/*! 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 16 In this problem we explore some ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 16

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 \(H_{3}.\)

Short Answer

Expert verified
H_3(\xi)=8\xi^3-12\xi; H_4=16\xi^4-48\xi^2+12; H_5=32\xi^5-160\xi^3+120\xi; H_6=64\xi^6-480\xi^4+720\xi^2-120. Derivatives verified. H_1=2\xi, H_2=4\xi^2-2, H_3=8\xi^3-12\xi from generating function.

Step by step solution

01

Compute Hermite Polynomial H3 using Rodrigues Formula

The Rodrigues formula for Hermite polynomials is given by \( H_n(\xi) = (-1)^n e^{\xi^2} \left(\frac{d}{d \xi}\right)^n e^{-\xi^2} \). To find \( H_3(\xi) \), we need to apply the third derivative to the function \( e^{-\xi^2} \): \[ H_3(\xi) = (-1)^3 e^{\xi^2} \left(\frac{d^3}{d \xi^3}\right) e^{-\xi^2} \] First, compute the first derivative: \( \frac{d}{d \xi} e^{-\xi^2} = -2\xi e^{-\xi^2} \).Second derivative: \( \frac{d^2}{d \xi^2} e^{-\xi^2} = 4\xi^2 e^{-\xi^2} - 2 e^{-\xi^2} \).Third derivative: \( \frac{d^3}{d \xi^3} e^{-\xi^2} = -8 \xi^3 e^{-\xi^2} + 12 \xi e^{-\xi^2} \). Substituting back: \( H_3(\xi) = -e^{\xi^2}(-8\xi^3 e^{-\xi^2} + 12\xi e^{-\xi^2}) = 8 \xi^3 - 12\xi \). Thus, \( H_3(\xi) = 8 \xi^3 - 12 \xi \).
02

Compute Hermite Polynomial H4 using Rodrigues Formula

Now we compute \( H_4(\xi) \) using the same formula:\[ H_4(\xi) = (-1)^4 e^{\xi^2} \left(\frac{d^4}{d \xi^4}\right) e^{-\xi^2} \] Start again from the third derivative:Fourth derivative: \( \frac{d^4}{d \xi^4} e^{-\xi^2} = 16 \xi^4 e^{-\xi^2} - 48 \xi^2 e^{-\xi^2} + 12 e^{-\xi^2} \).Substituting back gives:\( H_4(\xi) = e^{\xi^2}(16 \xi^4 e^{-\xi^2} - 48 \xi^2 e^{-\xi^2} + 12 e^{-\xi^2}) \)\( = 16\xi^4 - 48\xi^2 + 12 \).
03

Derive H5 Using Recursion Relation

The recursion relation for Hermite polynomials states: \( H_{n+1}(\xi) = 2\xi H_n(\xi) - 2n H_{n-1}(\xi) \).Using \( H_3(\xi) = 8\xi^3 - 12\xi \) and \( H_4(\xi) = 16\xi^4 - 48\xi^2 + 12 \), compute \( H_5(\xi) \):\[ H_5(\xi) = 2\xi H_4(\xi) - 8H_3(\xi) \]\( H_5(\xi) = 2\xi (16\xi^4 - 48\xi^2 + 12) - 8(8\xi^3 - 12\xi) \)= \( 32\xi^5 - 96\xi^3 + 24\xi - 64\xi^3 + 96\xi \)= \( 32\xi^5 - 160\xi^3 + 120\xi \).
04

Derive H6 Using Recursion Relation

Next, use the recursion relation for \( H_6(\xi) \):\[ H_6(\xi) = 2\xi H_5(\xi) - 10 H_4(\xi) \]From the earlier results: \( H_5(\xi) = 32\xi^5 - 160\xi^3 + 120\xi \) and \( H_4(\xi) = 16\xi^4 - 48\xi^2 + 12 \), calculate:\( H_6(\xi) = 2\xi (32\xi^5 - 160\xi^3 + 120\xi) - 10(16\xi^4 - 48\xi^2 + 12) \)= \( 64\xi^6 - 320\xi^4 + 240\xi^2 - 160\xi^4 + 480\xi^2 - 120 \)= \( 64\xi^6 - 480\xi^4 + 720\xi^2 - 120 \).
05

Differentiate H5 and H6 to Verify Differentiation Property

According to the property, \( \frac{d}{d\xi}H_n(\xi) = 2nH_{n-1}(\xi) \).Check \( H_5(\xi) \):\( \frac{d}{d\xi}(32\xi^5 - 160\xi^3 + 120\xi) = 160\xi^4 - 480\xi^2 + 120 \).\( 2\times5H_4(\xi) = 10(16\xi^4 - 48\xi^2 + 12) = 160\xi^4 - 480\xi^2 + 120 \), which matches.Check \( H_6(\xi) \):\( \frac{d}{d\xi}(64\xi^6 - 480\xi^4 + 720\xi^2 - 120) = 384\xi^5 - 1920\xi^3 + 1440\xi \).\( 2\times6H_5(\xi) = 12(32\xi^5 - 160\xi^3 + 120\xi) = 384\xi^5 - 1920\xi^3 + 1440\xi \), which verifies the derivative.
06

