rearranging this equation, we have: \[(\lambda_2 - \lambda_1) \left\langle v_1 \middle| v_2 \right\rangle = 0.\] Since we assumed that the eigenvalues are distinct, $\lambda_1 \neq \lambda_2$, which implies $\lambda_2 - \lambda_1 \neq 0$. Therefore, the only way for the equation to hold is if $\left\langle v_1 \middle| v_2 \right\rangle = 0$. Hence, eigenvectors corresponding to distinct eigenvalues of a normal operator are orthogonal. ◻

Spectral Decomposition and Diagonalization

Normal operators are significant because they are diagonalizable by a unitary transformation. This is formalized in the spectral theorem for normal operators, which states that a normal operator can be expressed in terms of its eigenvalues and eigenvectors.

Theorem 4 (Spectral Decomposition of Normal Operators). A normal operator $A$ on a finite-dimensional vector space can be expressed as a spectral decomposition: \[A = \sum_{i} \lambda_i P_i,\] where $\{\lambda_i\}$ are the distinct eigenvalues of $A$, and $P_i$ is the orthogonal projector onto the eigenspace $E_i$ corresponding to the eigenvalue $\lambda_i$. The projectors $P_i$ satisfy the properties:

$P_i P_j = 0$ for $i \neq j$ (orthogonality)
$P_i^2 = P_i$ (projection)
$P_i^\dagger = P_i$ (orthogonality of projection)
$\sum_{i} P_i = I$ (completeness relation), where $I$ is the identity operator.

If the eigenspaces are one-dimensional (i.e., eigenvalues are non-degenerate), and $\{\left| v_i \right\rangle\}$ are normalized eigenvectors corresponding to distinct eigenvalues $\lambda_i$, then $P_i = \left| v_i \right\rangle\left\langle v_i \right|$. In the case of degeneracy, $P_i = \sum_{j} \left| v_{ij} \right\rangle\left\langle v_{ij} \right|$, where $\{\left| v_{ij} \right\rangle\}$ is an orthonormal basis for the eigenspace $E_i$.

The spectral decomposition is also known as the eigenvalue decomposition. It provides a canonical form for normal operators, expressing them as a linear combination of projection operators, weighted by their eigenvalues. In a basis formed by orthonormal eigenvectors of $A$, the matrix representation of $A$ becomes diagonal, with the eigenvalues on the diagonal. This diagonalization simplifies many calculations and provides a clear understanding of the action of the operator.

The completeness relation $\sum_{i} P_i = I$ is crucial. It implies that the eigenspaces of a normal operator span the entire vector space. In other words, any vector in the space can be written as a linear combination of eigenvectors of $A$.

Degeneracy and Orthonormalization of Eigenvectors

Degeneracy occurs when an eigenvalue has more than one linearly independent eigenvector. In such cases, the eigenspace corresponding to the degenerate eigenvalue is multi-dimensional. While eigenvectors corresponding to distinct eigenvalues are guaranteed to be orthogonal, eigenvectors corresponding to the same degenerate eigenvalue are not necessarily orthogonal to each other. However, within each degenerate eigenspace, we can always construct an orthonormal basis using the Gram-Schmidt orthonormalization procedure.

For each eigenvalue $\lambda_i$, we find the set of linearly independent eigenvectors that span the eigenspace $E_i$. If this set is not already orthonormal, we apply the Gram-Schmidt process to obtain an orthonormal basis $\{\left| v_{ij} \right\rangle\}$ for $E_i$. By doing this for each eigenspace, we can construct a complete orthonormal set of eigenvectors for the normal operator $A$. This set of eigenvectors forms an orthonormal basis for the entire vector space, in which the operator $A$ is represented by a diagonal matrix.

In summary, normal operators are well-behaved in terms of their eigenvalue problem. They possess orthogonal eigenvectors (for distinct eigenvalues), and they are diagonalizable. These properties make normal operators, including Hermitian and unitary operators, indispensable tools in quantum mechanics and various branches of physics and mathematics.

Hermitian Operators

Hermitian operators hold a position of paramount importance in quantum mechanics. They are not merely a mathematical construct but are fundamental to the very formulation of the theory, serving as the mathematical representatives of physical observables. As a subclass of normal operators, they inherit all the favorable properties of normal operators, and possess additional crucial characteristics that make them indispensable in quantum theory.

Definition 3 (Hermitian Operator). An operator $H$ is termed Hermitian or self-adjoint if it is equal to its own adjoint. This defining property is mathematically expressed as: \[H = H^\dagger.\]

The self-adjoint nature of Hermitian operators has profound implications, particularly concerning their eigenvalues and eigenvectors, which are directly linked to measurable physical quantities in quantum systems.

Real Eigenvalues: A Cornerstone of Physical Observables

One of the most significant properties of Hermitian operators is that their eigenvalues are guaranteed to be real numbers. This is not just a mathematical curiosity; it is a necessity for the consistency of quantum mechanics with physical reality.

Theorem 5 (Real Eigenvalues of Hermitian Operators). All eigenvalues of a Hermitian operator are real.

Proof. Proof. Let $H$ be a Hermitian operator, and suppose $\lambda$ is an eigenvalue of $H$ with corresponding eigenvector $\left| v \right\rangle \neq \mathbf{0}$. By definition, this means $H\left| v \right\rangle = \lambda \left| v \right\rangle$. To prove that $\lambda$ is real, we start with the eigenvalue equation and use the Hermitian property.

Consider the inner product of $\left| v \right\rangle$ with $H\left| v \right\rangle$: \[\left\langle v \middle| H v \right\rangle = \left\langle v \middle| \lambda v \right\rangle = \lambda \left\langle v \middle| v \right\rangle.\] Now, let’s use the definition of the adjoint operator. By definition, $\left\langle v \middle| H v \right\rangle = \left\langle H^\dagger v \middle| v \right\rangle$. Since $H$ is Hermitian, $H^\dagger = H$, so we have: \[\left\langle v \middle| H v \right\rangle = \left\langle H v \middle| v \right\rangle.\] Substituting $H\left| v \right\rangle = \lambda \left| v \right\rangle$ into the right side, we get: \[\left\langle H v \middle| v \right\rangle = \left\langle \lambda v \middle| v \right\rangle = \lambda^* \left\langle v \middle| v \right\rangle,\] where we have used the property that when a scalar is pulled out of the first position in an inner product, it becomes its complex conjugate.

Equating the two expressions for $\left\langle v \middle| H v \right\rangle$, we have: \[\lambda \left\langle v \middle| v \right\rangle = \lambda^* \left\langle v \middle| v \right\rangle.\] Since $\left| v \right\rangle$ is an eigenvector, it is non-zero, so $\left\langle v \middle| v \right\rangle = \|\left| v \right\rangle\|^2 > 0$. We can divide both sides by $\left\langle v \middle| v \right\rangle$: \[\lambda = \lambda^*.\] A complex number $\lambda$ is equal to its complex conjugate $\lambda^*$ if and only if its imaginary part is zero, meaning $\lambda$ must be a real number. Therefore, the eigenvalues of a Hermitian operator are always real. ◻

Physical Observables and Measurement Outcomes

In quantum mechanics, the postulate that physical observables are represented by Hermitian operators is of paramount importance. This postulate directly connects the mathematical formalism to the physical world. The reason for this connection lies in the real nature of the eigenvalues of Hermitian operators.

Physical measurements must yield real numbers. Quantities like energy, momentum, position, and spin are all measurable and have real values. If the operators representing these observables had complex eigenvalues, it would imply that measurements could result in complex values, which is not physically meaningful for these quantities.

The eigenvalues of a Hermitian operator represent the possible outcomes of measuring the physical observable associated with that operator. When a measurement of an observable $H$ is performed on a quantum system in a state, the result will always be one of the eigenvalues $\{\lambda_i\}$ of $H$. For instance, if we measure the spin of an electron along the z-axis, the operator corresponding to this observable is related to the Pauli matrix $\sigma_z$, which is Hermitian. The eigenvalues of $\sigma_z$ are $+1$ and $-1$, representing the possible outcomes "spin up" and "spin down" along the z-axis, respectively. These outcomes are real, physically measurable values.

