Lecture Notes on Advanced Quantum Mechanics: Operators and Decompositions

Introduction

This lecture covers several key concepts in linear algebra and operator theory relevant to quantum mechanics. We begin by exploring the trace operator, its properties, and its basis independence. Following this, we discuss functions of operators, focusing on their definition via eigenbasis expansion and illustrating with the example of the exponential of the swap operator (\(\sigma_x\)). We then introduce the tensor product, a crucial tool for describing composite quantum systems, and examine its properties and matrix representations. Finally, we delve into matrix decompositions, including diagonalization, polar decomposition, and singular value decomposition (SVD), highlighting their importance in various applications, including quantum information processing. The lecture aims to provide a solid foundation in these mathematical tools, which are essential for advanced topics in quantum mechanics.

Trace Operator

Definition and Basis Independence

The trace operator is a fundamental operation in linear algebra, particularly useful in contexts where information about a subsystem is unknown or inaccessible.

Definition 1 (Trace Operator). The trace of an operator \(A\), denoted as \(\text{Tr}(A)\), is defined as the sum of the diagonal elements of the matrix representation of \(A\) in a chosen basis. If \(\{ \left|{j}\right\rangle \}\) is a basis, then \[\text{Tr}(A) = \sum_{j} \left\langle{j}\right| A \left|{j}\right\rangle.\] Let \(A_{ij} = \left\langle{i}\middle|{A\left|{j}\right\rangle}\right\rangle\) be the matrix elements of the operator \(A\) in a given basis. Then, the trace is given by \[\text{Tr}(A) = \sum_{i} A_{ii}.\]

For the trace to be a well-defined property of the operator itself, it must be independent of the choice of basis. We will now demonstrate this basis independence.

Consider an operator \(A\) expressed in its eigenbasis \(\{ \left|{d_i}\right\rangle \}\) as: \[A = \sum_{i} \lambda_i \left|{d_i}\middle\rangle\middle\langle{d_i}\right|,\] where \(\lambda_i\) are the eigenvalues and \(\left|{d_i}\right\rangle\) are the eigenvectors of \(A\).

In this eigenbasis, the trace of \(A\) is: \[\begin{aligned}\text{Tr}(A) &= \sum_{j} \left\langle{d_j}\right| A \left|{d_j}\right\rangle \\&= \sum_{j} \left\langle{d_j}\right| \left( \sum_{i} \lambda_i \left|{d_i}\middle\rangle\middle\langle{d_i}\right| \right) \left|{d_j}\right\rangle \\&= \sum_{j} \sum_{i} \lambda_i \left\langle{d_j}\middle|{d_i}\right\rangle \left\langle{d_i}\middle|{d_j}\right\rangle \\&= \sum_{j} \sum_{i} \lambda_i \delta_{ji} \left\langle{d_i}\middle|{d_j}\right\rangle \quad \text{since } \{ \left|{d_i}\right\rangle \} \text{ is an orthonormal basis} \\&= \sum_{i} \lambda_i \left\langle{d_i}\middle|{d_i}\right\rangle \\&= \sum_{i} \lambda_i.\end{aligned}\] Thus, in the eigenbasis, the trace of \(A\) is the sum of its eigenvalues.

Now, let’s consider a change of basis to a new orthonormal basis \(\{ \left|{w_j}\right\rangle \}\). Let \(A_v\) and \(A_w\) be the matrix representations of \(A\) in bases \(\{ \left|{v_i}\right\rangle \}\) and \(\{ \left|{w_j}\right\rangle \}\) respectively. There exists a change of basis matrix \(U\) such that \(A_w = U A_v U^{-1}\). The trace of a matrix is invariant under cyclic permutations, so \[\text{Tr}(A_w) = \text{Tr}(U A_v U^{-1}) = \text{Tr}(U^{-1} U A_v) = \text{Tr}(I A_v) = \text{Tr}(A_v).\] Thus, the trace is independent of the basis.

Trace and Eigenvalues

As shown earlier, in the eigenbasis representation, the trace of an operator \(A\) is indeed the sum of its eigenvalues: \[\text{Tr}(A) = \sum_{i} \lambda_i.\] This provides a basis-independent way to calculate the sum of eigenvalues of an operator.

Cyclic Property

The trace operator possesses a crucial cyclic property, which is highly useful in various manipulations.

Proposition 1 (Cyclic Property of Trace). For operators \(A\), \(B\), and \(C\), the trace is cyclic under permutations of the operators within the trace: \[\text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB).\] More generally, for a product of \(n\) operators \(A_1, A_2, \dots, A_n\): \[\text{Tr}(A_1 A_2 \dots A_n) = \text{Tr}(A_2 \dots A_n A_1) = \dots = \text{Tr}(A_n A_1 \dots A_{n-1}).\]

Proof. Proof. Let’s consider the matrix representation of the operators in some basis. The matrix element of the product \(ABC\) is given by: \[(ABC)_{ii} = \sum_{k} \sum_{l} A_{ik} B_{kl} C_{li}.\] The trace of \(ABC\) is then: \[\text{Tr}(ABC) = \sum_{i} (ABC)_{ii} = \sum_{i} \sum_{k} \sum_{l} A_{ik} B_{kl} C_{li}.\] Now consider \(\text{Tr}(CAB)\): \[\text{Tr}(CAB) = \sum_{i} (CAB)_{ii} = \sum_{i} \sum_{k} \sum_{l} C_{ik} A_{kl} B_{li}.\] By renaming the indices (\(i \rightarrow l, k \rightarrow i, l \rightarrow k\)), we get: \[\text{Tr}(CAB) = \sum_{l} \sum_{i} \sum_{k} C_{li} A_{ik} B_{kl} = \sum_{i} \sum_{k} \sum_{l} A_{ik} B_{kl} C_{li} = \text{Tr}(ABC).\] Similarly, we can show \(\text{Tr}(BCA) = \text{Tr}(ABC)\). This property extends to products of more than three operators by repeated application of this result. ◻

This cyclic property is a powerful tool when manipulating expressions involving traces of operator products.

