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Chapter 2: Problem 64

$$\left(1-x^{2}\right) \frac{d^{2} f}{d x^{2}}-2 x \frac{d f}{d x}+\ell(\ell+1) f=0$$ 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
The series recursion relation is \(a_{n+2} = \frac{(n-\ell)(n+\ell+1)}{(n+2)(n+1)} a_n\). For non-integer \(\ell\), it diverges at \(x=1\); for integer \(\ell\), the series truncates. Legendre polynomials found include \(P_0(x)=1\), \(P_1(x)=x\), \(P_2(x)=\frac{1}{2}(3x^2-1)\), and \(P_3(x)=\frac{1}{2}(5x^3-3x)\).

Step by step solution

01

Substitute the Power Series into the Differential Equation

Let's substitute the power series solution \( f(x) = \sum_{n=0}^{\infty} a_n x^n \) into the differential equation. We first need to compute the derivatives. The first derivative is \( \frac{df}{dx} = \sum_{n=1}^{\infty} a_n n x^{n-1} \) and the second derivative is \( \frac{d^2f}{dx^2} = \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} \). Substituting these into the differential equation, multiply through by \(1 - x^2\) and adjust indexes accordingly.
02

Reorganize Terms and Extract the Recurrence Relation

After substituting the derivatives into the equation, collect like terms of \(x^n\). This yields:\[(1-x^2)\left( \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} \right) - 2x\left(\sum_{n=1}^{\infty} a_n n x^{n-1}\right) + \ell(\ell+1) \sum_{n=0}^{\infty} a_n x^n = 0\]Expand the expressions involving the sums and rearrange them based on powers of \(x^n\). Matching the coefficients of \(x^n\) to be zero provides the recursion relation:\[a_{n+2} = \frac{(n-\ell)(n+\ell+1)}{(n+2)(n+1)} a_n\].
03

Termination and Divergence Arguments

To argue that the series diverges at \(x=1\) unless it truncates, note that for non-integer \(\ell\), the series has infinitely many terms due to the recursion relation. The terms grow rapidly and lead to a divergent series at \(x=1\). Alternatively, if \(\ell\) is an integer, the sequence \(n = \ell, \ell+1, \ldots\) leads to zero in the numerator of the recursion relation thus causing the series to terminate.
04

Find Specific Legendre Polynomials for Integer \(\ell\)

To find specific Legendre polynomials, use the recurrence relation for integer values of \(\ell\). Beginning with \(P_0\), if \(\ell=0\), the polynomial is simply \(P_0(x) = a_0 = a_0(1)\). For \(\ell=1\), with initial condition \(P_1(x) = a_1x\), \(\ell=2\) provides \(P_2(x) = a_0(\frac{3x^2-1}{2})\), and \(\ell=3\) has \(P_3(x) = a_1(\frac{5x^3-3x}{2})\). These are the first four normalized Legendre polynomials.

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Key Concepts

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

Power Series Solution
Understanding the concept of a power series solution can demystify how complex functions fit nicely into differential equations. A power series solution expresses a function, say \(f(x)\), as an infinite sum of terms involving powers of \(x\). It's like breaking the function down into simpler pieces:
  • The function \(f(x)\) becomes \( \sum_{n=0}^{\infty} a_n x^n \), where \(a_n\) are constants.
  • This expression allows each term in the series to approximate the behavior of \(f(x)\) at different values of \(x\).
Subbing this series into our differential equation lets us match terms of the same power of \(x\) on both sides. This helps to simplify the equation. Instead of solving one big mystery, we solve for each \(a_n\) term by term. This strategy provides a recurrence relation that interlinks successive coefficients \(a_n\). Simplifying power series makes complex differential equations more approachable.
Recurrence Relation
The recurrence relation is a powerful tool when dealing with series solutions, particularly for differential equations. It forms the backbone of our solutions:
  • A recurrence relation connects terms in a sequence using explicit formulae.
  • For Legendre polynomials, the relationship was found as \[a_{n+2} = \frac{(n-\ell)(n+\ell+1)}{(n+2)(n+1)} a_n\], linking coefficients \(a_n\) of higher powers back to those of lower powers.
  • This iterative approach eliminates the need to determine every \(a_n\) individually from scratch; rather, if you know how one starts, you can find them all!
This relation is key in determining whether the series converges or diverges. If \(\ell\) is an integer, the series truncates at a finite term since further terms become zero due to the recurrence formula having zero in its numerator, forming smooth and finite polynomials.
Differential Equations
At the heart of our exploration lies the differential equation, a fundamental concept encapsulating relationships between a function and its derivatives. Here's what makes them essential and intriguing:
  • A differential equation like \((1-x^{2}) \frac{d^{2} f}{d x^{2}}-2 x \frac{d f}{d x}+\ell(\ell+1) f=0\) dictates how \(f(x)\) behaves changewise, guided by \(x\)'s power.
  • This relates to how the function and its slope, curvature, or other changing factors interrelate; it's the essence of motion, growth, and decay models.
  • By using a power series, we broke down the complexity into manageable elements, allowing easier handling of the differential components and ultimately leading us to express the function as the neat series we see in Legendre polynomials.
Understanding differential equations through Legendre polynomials and their series solutions illuminates their utility in engineering, physics, and beyond. It turns abstract change rules into concrete, graspable concepts.