Real eigenvalues of Hermitian operators are not just a mathematical property; they are a fundamental requirement for the consistency of quantum mechanics with experimental observations. They ensure that the predicted outcomes of measurements are real, physically sensible values, aligning with the nature of physical observables. This is a cornerstone of the Copenhagen interpretation and the practical application of quantum mechanics.

In summary, the association of physical observables with Hermitian operators and the reality of their eigenvalues is a crucial link between the abstract mathematical framework of quantum mechanics and the concrete, measurable world of physical phenomena.

Unitary Operators

Definition and Role of Unitary Operators

Unitary operators are another class of operators of paramount importance in quantum mechanics, particularly in the context of transformations and evolution of quantum states. They are characterized by their norm-preserving nature, which ensures that probabilities in quantum mechanics are conserved under their action.

This definition has several important implications:

Invertibility: Unitary operators are always invertible, and their inverse is simply their adjoint, $U^{-1} = U^\dagger$.
Norm Preservation: Unitary operators preserve the norm (length) of vectors. For any vector $\left| v \right\rangle$, the norm of the transformed vector $U\left| v \right\rangle$ is the same as the norm of $\left| v \right\rangle$: \[\|U\left| v \right\rangle\| = \sqrt{\left\langle Uv \middle| Uv \right\rangle} = \sqrt{\left\langle v \middle| U^\dagger U v \right\rangle} = \sqrt{\left\langle v \middle| I v \right\rangle} = \sqrt{\left\langle v \middle| v \right\rangle} = \|\left| v \right\rangle\|.\] This norm preservation is crucial in quantum mechanics because the norm squared of a quantum state vector represents probability. Unitary evolution ensures that the total probability remains conserved over time.
Inner Product Preservation: Unitary operators also preserve the inner product between vectors. For any two vectors $\left| u \right\rangle$ and $\left| v \right\rangle$, \[\left\langle Uu \middle| Uv \right\rangle = \left\langle u \middle| U^\dagger U v \right\rangle = \left\langle u \middle| I v \right\rangle = \left\langle u \middle| v \right\rangle.\] This property implies that if a set of vectors is orthonormal, applying a unitary operator to all of them will result in another orthonormal set.

Unitary operators play a fundamental role in quantum mechanics in several contexts, including:

Eigenvalues of Unitary Operators: Phase Factors and Geometric Interpretation

The eigenvalues of unitary operators possess a distinctive property: they are always complex numbers with an absolute value of 1. This means they lie on the unit circle in the complex plane and are often referred to as phase factors.

Theorem 6 (Eigenvalues of Unitary Operators are Phase Factors). The eigenvalues of a unitary operator $U$ are complex numbers $\lambda$ such that $|\lambda| = 1$. Therefore, they can be expressed in the form $\lambda = e^{i\phi}$ for some real number $\phi$.

Proof. Proof. Let $U$ be a unitary operator, and let $\lambda$ be an eigenvalue of $U$ with a corresponding non-zero eigenvector $\left| v \right\rangle$, so $U\left| v \right\rangle = \lambda \left| v \right\rangle$. To find the constraint on $\lambda$, we consider the norm of the vector $U\left| v \right\rangle$.

Equating the two expressions for $\|U\left| v \right\rangle\|$, we have: \[\|\left| v \right\rangle\| = |\lambda| \|\left| v \right\rangle\|.\] Since $\left| v \right\rangle$ is an eigenvector, it is non-zero, so $\|\left| v \right\rangle\| \neq 0$. We can divide both sides by $\|\left| v \right\rangle\|$ to obtain: \[1 = |\lambda|.\] Thus, the absolute value of any eigenvalue $\lambda$ of a unitary operator $U$ must be 1. Complex numbers with absolute value 1 can be written in exponential form as $\lambda = e^{i\phi} = \cos(\phi) + i\sin(\phi)$, where $\phi$ is a real number representing the phase angle.

Geometrically, the eigenvalues of a unitary operator lie on the unit circle in the complex plane. Multiplying a vector by a phase factor $e^{i\phi}$ corresponds to rotating the vector in the complex plane by an angle $\phi$ without changing its magnitude. In the context of quantum mechanics, these phase factors are crucial. For instance, in quantum interference phenomena, the relative phases between different quantum amplitudes determine the interference pattern.

In summary, unitary operators are transformations that preserve probability and inner products, and their eigenvalues are phase factors, reflecting rotations in the complex plane. These properties are essential for describing the dynamics and symmetries of quantum systems.

◻

Trace of an Operator

Definition and Calculation of the Trace

The trace is a fundamental operation in linear algebra that maps a linear operator to a scalar. It provides valuable information about the operator and is particularly useful in quantum mechanics for various purposes, including the description of mixed states and subsystems.

Definition 4 (Trace of an Operator). The trace of a linear operator $A$ acting on a finite-dimensional vector space $V$ is defined as the sum of the diagonal elements of its matrix representation in any orthonormal basis $\{\left| e_i \right\rangle\}$ of $V$. Mathematically, the trace of $A$, denoted as $\text{Tr}(A)$, is given by: \[\text{Tr}(A) = \sum_{i} \left\langle e_i \middle| A e_i \right\rangle.\]

In matrix form, if the operator $A$ is represented by a matrix $[A]$ in an orthonormal basis, where $[A]_{ij} = \left\langle e_i \middle| A e_j \right\rangle$, then the trace is the sum of the diagonal entries of the matrix: \[\text{Tr}(A) = \sum_{i} [A]_{ii} = \sum_{i} A_{ii}.\] This definition appears to depend on the choice of basis $\{\left| e_i \right\rangle\}$. However, a crucial property of the trace is its basis independence, which we will discuss shortly.

Basis Independence of the Trace

A remarkable and essential property of the trace is that it is invariant under a change of orthonormal basis. This means that the value of the trace of an operator is the same regardless of which orthonormal basis is used to compute it. This basis independence makes the trace an intrinsic property of the operator itself, rather than an artifact of a particular representation.

Proposition 7 (Basis Independence of the Trace). The trace of a linear operator is independent of the choice of orthonormal basis used to compute it.

The proof of basis independence involves showing that if we change from one orthonormal basis $\{\left| e_i \right\rangle\}$ to another orthonormal basis $\{\left| f_j \right\rangle\}$, the calculated trace remains the same. This can be formally shown using the properties of unitary transformations that relate different orthonormal bases. (The detailed proof will be discussed in a subsequent lecture.)

The implication of basis independence is profound. It means that the trace provides a coordinate-free characterization of the operator. We can choose any orthonormal basis that simplifies our calculations, and the trace will always yield the same value, reflecting a fundamental property of the operator itself.

Trace and Eigenvalues

There is a direct and important relationship between the trace of an operator and its eigenvalues. The trace of an operator is equal to the sum of its eigenvalues, counted with their algebraic multiplicities.

Proposition 8 (Trace as Sum of Eigenvalues). The trace of a linear operator $A$ is equal to the sum of its eigenvalues $\{\lambda_i\}$: \[\text{Tr}(A) = \sum_{i} \lambda_i.\]

To understand this, consider the spectral decomposition of a normal operator $A = \sum_i \lambda_i P_i$, where $\lambda_i$ are the eigenvalues and $P_i$ are the projectors onto the corresponding eigenspaces. In an eigenbasis of $A$, the matrix representation of $A$ is diagonal, with the eigenvalues along the diagonal. The sum of these diagonal elements is precisely the sum of the eigenvalues. Since the trace is basis-independent, this relationship holds true in any basis.

This property provides a useful way to compute the sum of eigenvalues without explicitly solving for each eigenvalue individually, and conversely, it connects the abstract concept of eigenvalues to the more computationally accessible trace operation.

Importance of Trace in Quantum Mechanics: Mixed States and Subsystems

In quantum mechanics, the trace operation is indispensable, particularly when dealing with statistical ensembles, mixed states, and subsystems of composite quantum systems.