Functions of Operators

Definition and Eigenbasis Expansion

Functions of operators are defined in a manner consistent with functions of numbers, particularly through power series expansions.

Definition 2 (Function of an Operator). If a function \(f(x)\) has a power series expansion \(f(x) = \sum_{i=0}^{\infty} a_i x^i\), then for an operator \(A\), the function of the operator \(f(A)\) is defined by substituting the operator into the power series: \[f(A) = \sum_{i=0}^{\infty} a_i A^i = a_0 I + a_1 A + a_2 A^2 + \dots\] where \(I\) is the identity operator.

If the operator \(A\) is normal (or Hermitian), it has an eigenbasis expansion \(A = \sum_{i} \lambda_i \left|{d_i}\middle\rangle\middle\langle{d_i}\right|\). Applying the function \(f(A)\) to this expansion yields: \[\begin{aligned}f(A) &= f\left( \sum_{j} \lambda_j \left|{d_j}\middle\rangle\middle\langle{d_j}\right| \right) \\&= \sum_{i=0}^{\infty} a_i \left( \sum_{j} \lambda_j \left|{d_j}\middle\rangle\middle\langle{d_j}\right| \right)^i \\&= \sum_{i=0}^{\infty} a_i \left( \sum_{j} \lambda_j^i \left|{d_j}\middle\rangle\middle\langle{d_j}\right| \right) \quad \text{since } (\left|{d_j}\middle\rangle\middle\langle{d_j}\right|)^n = \left|{d_j}\middle\rangle\middle\langle{d_j}\right| \\&= \sum_{j} \left( \sum_{i=0}^{\infty} a_i \lambda_j^i \right) \left|{d_j}\middle\rangle\middle\langle{d_j}\right| \\&= \sum_{j} f(\lambda_j) \left|{d_j}\middle\rangle\middle\langle{d_j}\right|.\end{aligned}\] Thus, the function of an operator \(A\) can be computed by applying the function to its eigenvalues and keeping the same eigenvectors.

Example: Exponential of Swap Operator

Let’s consider the swap operator, often denoted as \(\sigma_x\) (Pauli-X operator), in the computational basis \(\{\left|{0}\right\rangle, \left|{1}\right\rangle\}\). Its matrix representation is: \[\sigma_x= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\] We want to compute \(e^{i \sigma_x}\).

Eigenvalues and Eigenvectors of \(\sigma_x\)

To find the eigenvalues, we solve the characteristic equation \(\det(\sigma_x- \lambda I) = 0\): \[\det \begin{pmatrix} -\lambda & 1 \\ 1 & -\lambda \end{pmatrix} = \lambda^2 - 1 = 0.\] The eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = -1\).

For \(\lambda_1 = 1\): \[(\sigma_x- I) \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.\] This gives \(-\alpha + \beta = 0\), so \(\alpha = \beta\). Normalizing the eigenvector \(\left|{v_1}\right\rangle = \begin{pmatrix} \alpha \\ \alpha \end{pmatrix}\), we get \(2|\alpha|^2 = 1\), so \(\alpha = \frac{1}{\sqrt{2}}\). Thus, \(\left|{v_1}\right\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} (\left|{0}\right\rangle + \left|{1}\right\rangle) = \left|{+}\right\rangle\).

For \(\lambda_2 = -1\): \[(\sigma_x+ I) \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.\] This gives \(\alpha + \beta = 0\), so \(\beta = -\alpha\). Normalizing \(\left|{v_2}\right\rangle = \begin{pmatrix} \alpha \\ -\alpha \end{pmatrix}\), we get \(2|\alpha|^2 = 1\), so \(\alpha = \frac{1}{\sqrt{2}}\). Thus, \(\left|{v_2}\right\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}} (\left|{0}\right\rangle - \left|{1}\right\rangle) = \left|{-}\right\rangle\).

Eigenvalues are \(\lambda_1 = 1, \lambda_2 = -1\) and corresponding eigenvectors are \(\left|{+}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle + \left|{1}\right\rangle), \left|{-}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle - \left|{1}\right\rangle)\).

Method 1: Power Series Expansion

We use the power series expansion of the exponential function \(e^{ix} = \sum_{n=0}^{\infty} \frac{(ix)^n}{n!}\). \[e^{i \sigma_x} = \sum_{n=0}^{\infty} \frac{(i \sigma_x)^n}{n!} = \sum_{n=0}^{\infty} \frac{i^n (\sigma_x)^n}{n!}.\] Since \(\sigma_x^2 = I\), we have \(\sigma_x^n = I\) if \(n\) is even, and \(\sigma_x^n = \sigma_x\) if \(n\) is odd. \[\begin{aligned}e^{i \sigma_x} &= \sum_{k=0}^{\infty} \frac{i^{2k} (\sigma_x)^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{i^{2k+1} (\sigma_x)^{2k+1}}{(2k+1)!} \\&= \sum_{k=0}^{\infty} \frac{(-1)^{k} I}{(2k)!} + \sum_{k=0}^{\infty} \frac{i (-1)^{k} \sigma_x}{(2k+1)!} \\&= I \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k)!} + i \sigma_x\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)!} \\&= I \cos(1) + i \sigma_x\sin(1) \\&= \cos(1) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + i \sin(1) \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\&= \begin{pmatrix} \cos(1) & i \sin(1) \\ i \sin(1) & \cos(1) \end{pmatrix}.\end{aligned}\]

