/*! 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 19 Evaluate the following commutato... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the following commutators: (a) \([\mathbf{L} \cdot \mathbf{S}, \mathbf{L}],(\mathbf{b})[\mathbf{L} \cdot \mathbf{S}, \mathbf{S}]\) \((\mathrm{c})[\mathbf{L} \cdot \mathbf{S}, \mathbf{J}],(\mathrm{d})\left[\mathbf{L} \cdot \mathbf{S}, L^{2}\right],(\mathrm{e})\left[\mathbf{L} \cdot \mathbf{S}, S^{2}\right],(\mathrm{f})\left[\mathbf{L} \cdot \mathbf{S}, J^{2}\right] .\) Hint: \(\mathbf{L}\) and \(\mathbf{S}\) satisfy. the fundamental commutation relations for angular momentum (Equations 4.99 and 4.134 ), but they commute with each other.

Short Answer

Expert verified
All commutators evaluate to zero: (a) 0, (b) 0, (c) 0, (d) 0, (e) 0, (f) 0.

Step by step solution

01

Understand the Problem

We need to evaluate the commutators between the dot product \(\mathbf{L} \cdot \mathbf{S}\) and other angular momentum operators.
02

Use Angular Momentum Commutation Relations

Recall that for angular momentum operators, \([L_i, L_j] = i \hbar \epsilon_{ijk} L_k\) and \([S_i, S_j] = i \hbar \epsilon_{ijk} S_k\). However, \([L_i, S_j] = 0\) because \(\mathbf{L}\) and \(\mathbf{S}\) commute.
03

Evaluate \([\mathbf{L} \cdot \mathbf{S}, \mathbf{L}]\)

Since \(\mathbf{L} \cdot \mathbf{S} = L_i S_i\), use \([L_i S_i, L_j] = L_i [S_i, L_j] + [L_i, L_j] S_i = 0 + i \hbar \epsilon_{ijk} L_k S_i\). Simplifying gives zero as \(\epsilon_{ijk} L_k S_i = 0\) due to the independent components.
04

Evaluate \([\mathbf{L} \cdot \mathbf{S}, \mathbf{S}]\)

Similarly to step 3, \([L_i S_i, S_j] = L_i [S_i, S_j] + [L_i, S_j] S_i = 0 + i \hbar \epsilon_{ijk} L_i S_k = 0\) due to independence of \(\mathbf{L}\) and \(\mathbf{S}\).
05

Evaluate \([\mathbf{L} \cdot \mathbf{S}, \mathbf{J}]\)

Using \(\mathbf{J} = \mathbf{L} + \mathbf{S}\), \([\mathbf{L} \cdot \mathbf{S}, \mathbf{J}] = [\mathbf{L} \cdot \mathbf{S}, \mathbf{L}] + [\mathbf{L} \cdot \mathbf{S}, \mathbf{S}]= 0 + 0 = 0\).
06

Evaluate \([\mathbf{L} \cdot \mathbf{S}, L^2]\)

Use \(L^2 = L_i L_i\). Consider \([\mathbf{L} \cdot \mathbf{S}, L^2] = [L_i S_i, L_j L_j]\). This yields \(L_j[L_i, L_j]S_i + [L_j, L_i]L_j S_i\), both are zero. Hence, \([\mathbf{L} \cdot \mathbf{S}, L^2] = 0\).
07

Evaluate \([\mathbf{L} \cdot \mathbf{S}, S^2]\)

Using \(S^2 = S_i S_i\), \([\mathbf{L} \cdot \mathbf{S}, S^2] = [L_i S_i, S_j S_j]\). This results in similar terms to step 6, giving 0. Hence, \([\mathbf{L} \cdot \mathbf{S}, S^2] = 0\).
08

Evaluate \([\mathbf{L} \cdot \mathbf{S}, J^2]\)

Use \(J^2 = \mathbf{J} \cdot \mathbf{J} = L^2 + S^2 + 2 \mathbf{L} \cdot \mathbf{S}\). Thus \([\mathbf{L} \cdot \mathbf{S}, J^2] = [\mathbf{L} \cdot \mathbf{S}, L^2] + [\mathbf{L} \cdot \mathbf{S}, S^2] + 2[\mathbf{L} \cdot \mathbf{S}, \mathbf{L} \cdot \mathbf{S}]\), simplifying to 0.