Density Matrices and Mixed States

In quantum mechanics, the state of a system is not always described by a single pure state vector. Often, we deal with statistical mixtures of states, known as mixed states. Mixed states are described by density operators (or density matrices). The trace operation plays a crucial role in defining and working with density operators. For instance, the probability of obtaining a particular outcome in a measurement on a system described by a density operator is calculated using the trace formula.

Partial Trace and Subsystems

When considering composite quantum systems, which are systems composed of multiple subsystems (e.g., multiple qubits), we often need to describe the state of a subsystem independently of the rest of the system. The partial trace is a generalization of the trace operation that allows us to "trace out" the degrees of freedom of one subsystem to obtain the reduced density operator describing the state of the remaining subsystem.

Imagine a bipartite quantum system composed of subsystems A and B. If the composite system is described by a density operator $\rho_{AB}$, and we are interested only in the state of subsystem A, we can compute the reduced density operator for subsystem A, $\rho_A$, by taking the partial trace over subsystem B: \[\rho_A = \text{Tr}_B(\rho_{AB}).\] The partial trace $\text{Tr}_B$ is a linear map that acts on operators on the composite Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$ and produces operators on $\mathcal{H}_A$. It effectively averages over the degrees of freedom of subsystem B, providing a description of subsystem A alone, irrespective of the state of subsystem B.

The partial trace is essential for:

The trace and partial trace are thus powerful tools for simplifying the description of quantum systems, especially when dealing with complexity arising from mixed states, composite systems, and interactions with environments.

The trace operation is not merely a mathematical tool but a central concept in quantum mechanics, enabling the description of mixed states, the extraction of physical expectation values, and the analysis of subsystems within larger quantum systems. Its basis independence and connection to eigenvalues further solidify its fundamental nature in linear algebra and quantum theory.

Conclusion

This lecture has provided a comprehensive overview of eigenvalues and eigenvectors, and explored their essential properties within the context of normal, Hermitian, and unitary operators. These mathematical constructs are not merely abstract tools but are fundamental to the formulation and interpretation of quantum mechanics.

Key Takeaways:

Eigenvalues and Eigenvectors as Fundamental Tools: Eigenvalues and eigenvectors are indispensable for analyzing linear operators, providing a basis in which the action of an operator is simplified to scalar multiplication. They reveal intrinsic properties of operators and are crucial for understanding their behavior.
Normal Operators and Spectral Decomposition: Normal operators, encompassing Hermitian and unitary operators, are characterized by commuting with their adjoints. This normality ensures the existence of orthogonal eigenvectors and allows for spectral decomposition, expressing operators in terms of their eigenvalues and projectors onto eigenspaces.
Hermitian Operators and Real Observables: Hermitian operators are distinguished by having real eigenvalues. This property is paramount in quantum mechanics, as Hermitian operators represent physical observables, and their real eigenvalues correspond to the possible real-valued outcomes of physical measurements.
Unitary Operators and Phase Factor Eigenvalues: Unitary operators, defined by preserving norms and inner products, have eigenvalues that are phase factors (complex numbers with magnitude 1). They are essential for describing time evolution and transformations in quantum systems, ensuring probability conservation.
Trace: Basis-Independent Sum of Eigenvalues: The trace of an operator, defined as the sum of its diagonal elements in any orthonormal basis, is a basis-independent quantity equal to the sum of its eigenvalues. It is a powerful tool, particularly in quantum mechanics, for dealing with mixed states, expectation values, and subsystems through the partial trace.

Remark. Remark 1. Important Remarks:

Postulate of Hermitian Observables: The fundamental postulate that physical observables are represented by Hermitian operators is a cornerstone of quantum mechanics. This ensures that measurement outcomes are real and that the theory aligns with physical reality.
Eigenvalues as Measurement Outcomes: The eigenvalues of a Hermitian operator are not just abstract numbers; they are the physically possible results of measuring the corresponding observable. This direct link between mathematical eigenvalues and experimental outcomes is a defining feature of quantum theory.
Unitary Evolution and Transformations: Unitary operators are indispensable for describing the dynamics of isolated quantum systems and for representing symmetry transformations. Their norm-preserving nature guarantees the conservation of probability, a fundamental requirement for any physically consistent quantum evolution.
Trace for Subsystems and Complexity Reduction: The trace operation, especially the partial trace, is a powerful tool for managing complexity in quantum systems. It allows us to focus on subsystems, describe mixed states, and analyze entanglement, making complex quantum systems more tractable.

In the upcoming lecture, we will further explore the properties of the trace, including a formal proof of its basis independence and delve into its applications in quantum mechanics. We will specifically examine the role of the trace in the context of density matrices, quantum statistical mechanics, and the crucial concept of the partial trace for describing entangled and composite quantum systems.

As you prepare for the next lecture, consider the implications of the basis independence of the trace. How does this property simplify calculations and provide a more fundamental, basis-invariant characterization of quantum operators and states? Reflect on how the trace, as a seemingly simple operation, becomes an indispensable tool in the quantum mechanical toolbox.

Arrivederci.

