91Ó°ÊÓ

Consider a three-dimensional ket space. If a certain set of orthonormal kets, say, \(|1\rangle\), \(|2\rangle\), and \(|3\rangle\), are used as the base kets, the operators \(A\) and \(B\) are represented by \(A \doteq\left(\begin{array}{ccc}a & 0 & 0 \\ 0 & -a & 0 \\ 0 & 0 & -a\end{array}\right), \quad B \doteq\left(\begin{array}{ccc}b & 0 & 0 \\ 0 & 0 & -i b \\ 0 & i b & 0\end{array}\right)\) with \(a\) and \(b\) both real. a. Obviously \(A\) exhibits a degenerate spectrum. Does \(B\) also exhibit a degenerate spectrum? b. Show that \(A\) and \(B\) commute. c. Find a new set of orthonormal kets which are simultaneous eigenkets of both \(A\) and \(B\). Specify the eigenvalues of \(A\) and \(B\) for each of the three eigenkets. Does your specification of eigenvalues completely characterize each eigenket?

Step by step solution

01

Determine the Eigenvalues of B

To determine if operator \(B\) exhibits a degenerate spectrum, we need to find its eigenvalues. The matrix \(B\) is given as:\[B = \begin{pmatrix} b & 0 & 0 \ 0 & 0 & -ib \ 0 & ib & 0 \end{pmatrix}\]The characteristic equation is found by setting \(\det(B - \lambda I) = 0\), where \(I\) is the identity matrix. Solving for the determinant, we have:\[\det \begin{pmatrix} b - \lambda & 0 & 0 \ 0 & -\lambda & -ib \ 0 & ib & -\lambda \end{pmatrix} = (b - \lambda)[-\lambda(-\lambda) - (-ib)(ib)].\]Simplifying gives \(\lambda_1 = b\), and the remaining quadratic gives \(\lambda_2 = \lambda_3 = 0\). Therefore, \(B\) has eigenvalues \(b, 0, 0\), which means it exhibits a degenerate spectrum.

02

Demonstrate that A and B Commute

To show the operators \(A\) and \(B\) commute, we must check if \([A, B] = AB - BA = 0\). First, calculate \(AB\):\[AB = \begin{pmatrix} a & 0 & 0 \ 0 & -a & 0 \ 0 & 0 & -a \end{pmatrix} \begin{pmatrix} b & 0 & 0 \ 0 & 0 & -ib \ 0 & ib & 0 \end{pmatrix} = \begin{pmatrix} ab & 0 & 0 \ 0 & 0 & iab \ 0 & -iab & 0 \end{pmatrix}\]Next, calculate \(BA\):\[BA = \begin{pmatrix} b & 0 & 0 \ 0 & 0 & -ib \ 0 & ib & 0 \end{pmatrix} \begin{pmatrix} a & 0 & 0 \ 0 & -a & 0 \ 0 & 0 & -a \end{pmatrix} = \begin{pmatrix} ab & 0 & 0 \ 0 & 0 & iab \ 0 & -iab & 0 \end{pmatrix}\]Since \(AB = BA\), we have \([A, B] = 0\), proving that \(A\) and \(B\) commute.

03

Find Simultaneous Eigenkets and Corresponding Eigenvalues

Since \(A\) and \(B\) commute, they share orthonormal eigenkets. Consider the eigenvalues and find simultaneous eigenkets for both:1. For eigenvalue \(a\) of \(A\) and \(b\) of \(B\), the eigenket is \(|1\rangle\).2. For eigenvalue \(-a\) of \(A\) and \(0\) of \(B\), the first eigenket can be \(|2\rangle \equiv \begin{pmatrix} 0 \ \frac{1}{\sqrt{2}} \ \frac{i}{\sqrt{2}} \end{pmatrix}\).3. For eigenvalue \(-a\) of \(A\) and \(0\) of \(B\), the second eigenket can be \(|3\rangle \equiv \begin{pmatrix} 0 \ \frac{1}{\sqrt{2}} \ -\frac{i}{\sqrt{2}} \end{pmatrix}\).By specifying the eigenvalues \(a, b, 0\) for both operators, each eigenket is completely characterized by its corresponding eigenvalues.

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Key Concepts

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

Degenerate Spectrum

In quantum mechanics, a degenerate spectrum refers to a situation where different eigenstates correspond to the same eigenvalue for a particular operator. This is crucial because the degeneracy of eigenvalues can have significant implications for the physical properties of a quantum system.

Let's consider the operator \(B\), represented by the matrix:

• \[ \begin{pmatrix} b & 0 & 0 \ 0 & 0 & -ib \ 0 & ib & 0 \end{pmatrix} \]

Analyzing the matrix, we solve the characteristic equation \( \det(B - \lambda I) = 0 \) to find its eigenvalues. Solving the determinant gives eigenvalues \(b, 0, 0\). This implies \(B\) has a degenerate spectrum with \( \lambda_2 = \lambda_3 = 0 \) having multiplicity of two.

Both \(A\) and \(B\) in the given problem have degenerate spectra. Understanding this concept is essential as it helps find simultaneous eigenkets, as discussed later.

Commuting Operators

Operators are said to commute if their order of application does not affect the result. Mathematically, for operators \(A\) and \(B\), we say they commute if \([A, B] = AB - BA = 0\).

In many quantum mechanical problems, finding commuting operators often aids in simplifying calculations and reveals symmetries in a system. In the provided exercise, operators \(A\) and \(B\) are represented by specific matrices. Calculations of their products \(AB\) and \(BA\) yield the same matrix:

• \[ \begin{pmatrix} ab & 0 & 0 \ 0 & 0 & iab \ 0 & -iab & 0 \end{pmatrix} \]

The equality \(AB = BA\) confirms that \(A\) and \(B\) commute.

This commuting property is significant because it implies that \(A\) and \(B\) can have a common set of eigenkets. This property is explored further when discussing simultaneous eigenkets.

Simultaneous Eigenkets

When two operators commute, they can share a set of orthonormal eigenkets, known as simultaneous eigenkets. These eigenkets are vital because they satisfy the eigenvalue equations for both operators simultaneously.

In the problem given, since operators \(A\) and \(B\) commute, we look for such eigenkets. The simultaneous eigenkets are constructed as linear combinations of the original base kets \(|1\rangle\), \(|2\rangle\), and \(|3\rangle\). For example:

• The first eigenket for eigenvalues \(a\) of \(A\) and \(b\) of \(B\) is simply \(|1\rangle\).
• The second and third simultaneous eigenkets have eigenvalues \(-a\) for \(A\) and \(0\) for \(B\). These can be formed as \(\begin{pmatrix} 0 \ \frac{1}{\sqrt{2}} \ \frac{i}{\sqrt{2}} \end{pmatrix}\) and \(\begin{pmatrix} 0 \ \frac{1}{\sqrt{2}} \ -\frac{i}{\sqrt{2}} \end{pmatrix}\) respectively.

These simultaneous eigenkets provide a comprehensive description of the state of the system described by both operators. The eigenvalues \(a, -a, b, 0, 0\) used alongside the kets, fully characterize each eigenket.