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Key Concepts

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

Angular Momentum
In quantum mechanics, angular momentum is a fundamental concept that plays a significant role in understanding atomic and subatomic systems. Angular momentum in this context is a vector operator, denoted by \( \mathbf{L} \), which is associated with rotational symmetry. This operator has three components, typically represented as \( L_x, L_y, \) and \( L_z \). The total angular momentum operator satisfies specific commutation relations, specifically \([L_i, L_j] = i \hbar \epsilon_{ijk} L_k\). These relations reflect the non-commutative nature of angular momentum in quantum mechanics.

The magnitude of the angular momentum vector is described by \( L^2 = L_x^2 + L_y^2 + L_z^2 \). Measurement of angular momentum leads to quantization, meaning its values are discrete, typically depending on quantum numbers \( l \) and \( m \). Understanding these operators and their behaviors is crucial when solving problems involving angular momentum, such as evaluating commutators in quantum systems.
Spin Operators
Spin is an intrinsic form of angular momentum carried by elementary particles, composite particles, and atomic nuclei. Different from \( \mathbf{L} \), spin angular momentum is a property inherent to particles themselves. Spin operators, denoted as \( \mathbf{S} \), have components similar to orbital angular momentum, \( S_x, S_y, \) and \( S_z \), and obey similar commutation relations: \([S_i, S_j] = i \hbar \epsilon_{ijk} S_k\).

Despite this similarity, spin differs because it does not rely on particles' motion through space. Particles can have half-integer spins, which have far-reaching implications in quantum mechanics, such as the Pauli exclusion principle governing the behavior of fermions. Spin operators are pivotal in understanding particle properties and behaviors in magnetic fields and require careful manipulation, especially when combined with other operators.
Fundamental Commutation Relations
Fundamental commutation relations play an essential role in ensuring the consistent application of quantum mechanical principles. They express how different quantum operators "commute" or do not commute with each other, informing us how measurement of one observable influences another. The relations \([L_i, L_j] = i \hbar \epsilon_{ijk} L_k\) and \([S_i, S_j] = i \hbar \epsilon_{ijk} S_k\) state that the measurements of angular momentum components cannot be simultaneously precise.

However, for mixed operators such as \( \mathbf{L} \) and \( \mathbf{S} \), the commutation relation becomes \([L_i, S_j] = 0\), indicating that measurements of orbital angular momentum and spin are independent of each other. These relations are not merely mathematical constructs; they have physical significance in understanding the nature of quantum systems, particularly in calculating the dynamics of particles.
Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It is characterized by principles such as wave-particle duality, quantization, and uncertainty. The mathematics of quantum mechanics involves operators, commutators, and complex functions that describe the probable states and behaviors of particles.

Operators like \( \mathbf{L} \) and \( \mathbf{S} \) are used to represent observables such as angular momentum and spin. Evaluating how these operators interact through commutators is crucial in solving quantum mechanical problems. Quantum mechanics transcends classical physics by accounting for phenomena like particle entanglement and superposition, further underpinning the fundamental structure of the universe.

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Most popular questions from this chapter

Consider a charged particle in the one-dimensional harmonic oscillator potential. Suppose we turn on a weak electric field \((E),\) so that the potential energy is shifted by an amount \(H^{\prime}=-q E x\) (a) Show that there is no first-order change in the energy levels, and calculate the second-order correction. Hint: See Problem 3.39 (b) The Schrödinger equation can be solved directly in this case, by a change of variables: \(x^{\prime} \equiv x-\left(q E / m \omega^{2}\right) .\) Find the exact energies, and show that they are consistent with the perturbation theory approximation.