--- title: "Lecture Notes on Eigenvalues, Eigenvectors, and Operators" author: "Your Name" date: "2025-03-28" format: html: toc: true # Table of Contents toc-depth: 2 code-tools: true theme: cosmo # Or "journal" for Distill-like minimalism --- # Introduction In quantum mechanics, the description of physical observables is fundamentally rooted in the concept of operators. A central postulate of quantum mechanics asserts that every observable quantity is represented by a Hermitian operator. While not every Hermitian operator necessarily corresponds to a directly observable quantity, this postulate establishes a crucial link between the mathematical formalism and physical reality. This lecture aims to revisit and refine the concepts of eigenvalues and eigenvectors, with a particular focus on their significance for normal and Hermitian operators. These operators are not merely mathematical abstractions; they are essential tools for understanding and describing quantum systems. The eigenvalue problem, at first glance, might appear to be an abstract mathematical exercise. However, it provides an indispensable framework for analyzing operators and vectors in a way that is profoundly useful in quantum mechanics. We will explore how the eigenvalue problem allows us to extract physically meaningful information from operators. Furthermore, we will extend our discussion to include normal and unitary operators, highlighting their unique properties and roles within the quantum mechanical framework. Finally, we will introduce the concept of the trace of an operator, an operation that becomes particularly important when we consider quantum subsystems and composite systems. This lecture will lay the groundwork for understanding how these mathematical constructs are used to describe quantum phenomena, measurements, and the evolution of quantum states. By the end of this lecture, you will gain a deeper appreciation for the role of eigenvalues, eigenvectors, and specific classes of operators in the language of quantum mechanics. # Eigenvalues and Eigenvectors ## Definition of Eigenvalues and Eigenvectors When a linear operator acts on a vector, it generally transforms it into a different vector, altering both its magnitude and direction. However, a special class of vectors, known as eigenvectors, exhibit a simpler behavior under the action of a specific operator. When an operator acts on an eigenvector, the resulting vector is merely a scaled version of the original eigenvector, maintaining its direction. This scaling factor is termed the eigenvalue. :::: tcolorbox ::: definition **Definition 1** (Eigenvalue and Eigenvector). *Let $A$ be a linear operator acting on a vector space $V$. A non-zero vector $\left| v \right\rangle \in V$ is called an **eigenvector** of $A$ if the application of $A$ to $\left| v \right\rangle$ results in a vector collinear with $\left| v \right\rangle$. Mathematically, this is expressed as: $$A\left| v \right\rangle = \lambda \left| v \right\rangle,$$ where $\lambda \in \mathbb{C}$ is a complex number called the **eigenvalue** associated with the eigenvector $\left| v \right\rangle$. The set of all eigenvalues of an operator is referred to as its **spectrum**.* ::: :::: The equation $A\left| v \right\rangle = \lambda \left| v \right\rangle$ is the fundamental **eigenvalue equation**. It signifies that the operator $A$, when applied to the eigenvector $\left| v \right\rangle$, acts simply as a scalar multiplication by the eigenvalue $\lambda$. This concept is central to understanding the behavior of linear operators and has profound implications in quantum mechanics, where operators represent physical observables. ## Matrix Formulation of the Eigenvalue Problem To practically solve the eigenvalue equation, especially in finite-dimensional vector spaces, we transition to a matrix representation. Choosing a basis $\{\left| e_i \right\rangle\}$ for the vector space $V$, we can represent the operator $A$ as a matrix $[A]$ and the eigenvector $\left| v \right\rangle$ as a column vector $\mathbf{v}$. In this basis, the eigenvalue equation $A\left| v \right\rangle = \lambda \left| v \right\rangle$ is transformed into a matrix equation. Let $A_{ij} = \left\langle e_i \middle| A e_j \right\rangle$ be the matrix elements of $A$ and $v_j = \left\langle e_j \middle| v \right\rangle$ be the components of $\left| v \right\rangle$ in the chosen basis. The eigenvalue equation in component form becomes: $$\sum_{j} A_{ij} v_j = \lambda v_i.$$ Rearranging terms to bring all terms to one side, we get: $$\sum_{j} A_{ij} v_j - \lambda v_i = 0.$$ To express this as a matrix equation, we introduce the identity matrix $I$, whose matrix elements are given by the Kronecker delta $\delta_{ij}$ (i.e., $\delta_{ij} = 1$ if $i=j$ and $0$ if $i \neq j$). We can rewrite $\lambda v_i$ as $\lambda \sum_{j} \delta_{ij} v_j = \sum_{j} \lambda \delta_{ij} v_j$. Substituting this back into the equation: $$\sum_{j} A_{ij} v_j - \sum_{j} \lambda \delta_{ij} v_j = 0,$$ $$\sum_{j} (A_{ij} - \lambda \delta_{ij}) v_j = 0.$$ In matrix notation, this system of linear equations is compactly written as: $$(A - \lambda I) \mathbf{v} = \mathbf{0},$$ where $I$ is the identity matrix of the same dimension as $A$, and $\mathbf{v}$ is the column vector representing the eigenvector $\left| v \right\rangle$. This equation is a system of homogeneous linear equations for the components of $\mathbf{v}$. ## The Characteristic Equation and the Eigenvalues The equation $(A - \lambda I) \mathbf{v} = \mathbf{0}$ has a non-trivial solution (i.e., $\mathbf{v} \neq \mathbf{0}$) if and only if the matrix $(A - \lambda I)$ is singular. A matrix is singular if and only if its determinant is zero. This condition leads to the characteristic equation, which is used to find the eigenvalues. :::: tcolorbox ::: proposition **Proposition 1** (Characteristic Equation). *The eigenvalues $\lambda$ of an operator $A$ are precisely the solutions to the **characteristic equation**: $$\det(A - \lambda I) = 0.$$* ::: :::: The determinant $\det(A - \lambda I)$ is a polynomial in $\lambda$ of degree $n$, where $n$ is the dimension of the matrix $A$. This polynomial, $p(\lambda) = \det(A - \lambda I)$, is called the **characteristic polynomial** of $A$. Setting the characteristic polynomial to zero, $p(\lambda) = 0$, yields the characteristic equation. According to the fundamental theorem of algebra, a polynomial equation of degree $n$ has exactly $n$ roots in the complex numbers, counting multiplicities. These roots are the eigenvalues of the operator $A$. Therefore, an $n \times n$ matrix (or an operator in an $n$-dimensional vector space) has exactly $n$ eigenvalues. These eigenvalues may be real or complex, and they may not all be distinct (degeneracy). ## Procedure for Finding Eigenvalues and Eigenvectors The process of finding eigenvalues and their corresponding eigenvectors is algorithmic and can be summarized in the following steps: ::::: tcolorbox :::: algorithm ::: algorithmic **Form the Characteristic Matrix:** Given an operator $A$, construct the characteristic matrix $(A - \lambda I)$, where $I$ is the identity matrix and $\lambda$ is an unknown scalar. **Calculate the Characteristic Polynomial:** Compute the determinant of the characteristic matrix: $p(\lambda) = \det(A - \lambda I)$. This will result in a polynomial in $\lambda$ of degree $n$, where $n$ is the dimension of $A$. **Solve the Characteristic Equation for Eigenvalues:** Solve the characteristic equation $p(\lambda) = 0$ for $\lambda$. The roots $\lambda_1, \lambda_2, \dots, \lambda_n$ of this equation are the eigenvalues of the operator $A$. **Determine Eigenvectors for Each Eigenvalue:** For each eigenvalue $\lambda_i$ ($i = 1, 2, \dots, n$), substitute $\lambda_i$ back into the homogeneous system of linear equations $(A - \lambda_i I) \mathbf{v} = \mathbf{0}$. **Solve for Eigenvectors:** Solve the system of linear equations $(A - \lambda_i I) \mathbf{v} = \mathbf{0}$ to find the eigenvector(s) $\mathbf{v}_i$ corresponding to the eigenvalue $\lambda_i$. Since $\det(A - \lambda_i I) = 0$, this system will have at least one non-trivial solution. In case of degeneracy (multiple linearly independent eigenvectors for the same eigenvalue), find a basis for the null space of $(A - \lambda_i I)$. **Normalize Eigenvectors (Optional but Recommended):** Normalize each eigenvector $\mathbf{v}_i$ so that it has unit length, i.e., $\left\langle v_i \middle| v_i \right\rangle = 1$. This is often convenient, especially in quantum mechanics. ::: **Complexity Analysis:** - Step 2 (Determinant Calculation): For an $n \times n$ matrix, determinant calculation using cofactor expansion is $O(n!)