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Most popular questions from this chapter

Consider the double delta-function potential \(V(x)=-\alpha[\delta(x+a)+\delta(x-a)]\) (a) Sketch this potential. (b) How many bound states does it possess? Find the allowed energies, for \(\alpha=\hbar^{2} / m a\) and for \(\alpha=\hbar^{2} / 4 m a,\) and sketch the wave functions. (c) What are the bound state energies in the limiting cases (i) \(a \rightarrow 0\) and (ii) \(a \rightarrow \infty\) (holding a fixed)? Explain why your answers are reasonable, by comparison with the single delta-function well.

Consider the potential $$V(x)=-\frac{\hbar^{2} a^{2}}{m} \operatorname{sech}^{2}(a x),$$ where \(a\) is a positive constant, and "sech" stands for the hyperbolic secant. (a) Graph this potential. (b) Check that this potential has the ground state $$\psi_{0}(x)=A \operatorname{sech}(a x),$$ and find its energy. Normalize \(\psi_{0},\) and sketch its graph. (c) Show that the function $$\psi_{k}(x)=A\left(\frac{i k-a \tanh (a x)}{i k+a}\right) e^{i k x},$$ (where \(k \equiv \sqrt{2 m E} / \hbar,\) as usual) solves the Schr枚dinger equation for any (positive) energy \(E .\) since \(\tanh z \rightarrow-1\) as \(z \rightarrow-\infty,\) \(\psi_{k}(x) \approx A e^{i k x}, \quad\) for large negative \(x.\) This represents, then, a wave coming in from the left with no accompanying reflected wave (i.e. no term \(\exp (-i k x)\) ). What is the asymptotic form of \(\psi_{k}(x)\) at large positive \(x\) ? What are \(R\) and \(T\), for this potential? Comment: This is a famous example of a reflectionless potential-every incident particle, regardless its energy, passes right through. 62.

Consider a particle of mass \(m\) in the potential $$V(x)=\left\\{\begin{array}{ll}\infty & x<0, \\ -32 \hbar^{2} / m a^{2} & 0 \leq x \leq a, \\\0 & x>a.\end{array}\right.$$ (a) How many bound states are there? (b) In the highest-energy bound state, what is the probability that the particle would be found outside the well \((x>a)\) ? Answer: 0.542 , so even though it is "bound" by the well, it is more likely to be found outside than inside!

In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that $$H_{n}(\xi)=(-1)^{n} e^{\xi^{2}}\left(\frac{d}{d \xi}\right)^{n} e^{-\xi^{2}}.$$ Use it to derive \(H_{3}\) and \(H_{4}.\) (b) The following recursion relation gives you \(H_{n+1}\) in terms of the two preceding Hermite polynomials: $$H_{n+1}(\xi)=2 \xi H_{n}(\xi)-2 n H_{n-1}(\xi).$$ Use it, together with your answer in (a), to obtain \(H_{5}\) and \(H_{6}\). (c) If you differentiate an \(n\) th-order polynomial, you get a polynomial of order \((n-1) .\) For the Hermite polynomials, in fact, $$\frac{d H_{n}}{d \xi}=2 n H_{n-1}(\xi).$$ Check this, by differentiating \(H_{5}\) and \(H_{6}\). (d) \(H_{n}(\xi)\) is the \(n\) th \(z\) -derivative, at \(z=0\), of the generating function \(\exp \left(-z^{2}+2 z \xi\right) ;\) or, to put it another way, it is the coefficient of \(z^{n} / n !\) in the Taylor series expansion for this function: $$e^{-z^{2}+2 z \xi}=\sum_{n=0}^{\infty} \frac{z^{n}}{n !} H_{n}(\xi).$$ Use this to obtain \(H_{1}, H_{2},\) and \(H_{3}.\)

(a) Show that the wave function of a particle in the infinite square well returns to its original form after a quantum revival time \(T=4 m a^{2} / \pi \hbar\) That is: \(\Psi(x, T)=\Psi(x, 0)\) for any state (not just a stationary state). (b) What is the classical revival time, for a particle of energy \(E\) bouncing back and forth between the walls? (c) For what energy are the two revival times equal?
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