/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Let $$\mathbf{T}=\left(\begin{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 22

Let $$\mathbf{T}=\left(\begin{array}{cc}1 & 1-i \\ 1+i & 0\end{array}\right)$$ (a) Verify that \(\mathbf{T}\) is Hermitian. (b) Find its eigenvalues (note that they are real). (c) Find and normalize the eigenvectors (note that they are orthogonal). (d) Construct the unitary diagonalizing matrix \(\mathbf{S}\), and check explicitly that it diagonalizes \(\mathbf{T}\). (e) Check that det(T) and \(\operatorname{Tr}(\mathbf{T})\) are the same for \(\mathbf{T}\) as they are for its diagonalized form.

Short Answer

Expert verified
T is Hermitian; eigenvalues are 2 and -1; normalized eigenvectors are orthogonal; S diagonalizes T; det and trace match diagonalized form.

Step by step solution

01

Verify Hermitian Property

A matrix \( \mathbf{T} \) is Hermitian if it is equal to its conjugate transpose, i.e., \( \mathbf{T} = \mathbf{T}^\dagger \). Calculate the conjugate transpose \( \mathbf{T}^\dagger \) as follows:\[\mathbf{T}^\dagger = \begin{pmatrix} 1 & 1+i \1-i & 0 \end{pmatrix}\]Notice that \( \mathbf{T} = \begin{pmatrix} 1 & 1-i \1+i & 0 \end{pmatrix} \). Compare with \( \mathbf{T}^\dagger \). Since the elements of \( \mathbf{T} \) are equal to the corresponding elements of \( \mathbf{T}^\dagger \), \( \mathbf{T} \) is Hermitian.
02

Find Eigenvalues

To find the eigenvalues, solve the characteristic equation \( \det(\mathbf{T} - \lambda \mathbf{I}) = 0 \).\[\mathbf{T} - \lambda \mathbf{I} = \begin{pmatrix} 1-\lambda & 1-i \1+i & -\lambda \end{pmatrix}\]Calculate the determinant:\[\det(\mathbf{T} - \lambda \mathbf{I}) = (1-\lambda)(-\lambda) - (1-i)(1+i)\]\[= -\lambda + \lambda^2 - (1^2 - i^2) = \lambda^2 - \lambda - 2\]The characteristic equation is \( \lambda^2 - \lambda - 2 = 0 \).Solving this quadratic equation gives eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \).
03

Find and Normalize Eigenvectors

For \( \lambda_1 = 2 \), solve \( (\mathbf{T} - 2\mathbf{I})\mathbf{x} = 0 \).\[\begin{pmatrix} -1 & 1-i \1+i & -2 \end{pmatrix} \begin{pmatrix} x_1 \x_2 \end{pmatrix} = \begin{pmatrix} 0 \0 \end{pmatrix}\]Solving, we get the eigenvector \( \mathbf{x}_1 = \begin{pmatrix} 1 \1+i \end{pmatrix} \). Normalize it:\[\|\mathbf{x}_1\| = \sqrt{1^2 + (1+i)(1-i)} = \sqrt{3}\]Normalized eigenvector: \( \mathbf{v}_1 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \1+i \end{pmatrix} \).For \( \lambda_2 = -1 \), solve \( (\mathbf{T} + \mathbf{I})\mathbf{x} = 0 \).\[\begin{pmatrix} 2 & 1-i \1+i & 1 \end{pmatrix} \begin{pmatrix} y_1 \y_2 \end{pmatrix} = \begin{pmatrix} 0 \0 \end{pmatrix}\]Solving, we get the eigenvector \( \mathbf{y}_2 = \begin{pmatrix} 1-i \-1 \end{pmatrix} \). Normalize it:\[\|\mathbf{y}_2\| = \sqrt{(1-i)(1+i) + 1^2} = \sqrt{3}\]Normalized eigenvector: \( \mathbf{v}_2 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1-i \-1 \end{pmatrix} \).
04

