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Chapter 11: Problem 15

Obtain the selection rule for \(\Delta \ell\) as follows: (a) Derive the commutation relation $$\left[L^{2},\left[L^{2}, \mathbf{r}\right]\right]=2 \hbar^{2}\left(\mathbf{r} L^{2}+L^{2} \mathbf{r}\right).$$ Hint: First show that $$\left[L^{2}, z\right]=2 i \hbar\left(x L_{y}-y L_{x}-i \hbar z\right)$$ Use this, and (in the final step) the fact that \(\mathbf{r} \cdot \mathbf{L}=\mathbf{r} \cdot(\mathbf{r} \times \mathbf{p})=\mathbf{0},\) to demonstrate that $$\left[L^{2},\left[L^{2}, z\right]\right]=2 \hbar^{2}\left(z L^{2}+L^{2} z\right)$$ The generalization from \(z\) to \(\mathbf{r}\) is trivial. (b) Sandwich this commutator between \(\left\langle n^{\prime} \ell^{\prime} m^{\prime}\right|\) and \(|n \ell m\rangle,\) and work out the implications.

Short Answer

Expert verified
The selection rule for \( \Delta \ell \) is that it can change by \( \pm 1 \).

Step by step solution

01

Understand the Problem

We are tasked with deriving the commutation relation of operators which is essential in quantum mechanics, specifically for angular momentum operators. The goal is to ultimately find the selection rule for changes in the orbital angular momentum quantum number \( \Delta \ell \).
02

Establish the Commutation Relation [L虏, z]

Use the hint to show that \( [L^2, z] = 2i \hbar (xL_y - yL_x - i \hbar z) \). Recall that \( L^2 = L_x^2 + L_y^2 + L_z^2 \) and apply commutation rules to each term, simplifying to the required form.
03

Calculate [L虏, [L虏, z]]

Using the established relation from Step 2, compute \( [L^2, [L^2, z]] \). Apply the commutator property \( [A, [B, C]] = [A, B]C + B[A, C] \). Simplify to achieve \( 2 \hbar^2 (z L^2 + L^2 z) \).
04

Generalize from z to r

Recognize that the relation \( [L^2, [L^2, z]] = 2\hbar^2 (zL^2 + L^2z) \) holds for any cartesian component of \( \mathbf{r} \); thus, it generalizes to \( [L^2, [L^2, \mathbf{r}]] = 2\hbar^2 (\mathbf{r}L^2 + L^2\mathbf{r}) \).
05

Use Quantum State Matrix Elements

Sandwich the commutation relation between the quantum states \( \langle n' \ell' m' | \) and \( |n \ell m \rangle \). This implies \( \langle n' \ell' m' | [L^2, [L^2, r]] | n \ell m \rangle = 0 \), derived from the orthogonality conditions and symmetry of states.
06

Derive Selection Rule for 螖鈩�

Consider the implication of \( [L^2, \mathbf{r} ] \) not affecting \( \langle n' \ell' m' |n \ell m \rangle \). It implies that any transitions must comply with \( \Delta \ell = \pm 1 \) for there to be a non-zero matrix element.

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Key Concepts

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

Commutation Relations
In quantum mechanics, commutation relations are fundamental to understanding how different operators interact. Operators that initialize and manage observables like position and momentum must adhere to a specific set of rules, known as commutation relations. For angular momentum operators, commutation relations define how these operators combine and influence quantum states.

The commutator of two operators \( A \) and \( B \) is typically expressed as \( [A, B] = AB - BA \). In the context of the exercise, the focus is on the commutation relation \( [L^2, z] \), where \( L^2 \) represents the total angular momentum squared, and \( z \) is the z-component of the position vector. By applying the general commutation rules for angular momentum, one derives \([L^2, z] = 2i\hbar(xL_y - yL_x - i\hbar z)\).

This relation demonstrates how the operations of \( L^2 \) and the position components influence each other, which is crucial in establishing how quantum states behave under these operations.
Quantum State Transitions
Quantum state transitions are changes that occur when a quantum system moves from one energy state to another. The probability of such transitions is determined by the matrix elements of operators between initial and final states.

