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Chapter 2: Problem 64

Reads $$\left(1-x^{2}\right) \frac{d^{2} f}{d x^{2}}-2 x \frac{d f}{d x}+\ell(\ell+1) f=0,$$ where \(\ell\) is some (non-negative) real number. (a) Assume a power series solution, $$f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$$ and obtain a recursion relation for the constants \(a_{n}\) (b) Argue that unless the series truncates (which can only happen if \(\ell\) is an integer \(),\) the solution will diverge at \(x=1\) (c) When \(\ell\) is an integer, the series for one of the two linearly independent solutions (either \(f\) even or \(f\) odd depending on whether \(\ell\) is even or odd) will truncate, and those solutions are called Legendre polynomials \(P_{\ell}(x)\) Find \(P_{0}(x), P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) from the recursion relation. Leave your answer in terms of either \(a_{0}\) or \(a_{1} .^{69}\)

Short Answer

Expert verified
Use a recursion relation for coefficients to derive Legendre polynomials.

Step by step solution

01

Assume a Power Series Solution

Assume the power series solution is given by \( f(x) = \sum_{n=0}^{\infty} a_n x^n \). This means we can express \( f(x) \) as an infinite series where each term is a coefficient \( a_n \) multiplied by \( x^n \).
02

Compute the Derivatives

Calculate the first and second derivatives of the power series:- First derivative: \( \frac{df}{dx} = \sum_{n=1}^{\infty} n a_n x^{n-1} \)- Second derivative: \( \frac{d^2 f}{dx^2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \).
03

Substitute into Original Equation

Substitute these derivatives into the original differential equation:\[ (1-x^2) \frac{d^2 f}{dx^2} - 2x \frac{d f}{dx} + \ell(\ell+1)f = 0 \]Distribute and simplify each term to obtain:\[ (1-x^2) \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - 2x \sum_{n=1}^{\infty} n a_n x^{n-1} + \ell(\ell+1) \sum_{n=0}^{\infty} a_n x^n = 0 \].
04

Simplify and Align Power Series Terms

To express everything in terms of powers of \( x \), manage the series indices so all sums start from the same term. Adjust each series such that they contribute terms of the same degree \( x^n \). This typically involves changing indices and combining similar terms.
05

Obtain the Recurrence Relation

Combine the series from the earlier step so that all coefficients of \( x^n \) equal zero. Compare coefficients for every power of \( x \) to derive a recursive relation: \[ a_{n+2} = \frac{\ell (\ell + 1) - n (n+1)}{(n+2)(n+1)} a_n \].
06

Discuss Divergence for Non-Truncated Series

For non-integer \( \ell \), the series does not truncate naturally, which can cause the series to diverge at \( x = 1 \) because terms will continue to add indefinitely, leading towards infinity at the endpoint.
07

Consider Integer \( \ell \) and Calculate Legendre Polynomials

When \( \ell \) is an integer, select terms so the series truncates. Begin with different initial conditions (e.g., \( a_0 \) and \( a_1 \)) to compute:- \( P_0(x) = 1 \)- \( P_1(x) = x \)- \( P_2(x) = \frac{1}{2}(3x^2 - 1) \)- \( P_3(x) = \frac{1}{2}(5x^3 - 3x) \)These are the Legendre polynomials, which occur when \( \ell \) is a whole number.

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Key Concepts

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

Power Series Solution
A power series solution involves representing a function as an infinite sum of terms. In this context, we assume that the function \( f(x) \) can be expressed as a power series:
  • \( f(x) = \sum_{n=0}^{\infty} a_n x^n \)
The power series essentially allows us to approximate functions with a sequence of terms involving coefficients \( a_n \) and powers of \( x^n \). By assuming such a representation, we can explore complex differential equations and find solutions by analyzing each term independently.

This technique is especially useful when dealing with differential equations that cannot be simplified by straightforward methods. By transforming the function into a series, the derivatives become simpler in theory as they follow predictable patterns. Such transformations enable us to work systematically toward a solution.
Recurrence Relation
A recurrence relation is an equation that expresses each term in a series as a function of the preceding terms. In the context of this problem, deriving a recurrence relation for the power series allows us to systematically compute each coefficient \( a_n \), step by step, using the previously found terms.