\mathrm{A}\( free particle of mass \)m\( is confined to a ring of circumference \)L\( such that \)\psi(x+L)=\psi(x) .\( The unperturbed Hamiltonian is \\[ H^{0}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}} \\] to which we add a perturbation \\[ H^{\prime}=V_{0} \cos \left(2 \pi \frac{x}{L}\right) \\] (a) Show that the unperturbed states may be written \\[ \psi_{n}^{0}(x)=\frac{1}{\sqrt{L}} e^{i 2 \pi n x / L} \\] for \)n=0,\pm 1,\pm 2\( and that, apart from \)n=0,\( all of these states are two-fold degenerate. (b) Find a general expression for the matrix elements of the perturbation: \)H_{m n}^{\prime}=\left\langle\psi_{m}^{0}\left|H^{\prime}\right| \psi_{n}^{0}\right\rangle\( (c) Consider the degenerate pair of states with \)n=\pm 1\(. Construct the matrix \)W\( and calculate the first-order energy corrections, \)E^{1}\(. Note that the degeneracy does not liff at first order. Therefore, diagonalizing \)\mathrm{W}\( does not tell us what the "good" states are. (d) Construct the matrix \)\mathrm{W}^{2}(\text { Problem } 7.40)\( for the states \)n=\pm 1,\( and show that the degeneracy lifts at second order. What are the good linear combinations of the states with \)n=\pm 1 ?$ (e) What are the energies, accurate to second order, for these states? 36

Consider a three-level system with the unperturbed Hamiltonian $$\mathrm{H}^{0}=\left(\begin{array}{ccc} \epsilon_{a} & 0 & 0 \\ 0 & \epsilon_{a} & 0 \\ 0 & 0 & \epsilon_{c} \end{array}\right)$$ \(\left(\epsilon_{a} > \epsilon_{c}\right)\) and the perturbation $$\mathrm{H}^{\prime}=\left(\begin{array}{ccc} 0 & 0 & V \\ 0 & 0 & V \\ V^{*} & V^{*} & 0 \end{array}\right)$$ since the \((2 \times 2)\) matrix \(W\) is diagonal (and in fact identically 0 ) in the basis of states (1,0,0) and \((0,1,0),\) you might assume they are the good states, but they're not. To see this: (a) Obtain the exact eigenvalues for the perturbed Hamiltonian \(\mathrm{H}=\mathrm{H}^{0}+\mathrm{H}^{\prime}\) (b) Expand your results from part (a) as a power series in \(|V|\) up to second order. (c) What do you obtain by applying nondegenerate perturbation theory to find the energies of all three states (up to second order)? This would work if the assumption about the good states above were correct. Moral: If any of the eigenvalues of \(\mathrm{W}\) are equal, the states that diagonalize \(\mathrm{W}\) are not unique, and diagonalizing \(W\) does not determine the "good" states. When this happens (and it's not uncommon), you need to use second-order degenerate perturbation theory (see Problem 7.40 ).

A free particle of mass \(m\) is confined to a ring of circumference \(L\) such that \(\psi(x+L)=\psi(x) .\) The unperturbed Hamiltonian is $$H^{0}=-\frac{\hbar^{2}}{2 m} \frac{d^{2}}{d x^{2}}$$ to which we add a perturbation $$H^{\prime}=V_{0} \cos \left(2 \pi \frac{x}{L}\right)$$ (a) Show that the unperturbed states may be written $$\psi_{n}^{0}(x)=\frac{1}{\sqrt{L}} e^{i 2 \pi n x / L}$$ for \(n=0,\pm 1,\pm 2\) and that, apart from \(n=0,\) all of these states are two- fold degenerate. (b) Find a general expression for the matrix elements of the perturbation: $$H_{m n}^{\prime}=\left\langle\psi_{m}^{0}\left|H^{\prime}\right| \psi_{n}^{0}\right\rangle$$ (c) Consider the degenerate pair of states with \(n=\pm 1\). Construct the matrix \(W\) and calculate the first-order energy corrections, \(E^{1}\). Note that the degeneracy does not lift at first order. Therefore, diagonalizing W does not tell us what the "good" states are. (d) Construct the matrix \(\mathrm{W}^{2}\) (Problem 7.40 ) for the states \(n=\pm 1\), and show that the degeneracy lifts at second order. What are the good linear combinations of the states with \(n=\pm 1\) ? (e) What are the energies, accurate to second order, for these states? 36

In the text I asserted that the first-order corrections to an \(n\) -fold degenerate energy are the eigenvalues of the \(W\) matrix, and I justified this claim as the "natural" generalization of the case \(n=2 .\) Prove it, by reproducing the steps in Section \(7.2 .1,\) starting with $$\psi^{0}=\sum_{j=1}^{n} \alpha_{j} \psi_{j}^{0}$$ (generalizing Equation 7.17 ), and ending by showing that the analog to Equation 7.27 can be interpreted as the eigenvalue equation for the matrix \(\mathrm{W}\)

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