In the exercise, we examine transitions using the matrix element \( \langle n' \ell' m' | [L^2, [L^2, \mathbf{r}]] | n \ell m \rangle \). Because commutators like \( [L^2, \mathbf{r}] \) strongly affect how states interact, they help define which transitions are allowed or forbidden. In particular, the orthogonality and symmetry of quantum states impose constraints through selection rules.

The selection rule derived from the commutation relation indicates that for any transition to have a non-zero probability, the change in orbital angular momentum, denoted as \( \Delta \ell \), must satisfy certain conditions like \( \Delta \ell = \pm 1 \). These rules ensure that only specific transitions happen, maintaining the principles of quantum mechanics.
Orbital Angular Momentum
Orbital angular momentum is a fundamental aspect of quantum mechanics, describing the rotation of particles around a central point or axis. It is quantized and associated with quantum numbers that define allowable energy states.

The orbital angular momentum is represented by \( \mathbf{L} \), consisting of the operators \( L_x, L_y, \) and \( L_z \). These operators govern how particles rotate in space and how their states are interconnected. In the exercise, \( L^2 = L_x^2 + L_y^2 + L_z^2 \) refers to the total angular momentum squared, a crucial element when understanding rotational dynamics and energy levels.

Angular momentum's quantized nature dictates that particles can only rotate in discrete steps, defined by the related quantum numbers \( \ell \). These steps translate to possible value ranges for \( \ell \), normally integers that determine the particle's angular behavior and interaction potential.
Angular Momentum Operators
Angular momentum operators are mathematical constructs that describe the rotational properties of particles in quantum mechanics. These operators are fundamental for predicting how particles behave in angular momentum space.

There are several key operators related to angular momentum: \( L_x, L_y, \) and \( L_z \) correspond to the angular momentum components over each spatial direction, while \( L^2 \) represents the total angular momentum squared. These operators are linked through commutation relations, which define their interdependencies and the conservation of angular momentum.

In quantum mechanics, these operators are applied to wave functions to extract information about the system's rotational symmetry. The exercise leverages these operators to derive selection rules, illustrating the step-by-step influence of angular momentum changes on the quantum states. Mastery of these operators and their relations allows for a robust understanding of rotational dynamics within quantum systems.

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Most popular questions from this chapter

A particle is in the ground state of the harmonic oscillator with classical frequency \(\omega,\) when suddenly the spring constant quadruples, so \(\omega^{\prime}=2 \omega\) without initially changing the wave function (of course, \(\Psi\) will now evolve differently, because the Hamiltonian has changed). What is the probability that a measurement of the energy would still return the value \(\hbar \omega / 2\) ? What is the probability of getting \(\hbar \omega^{2}\) Answer: 0.943

Suppose the perturbation takes the form of a delta function (in time): \\[ \hat{H}^{\prime}=\hat{U} \delta(t) \\] assume that \(U_{a a}=U_{b b}=0,\) and let \(U_{a b}=U_{b a}^{*} \equiv \alpha,\) If \(c_{a}(-\infty)=1\) and \(c_{b}(-\infty)=0,\) find \(c_{a}(t)\) and \(c_{b}(t),\) and check that \(\left|c_{a}(t)\right|^{2}+\left|c_{b}(t)\right|^{2}=1\) What is the net probability \((P_{a \rightarrow b} \text { for } t \rightarrow \infty\) ) that a transition occurs? Hint: \right. You might want to treat the delta function as the limit of a sequence of rectangles.