$, but using Gaussian elimination is $O(n^3)$. - Step 3 (Solving Polynomial Equation): Finding roots of a polynomial of degree $n$ can be computationally intensive and depends on the desired precision and method used. For small $n$, analytical solutions may exist; for larger $n$, numerical methods are typically employed. - Step 5 (Solving Linear System): Solving a system of linear equations using Gaussian elimination is $O(n^3)$. This step is repeated for each eigenvalue. The overall complexity is dominated by determinant and root-finding, and solving linear systems, making it computationally feasible for matrices of moderate size. :::: ::::: :::: tcolorbox ::: example **Example 1**. *Let's find the eigenvalues and eigenvectors of the Pauli-Z matrix $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.* ***Step 1 & 2: Characteristic Equation.** Form the characteristic matrix $(\sigma_z - \lambda I)$ and compute its determinant: $$\det(\sigma_z - \lambda I) = \det \left( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) = \det \begin{pmatrix} 1-\lambda & 0 \\ 0 & -1-\lambda \end{pmatrix}.$$ The determinant is $(1-\lambda)(-1-\lambda) - (0)(0) = (1-\lambda)(-1-\lambda)$.* ***Step 3: Solve for Eigenvalues.** Set the determinant to zero to find the eigenvalues: $$(1-\lambda)(-1-\lambda) = 0.$$ The solutions are $\lambda_1 = 1$ and $\lambda_2 = -1$. Thus, the eigenvalues of $\sigma_z$ are $1$ and $-1$.* ***Step 4 & 5: Solve for Eigenvectors.** For $\lambda_1 = 1$, we solve $(\sigma_z - 1 I) \mathbf{v}_1 = \mathbf{0}$: $$\left( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - 1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.$$ This system of equations is: $$\begin{aligned} 0\alpha + 0\beta &= 0 \\ 0\alpha - 2\beta &= 0 \end{aligned}$$ From the second equation, we get $\beta = 0$, while $\alpha$ is undetermined and can be any non-zero complex number. Choosing $\alpha = 1$, we get an eigenvector $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.* *For $\lambda_2 = -1$, we solve $(\sigma_z - (-1) I) \mathbf{v}_2 = \mathbf{0}$: $$\left( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - (-1) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.$$ This system of equations is: $$\begin{aligned} 2\alpha + 0\beta &= 0 \\ 0\alpha + 0\beta &= 0 \end{aligned}$$ From the first equation, we get $\alpha = 0$, while $\beta$ is undetermined. Choosing $\beta = 1$, we get an eigenvector $\mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.* ***Step 6: Normalize Eigenvectors.** The eigenvectors $\left| v_1 \right\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\left| v_2 \right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are already normalized as $\sqrt{|1|^2 + |0|^2} = 1$ and $\sqrt{|0|^2 + |1|^2} = 1$.* *Thus, the eigenvalues of $\sigma_z$ are $\lambda_1 = 1$ and $\lambda_2 = -1$, with corresponding normalized eigenvectors $\left| v_1 \right\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\left| v_2 \right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.* ::: :::: ## Normalization and Phase Freedom of Eigenvectors Eigenvectors are determined by the eigenvalue equation up to a multiplicative constant. If $\left| v \right\rangle$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $c\left| v \right\rangle$ is also an eigenvector for any non-zero complex number $c$, since $A(c\left| v \right\rangle) = c(A\left| v \right\rangle) = c(\lambda\left| v \right\rangle) = \lambda(c\left| v \right\rangle)$. This scalar multiplication freedom allows us to normalize eigenvectors. **Normalization** is the process of choosing a specific representative eigenvector by scaling it such that its norm is unity. For an eigenvector $\left| v \right\rangle$, the normalized eigenvector $\left| v \right\rangle_{\text{norm}}$ is obtained by dividing $\left| v \right\rangle$ by its norm $\|\left| v \right\rangle\| = \sqrt{\left\langle v \middle| v \right\rangle}$: $$\left| v \right\rangle_{\text{norm}} = \frac{\left| v \right\rangle}{\|\left| v \right\rangle\|} = \frac{\left| v \right\rangle}{\sqrt{\left\langle v \middle| v \right\rangle}}.$$ Normalized eigenvectors are particularly useful in quantum mechanics, where probabilities are calculated using inner products and wave functions are often normalized. Even after normalization, there remains a **phase freedom**. If $\left| v \right\rangle_{\text{norm}}$ is a normalized eigenvector, then $e^{i\phi} \left| v \right\rangle_{\text{norm}}$ is also a normalized eigenvector for any real number $\phi$, because $|e^{i\phi}| = 1$ and the eigenvalue equation remains satisfied: $A(e^{i\phi} \left| v \right\rangle_{\text{norm}}) = e^{i\phi} (A\left| v \right\rangle_{\text{norm}}) = e^{i\phi} (\lambda \left| v \right\rangle_{\text{norm}}) = \lambda (e^{i\phi} \left| v \right\rangle_{\text{norm}})$. This phase factor $e^{i\phi}$ does not change the physical state represented by the eigenvector in quantum mechanics, as physical observables are typically independent of the global phase. # Normal Operators ## Definition and Significance of Normal Operators Normal operators constitute a particularly important class of linear operators in linear algebra and quantum mechanics. Their significance stems from their well-behaved spectral properties, which generalize those of both Hermitian and unitary operators. :::: tcolorbox ::: definition **Definition 2** (Normal Operator). *A linear operator $A$ on a vector space is defined as **normal** if it commutes with its adjoint operator $A^\dagger$. Mathematically, the condition for normality is: $$A A^\dagger = A^\dagger A.$$* ::: :::: This commutativity condition is the defining characteristic of normal operators. It ensures that $A$ and $A^\dagger$ share a common set of eigenvectors, a property that greatly simplifies their analysis and application. It is crucial to recognize that both Hermitian and unitary operators are special cases of normal operators. - **Hermitian Operators:** By definition, a Hermitian operator $H$ is self-adjoint, meaning $H^\dagger = H$. Consequently, $H H^\dagger = H H = H^\dagger H$, satisfying the normality condition. Hermitian operators are fundamental in quantum mechanics as they represent physical observables. - **Unitary Operators:** A unitary operator $U$ is defined by the property $U^\dagger U = U U^\dagger = I$, where $I$ is the identity operator. From this definition, it is evident that $U U^\dagger = U^\dagger U$, thus unitary operators are also normal. Unitary operators are essential for describing time evolution and transformations in quantum systems. The class of normal operators is broader than just Hermitian and unitary operators, encompassing any operator that satisfies the commutativity condition $A A^\dagger = A^\dagger A$. ## Eigenvalues and Eigenvectors of Normal Operators Normal operators exhibit several key properties concerning their eigenvalues and eigenvectors that make them particularly tractable and important in applications. ### Eigenvalues of the Adjoint Operator A fundamental property of normal operators is the relationship between the eigenvalues of a normal operator and its adjoint. :::: tcolorbox ::: proposition **Proposition 2** (Eigenvalues of the Adjoint of a Normal Operator). *If $A$ is a normal operator, and $\left| v \right\rangle$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, i.e., $A\left| v \right\rangle = \lambda \left| v \right\rangle$, then $\left| v \right\rangle$ is also an eigenvector of the adjoint operator $A^\dagger$, but with eigenvalue $\lambda^*$, the complex conjugate of $\lambda$. That is, $A^\dagger\left| v \right\rangle = \lambda^* \left| v \right\rangle$.* ::: :::: ::: proof *Proof.* To prove this, we consider the operator $B = A - \lambda I$, where $I$ is the identity operator. If $A$ is normal, then $B$ is also normal. To see this, we compute $BB^\dagger$ and $B^\dagger B$: $$\begin{aligned} BB^\dagger &= (A - \lambda I)(A - \lambda I)^\dagger = (A - \lambda I)(A^\dagger - \lambda^* I) \\ &= AA^\dagger - \lambda^* A - \lambda A^\dagger + \lambda \lambda^* I \\ B^\dagger B &= (A - \lambda I)^\dagger (A - \lambda I) = (A^\dagger - \lambda^* I)(A - \lambda I) \\ &= A^\dagger A - \lambda A^\dagger - \lambda^* A + \lambda^* \lambda I \end{aligned}$$ Since $A$ is normal, $AA^\dagger = A^\dagger A$, and thus $BB^\dagger = B^\dagger B$, meaning $B$ is also normal. Now, if $\left| v \right\rangle$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A\left| v \right\rangle = \lambda \left| v \right\rangle$, which implies $B\left| v \right\rangle = (A - \lambda I)\left| v \right\rangle = A\left| v \right\rangle - \lambda \left| v \right\rangle = \mathbf{0}$. For any vector $\left| u \right\rangle$, we have $\|B\left| u \right\rangle\|^2 = \left\langle B\left| u \right\rangle \middle| B\left| u \right\rangle \right\rangle = \left\langle u \middle| B^\dagger B \left| u \right\rangle \right\rangle$. If $B\left| v \right\rangle = \mathbf{0}$, then $\|B\left| v \right\rangle\|^2 = \left\langle v \middle| B^\dagger B \left| v \right\rangle \right\rangle = 0$. This implies $\left\langle v \middle| B^\dagger B \left| v \right\rangle \right\rangle = \|B^\dagger \left| v \right\rangle\|^2 = 0$ because $B^\dagger B$ is a positive semi-definite operator (for normal $B$, $BB^\dagger = B^\dagger B$, and both are positive semi-definite). Therefore, $\|B^\dagger \left| v \right\rangle\|^2 = 0$ implies $B^\dagger \left| v \right\rangle = \mathbf{0}$. Substituting back $B^\dagger = (A - \lambda I)^\dagger = A^\dagger - \lambda^* I$, we have $(A^\dagger - \lambda^* I)\left| v \right\rangle = \mathbf{0}$, which gives $A^\dagger \left| v \right\rangle = \lambda^* \left| v \right\rangle$. Thus, if $\left| v \right\rangle$ is an eigenvector of a normal operator $A$ with eigenvalue $\lambda$, it is also an eigenvector of $A^\dagger$ with eigenvalue $\lambda^*$. ◻ ::: ### Orthogonality of Eigenvectors Another crucial property of normal operators is the orthogonality of their eigenvectors corresponding to distinct eigenvalues. :::: tcolorbox ::: theorem **Theorem 3** (Orthogonality of Eigenvectors of Normal Operators). *Eigenvectors of a normal operator corresponding to distinct eigenvalues are mutually orthogonal.