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Chapter 1: Problem 23

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 that are simultancous 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?

Short Answer

Expert verified
(a) No, B is not degenerate. (b) A and B commute. (c) New kets: |1鉄�, |2'鉄�, |3'鉄�; eigenvalues: (a, b), (-a, b), (-a, -b).

Step by step solution

01

Determine if B has a degenerate spectrum

The spectrum is degenerate if there are repeated eigenvalues. To find the eigenvalues of operator \(B\), we solve the characteristic equation: \[\text{det}(B - \lambda I) = 0.\] This implies: \[\begin{vmatrix} b - \lambda & 0 & 0 \ 0 & -\lambda & -ib \ 0 & ib & -\lambda \end{vmatrix} = 0.\] \(\lambda_1 = b\) and the submatrix for the last two rows and columns is: \[ |-\lambda|^2 - (ib)(-ib) = \lambda^2 - b^2 = 0. \] Solving, we get \(\lambda_2 = b\) and \(\lambda_3 = -b\). Therefore, \(B\) does not have a degenerate spectrum because the eigenvalues \(b, -b, b\) are distinct in only two cases: \(b\) repeats once.
02

Verify if A and B commute

Two operators commute if their product in either order is the same. We calculate \([A, B] = AB - BA\) using matrix multiplication. Compute matrix \(AB\) and then \(BA\):\[ AB = \begin{pmatrix} ab & 0 & 0 \ 0 & 0 & aib \ 0 & -aib & 0 \end{pmatrix}, \] and\[ BA = \begin{pmatrix} ab & 0 & 0 \ 0 & 0 & -aib \ 0 & aib & 0 \end{pmatrix}. \] Both produce \([A, B] = 0\). Therefore, \(A\) and \(B\) commute.
03

Find simultaneous eigenkets of A and B

For simultaneous eigenkets, both operators should act in a way that scales the vectors by scalars (the eigenvalues). From Step 2, it's verified that they can share eigenkets. Start with eigenkets from \(A\):\[ |1\rangle \rightarrow a, \quad |2\rangle \rightarrow -a, \quad |3\rangle \rightarrow -a. \] From \(B\) eigenvectors, solve \([B - bI] |2\rangle\) and \([B - (-b)I] |3\rangle\) to yield:\[ |2'\rangle = \frac{1}{\sqrt{2}}(|2\rangle + i|3\rangle), \quad |3'\rangle = \frac{1}{\sqrt{2}}(|2\rangle - i|3\rangle). \] Confirm each of these kets yields respective eigenvalue actions from both \(A\) and \(B\).
04

Specify eigenvalues and confirm uniqueness

The simultaneous eigenvectors \(|1\rangle\), \(|2'\rangle\), \(|3'\rangle\) are given by eigenvalues as follows:\(|1\rangle\) has eigenvalues \(a, b\), and \(|2'\rangle\), \(|3'\rangle\) both have \(-a, b\) and \(-a, -b\). This eigenvalue specification characterizes the eigenkets sufficiently because each is associated with a unique pair of eigenvalues, which uniquely represents its state within the space.

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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, the term "degenerate spectrum" refers to the scenario where an operator has repeated eigenvalues. Operators, which are mathematical entities representing physical observables, have associated spectra鈥攕ets of eigenvalues that define their behavior. When an operator has a repeated eigenvalue, we describe its spectrum as degenerate.

Here's how this works: when we solve the operator's characteristic equation, if there are multiple kets (vectors in the ket space) corresponding to the same eigenvalue, this implies degeneracy. It's important because degeneracy can lead to multiple quantum states sharing the same energy level, which impacts the system's symmetry and potential behaviors.

For the operator \(A\) given in the exercise, its eigenvalues include \(a\) and \(-a\), with \(-a\) repeating. Thus, \(A\) has a degenerate spectrum because it has more than one vector associated with the same eigenvalue \(-a\). On the other hand, when checking the operator \(B\) from our exercise, we found that while \(b\) appears twice, \(-b\) also exists, which means \(B\) does not have a completely degenerate spectrum. It has variability since at least one eigenvalue is different.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are central to understanding quantum operators and their effects on quantum states. An eigenvector of an operator remains in the same direction even after the operator acts on it, although its magnitude may change, scaled by a factor known as the eigenvalue.

This property helps simplify the analysis of quantum systems by reducing the complexity involved in predicting the behavior of quantum states. For any linear operator \(A\), if \(|v\rangle\) is an eigenvector and \(\lambda\) is the corresponding eigenvalue, then the relation \(A|v\rangle = \lambda|v\rangle\) holds true. This relationship is foundational in quantum mechanics, often utilized in solving the Schr枚dinger equation to determine allowed energy levels of quantum systems.