Use Generating Function to Derive H1, H2, H3

The generating function is given by \( e^{-z^2 + 2z\xi} = \sum_{n=0}^{\infty} \frac{z^n}{n!} H_n(\xi) \).For \( H_1(\xi) \), expand up to the linear term: \( -z^2 + 2z\xi = 2z\xi \). The coefficient of \( z \) is \( 2\xi \). Thus, \( H_1(\xi) = 2\xi \).For \( H_2(\xi) \), expand to the quadratic term: \( (-z^2 + 2z\xi)^2/2! = (4z^2\xi^2 - 2z^2)/2 = 2\xi^2 - 1 \). Hence, \( H_2(\xi) = 4\xi^2 - 2 \).Verification for \( H_3(\xi) \): from Rodrigues computation, \( H_3(\xi)=8\xi^3-12\xi \).

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Key Concepts

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

Rodrigues Formula
Rodrigues' formula is a powerful tool for generating Hermite polynomials as it allows for a systematic way to calculate any Hermite polynomial. The formula is expressed as:\[ H_n(\xi) = (-1)^n e^{\xi^2} \left(\frac{d}{d \xi}\right)^n e^{-\xi^2} \] This formula involves
  • Taking the exponential function \(e^{-\xi^2}\)
  • Applying the \(n\)-th derivative
  • Finally, multiplying by \((-1)^n e^{\xi^2}\)
For example, to find \(H_3(\xi)\), we apply the third derivative to \(e^{-\xi^2}\). Following this, we get:
  • First derivative: \(-2\xi e^{-\xi^2}\)
  • Second derivative: \(4\xi^2 e^{-\xi^2} - 2 e^{-\xi^2}\)
  • Third derivative: \(-8 \xi^3 e^{-\xi^2} + 12 \xi e^{-\xi^2}\)
Substituting back, we find \(H_3(\xi) = 8\xi^3 - 12\xi\). This step-by-step differentiation illustrates how the complexity of derivatives transforms into a polynomial form using Rodrigues' formula.
Recursion Relation
The recursion relation provides a straightforward method for calculating higher-order Hermite polynomials from known lower-order ones. The relation is given by:\[ H_{n+1}(\xi) = 2 \xi H_n(\xi) - 2n H_{n-1}(\xi) \]Using this relation:
  • We can compute \(H_5(\xi)\) using \(H_4(\xi)\) and \(H_3(\xi)\)
  • Then compute \(H_6(\xi)\) using \(H_5(\xi)\) and \(H_4(\xi)\)
For instance, using \(H_3(\xi) = 8\xi^3 - 12\xi\) and \(H_4(\xi) = 16\xi^4 - 48\xi^2 + 12\), the relation allows us to find:
  • \(H_5(\xi) = 32\xi^5 - 160\xi^3 + 120\xi\)
  • \(H_6(\xi) = 64\xi^6 - 480\xi^4 + 720\xi^2 - 120\)
The recursion relation simplifies the calculation process significantly by reducing the need for direct derivative computation for higher-order polynomials.
Generating Function
The generating function is an elegant mathematical expression that encodes all Hermite polynomials into a single formula. It is written as:\[ e^{-z^2 + 2z\xi} = \sum_{n=0}^{\infty} \frac{z^n}{n!} H_n(\xi) \]Here's how it works:
  • The coefficient of each \(z^n/n!\) in the series corresponds to the \(n\)-th Hermite polynomial \(H_n(\xi)\)
  • This allows us to find Hermite polynomials by expanding the exponential function
For example, to derive \(H_1(\xi)\), \(H_2(\xi)\), and \(H_3(\xi)\):
  • Expand the expression up to its linear term for \(H_1(\xi)\): \(2z\xi\), resulting in \(H_1(\xi) = 2 \xi\)
  • For \(H_2(\xi)\), expand to quadratic terms: the terms offer \(H_2(\xi) = 4\xi^2 - 2\)
  • \(H_3(\xi)\) coincides with previous findings: \(8\xi^3 - 12\xi\)
This approach is particularly appreciated for its theoretical depth and practical simplicity.
Differentiation Property
The differentiation property of Hermite polynomials states that differentiating a Hermite polynomial yields a multiple of another Hermite polynomial with a degree lower by one:\[ \frac{d}{d \xi} H_n(\xi) = 2n H_{n-1}(\xi) \]This property ensures that:
  • The derivative of a Hermite polynomial retains the polynomial structure but of lower degree
  • The expression above facilitates calculations involving derivatives
Let's verify this by differentiating \(H_5(\xi)\) and \(H_6(\xi)\):
  • \(H_5(\xi) = 32\xi^5 - 160\xi^3 + 120\xi\)
  • Derivative: \(\frac{d}{d \xi}H_5(\xi) = 160\xi^4 - 480\xi^2 + 120\)
  • The relation \(2\times5H_4(\xi) = 10(16\xi^4 - 48\xi^2 + 12)\) is confirmed
Similarly, verifying for \(H_6(\xi)\) gives consistent results, demonstrating the property鈥檚 reliability in Hermite polynomial theory.