* ::: :::: ::: proof *Proof.* Let $A$ be a normal operator, and let $\left| v_1 \right\rangle$ and $\left| v_2 \right\rangle$ be eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_1$ and $\lambda_2$, respectively, so $A\left| v_1 \right\rangle = \lambda_1 \left| v_1 \right\rangle$ and $A\left| v_2 \right\rangle = \lambda_2 \left| v_2 \right\rangle$, with $\lambda_1 \neq \lambda_2$. We want to show that $\left\langle v_1 \middle| v_2 \right\rangle = 0$. Consider the inner product $\left\langle v_1 \middle| A v_2 \right\rangle$. Using the eigenvalue equation for $\left| v_2 \right\rangle$, we have: $$\left\langle v_1 \middle| A v_2 \right\rangle = \left\langle v_1 \middle| \lambda_2 v_2 \right\rangle = \lambda_2 \left\langle v_1 \middle| v_2 \right\rangle.$$ Alternatively, we can use the adjoint operator $A^\dagger$ and the property of the adjoint of an operator acting on the bra vector: $$\left\langle v_1 \middle| A v_2 \right\rangle = \left\langle A^\dagger v_1 \middle| v_2 \right\rangle.$$ Since $A$ is normal and $A\left| v_1 \right\rangle = \lambda_1 \left| v_1 \right\rangle$, from Proposition [\[proposition:Eigenvalues of Adjoint of Normal Operator\]](#proposition:Eigenvalues of Adjoint of Normal Operator){reference-type="ref" reference="proposition:Eigenvalues of Adjoint of Normal Operator"}, we know that $A^\dagger \left| v_1 \right\rangle = \lambda_1^* \left| v_1 \right\rangle$. Thus, $$\left\langle A^\dagger v_1 \middle| v_2 \right\rangle = \left\langle \lambda_1^* v_1 \middle| v_2 \right\rangle = \lambda_1 \left\langle v_1 \middle| v_2 \right\rangle.$$ Equating the two expressions for $\left\langle v_1 \middle| A v_2 \right\rangle$, we get: $$\lambda_2 \left\langle v_1 \middle| v_2 \right\rangle = \lambda_1 \left\langle v_1 \middle| v_2 \right\rangle.$$ Rearranging this equation, we have: $$(\lambda_2 - \lambda_1) \left\langle v_1 \middle| v_2 \right\rangle = 0.$$ Since we assumed that the eigenvalues are distinct, $\lambda_1 \neq \lambda_2$, which implies $\lambda_2 - \lambda_1 \neq 0$. Therefore, the only way for the equation to hold is if $\left\langle v_1 \middle| v_2 \right\rangle = 0$. Hence, eigenvectors corresponding to distinct eigenvalues of a normal operator are orthogonal. ◻ ::: ### Spectral Decomposition and Diagonalization Normal operators are significant because they are diagonalizable by a unitary transformation. This is formalized in the spectral theorem for normal operators, which states that a normal operator can be expressed in terms of its eigenvalues and eigenvectors. :::: tcolorbox ::: theorem **Theorem 4** (Spectral Decomposition of Normal Operators). *A normal operator $A$ on a finite-dimensional vector space can be expressed as a spectral decomposition: $$A = \sum_{i} \lambda_i P_i,$$ where $\{\lambda_i\}$ are the distinct eigenvalues of $A$, and $P_i$ is the orthogonal projector onto the eigenspace $E_i$ corresponding to the eigenvalue $\lambda_i$. The projectors $P_i$ satisfy the properties:* 1. *$P_i P_j = 0$ for $i \neq j$ (orthogonality)* 2. *$P_i^2 = P_i$ (projection)* 3. *$P_i^\dagger = P_i$ (orthogonality of projection)* 4. *$\sum_{i} P_i = I$ (completeness relation), where $I$ is the identity operator.* *If the eigenspaces are one-dimensional (i.e., eigenvalues are non-degenerate), and $\{\left| v_i \right\rangle\}$ are normalized eigenvectors corresponding to distinct eigenvalues $\lambda_i$, then $P_i = \left| v_i \right\rangle\left\langle v_i \right|$. In the case of degeneracy, $P_i = \sum_{j} \left| v_{ij} \right\rangle\left\langle v_{ij} \right|$, where $\{\left| v_{ij} \right\rangle\}$ is an orthonormal basis for the eigenspace $E_i$.* ::: :::: The spectral decomposition is also known as the eigenvalue decomposition. It provides a canonical form for normal operators, expressing them as a linear combination of projection operators, weighted by their eigenvalues. In a basis formed by orthonormal eigenvectors of $A$, the matrix representation of $A$ becomes diagonal, with the eigenvalues on the diagonal. This diagonalization simplifies many calculations and provides a clear understanding of the action of the operator. The completeness relation $\sum_{i} P_i = I$ is crucial. It implies that the eigenspaces of a normal operator span the entire vector space. In other words, any vector in the space can be written as a linear combination of eigenvectors of $A$. ### Degeneracy and Orthonormalization of Eigenvectors **Degeneracy** occurs when an eigenvalue has more than one linearly independent eigenvector. In such cases, the eigenspace corresponding to the degenerate eigenvalue is multi-dimensional. While eigenvectors corresponding to distinct eigenvalues are guaranteed to be orthogonal, eigenvectors corresponding to the same degenerate eigenvalue are not necessarily orthogonal to each other. However, within each degenerate eigenspace, we can always construct an orthonormal basis using the Gram-Schmidt orthonormalization procedure. For each eigenvalue $\lambda_i$, we find the set of linearly independent eigenvectors that span the eigenspace $E_i$. If this set is not already orthonormal, we apply the Gram-Schmidt process to obtain an orthonormal basis $\{\left| v_{ij} \right\rangle\}$ for $E_i$. By doing this for each eigenspace, we can construct a complete orthonormal set of eigenvectors for the normal operator $A$. This set of eigenvectors forms an orthonormal basis for the entire vector space, in which the operator $A$ is represented by a diagonal matrix. In summary, normal operators are well-behaved in terms of their eigenvalue problem. They possess orthogonal eigenvectors (for distinct eigenvalues), and they are diagonalizable. These properties make normal operators, including Hermitian and unitary operators, indispensable tools in quantum mechanics and various branches of physics and mathematics. # Hermitian Operators Hermitian operators hold a position of paramount importance in quantum mechanics. They are not merely a mathematical construct but are fundamental to the very formulation of the theory, serving as the mathematical representatives of physical observables. As a subclass of normal operators, they inherit all the favorable properties of normal operators, and possess additional crucial characteristics that make them indispensable in quantum theory. :::: tcolorbox ::: definition **Definition 3** (Hermitian Operator). *An operator $H$ is termed **Hermitian** or **self-adjoint** if it is equal to its own adjoint. This defining property is mathematically expressed as: $$H = H^\dagger.$$* ::: :::: The self-adjoint nature of Hermitian operators has profound implications, particularly concerning their eigenvalues and eigenvectors, which are directly linked to measurable physical quantities in quantum systems. ## Real Eigenvalues: A Cornerstone of Physical Observables One of the most significant properties of Hermitian operators is that their eigenvalues are guaranteed to be real numbers. This is not just a mathematical curiosity; it is a necessity for the consistency of quantum mechanics with physical reality. :::: tcolorbox ::: theorem **Theorem 5** (Real Eigenvalues of Hermitian Operators). *All eigenvalues of a Hermitian operator are real.* ::: :::: ::: proof *Proof.* Let $H$ be a Hermitian operator, and suppose $\lambda$ is an eigenvalue of $H$ with corresponding eigenvector $\left| v \right\rangle \neq \mathbf{0}$. By definition, this means $H\left| v \right\rangle = \lambda \left| v \right\rangle$. To prove that $\lambda$ is real, we start with the eigenvalue equation and use the Hermitian property. Consider the inner product of $\left| v \right\rangle$ with $H\left| v \right\rangle$: $$\left\langle v \middle| H v \right\rangle = \left\langle v \middle| \lambda v \right\rangle = \lambda \left\langle v \middle| v \right\rangle.$$ Now, let's use the definition of the adjoint operator. By definition, $\left\langle v \middle| H v \right\rangle = \left\langle H^\dagger v \middle| v \right\rangle$. Since $H$ is Hermitian, $H^\dagger = H$, so we have: $$\left\langle v \middle| H v \right\rangle = \left\langle H v \middle| v \right\rangle.