In our exercise, each vector \(|1\rangle, |2\rangle,\) and \(|3\rangle\) are acted upon by \(A\) and \(B\) to determine their eigenvalues. Consequently, we identify the eigenvalues for both operators across these kets, ensuring that each ket satisfies the eigenvalue equations for \(A\) and \(B\). The process involves solving for eigenvectors that can satisfy both operators simultaneously, which we did by adjusting existing kets into combinations \(|2'\rangle\) and \(|3'\rangle\).
Commutativity of Operators
The principle of the commutativity of operators in quantum mechanics specifies that two operators \(A\) and \(B\) commute if their order of application does not affect the outcome. Mathematically, this is expressed as \([A, B] = AB - BA = 0\).

This property is significant because when two observables have commuting operators, it means they have a common set of eigenvectors. Thus, these observables can be measured simultaneously with precision, revealing important insights about the quantum system's state.

Examining the exercise, we computed the matrix products \(AB\) and \(BA\) as applied to the given matrices for operators \(A\) and \(B\). Both matrices resulted in identical matrices, verifying that \([A, B] = 0\). Therefore, any ket in the space could be reshuffled into new kets that are simultaneous eigenvectors of \(A\) and \(B\). These transformations affirm that the operators commute, allowing shared eigenvectors to simultaneously determine the eigenvalues for different quantum states.

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Most popular questions from this chapter

The Hamiltonian operator for a two-state system is given by \\[H=a(|1\rangle\langle 1|-| 2\rangle\langle 2|+| 1\rangle\langle 2|+| 2\rangle\langle 1|)\\] where \(a\) is a number with the dimension of energy. Find the energy eigenvalues and the corresponding energy eigenkets (as linear combinations of |1\rangle and |2\rangle ).

Using the rules of bra-ket algebra, prove or evaluate the following: (a) \(\operatorname{tr}(X Y)=\operatorname{tr}(Y X),\) where \(X\) and \(Y\) are operators. (b) \((X Y)^{\dagger}=Y^{\prime} X^{\dagger},\) where \(X\) and \(Y\) are operators. (c) \(\exp [i f(A)]=?\) in ket-bra form, where \(A\) is a Hermitian operator whose eigenvalues are known. (d) \(\sum_{a^{\prime}} \psi_{a^{\prime}}^{*}\left(\mathbf{x}^{\prime}\right) \psi_{a^{\prime}}\left(\mathbf{x}^{\prime \prime}\right),\) where \(\psi_{a^{\prime}}\left(\mathbf{x}^{\prime}\right)=\left\langle\mathbf{x}^{\prime} | a^{\prime}\right\rangle\).

(a) Consider two kets \(|\alpha\rangle\) and \(|\beta\rangle .\) Suppose \(\left\langle a^{\prime} | \alpha\right\rangle,\left\langle a^{\prime \prime} | \alpha\right\rangle, \ldots\) and \(\left\langle a^{\prime} | \beta\right\rangle\) \(\left\langle a^{\prime \prime} | \beta\right\rangle, \ldots\) are all known, where \(\left|a^{\prime}\right\rangle,\left|a^{\prime \prime}\right\rangle, \ldots\) form a complete set of base kets. Find the matrix representation of the operator \(|\alpha\rangle\langle\beta|\) in that basis. (b) We now consider a spin \(\frac{1}{2}\) system and let \(| \alpha\) ) and \(|\beta\rangle\) be \(| s_{z}=h / 2\) ) and \(\left|s_{x}=h / 2\right\rangle,\) respectively. Write down explicitly the square matrix that corresponds to \(|\alpha\rangle\langle\beta|\) in the usual \((s_{z}\) diagonal) basis..

Let \(A\) and \(B\) be observables. Suppose the simultancous eigenkets of \(A\) and \(B\) \(\left\\{\left|a^{\prime}, b^{\prime}\right\rangle\right\\}\) form a complete orthonormal set of base kets. Can we always conclude that \\[[A, B]=0 ?\\] If your answer is yes, prove the assertion. If your answer is no, give a counterexample.

Using the orthonormality of \(|+\rangle\) and \(|-\rangle,\) prove \\[\left[S_{i}, S_{j}\right]=i \varepsilon_{i j k} \hbar S_{k}, \quad\left[S_{i}, S_{j}\right]=\left(\frac{\hbar^{2}}{2}\right) \delta_{i j}\\] where \\[\begin{array}{l}\left.S_{x}=\frac{\hbar}{2}(|+)\langle-|+|-\rangle\langle+|\right), \quad S_{y}=\frac{i \hbar}{2}(-|+\rangle\langle-|+|-\rangle\langle+|) \\\S_{z}=\frac{\hbar}{2}(|+\rangle\langle+|-|-\rangle\langle-|)\end{array}\\]
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