Method 2: Eigenbasis Decomposition

Using the eigenbasis decomposition formula \(f(A) = \sum_{j} f(\lambda_j) \left|{v_j}\middle\rangle\middle\langle{v_j}\right|\), we have: \[e^{i \sigma_x} = e^{i \lambda_1} \left|{v_1}\middle\rangle\middle\langle{v_1}\right| + e^{i \lambda_2} \left|{v_2}\middle\rangle\middle\langle{v_2}\right| = e^{i} \left|{+}\middle\rangle\middle\langle{+}\right| + e^{-i} \left|{-}\middle\rangle\middle\langle{-}\right|.\] We need to express \(\left|{+}\middle\rangle\middle\langle{+}\right|\) and \(\left|{-}\middle\rangle\middle\langle{-}\right|\) in the computational basis. \[\begin{aligned}\left|{+}\middle\rangle\middle\langle{+}\right| &= \frac{1}{\sqrt{2}} (\left|{0}\right\rangle + \left|{1}\right\rangle) \frac{1}{\sqrt{2}} (\left\langle{0}\right| + \left\langle{1}\right|) = \frac{1}{2} (\left|{0}\middle\rangle\middle\langle{0}\right| + \left|{0}\middle\rangle\middle\langle{1}\right| + \left|{1}\middle\rangle\middle\langle{0}\right| + \left|{1}\middle\rangle\middle\langle{1}\right|) = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \\\left|{-}\middle\rangle\middle\langle{-}\right| &= \frac{1}{\sqrt{2}} (\left|{0}\right\rangle - \left|{1}\right\rangle) \frac{1}{\sqrt{2}} (\left\langle{0}\right| - \left\langle{1}\right|) = \frac{1}{2} (\left|{0}\middle\rangle\middle\langle{0}\right| - \left|{0}\middle\rangle\middle\langle{1}\right| - \left|{1}\middle\rangle\middle\langle{0}\right| + \left|{1}\middle\rangle\middle\langle{1}\right|) = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}.\end{aligned}\] Therefore, \[\begin{aligned}e^{i \sigma_x} &= e^{i} \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} + e^{-i} \frac{1}{2} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \\&= \frac{1}{2} \begin{pmatrix} e^i + e^{-i} & e^i - e^{-i} \\ e^i - e^{-i} & e^i + e^{-i} \end{pmatrix} \\&= \begin{pmatrix} \frac{e^i + e^{-i}}{2} & \frac{e^i - e^{-i}}{2} \\ \frac{e^i - e^{-i}}{2} & \frac{e^i + e^{-i}}{2} \end{pmatrix} \\&= \begin{pmatrix} \cos(1) & i \sin(1) \\ i \sin(1) & \cos(1) \end{pmatrix}.\end{aligned}\] Both methods yield the same result.

Tensor Product

Definition and Properties

The tensor product is a way to combine vector spaces and operators to describe composite systems. If we have two vector spaces \(U\) and \(V\), their tensor product space is denoted as \(U \otimes V\). If \(\{\left|{u_i}\right\rangle\}\) is a basis for \(U\) and \(\{\left|{v_j}\right\rangle\}\) is a basis for \(V\), then \(\{\left|{u_i}\right\rangle \otimes \left|{v_j}\right\rangle\}\) forms a basis for \(U \otimes V\). For vectors \(\left|{u}\right\rangle \in U\) and \(\left|{v}\right\rangle \in V\), their tensor product is denoted as \(\left|{u}\right\rangle \otimes \left|{v}\right\rangle\) or simply \(\left|{u}\right\rangle\left|{v}\right\rangle\) by positional notation.

Bilinearity and Distributivity

The tensor product is bilinear: \[\begin{aligned}(c_1 \left|{u_1}\right\rangle + c_2 \left|{u_2}\right\rangle) \otimes \left|{v}\right\rangle &= c_1 (\left|{u_1}\right\rangle \otimes \left|{v}\right\rangle) + c_2 (\left|{u_2}\right\rangle \otimes \left|{v}\right\rangle) \\\left|{u}\right\rangle \otimes (d_1 \left|{v_1}\right\rangle + d_2 \left|{v_2}\right\rangle) &= d_1 (\left|{u}\right\rangle \otimes \left|{v_1}\right\rangle) + d_2 (\left|{u}\right\rangle \otimes \left|{v_2}\right\rangle) \\c (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) &= (c \left|{u}\right\rangle) \otimes \left|{v}\right\rangle = \left|{u}\right\rangle \otimes (c \left|{v}\right\rangle).\end{aligned}\] and distributive: \[\begin{aligned}(\left|{u_1}\right\rangle + \left|{u_2}\right\rangle) \otimes \left|{v}\right\rangle &= \left|{u_1}\right\rangle \otimes \left|{v}\right\rangle + \left|{u_2}\right\rangle \otimes \left|{v}\right\rangle \\\left|{u}\right\rangle \otimes (\left|{v_1}\right\rangle + \left|{v_2}\right\rangle) &= \left|{u}\right\rangle \otimes \left|{v_1}\right\rangle + \left|{u}\right\rangle \otimes \left|{v_2}\right\rangle.\end{aligned}\]

Inner Product

The inner product in the tensor product space is defined using the inner products of the individual spaces. If \(\left|{u_1}\right\rangle, \left|{u_2}\right\rangle \in U\) and \(\left|{v_1}\right\rangle, \left|{v_2}\right\rangle \in V\), then \[\left\langle{\left|{u_1}\right\rangle \otimes \left|{v_1}\right\rangle}\middle|{\left|{u_2}\right\rangle \otimes \left|{v_2}\right\rangle}\right\rangle = \left\langle{u_1}\middle|{u_2}\right\rangle_U \left\langle{v_1}\middle|{v_2}\right\rangle_V.\] If \(U\) and \(V\) are Hilbert spaces, then \(U \otimes V\) is also a Hilbert space with this inner product.