$$ Substituting $H\left| v \right\rangle = \lambda \left| v \right\rangle$ into the right side, we get: $$\left\langle H v \middle| v \right\rangle = \left\langle \lambda v \middle| v \right\rangle = \lambda^* \left\langle v \middle| v \right\rangle,$$ where we have used the property that when a scalar is pulled out of the first position in an inner product, it becomes its complex conjugate. Equating the two expressions for $\left\langle v \middle| H v \right\rangle$, we have: $$\lambda \left\langle v \middle| v \right\rangle = \lambda^* \left\langle v \middle| v \right\rangle.$$ Since $\left| v \right\rangle$ is an eigenvector, it is non-zero, so $\left\langle v \middle| v \right\rangle = \|\left| v \right\rangle\|^2 > 0$. We can divide both sides by $\left\langle v \middle| v \right\rangle$: $$\lambda = \lambda^*.$$ A complex number $\lambda$ is equal to its complex conjugate $\lambda^*$ if and only if its imaginary part is zero, meaning $\lambda$ must be a real number. Therefore, the eigenvalues of a Hermitian operator are always real. ◻ ::: ## Physical Observables and Measurement Outcomes In quantum mechanics, the postulate that physical observables are represented by Hermitian operators is of paramount importance. This postulate directly connects the mathematical formalism to the physical world. The reason for this connection lies in the real nature of the eigenvalues of Hermitian operators. Physical measurements must yield real numbers. Quantities like energy, momentum, position, and spin are all measurable and have real values. If the operators representing these observables had complex eigenvalues, it would imply that measurements could result in complex values, which is not physically meaningful for these quantities. The eigenvalues of a Hermitian operator represent the possible outcomes of measuring the physical observable associated with that operator. When a measurement of an observable $H$ is performed on a quantum system in a state, the result will always be one of the eigenvalues $\{\lambda_i\}$ of $H$. For instance, if we measure the spin of an electron along the z-axis, the operator corresponding to this observable is related to the Pauli matrix $\sigma_z$, which is Hermitian. The eigenvalues of $\sigma_z$ are $+1$ and $-1$, representing the possible outcomes \"spin up\" and \"spin down\" along the z-axis, respectively. These outcomes are real, physically measurable values. ::: tcolorbox **Real eigenvalues of Hermitian operators are not just a mathematical property; they are a fundamental requirement for the consistency of quantum mechanics with experimental observations.** They ensure that the predicted outcomes of measurements are real, physically sensible values, aligning with the nature of physical observables. This is a cornerstone of the Copenhagen interpretation and the practical application of quantum mechanics. ::: In summary, the association of physical observables with Hermitian operators and the reality of their eigenvalues is a crucial link between the abstract mathematical framework of quantum mechanics and the concrete, measurable world of physical phenomena. # Unitary Operators ## Definition and Role of Unitary Operators Unitary operators are another class of operators of paramount importance in quantum mechanics, particularly in the context of transformations and evolution of quantum states. They are characterized by their norm-preserving nature, which ensures that probabilities in quantum mechanics are conserved under their action. This definition has several important implications: - **Invertibility:** Unitary operators are always invertible, and their inverse is simply their adjoint, $U^{-1} = U^\dagger$. - **Norm Preservation:** Unitary operators preserve the norm (length) of vectors. For any vector $\left| v \right\rangle$, the norm of the transformed vector $U\left| v \right\rangle$ is the same as the norm of $\left| v \right\rangle$: $$\|U\left| v \right\rangle\| = \sqrt{\left\langle Uv \middle| Uv \right\rangle} = \sqrt{\left\langle v \middle| U^\dagger U v \right\rangle} = \sqrt{\left\langle v \middle| I v \right\rangle} = \sqrt{\left\langle v \middle| v \right\rangle} = \|\left| v \right\rangle\|.$$ This norm preservation is crucial in quantum mechanics because the norm squared of a quantum state vector represents probability. Unitary evolution ensures that the total probability remains conserved over time. - **Inner Product Preservation:** Unitary operators also preserve the inner product between vectors. For any two vectors $\left| u \right\rangle$ and $\left| v \right\rangle$, $$\left\langle Uu \middle| Uv \right\rangle = \left\langle u \middle| U^\dagger U v \right\rangle = \left\langle u \middle| I v \right\rangle = \left\langle u \middle| v \right\rangle.$$ This property implies that if a set of vectors is orthonormal, applying a unitary operator to all of them will result in another orthonormal set. Unitary operators play a fundamental role in quantum mechanics in several contexts, including: ## Eigenvalues of Unitary Operators: Phase Factors and Geometric Interpretation The eigenvalues of unitary operators possess a distinctive property: they are always complex numbers with an absolute value of 1. This means they lie on the unit circle in the complex plane and are often referred to as phase factors. :::: tcolorbox ::: theorem **Theorem 6** (Eigenvalues of Unitary Operators are Phase Factors). *The eigenvalues of a unitary operator $U$ are complex numbers $\lambda$ such that $|\lambda| = 1$. Therefore, they can be expressed in the form $\lambda = e^{i\phi}$ for some real number $\phi$.* ::: :::: :::: proof *Proof.* Let $U$ be a unitary operator, and let $\lambda$ be an eigenvalue of $U$ with a corresponding non-zero eigenvector $\left| v \right\rangle$, so $U\left| v \right\rangle = \lambda \left| v \right\rangle$. To find the constraint on $\lambda$, we consider the norm of the vector $U\left| v \right\rangle$. Since $U$ is unitary, it preserves the norm of vectors. Thus, the norm of $U\left| v \right\rangle$ must be equal to the norm of $\left| v \right\rangle$: $$\|U\left| v \right\rangle\| = \|\left| v \right\rangle\|.$$ On the other hand, from the eigenvalue equation $U\left| v \right\rangle = \lambda \left| v \right\rangle$, we can also express the norm of $U\left| v \right\rangle$ in terms of $\lambda$ and $\left| v \right\rangle$: $$\|U\left| v \right\rangle\| = \|\lambda \left| v \right\rangle\| = |\lambda| \|\left| v \right\rangle\|,$$ where $|\lambda|$ is the absolute value (or modulus) of the complex number $\lambda$. Equating the two expressions for $\|U\left| v \right\rangle\|$, we have: $$\|\left| v \right\rangle\| = |\lambda| \|\left| v \right\rangle\|.$$ Since $\left| v \right\rangle$ is an eigenvector, it is non-zero, so $\|\left| v \right\rangle\| \neq 0$. We can divide both sides by $\|\left| v \right\rangle\|$ to obtain: $$1 = |\lambda|.$$ Thus, the absolute value of any eigenvalue $\lambda$ of a unitary operator $U$ must be 1. Complex numbers with absolute value 1 can be written in exponential form as $\lambda = e^{i\phi} = \cos(\phi) + i\sin(\phi)$, where $\phi$ is a real number representing the phase angle. Geometrically, the eigenvalues of a unitary operator lie on the unit circle in the complex plane. Multiplying a vector by a phase factor $e^{i\phi}$ corresponds to rotating the vector in the complex plane by an angle $\phi$ without changing its magnitude. In the context of quantum mechanics, these phase factors are crucial. For instance, in quantum interference phenomena, the relative phases between different quantum amplitudes determine the interference pattern. ::: mdframed **In summary, unitary operators are transformations that preserve probability and inner products, and their eigenvalues are phase factors, reflecting rotations in the complex plane. These properties are essential for describing the dynamics and symmetries of quantum systems.** ::: ◻ :::: # Trace of an Operator ## Definition and Calculation of the Trace The trace is a fundamental operation in linear algebra that maps a linear operator to a scalar. It provides valuable information about the operator and is particularly useful in quantum mechanics for various purposes, including the description of mixed states and subsystems. :::: tcolorbox ::: definition **Definition 4** (Trace of an Operator). *The **trace** of a linear operator $A$ acting on a finite-dimensional vector space $V$ is defined as the sum of the diagonal elements of its matrix representation in any orthonormal basis $\{\left| e_i \right\rangle\}$ of $V$. Mathematically, the trace of $A$, denoted as $\text{Tr}(A)$, is given by: $$\text{Tr}(A) = \sum_{i} \left\langle e_i \middle| A e_i \right\rangle.$$* ::: :::: In matrix form, if the operator $A$ is represented by a matrix $[A]$ in an orthonormal basis, where $[A]_{ij} = \left\langle e_i \middle| A e_j \right\rangle$, then the trace is the sum of the diagonal entries of the matrix: $$\text{Tr}(A) = \sum_{i} [A]_{ii} = \sum_{i} A_{ii}.