When we substitute the power series representation of \( f(x) \) and its derivatives into the original differential equation, we equate coefficients of like powers of \( x \):
  • Each term by itself should be zero because they collectively describe the entire equation.
  • Align the series terms to have matching powers of \( x \), facilitating comparison and simplification.
  • Upon simplification, we arrive at: \( a_{n+2} = \frac{\ell (\ell + 1) - n (n+1)}{(n+2)(n+1)} a_n \)
This recurrence relation is pivotal because it allows us to compute successive terms of the series based solely on initial conditions, like \( a_0 \) or \( a_1 \).

In the special case of integer \( \ell \), the relation ensures that the series truncates, which is crucial for forming the Legendre polynomials鈥攕olutions that remain finite and well-behaved.
Differential Equations
Differential equations involve functions and their derivatives and are fundamental in describing how changes occur in a system. The complexity of these equations often requires innovative methods for solving them, especially when traditional algebraic techniques do not apply.

In this problem, the given differential equation:
  • \( (1-x^2) \frac{d^2 f}{dx^2} - 2x \frac{df}{dx} + \ell(\ell+1)f = 0 \)
demonstrates a classic structure where modifications to a function鈥檚 behavior are related back to its derivatives. This structure is common in physical systems, such as those described by Legendre polynomials in spherical harmonics.

Differential equations like these are where we see why techniques such as power series solutions become invaluable. They enable us to transcribe a seemingly complicated function and slowly unravel it through recursive steps to reach a viable鈥攁nd analytically satisfying鈥攕olution.
  • It shows how differential equations can be tackled using a combined strategy of approximation (power series) and recursion (recurrence relation).
  • Highlights the critical value of integer \( \ell \) for solutions that define Legendre polynomials, a central result in many physics and engineering applications.
Understanding the linkage between these mathematical concepts and their application culminates in better comprehension of natural phenomena.

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Most popular questions from this chapter

In this problem you will show that the number of nodes of the stationary states of a one-dimensional potential always increases with energy. Consider two (real, normalized) solutions \(\left(\psi_{n} \text { and } \psi_{m}\) ) to the time- \right. independent Schr枚dinger equation (for a given potential \(V(x)\) ), with energies \(E_{n}>E_{m}\) (a) Show that $$\frac{d}{d x}\left(\frac{d \psi_{m}}{d x} \psi_{n}-\psi_{m} \frac{d \psi_{n}}{d x}\right)=\frac{2 m}{\hbar^{2}}\left(E_{n}-E_{m}\right) \psi_{m} \psi_{n}$$ (b) Let \(x_{1}\) and \(x_{2}\) be two adjacent nodes of the function \(\psi_{m}(x) .\) Show that $$\psi_{m}^{\prime}\left(x_{2}\right) \psi_{n}\left(x_{2}\right)-\psi_{m}^{\prime}\left(x_{1}\right) \psi_{n}\left(x_{1}\right)=\frac{2 m}{\hbar^{2}}\left(E_{n}-E_{m}\right) \int_{x_{1}}^{x_{2}} \psi_{m} \psi_{n} d x$$ (c) If \(\psi_{n}(x)\) has no nodes between \(x_{1}\) and \(x_{2},\) then it must have the same sign everywhere in the interval. Show that (b) then leads to a contradiction. Therefore, between every pair of nodes of \(\psi_{m}(x), \psi_{n}(x)\) must have at least one node, and in particular the number of nodes increases with energy.