Operators on Tensor Product Spaces

If \(A\) is an operator on \(U\) and \(B\) is an operator on \(V\), their tensor product operator \(A \otimes B\) acts on \(U \otimes V\) such that: \[(A \otimes B) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = (A \left|{u}\right\rangle) \otimes (B \left|{v}\right\rangle).\] If an operator acts only on subspace \(U\), it is implicitly considered as \(A \otimes I_V\), where \(I_V\) is the identity operator on \(V\). Similarly, an operator acting on \(V\) is \(I_U \otimes B\).

Tensor Operators and Commutators

Operators acting on different tensor product spaces commute. Let \(A\) act on \(U\) and \(B\) act on \(V\). Then effectively we are considering \(A \otimes I_V\) and \(I_U \otimes B\). Their commutator is: \[\begin{aligned}[A \otimes I_V, I_U \otimes B] &= (A \otimes I_V)(I_U \otimes B) - (I_U \otimes B)(A \otimes I_V) \\&= (A I_U) \otimes (I_V B) - (I_U A) \otimes (B I_V) \\&= A \otimes B - A \otimes B = 0.\end{aligned}\] Thus, operators acting on different subspaces commute.

Matrix Representation

Vectors and Operators

If \(\left|{u}\right\rangle = \sum_i u_i \left|{e_i}\right\rangle\) in basis \(\{\left|{e_i}\right\rangle\}\) of \(U\) and \(\left|{v}\right\rangle = \sum_j v_j \left|{f_j}\right\rangle\) in basis \(\{\left|{f_j}\right\rangle\}\) of \(V\), then \(\left|{u}\right\rangle \otimes \left|{v}\right\rangle\) in the basis \(\{\left|{e_i}\right\rangle \otimes \left|{f_j}\right\rangle\}\) of \(U \otimes V\) is represented by a vector whose components are \(u_i v_j\). Conventionally, the index \(j\) (from the right tensor factor) is the fast-changing index, and \(i\) (from the left tensor factor) is the slow-changing index.

If \(A\) is a matrix representing an operator on \(U\) and \(B\) is a matrix representing an operator on \(V\), then the matrix representing \(A \otimes B\) is the Kronecker product of matrices \(A\) and \(B\). If \(A = (A_{ik})\) and \(B = (B_{jl})\), then the matrix for \(A \otimes B\) has blocks \(A_{ik} B\), where \(B\) is the entire matrix \(B\).

Example: \(\sigma_x\otimes \sigma_z\)

Let’s find the matrix representation of \(\sigma_x\otimes \sigma_z\). \[\sigma_x= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_z= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.\] Then \(\sigma_x\otimes \sigma_z\) is given by: \[\sigma_x\otimes \sigma_z= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 \cdot \sigma_z& 1 \cdot \sigma_z\\ 1 \cdot \sigma_z& 0 \cdot \sigma_z\end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}.\] Here, the basis is ordered as \(\{\left|{00}\right\rangle, \left|{01}\right\rangle, \left|{10}\right\rangle, \left|{11}\right\rangle\}\).

Matrix Decomposition

Matrix decomposition is a fundamental concept in linear algebra and numerical analysis, involving the factorization of a matrix into a product of matrices with specific structures and properties. These decompositions reveal the intrinsic geometric and algebraic characteristics of linear transformations, providing powerful tools for theoretical analysis and practical computations in quantum mechanics and quantum computing.

The study of matrix decompositions offers multiple advantages:

• Computational efficiency: Matrix decompositions often simplify complex calculations such as solving linear systems, computing determinants, finding inverses, and determining eigenvalues.

• Geometric interpretation: They provide insights into the geometric actions of matrices (rotations, reflections, scaling).

• Dimensional reduction: Some decompositions allow for approximation by lower-dimensional representations, preserving the most significant features of the data.

• Analysis of quantum systems: In quantum mechanics, decompositions help analyze states, operations, and especially entanglement properties.

In this section, we will explore three important matrix decompositions:

1. Diagonalization and Spectral Decomposition: Representing an operator in terms of its eigenvalues and eigenvectors

2. Polar Decomposition: Factoring a matrix into a product of a unitary operator and a positive semi-definite operator

3. Singular Value Decomposition (SVD): Generalizing diagonalization to arbitrary matrices, with powerful applications in entanglement detection

Diagonalization and Spectral Decomposition

Diagonalization is a specific type of matrix decomposition that transforms a square matrix into a diagonal form. It is particularly relevant for normal matrices (matrices that commute with their adjoints, \(AA^\dagger = A^\dagger A\)), which include Hermitian (self-adjoint) and unitary matrices—types frequently encountered in quantum mechanics.