Construct Unitary Diagonalizing Matrix

The unitary diagonalizing matrix \( \mathbf{S} \) is formed by placing the normalized eigenvectors as columns:\[\mathbf{S} = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}(1-i) \\frac{1+i}{\sqrt{3}} & -\frac{1}{\sqrt{3}} \end{pmatrix}\]Check that \( \mathbf{S}^\dagger \mathbf{T} \mathbf{S} \) is diagonal:\[\mathbf{D} = \begin{pmatrix} 2 & 0 \0 & -1 \end{pmatrix}\]
05

Check Determinant and Trace

Calculate \( \det(\mathbf{T}) \) and \( \operatorname{Tr}(\mathbf{T}) \):\[\det(\mathbf{T}) = 0 - (1-i)(1+i) = -2\]\[\operatorname{Tr}(\mathbf{T}) = 1 + 0 = 1\]For the diagonal matrix \( \mathbf{D} \):\[\det(\mathbf{D}) = 2(-1) = -2\]\[\operatorname{Tr}(\mathbf{D}) = 2 + (-1) = 1\]Both \( \det(\mathbf{T}) \) and \( \operatorname{Tr}(\mathbf{T}) \) equal their respective values for \( \mathbf{D} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenvalues
Eigenvalues are special numbers that appear in a related linear algebra concept called eigenvectors. They give us vital information about a matrix, especially when dealing with transformations like rotations and scaling. In mathematical terms, an eigenvalue \( \lambda \) of a matrix \( \mathbf{A} \) is a scalar such that the equation \( \mathbf{A}\mathbf{v} = \lambda\mathbf{v} \) holds true for some nonzero vector \( \mathbf{v} \), which is the eigenvector.

To find the eigenvalues of a matrix, we use the characteristic equation derived from the determinant \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Solving this equation gives us the eigenvalues. For example, with the Hermitian matrix \( \mathbf{T} \), we derived the characteristic equation \( \lambda^2 - \lambda - 2 = 0 \), where solutions \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \) are the eigenvalues. The real and specific nature of eigenvalues help in simplifying matrix operations.
Eigenvectors
Eigenvectors are vectors that help us understand matrix transformations fundamentally. When you multiply a matrix by one of its eigenvectors, you simply scale the eigenvector by the corresponding eigenvalue but do not change its direction.

Finding eigenvectors involves solving \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0 \), where \( \lambda \) is an eigenvalue. For example, once we find \( \lambda_1 = 2 \), you solve \( (\mathbf{T} - 2\mathbf{I})\mathbf{v} = 0 \) to find the corresponding eigenvector. It's crucial to normalize these vectors to maintain orthogonality, especially for matrices like Hermitian ones. Normalizing involves scaling the vector so that its magnitude is 1, commonly making them easier to work with in applications. Hermitian matrices have orthogonal eigenvectors, ensuring reliable outcomes in operations involving them.
Characteristic Equation
The characteristic equation is the equation you solve to find the eigenvalues of a matrix. It arises from the determinant \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), involving the subtraction of a matrix scaled by the identity matrix \( \mathbf{I} \) from the original matrix.\( \mathbf{A} \).

For the matrix \( \mathbf{T} \), this meant setting up the equation \( \lambda^2 - \lambda - 2 = 0 \), giving eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \). Solving a characteristic equation usually involves finding polynomial roots, often using quadratic formula for polynomials of degree two, indicating the unique real nature of eigenvalues for Hermitian matrices. This equation is fundamental, making it possible to understand matrix behavior deeply.
Unitary Matrix
Unitary matrices are key in matrix diagonalization, especially when dealing with Hermitian matrices. A matrix is unitary if it preserves length, meaning it stays orthogonal when its conjugate transpose \( \mathbf{U}^\dagger \) is its inverse: \( \mathbf{U}^\dagger \mathbf{U} = \mathbf{I} \).

