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Chapter 6: Problem 2

In terms of the variable \(\theta\), Legendre's equation is \\[\sin \theta \frac{d}{d \theta}\left(\sin \theta \frac{d \Theta(\theta)}{d \theta}\right)+\left(\beta^{2} \sin ^{2} \theta-m^{2}\right) \Theta(\theta)=0\\] Let \(x=\cos \theta\) and \(P(x)=\Theta(\theta)\) and show that \\[\left(1-x^{2}\right) \frac{d^{2} P(x)}{d x^{2}}-2 x \frac{d P(x)}{d x}+\left[\beta-\frac{m^{2}}{1-x^{2}}\right] P(x)=0\\]

Short Answer

Expert verified
Substitute \(x=\cos \theta\) and simplify the derivatives to match Legendre's standard form.

Step by step solution

01

Substitute Variables

We start by substituting \( x = \cos \theta \) into the given Legendre's equation. This implies that we also need to express \( \sin \theta \) in terms of \( x \). Since \( x = \cos \theta \), we have \( \sin \theta = \sqrt{1-x^2} \).
02

Express Derivatives

Express the derivative \( \frac{d}{d\theta} \) in terms of \( \frac{d}{dx} \). By the chain rule, \( \frac{d}{d\theta} = \frac{dx}{d\theta} \frac{d}{dx} \). Since \( x = \cos \theta \), we have \( \frac{dx}{d\theta} = -\sin \theta \). Thus, \( \frac{d}{d\theta} = -\sqrt{1-x^2}\frac{d}{dx} \).
03

Substitute Derivatives into the Equation

Substitute \( \sin \theta = \sqrt{1-x^2} \) and \( \frac{d}{d\theta} = -\sqrt{1-x^2}\frac{d}{dx} \) into the original equation. The term \( \sin \theta \frac{d\Theta}{d\theta} \) becomes \( (1-x^2) \frac{dP}{dx} \). Therefore, \( \sin \theta \frac{d}{d\theta}(\sin \theta \frac{d\Theta}{d\theta}) \) becomes \( \frac{d}{dx}((1-x^2) \frac{dP}{dx}) \).
04

Derive \( \sin \theta \frac{d}{d\theta}(\sin \theta \frac{d\Theta}{d\theta}) \)

For the second derivative term, we have: \[ \frac{d}{dx}((1-x^2) \frac{dP}{dx}) = \frac{d}{dx}((1-x^2) \frac{dP}{dx}) = (1-x^2)\frac{d^2 P}{dx^2} - 2x \frac{dP}{dx} \] by applying the product rule.
05

Substitute Everything Back into the Equation

Now substitute this back into the Legendre's differential equation. The equation turns into:\[(1-x^2)\frac{d^2 P}{dx^2} - 2x \frac{dP}{dx} + \left(\beta^2 (1-x^2) - m^2\right)P(x) = 0\].
06

Simplify the Equation

You can expand \( \beta^2 (1-x^2) - m^2 \) to get \( \beta^2 - \beta^2 x^2 - m^2 \). Simplifying further by collecting like terms, we arrive at:\[(1-x^2)\frac{d^2 P}{dx^2} - 2x \frac{dP}{dx} + \left[ \beta^2 - \frac{m^2}{1-x^2} \right] P(x) = 0 \].

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Key Concepts

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

Differential Equations
Differential equations are equations that involve a function and its derivatives. They are essential in describing various physical phenomena, such as motion, heat, and electricity. The equation given in the original exercise is a type of differential equation known as Legendre's differential equation.

This particular form of a differential equation is a second-order linear differential equation, which means it involves the second derivative of the function. Specifically, the Legendre differential equation plays a significant role in physics, particularly in solving problems involving spherical symmetry.

In the context of the given problem, Legendre's equation is transformed into a more workable form by changing the variable from \( \theta \) to \( x \), simplifying the solution process while maintaining the essence of the original problem.
Legendre Polynomials
Legendre polynomials are solutions to Legendre's differential equation. They are a sequence of orthogonal polynomials and play an essential role in mathematical physics. These polynomials appear in contexts like the gravitational potential and electrostatic potential in spherical coordinates.

The solution typically involves finding these polynomials based on boundary conditions or constraints given in the problem. The beauty of Legendre polynomials lies in their orthogonality property over the interval \( [-1,1] \), which allows them to be particularly useful in solving differential equations, especially when dealing with expansions of functions in terms of a series.

In the exercise, by setting \( P(x) = \Theta(\theta) \), the goal is to express the solutions in terms of these polynomials, which simplifies subsequent calculations due to their well-established properties.
Mathematical Transformation
Mathematical transformation is a fundamental technique in solving differential equations. By transforming a given equation, we make it easier to solve by changing its form or variables. In the exercise, the substitution \( x = \cos \theta \) is a perfect example of such a transformation.