Theorem 2 (Spectral Theorem for Normal Matrices). For any normal matrix \(A\) of size \(n \times n\), there exists a unitary matrix \(R\) of size \(n \times n\) and a diagonal matrix \(D\) of size \(n \times n\) such that \[R^\dagger A R = D,\] where \(D\) is a diagonal matrix whose diagonal entries are the eigenvalues of \(A\). Equivalently, this decomposition can be written as \[A = R D R^\dagger.\] This is known as the spectral decomposition or eigendecomposition of \(A\). The columns of \(R\) are the eigenvectors of \(A\), and the diagonal matrix \(D\) contains the corresponding eigenvalues.

If we explicitly write out the spectral decomposition, we get: \[A = \sum_{i} \lambda_i \left|{d_i}\middle\rangle\middle\langle{d_i}\right|\] where \(\lambda_i\) are the eigenvalues and \(\left|{d_i}\right\rangle\) are the corresponding eigenvectors (columns of \(R\)). Each term \(\left|{d_i}\middle\rangle\middle\langle{d_i}\right|\) represents a projection operator onto the eigenspace corresponding to \(\lambda_i\).

This form is particularly illuminating in quantum mechanics, as it expresses the operator as a sum of projection operators weighted by the eigenvalues. For a Hermitian operator representing an observable, these eigenvalues correspond to possible measurement outcomes, and the projectors indicate the quantum states that yield those outcomes.

Spectral decomposition of a normal matrix (A)

Geometric Interpretation

The spectral decomposition has a clear geometric interpretation:

1. \(R\) represents a change of basis to the eigenvector basis

2. \(D\) represents a stretching or scaling along the eigenvector directions

3. \(R^\dagger\) represents a change back to the original basis

Therefore, any normal operator can be understood as a rotation to a special basis (the eigenbasis), followed by stretching along the principal axes, and then rotating back to the original basis.

Applications in Quantum Mechanics

In quantum mechanics, diagonalization is intrinsically linked to finding the energy spectrum and eigenstates of a system. If \(A\) represents a Hamiltonian operator \(H\), the eigenvalues in \(D\) correspond to the energy levels of the system, and the columns of \(R\) are the corresponding energy eigenstates. The spectral decomposition is used to:

• Solve the time-independent Schrödinger equation: Finding the eigenvalues and eigenvectors of the Hamiltonian directly gives the stationary states and their energies.

• Calculate time evolution: The time evolution operator \(U(t) = e^{-iHt/\hbar}\) can be easily computed using the spectral decomposition of \(H\). If \(H = R D R^\dagger\), then \(U(t) = R e^{-iDt/\hbar} R^\dagger\), where \(e^{-iDt/\hbar}\) is easily computed by exponentiating the diagonal elements of \(D\).

• Analyze quantum observables: For any observable represented by a Hermitian operator, its spectral decomposition provides the possible outcomes of a measurement (eigenvalues) and the corresponding states (eigenvectors).

• Function evaluation: As shown earlier in our discussion of functions of operators, the spectral decomposition allows us to easily evaluate functions of operators: \(f(A) = \sum_i f(\lambda_i)\left|{d_i}\middle\rangle\middle\langle{d_i}\right|\).

Example 1. Consider the Pauli-X matrix \(\sigma_X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) that we analyzed earlier when discussing functions of operators.

We found that its eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = -1\) with corresponding eigenvectors \(\left|{+}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle + \left|{1}\right\rangle)\) and \(\left|{-}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle - \left|{1}\right\rangle)\).

The spectral decomposition of \(\sigma_X\) is: \[\sigma_X = 1 \cdot \left|{+}\middle\rangle\middle\langle{+}\right| + (-1) \cdot \left|{-}\middle\rangle\middle\langle{-}\right|\]

This expresses the bit-flip operation as a difference between projections onto the \(\left|{+}\right\rangle\) and \(\left|{-}\right\rangle\) states, providing insight into its action in the Hadamard basis.

Polar Decomposition

Polar decomposition provides a way to express any linear operator in terms of a product of a unitary operator and a positive semi-definite operator. This decomposition is analogous to the polar form of complex numbers, \(z = re^{i\theta}\), where \(r = |z| \ge 0\) is a non-negative real number and \(e^{i\theta}\) is a complex number on the unit circle (unitary in 1D). In matrix terms, the unitary part represents a rotation (or change of basis), and the positive semi-definite part represents a scaling or magnitude.

Positive Semi-Definite Operators

Before discussing polar decomposition, we formally define positive semi-definite operators.

Definition 3 (Positive Semi-Definite Operator). A Hermitian operator \(J\) on a Hilbert space \(H\) is called positive semi-definite (or positive for short) if for all vectors \(\left|{\psi}\right\rangle \in H\), \[\left\langle{\psi}\middle|{J\left|{\psi}\right\rangle}\right\rangle \ge 0.\] If, in addition, \(\left\langle{\psi}\middle|{J\left|{\psi}\right\rangle}\right\rangle > 0\) for all non-zero vectors \(\left|{\psi}\right\rangle \neq 0\), then \(J\) is called positive definite.

For any linear operator \(A\), the operator \(A^\dagger A\) is always positive semi-definite. This is because for any vector \(\left|{\psi}\right\rangle\), \[\left\langle{\psi}\middle|{A^\dagger A \left|{\psi}\right\rangle}\right\rangle = \left\langle{A\left|{\psi}\right\rangle}\middle|{A\left|{\psi}\right\rangle}\right\rangle = \|A\left|{\psi}\right\rangle\|^2 \ge 0,\] where \(\|A\left|{\psi}\right\rangle\|\) is the norm of the vector \(A\left|{\psi}\right\rangle\), which is always non-negative. This property is essential for the construction of the positive semi-definite part in the polar decomposition.