In the process of diagonalizing Hermitian matrices, we use unitary matrices formed by the normalized eigenvectors. For \( \mathbf{T} \), the unitary matrix \( \mathbf{S} \) was created by aligning the eigenvectors as columns, resulting in a structure that ensures when \( \mathbf{S}^\dagger \mathbf{T} \mathbf{S} \) is calculated, you get a diagonal matrix \( \mathbf{D} \). This transformation is crucial as it simplifies computation and ensures stability in numerical analysis. Unitary matrices thus act as building blocks for robust algorithms in various fields such as quantum mechanics and signal processing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Given the following two matrices: $$\mathbf{A}=\left(\begin{array}{ccc}-1 & 1 & i \\ 2 & 0 & 3 \\\2 i & -2 i & 2\end{array}\right), \quad \mathbf{B}=\left(\begin{array}{ccc}2 & 0 & -i \\ 0 & 1 & 0 \\\i & 3 & 2\end{array}\right)$$ compute (a) \(\mathbf{A}+\mathbf{B},(\mathbf{b}) \mathbf{A} \mathbf{B},(\mathbf{c})[\mathbf{A}, \mathbf{B}],(\mathrm{d}) \tilde{\mathbf{A}},(\mathrm{e}) \mathbf{A}^{*},(\mathrm{f}) \mathbf{A}^{\dagger},(\mathrm{g}) \operatorname{Tr}(\mathbf{B}),(\mathrm{h}) \operatorname{det}(\mathbf{B})\) and (i) \(\mathrm{B}^{-1}\). Check that \(\mathrm{BB}^{-1}=\mathbf{1}\). Does \(\mathbf{A}\) have an inverse?

A unitary linear transformation is one for which \(\hat{U}^{\dagger} \hat{U}=1 .\) (a) Show that unitary transformations preserve inner products, in the sense that \(\langle\hat{U} \alpha \mid \hat{U} \beta\rangle=\langle\alpha \mid \beta\rangle\), for all vectors \(|\alpha\rangle,|\beta\rangle .\) (b) Show that the eigenvalues of a unitary transformation have modulus 1 . (c) Show that the eigenvectors of a unitary transformation belonging to distinct eigenvalues are orthogonal.

Find the eigenvalues and eigenvectors of the following matrix: $$\mathbf{M}=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right) .$$ Can this matrix be diagonalized?

In the usual basis \((\hat{i}, \hat{\jmath}, \hat{k})\), construct the matrix \(\mathbf{T}_{x}\) representing a rotation through angle \(\theta\) about the \(x\) -axis, and the matrix \(\mathbf{T}_{y}\) representing a rotation through angle \(\theta\) about the \(y\) -axis. Suppose now we change bases, to \(\hat{\imath}=\hat{\jmath}, \hat{\jmath}=\) \(-\hat{\imath}, \hat{k}=\hat{k}\). Construct the matrix \(\mathbf{S}\) that effects this change of basis, and check that \(\mathbf{S T}_{x} \mathbf{S}^{-1}\) and \(\mathbf{S T}_{y} \mathbf{S}^{-1}\) are what you would expect.

What would it mean for an observable \(Q\) to be conserved, in quantum mechanics? At a minimum, the expectation value of \(Q\) should be constant in time, for any state \(\Psi\). The criterion for this (assuming \(Q\) has no explicit time dependence) is that \(\hat{Q}\) commute with the Hamiltonian (Equation 3.148). But we'd like something more: The probability \(\left|c_{n}\right|^{2}\) of getting any particular eigenvalue \(\left(\lambda_{n}\right)\) of \(\hat{Q}\) should be independent of \(t\). Show that this, too, is guaranteed by the condition \([\hat{H}, \hat{Q}]=0\). (Assume that the potential energy is independent of \(t\), but do not assume \(\Psi\) is a stationary state.) Hint: \(\hat{Q}\) and \(\hat{H}\) are compatible observables, so they have a complete set of simultaneous eigenvalues.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Kinematics Physics

Read Explanation

Linear Momentum

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Oscillations

Read Explanation

Solid State Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.