$$ This definition appears to depend on the choice of basis $\{\left| e_i \right\rangle\}$. However, a crucial property of the trace is its basis independence, which we will discuss shortly. ## Basis Independence of the Trace A remarkable and essential property of the trace is that it is invariant under a change of orthonormal basis. This means that the value of the trace of an operator is the same regardless of which orthonormal basis is used to compute it. This basis independence makes the trace an intrinsic property of the operator itself, rather than an artifact of a particular representation. :::: tcolorbox ::: proposition **Proposition 7** (Basis Independence of the Trace). *The trace of a linear operator is independent of the choice of orthonormal basis used to compute it.* ::: :::: The proof of basis independence involves showing that if we change from one orthonormal basis $\{\left| e_i \right\rangle\}$ to another orthonormal basis $\{\left| f_j \right\rangle\}$, the calculated trace remains the same. This can be formally shown using the properties of unitary transformations that relate different orthonormal bases. (The detailed proof will be discussed in a subsequent lecture.) The implication of basis independence is profound. It means that the trace provides a coordinate-free characterization of the operator. We can choose any orthonormal basis that simplifies our calculations, and the trace will always yield the same value, reflecting a fundamental property of the operator itself. ## Trace and Eigenvalues There is a direct and important relationship between the trace of an operator and its eigenvalues. The trace of an operator is equal to the sum of its eigenvalues, counted with their algebraic multiplicities. :::: tcolorbox ::: proposition **Proposition 8** (Trace as Sum of Eigenvalues). *The trace of a linear operator $A$ is equal to the sum of its eigenvalues $\{\lambda_i\}$: $$\text{Tr}(A) = \sum_{i} \lambda_i.$$* ::: :::: To understand this, consider the spectral decomposition of a normal operator $A = \sum_i \lambda_i P_i$, where $\lambda_i$ are the eigenvalues and $P_i$ are the projectors onto the corresponding eigenspaces. In an eigenbasis of $A$, the matrix representation of $A$ is diagonal, with the eigenvalues along the diagonal. The sum of these diagonal elements is precisely the sum of the eigenvalues. Since the trace is basis-independent, this relationship holds true in any basis. This property provides a useful way to compute the sum of eigenvalues without explicitly solving for each eigenvalue individually, and conversely, it connects the abstract concept of eigenvalues to the more computationally accessible trace operation. ## Importance of Trace in Quantum Mechanics: Mixed States and Subsystems In quantum mechanics, the trace operation is indispensable, particularly when dealing with statistical ensembles, mixed states, and subsystems of composite quantum systems. ### Density Matrices and Mixed States In quantum mechanics, the state of a system is not always described by a single pure state vector. Often, we deal with statistical mixtures of states, known as mixed states. Mixed states are described by density operators (or density matrices). The trace operation plays a crucial role in defining and working with density operators. For instance, the probability of obtaining a particular outcome in a measurement on a system described by a density operator is calculated using the trace formula. ### Partial Trace and Subsystems When considering composite quantum systems, which are systems composed of multiple subsystems (e.g., multiple qubits), we often need to describe the state of a subsystem independently of the rest of the system. The **partial trace** is a generalization of the trace operation that allows us to \"trace out\" the degrees of freedom of one subsystem to obtain the reduced density operator describing the state of the remaining subsystem. Imagine a bipartite quantum system composed of subsystems A and B. If the composite system is described by a density operator $\rho_{AB}$, and we are interested only in the state of subsystem A, we can compute the reduced density operator for subsystem A, $\rho_A$, by taking the partial trace over subsystem B: $$\rho_A = \text{Tr}_B(\rho_{AB}).$$ The partial trace $\text{Tr}_B$ is a linear map that acts on operators on the composite Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$ and produces operators on $\mathcal{H}_A$. It effectively averages over the degrees of freedom of subsystem B, providing a description of subsystem A alone, irrespective of the state of subsystem B. The partial trace is essential for: The trace and partial trace are thus powerful tools for simplifying the description of quantum systems, especially when dealing with complexity arising from mixed states, composite systems, and interactions with environments. ::: tcolorbox **The trace operation is not merely a mathematical tool but a central concept in quantum mechanics, enabling the description of mixed states, the extraction of physical expectation values, and the analysis of subsystems within larger quantum systems. Its basis independence and connection to eigenvalues further solidify its fundamental nature in linear algebra and quantum theory.** ::: # Conclusion This lecture has provided a comprehensive overview of eigenvalues and eigenvectors, and explored their essential properties within the context of normal, Hermitian, and unitary operators. These mathematical constructs are not merely abstract tools but are fundamental to the formulation and interpretation of quantum mechanics. **Key Takeaways:** - **Eigenvalues and Eigenvectors as Fundamental Tools:** Eigenvalues and eigenvectors are indispensable for analyzing linear operators, providing a basis in which the action of an operator is simplified to scalar multiplication. They reveal intrinsic properties of operators and are crucial for understanding their behavior. - **Normal Operators and Spectral Decomposition:** Normal operators, encompassing Hermitian and unitary operators, are characterized by commuting with their adjoints. This normality ensures the existence of orthogonal eigenvectors and allows for spectral decomposition, expressing operators in terms of their eigenvalues and projectors onto eigenspaces. - **Hermitian Operators and Real Observables:** Hermitian operators are distinguished by having real eigenvalues. This property is paramount in quantum mechanics, as Hermitian operators represent physical observables, and their real eigenvalues correspond to the possible real-valued outcomes of physical measurements. - **Unitary Operators and Phase Factor Eigenvalues:** Unitary operators, defined by preserving norms and inner products, have eigenvalues that are phase factors (complex numbers with magnitude 1). They are essential for describing time evolution and transformations in quantum systems, ensuring probability conservation. - **Trace: Basis-Independent Sum of Eigenvalues:** The trace of an operator, defined as the sum of its diagonal elements in any orthonormal basis, is a basis-independent quantity equal to the sum of its eigenvalues. It is a powerful tool, particularly in quantum mechanics, for dealing with mixed states, expectation values, and subsystems through the partial trace. :::: tcolorbox ::: remark **Remark 1**. ***Important Remarks:*** - ***Postulate of Hermitian Observables:** The fundamental postulate that physical observables are represented by Hermitian operators is a cornerstone of quantum mechanics. This ensures that measurement outcomes are real and that the theory aligns with physical reality.* - ***Eigenvalues as Measurement Outcomes:** The eigenvalues of a Hermitian operator are not just abstract numbers; they are the physically possible results of measuring the corresponding observable. This direct link between mathematical eigenvalues and experimental outcomes is a defining feature of quantum theory.* - ***Unitary Evolution and Transformations:** Unitary operators are indispensable for describing the dynamics of isolated quantum systems and for representing symmetry transformations. Their norm-preserving nature guarantees the conservation of probability, a fundamental requirement for any physically consistent quantum evolution.* - ***Trace for Subsystems and Complexity Reduction:** The trace operation, especially the partial trace, is a powerful tool for managing complexity in quantum systems. It allows us to focus on subsystems, describe mixed states, and analyze entanglement, making complex quantum systems more tractable.* ::: :::: In the upcoming lecture, we will further explore the properties of the trace, including a formal proof of its basis independence and delve into its applications in quantum mechanics. We will specifically examine the role of the trace in the context of density matrices, quantum statistical mechanics, and the crucial concept of the partial trace for describing entangled and composite quantum systems. As you prepare for the next lecture, consider the implications of the basis independence of the trace. How does this property simplify calculations and provide a more fundamental, basis-invariant characterization of quantum operators and states? Reflect on how the trace, as a seemingly simple operation, becomes an indispensable tool in the quantum mechanical toolbox. Arrivederci.