Polar Decomposition Theorem

Theorem 3 (Polar Decomposition Theorem). Let \(A\) be a linear operator on a finite-dimensional Hilbert space. Then there exists a unitary operator \(U\) and a positive semi-definite operator \(J\) such that \[A = UJ.\] This is called the left polar decomposition. Similarly, there exists a positive semi-definite operator \(K\) and a unitary operator \(V\) such that \[A = KV.\] This is called the right polar decomposition. Furthermore, the positive semi-definite operators are uniquely determined as \(J = \sqrt{A^\dagger A}\) and \(K = \sqrt{A A^\dagger}\). If \(A\) is invertible, then the unitary operators \(U\) and \(V\) are also unique.

To understand the polar decomposition more intuitively, let’s break down its components:

Geometric interpretation of polar decomposition

Proof Sketch for the Polar Decomposition

Let’s sketch a proof of the left polar decomposition \(A = UJ\):

1. Define \(J = \sqrt{A^\dagger A}\), which exists because \(A^\dagger A\) is positive semi-definite.

2. We need to find a unitary operator \(U\) such that \(A = UJ\). If \(J\) is invertible, then \(U = AJ^{-1}\) would be the candidate.

3. To prove that this \(U\) is indeed unitary, we need to show that \(U^\dagger U = I\): \[\begin{aligned} U^\dagger U &= (AJ^{-1})^\dagger (AJ^{-1}) \\ &= (J^{-1})^\dagger A^\dagger A J^{-1} \\ &= J^{-1} A^\dagger A J^{-1} \quad \text{(since $J$ is Hermitian)} \\ &= J^{-1} J^2 J^{-1} \quad \text{(since $J^2 = A^\dagger A$)} \\ &= I \end{aligned}\]

4. If \(J\) is not invertible, the construction is more complex. One approach is to:

   • Consider the orthogonal decomposition of the Hilbert space \(H = \text{Ker}(J) \oplus \text{Im}(J)\) where \(\text{Ker}(J)\) is the kernel (nullspace) of \(J\) and \(\text{Im}(J)\) is its image (range).

   • Note that \(\text{Ker}(J) = \text{Ker}(A)\) since \(J^2 = A^\dagger A\).

   • Define \(U\) to act as \(AJ^{-1}\) on \(\text{Im}(J)\) and as any unitary mapping on \(\text{Ker}(J)\).

Relationship with Eigenbasis

To understand the connection between polar decomposition and the spectral decomposition, let’s express \(J = \sqrt{A^\dagger A}\) in terms of the eigenbasis of \(A^\dagger A\).

Let \(A^\dagger A = \sum_j \lambda_j^2 \left|{j}\middle\rangle\middle\langle{j}\right|\) where \(\lambda_j^2\) are the eigenvalues of \(A^\dagger A\) and \(\left|{j}\right\rangle\) are the corresponding eigenvectors. The square root of \(A^\dagger A\) is: \[J = \sqrt{A^\dagger A} = \sum_j \lambda_j \left|{j}\middle\rangle\middle\langle{j}\right|\]

For each eigenvector \(\left|{j}\right\rangle\) with non-zero eigenvalue, define \(\left|{\psi_j}\right\rangle = \frac{1}{\lambda_j}A\left|{j}\right\rangle\). We can verify that these vectors form an orthonormal set (for non-zero \(\lambda_j\)).

The unitary operator \(U\) can then be constructed as: \[U = \sum_j \left|{\psi_j}\middle\rangle\middle\langle{j}\right|\] extended to a complete unitary operator by defining its action on the kernel of \(A\).

Applications of Polar Decomposition

Polar decomposition has various applications in quantum mechanics and quantum information:

• Analysis of quantum channels: Quantum operations can be decomposed into a unitary component and a positive component, helping to distinguish between reversible and irreversible parts of the dynamics.

• Quantum state preparation: Determining how to transform one state into another by separating the transformation into rotation and scaling components.

• Connection to SVD: Polar decomposition is closely related to the Singular Value Decomposition (SVD), which we’ll explore next, providing insights into bipartite quantum systems.

• Quantum circuit decomposition: Breaking down complex quantum gates into simpler components for implementation.

Singular Value Decomposition (SVD)

Singular Value Decomposition (SVD) is an exceptionally versatile and powerful matrix decomposition technique that generalizes diagonalization to non-square matrices and provides valuable insights into the structure and properties of linear transformations. Unlike diagonalization, SVD applies to any \(M \times N\) matrix, regardless of whether it is square or normal.

Singular Value Decomposition Theorem

Theorem 4 (Singular Value Decomposition (SVD) Theorem). For any \(M \times N\) matrix \(A\), there exist:

• A unitary matrix \(U\) of size \(M \times M\).

• A diagonal matrix \(\Sigma\) of size \(M \times N\) with non-negative real numbers on the diagonal, called singular values.

• A unitary matrix \(V\) of size \(N \times N\).

such that \[A = U \Sigma V^\dagger.\] The diagonal entries \(\sigma_i = \Sigma_{ii}\) for \(i = 1, 2, \dots, p = \min(M, N)\) are the singular values of \(A\), and are typically arranged in descending order: \(\sigma_1 \ge \sigma_2 \ge \dots \ge \sigma_p \ge 0\). The columns of \(U\) are called left singular vectors, and the columns of \(V\) are called right singular vectors.

Singular Value Decomposition of an (M N) matrix

Relationship to Polar Decomposition

The SVD and polar decomposition are closely related. If \(A = U \Sigma V^\dagger\) is the SVD of \(A\), then:

• \(A = (U V^\dagger)(V \Sigma V^\dagger)\) gives the right polar decomposition, where \(K = V \Sigma V^\dagger = \sqrt{A A^\dagger}\) and the unitary component is \(U V^\dagger\).