This transformation takes advantage of trigonometric identities, converting the Legendre differential equation into a more manageable form. The substitution effectively changes the angle-dependent function to one dependent on a simpler variable, making computations more straightforward.

This approach not only simplifies the structure of the differential equation but also reveals the underlying relationship between different variables, which can be crucial to obtaining an analytic solution. Such transformations are a powerful method in both pure and applied mathematics, aiding in the understanding of complex problems.

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Most popular questions from this chapter

\- Let \(\alpha_{\max }\) be the largest possible value of \(\alpha \pm m \hbar .\) By definition then, we have that \(\hat{L}_{z} \psi_{\alpha_{n u} \beta}=\alpha_{\max } \psi_{\alpha_{\max } \beta}\) \(\hat{L}^{2} \psi_{\alpha_{\max }^{p}}=\beta^{2} \psi_{\alpha_{\max } \beta}\) and $$ \hat{L}_{+} \psi_{\alpha_{\max } \beta}=0 $$ Operate on the last equation with \(\hat{L}_{-}\)to obtain $$ \begin{aligned} \hat{L}_{-} \hat{L}_{+} \psi_{\alpha_{\max }^{\beta}} &=0 \\ &=\left(\hat{L}^{2}-\hat{L}_{z}^{2}-\hbar \hat{L}_{z}\right) \psi_{\alpha_{\max } \beta} \end{aligned} $$ and $$ \beta^{2}=\alpha_{\max }^{2}+\hbar \alpha_{\max } $$Use a parallel procedure on \(\psi_{\alpha_{\min }} \beta\) to obtain $$ \beta^{2}=\alpha_{\min }^{2}-h \alpha_{\min } $$ Now show that \(\alpha_{\max }=-\alpha_{\min }\), and then argue that the possible values of the eigenvalues \(\alpha\) of \(\hat{L}_{z}\) extend from \(+\alpha_{\max }\) to \(-\alpha_{\max }\) in steps of magnitude \(h\). This is possible only if \(\alpha_{\max }\) is itself an integer (or perhaps a half- integer) times \(\hbar\). Finally, show that this last result leads to $$ \beta^{2}=l(l+1) \hbar^{2} \quad l=0,1,2, \ldots $$ and $$ \alpha=m \hbar \quad m=0, \pm 1, \pm 2, \ldots, \pm l $$

Given the first equality, show that the ground-state energy of a hydrogen atom can be written as \\[E_{0}=-\frac{\hbar^{2}}{2 m_{\mathrm{e}} a_{0}^{2}}=-\frac{e^{2}}{8 \pi \varepsilon_{0} a_{0}}=-\frac{m_{\mathrm{e}} e^{4}}{32 \pi^{2} \varepsilon_{0}^{2} \hbar^{2}}=-\frac{m_{\mathrm{e}} e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\\]

7\. Consider a transition between the \(l=2\) and the \(l=3\) states of atomic hydrogen. What is the total number of conceivable transitions between these two states in an external magnetic field? For light whose electric field vector is parallel to the direction of the external magnetic field, the selection rule is \(\Delta m=0\). For light whose electric field vector is perpendicular to the direction of the external magnetic field, the selection rule is \(\Delta m=\pm 1\). In each case, how many of the possible transitions are allowed?

The average values of \(1 / r, 1 / r^{2},\) and \(1 / r^{3}\) for a hydrogenlike atom can be evaluated in general and are given by \\[\begin{array}{c} \left\langle\frac{1}{r}\right\rangle_{n l}=\frac{Z}{a_{0} n^{2}} \\ \left(\frac{1}{r^{2}}\right)_{n l}=\frac{Z^{2}}{a_{0}^{2} n^{3}\left(l+\frac{1}{2}\right)} \end{array}\\] and \\[\left\langle\frac{1}{r^{3}}\right\rangle_{n l}=\frac{Z^{3}}{a_{0}^{3} n^{3} l\left(l+\frac{1}{2}\right)(l+1)}\\] Verify these formulas explicitly for the \(\psi_{210}\) orbital.

7\. Consider a transition between the \(l=2\) and the \(l=3\) states of atomic hydrogen. What is the total number of conceivable transitions between these two states in an external magnetic field? For light whose electric field vector is parallel to the direction of the external magnetic field, the selection rule is \(\Delta m=0\). For light whose electric field vector is perpendicular to the direction of the external magnetic field, the selection rule is \(\Delta m=\pm 1\). In each case, how many of the possible transitions are allowed?
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