Suppose $$V(x)=\left\\{\begin{array}{ll} -\alpha / x^{2}, & x>0 \\ \infty, & x \leq 0 \end{array}\right.$$ where \(a\) is some positive constant with the appropriate dimensions. We'd like to find the bound states-solutions to the time-independent Schr枚dinger equation $$-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}-\frac{\alpha}{x^{2}} \psi=E \psi$$ with negative energy \((E<0)\) (a) Let's first go for the ground state energy, \(E_{0}\). Prove, on dimensional grounds, that there is no possible formula for \(E_{0}\) no way to construct (from the available constants \(m, \hbar,\) and \(\alpha\) ) a quantity with the units of energy. That's weird, but it gets worse \(\ldots .\) (b) For convenience, rewrite Equation 2.190 as $$\frac{d^{2} \psi}{d x^{2}}+\frac{\beta}{x^{2}} \psi=\kappa^{2} \psi, \text { where } \beta \equiv \frac{2 m \alpha}{\hbar^{2}} \text { and } \kappa \equiv \frac{\sqrt{-2 m E}}{\hbar}$$ Show that if \(\psi(x)\) satisfies this equation with energy \(E\), then so too does \(\psi(\lambda x),\) with energy \(E^{\prime}=\lambda^{2} E,\) for any positive number \(\lambda .\) [This is a catastrophe: if there exists any solution at all, then there's a solution for every (negative) energy! Unlike the square well, the harmonic oscillator, and every other potential well we have encountered, there are no discrete allowed states- and no ground state. A system with no ground state-no lowest allowed energy-would be wildly unstable, cascading down to lower and lower levels, giving off an unlimited amount of energy as it falls. It might solve our energy problem, but we'd all be fried in the process. \(]\) Well, perhaps there simply are no solutions at all\(\ldots .\) (c) (Use a computer for the remainder of this problem.) Show that $$\psi_{\kappa}(x)=A \sqrt{x} K_{i g}(\kappa x)$$ satisfies Equation 2.191 (here \(K_{i g}\) is the modified Bessel function of order \(i g,\) and \(g \equiv \sqrt{\beta-1 / 4}\) ). Plot this function, for \(g=4\) (you might as well let \(\kappa=1\) for the graph; this just sets the scale of length). Notice that it goes to 0 as \(x \rightarrow 0\) and as \(x \rightarrow \infty\). And it's normalizable: determine \(A\). 66 wout the old rule that the number of nodes counts the number of lower-energy states? This function has an infinite number of nodes, regardless of the energy (i.e. of \(\kappa\) ). I guess that's consistent, since for any \(E\) there are always an infinite number of states with even lower energy. (d) This potential confounds practically everything we have come to expect. The problem is that it blows up too violently as \(x \rightarrow 0 .\) If you move the "brick wall" over a hair, $$V(x)=\left\\{\begin{array}{ll} -\alpha / x^{2}, & x > \epsilon > 0 \\ \infty, & x \leq \epsilon \end{array}\right.$$ it's suddenly perfectly normal. Plot the ground state wave function, for \(g=4\) and \(\epsilon=1\) (you'll first need to determine the appropriate value of \(\kappa\) ), from \(x=0\) to \(x=6 .\) Notice that we have introduced a new parameter \((\epsilon),\) with the dimensions of length, so the argument in (a) is out the window. Show that the ground state energy takes the form $$E_{0}=-\frac{\alpha}{\epsilon^{2}} f(\beta)$$ for some function \(f\) of the dimensionless quantity \(\beta\)

Find the allowed energies of the balf harmonic oscillator $$V(x)=\left\\{\begin{array}{ll} (1 / 2) m \omega^{2} x^{2}, & x >0 \\ \infty, & x < 0 \end{array}\right.$$ (This represents, for example, a spring that can be stretched, but not compressed.) Hint: This requires some careful thought, but very little actual calculation.

If two (or more) distinct \(^{56}\) solutions to the (time-independent) Schr枚dinger equation have the same energy \(E,\) these states are said to be degenerate. For example, the free particle states are doubly degenerate- one solution representing motion to the right, and the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension \(^{57}\) \((-\infty

A free particle has the initial wave function $$\Psi(x, 0)=A e^{-a x^{2}}, $$ where \(A\) and \(a\) are (real and positive) constants. (a) Normalize \(\Psi(x, 0)\) (b) Find \(\Psi(x, t) .\) Hint: Integrals of the form $$\int_{-\infty}^{+\infty} e^{-\left(a x^{2}+b x\right)} d x$$ can be handled by "completing the square": Let \(y \equiv \sqrt{a}[x+(b / 2 a)]\) and note that \(\left(a x^{2}+b x\right)=y^{2}-\left(b^{2} / 4 a\right) .\) (c) Find \(|\Psi(x, t)|^{2}\). Express your answer in terms of the quantity $$w \equiv \sqrt{a /\left[1+(2 \hbar a t / m)^{2}\right]}.$$ Sketch \(|\Psi|^{2}(\text { as a function of } x)\) at \(t=0,\) and again for some very large \(t\) Qualitatively, what happens to \(|\Psi|^{2},\) as time goes on? (d) \(\quad\) Find \(\langle x\rangle,\langle p\rangle,\left\langle x^{2}\right\rangle,\left\langle p^{2}\right\rangle, \sigma_{x},\) and \(\sigma_{p .}\) Partial answer: \(\left\langle p^{2}\right\rangle=a \hbar^{2},\) but it may take some algebra to reduce it to this simple form. (e) Does the uncertainty principle hold? At what time \(t\) does the system come closest to the uncertainty limit?
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