• \(A = (U \Sigma U^\dagger)(U V^\dagger)\) gives the left polar decomposition, where \(J = U \Sigma U^\dagger = \sqrt{A^\dagger A}\) and the unitary component is \(U V^\dagger\).

Construction of SVD from Polar Decomposition

Given the polar decomposition \(A = UJ\), where \(J = \sqrt{A^\dagger A}\), we can derive the SVD as follows:

1. Diagonalize \(J = R S R^\dagger\), where \(S\) is diagonal containing the singular values, and \(R\) is unitary.

2. Define \(V = R\) (the right singular vectors).

3. Substitute into the polar decomposition: \(A = U R S R^\dagger = (U R) S R^\dagger\).

4. Define \(U' = U R\) (which is unitary as a product of unitaries).

5. The SVD is then \(A = U' S V^\dagger\), where \(U'\) contains the left singular vectors.

This construction highlights why the singular values of \(A\) are the square roots of the eigenvalues of \(A^\dagger A\) (or equivalently, \(A A^\dagger\)).

Geometric Interpretation of SVD

The SVD provides a powerful geometric interpretation of how a matrix transforms vectors:

1. \(V^\dagger\) rotates the input space to align with the principal directions.

2. \(\Sigma\) scales each principal direction by the corresponding singular value.

3. \(U\) rotates the scaled vectors to the output orientation.

This interpretation makes clear that a matrix stretches space along orthogonal directions (the right singular vectors) by amounts equal to the singular values, and then rotates these stretched directions to new orientations (the left singular vectors).

Geometric interpretation of SVD as sequence of transformations

Algorithm for Computing SVD

Input: An \(M \times N\) matrix \(A\)
Output: Unitary matrices \(U\), \(V\) and diagonal matrix \(\Sigma\) such that \(A = U \Sigma V^\dagger\)

Compute the matrix \(B = A^\dagger A\) Find the eigenvalues \(\lambda_1, \lambda_2, \dots, \lambda_N\) and eigenvectors \(\left|{v_1}\right\rangle, \left|{v_2}\right\rangle, \dots, \left|{v_N}\right\rangle\) of \(B\) Sort the eigenvalues in descending order: \(\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_N \geq 0\) Rearrange the eigenvectors to correspond to the sorted eigenvalues Calculate the singular values \(\sigma_i = \sqrt{\lambda_i}\) for \(i = 1, 2, \dots, N\) Form the unitary matrix \(V\) with columns as the eigenvectors \(\left|{v_1}\right\rangle, \left|{v_2}\right\rangle, \dots, \left|{v_N}\right\rangle\) Construct the \(M \times N\) diagonal matrix \(\Sigma\) with singular values \(\sigma_1, \sigma_2, \dots, \sigma_{\min(M,N)}\) on the diagonal For each non-zero singular value \(\sigma_i\), compute the left singular vector \(\left|{u_i}\right\rangle = \frac{1}{\sigma_i} A \left|{v_i}\right\rangle\) Orthonormally complete the set \(\{\left|{u_i}\right\rangle\}\) if necessary to form a basis for \(\mathbb{C}^M\) Form the unitary matrix \(U\) with columns as the vectors \(\left|{u_1}\right\rangle, \left|{u_2}\right\rangle, \dots, \left|{u_M}\right\rangle\) return \(U, \Sigma, V\)

The complexity of computing the SVD is dominated by the eigenvalue decomposition of \(A^\dagger A\), which is typically \(O(N^3)\) for an \(N \times N\) matrix. The algorithm above is conceptual; in practice, more efficient and numerically stable algorithms such as QR iteration or divide-and-conquer strategies are used.

Low-Rank Approximation and Dimensionality Reduction

One of the most powerful applications of SVD is in low-rank approximation and dimensionality reduction. If we truncate the SVD to keep only the first \(k < p = \min(M,N)\) singular values and vectors, we get an approximation: \[\begin{aligned}A_k = \sum_{i=1}^k \sigma_i \left|{u_i}\right\rangle\left\langle{v_i}\right| = U_k \Sigma_k V_k^\dagger\end{aligned}\] where \(U_k\) consists of the first \(k\) columns of \(U\), \(\Sigma_k\) is a \(k \times k\) diagonal matrix with the top \(k\) singular values, and \(V_k\) consists of the first \(k\) columns of \(V\).

Theorem 5 (Eckart-Young-Mirsky Theorem). The matrix \(A_k\) obtained by truncating the SVD to the \(k\) largest singular values is the best rank-\(k\) approximation to \(A\) in both the Frobenius norm and the spectral norm: \[\begin{aligned}A_k = \arg\min_{rank(B) \leq k} \|A - B\|_F = \arg\min_{rank(B) \leq k} \|A - B\|_2\end{aligned}\]

This optimal approximation property makes SVD extremely valuable for dimensionality reduction, data compression, noise filtering, and feature extraction.