Problem 11.5 Suppose you don't assume \(H_{a a}^{\prime}=H_{b b}^{\prime}=0\) (a) Find \(c_{a}(t)\) and \(c_{b}(t)\) in first-order perturbation theory, for the case \(c_{a}(0)=1, c_{b}(0)=0 .\) Show that \(\left|c_{a}^{(1)}(t)\right|^{2}+\left|c_{b}^{(1)}(t)\right|^{2}=1,\) to first order in \(\hat{H}^{\prime}\) (b) There is a nicer way to handle this problem. Let \\[ d_{a} \equiv e^{\frac{1}{n} \int_{0}^{t} H_{a a}^{\prime}\left(t^{\prime}\right) d t^{\prime}} c_{a}, \quad d_{b} \equiv e^{\frac{1}{h} \int_{0}^{t} H_{b b}^{\prime}\left(t^{\prime}\right) d t^{\prime}} c_{b} \\] Show that \\[ \dot{d}_{a}=-\frac{i}{\hbar} e^{i \phi} H_{a b}^{\prime} e^{-i \omega_{0} t} d_{b} ; \quad \dot{d}_{b}=-\frac{i}{\hbar} e^{-i \phi} H_{b a}^{\prime} e^{i \omega_{0} t} d_{a} \\] where \\[ \phi(t) \equiv \frac{1}{\hbar} \int_{0}^{l}\left[H_{a a}^{\prime}\left(t^{\prime}\right)-H_{b b}^{\prime}\left(t^{\prime}\right)\right] d t^{\prime} \\] So the equations for \(d_{a}\) and \(d_{b}\) are identical in structure to Equation 11.17 (with an extra factor \(e^{i \phi}\) tacked onto \(\hat{H}^{\prime}\) ). (c) Use the method in part (b) to obtain \(c_{a}(t)\) and \(c_{b}(t)\) in first- order perturbation theory, and compare your answer to (a). Comment on any discrepancies.

We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption?

The case of an infinite square well whose right wall expands at a constant velocity \((v)\) can be solved exactly. 45 . A complete set of solutions is $$\Phi_{n}(x, t) \equiv \sqrt{\frac{2}{w}} \sin \left(\frac{n \pi}{w} x\right) e^{i\left(m v x^{2}-2 E_{n}^{i} a t\right) / 2 \hbar w}$$ where \(w(t) \equiv a+v t\) is the width of the well and \(E_{n}^{i} \equiv n^{2} \pi^{2} \hbar^{2} / 2 m a^{2}\) is the \(n\) th allowed energy of the original well (width \(a\) ). The general solution is a linear combination of the \(\Phi^{\prime}\) s: $$\Psi(x, t)=\sum_{n=1}^{\infty} c_{n} \Phi_{n}(x, t)$$ whose coefficients \(c_{n}\) are independent of \(t\) (a) Check that Equation 11.136 satisfies the time-dependent Schr枚dinger equation, with the appropriate boundary conditions. (b) Suppose a particle starts out \((t=0)\) in the ground state of the initial well: $$\Psi(x, 0)=\sqrt{\frac{2}{a}} \sin \left(\frac{\pi}{a} x\right)$$ Show that the expansion coefficients can be written in the form $$c_{n}=\frac{2}{\pi} \int_{0}^{\pi} e^{-i \alpha z^{2}} \sin (n z) \sin (z) d z$$ where \(\alpha \equiv m v a / 2 \pi^{2} \hbar\) is a dimensionless measure of the speed with which the well expands. (Unfortunately, this integral cannot be evaluated in terms of elementary functions.) (c) Suppose we allow the well to expand to twice its original width, so the "external" time is given by \(w\left(T_{e}\right)=2 a\). The "internal" time is the period of the time-dependent exponential factor in the (initial) ground state. Determine \(T_{e}\) and \(T_{i}\), and show that the adiabatic regime corresponds to \(\alpha \ll 1,\) so that \(\exp \left(-i \alpha z^{2}\right) \approx 1\) over the domain of integration. Use this to determine the expansion coefficients, \(c_{n}\). Construct \(\Psi(x, t),\) and confirm that it is consistent with the adiabatic theorem. (d) Show that the phase factor in \(\Psi(x, t)\) can be written in the form $$\theta(t)=-\frac{1}{\hbar} \int_{0}^{t} E_{1}\left(t^{\prime}\right) d t^{\prime}$$ where \(E_{n}(t) \equiv n^{2} \pi^{2} \hbar^{2} / 2 m w^{2}\) is the \(n\) th instantaneous eigenvalue, at time \(t .\) Comment on this result. What is the geometric phase? If the well now contracts back to its original size, what is Berry's phase for the cycle?
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