Low-rank approximation using truncated SVD

Applications of SVD in Quantum Information Theory

SVD has numerous applications in quantum information theory and quantum computing:

• Schmidt Decomposition: For a pure bipartite quantum state \(\left|{\psi}\right\rangle \in H_A \otimes H_B\), the Schmidt decomposition is a direct application of SVD to the coefficient matrix of the state. If we write \[\left|{\psi}\right\rangle = \sum_{i,j} c_{ij} \left|{i}\right\rangle_A \otimes \left|{j}\right\rangle_B\] where \(\{\left|{i}\right\rangle_A\}\) and \(\{\left|{j}\right\rangle_B\}\) are orthonormal bases for \(H_A\) and \(H_B\), then the SVD of matrix \(C = (c_{ij})\) gives us \[\left|{\psi}\right\rangle = \sum_{k=1}^r \sigma_k \left|{u_k}\right\rangle_A \otimes \left|{v_k}\right\rangle_B\] where \(r\) is the Schmidt rank of \(\left|{\psi}\right\rangle\), \(\sigma_k\) are the Schmidt coefficients (singular values), and \(\{\left|{u_k}\right\rangle_A\}\) and \(\{\left|{v_k}\right\rangle_B\}\) are orthonormal bases for subsystems \(A\) and \(B\).

• Entanglement Detection: A pure bipartite state is entangled if and only if it has Schmidt rank greater than 1 (equivalently, if it has more than one non-zero singular value). The degree of entanglement can be quantified by measures derived from the singular values, such as the entanglement entropy \(S = -\sum_k \lambda_k \log_2 \lambda_k\) where \(\lambda_k = \sigma_k^2\) are the eigenvalues of the reduced density matrix.

• Quantum Process Tomography: SVD is used in process tomography to reconstruct quantum operations from experimental data, helping identify the principal components of quantum channels.

• Quantum Machine Learning: In quantum machine learning algorithms, SVD is used for dimensionality reduction, feature extraction, and data compression, similar to its role in classical machine learning.

• Tensor Network Contractions: SVD is a key component in tensor network methods such as Matrix Product States (MPS) and the Density Matrix Renormalization Group (DMRG), which are widely used for simulating many-body quantum systems.

Example 2 (Schmidt Decomposition of an Entangled State). Consider the two-qubit Bell state \(\left|{\psi}\right\rangle = \frac{1}{\sqrt{2}}(\left|{00}\right\rangle + \left|{11}\right\rangle)\). We can write this as \[\left|{\psi}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle \otimes \left|{0}\right\rangle + \left|{1}\right\rangle \otimes \left|{1}\right\rangle)\] The coefficient matrix is \[C = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\] The SVD of \(C\) is trivial since it’s already diagonal (up to a constant factor): \(C = U \Sigma V^\dagger\) where \(U = V = I\) and \(\Sigma = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

Thus, the Schmidt decomposition of \(\left|{\psi}\right\rangle\) is \[\left|{\psi}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle \otimes \left|{0}\right\rangle + \left|{1}\right\rangle \otimes \left|{1}\right\rangle)\] with Schmidt coefficients \(\sigma_1 = \sigma_2 = \frac{1}{\sqrt{2}}\).

The fact that there are two non-zero Schmidt coefficients confirms that this is an entangled state, and since the coefficients are equal, it’s maximally entangled.

Remark. Remark 1 (Summary of Matrix Decompositions). We have examined three fundamental matrix decompositions, each with distinct properties and applications:

• Diagonalization (Spectral Decomposition):

  • Form: \(A = R D R^\dagger\), where \(D\) is diagonal, \(R\) is unitary

  • Applicable to: Normal matrices (including Hermitian, Unitary matrices)

  • Components: Eigenvalues (in \(D\)), Eigenvectors (columns of \(R\))

  • Key property: Expresses an operator in terms of its eigenbasis

  • Applications: Solving the Schrödinger equation, computing time evolution, evaluating functions of operators

• Polar Decomposition:

  • Form: \(A = UJ\) (left) or \(A = KV\) (right), where \(U, V\) are unitary, \(J, K\) are positive semi-definite

  • Applicable to: Any square matrix

  • Components: Unitary part (rotation), Positive semi-definite part (magnitude)

  • Key property: Separates rotational and scaling aspects of a transformation

  • Applications: Analogous to polar form of complex numbers, analyzing quantum channels

• Singular Value Decomposition (SVD):

  • Form: \(A = U \Sigma V^\dagger\), where \(U, V\) are unitary, \(\Sigma\) is diagonal with non-negative entries

  • Applicable to: Any matrix (square or rectangular)

  • Components: Singular values (in \(\Sigma\)), Left singular vectors (columns of \(U\)), Right singular vectors (columns of \(V\))

  • Key property: Provides optimal low-rank approximations

  • Applications: Dimensionality reduction, Schmidt decomposition of quantum states, entanglement detection, quantum process tomography

These decompositions offer complementary perspectives on linear transformations and enable a deeper understanding of the structure and properties of operators, particularly in quantum mechanics and quantum information theory.

Conclusion

This lecture has provided an overview of essential mathematical tools and concepts related to operators and matrix decompositions, which are fundamental in quantum mechanics and quantum computing. We discussed the trace operator and its cyclic properties, functions of operators defined through eigenbasis expansion, and the tensor product for composite systems. Furthermore, we explored matrix decomposition techniques, including diagonalization, polar decomposition, and singular value decomposition (SVD). These tools are crucial for understanding and manipulating quantum systems, analyzing quantum algorithms, and exploring advanced topics in quantum information theory.

In the next sessions, we will engage in exercises to solidify these concepts and further explore their applications in quantum mechanics and quantum computing. We will also delve into how singular value decomposition can be applied to detect and quantify entanglement in quantum systems, a key concept in quantum information science.

Follow-up Questions and Topics for Next Lecture:

• How can the trace operator be used to calculate expectation values in quantum mechanics?

• Can we extend the concept of functions of operators to more complex functions beyond power series?

• How does singular value decomposition help in understanding entanglement in bipartite quantum states?

• What are some practical algorithms for computing singular value decomposition, and what are their computational complexities?