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Most popular questions from this chapter

Delta functions live under integral signs, and two expressions \(\left(D_{1}(x) \text { and } D_{2}(x)\right)\) involving delta functions are said to be equal if $$\int_{-\infty}^{+\infty} f(x) D_{1}(x) d x=\int_{-\infty}^{+\infty} f(x) D_{2}(x) d x,$$ for every (ordinary) function \(f(x).\) (a) Show that $$\delta(c x)=\frac{1}{|c|} \delta(x).$$ where \(c\) is a real constant. (Be sure to check the case where \(c\) is negative.) (b) \(\operatorname{Let} \theta(x)\) be the step function: $$\theta(x) \equiv\left\\{\begin{array}{ll}1, & x>0 \\\0, & x<0\end{array}\right.$$ (In the rare case where it actually matters, we define \(\theta(0)\) to be \(1 / 2\).) Show that \(d \theta / d x=\delta(x).\)

Consider the double delta-function potential $$V(x)=-\alpha[\delta(x+a)+\delta(x-a)],$$ where \(a\) and \(a\) are positive constants. (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.

The Boltzmann equation \(^{68}\) $$P(n)=\frac{1}{Z} e^{-\beta E_{n}}, \quad Z \equiv \sum_{n} e^{-\beta E_{n}}, \quad \beta \equiv \frac{1}{k_{B} T}$$ gives the probability of finding a system in the state \(n\) (with energy \(E_{n}\) ), at temperature \(T (k_{B}\) is Boltzmann's constant). Note: The probability here refers to the random thermal distribution, and has nothing to do with quantum indeterminacy. Quantum mechanics will only enter this problem through quantization of the energies \(E_{n}\) (a) Show that the thermal average of the system's energy can be written as $$\bar{E}=\sum_{n} E_{n} P(n)=-\frac{\partial}{\partial \beta} \ln (Z)$$ (b) For a quantum simple harmonic oscillator the index \(n\) is the familiar quantum number, and \(E_{n}=(n+1 / 2) \hbar \omega .\) Show that in this case the partition function \(Z\) is $$Z=\frac{e^{-\beta \hbar \omega / 2}}{1-e^{-\beta \hbar \omega}}$$ You will need to sum a geometric series. Incidentally, for a classical simple harmonic oscillator it can be shown that \(Z\) classical \(=2 \pi /(\omega \beta)\) (c) Use your results from parts (a) and (b) to show that for the quantum oscillator $$\bar{E}=\left(\frac{\hbar \omega}{2}\right) \frac{1+e^{-\beta \hbar \omega}}{1-e^{-\beta \hbar \omega}}$$ For a classical oscillator the same reasoning would give \(\bar{E}_{\text {classical }}=1 / \beta=k_{B} T\) (d) Acrystal consisting of \(N\) atoms can be thought of as a collection of \(3 N\) oscillators (each atom is attached by springs to its 6 nearest neighbors, along the \(x, y,\) and \(z\) directions, but those springs are shared by the atoms at the two ends). The heat capacity of the crystal (per atom) will therefore be $$C=3 \frac{\partial \bar{E}}{\partial T}$$ Show that (in this model) $$C=3 k_{B}\left(\frac{\theta_{E}}{T}\right)^{2} \frac{e^{\theta_{E} / T}}{\left(e^{\theta_{E} / T}-1\right)^{2}}$$ where \(\theta_{E} \equiv \hbar \omega / k_{B}\) is the so-called Einstein temperature. The same reasoning using the classical expression for \(\bar{E}\) yields \(C_{\text {classical }}=3 k_{B}\) independent of temperature. (e) Sketch the graph of \(C / k_{B}\) versus \(T / \theta_{E} .\) Your result should look something like the data for diamond in Figure 2.24 , and nothing like the classical prediction.

Suppose $$V(x)=\left\\{\begin{array}{ll} m g x, & x>0 \\ \infty, & x \leq 0 \end{array}\right.$$ (a) Solve the (time-independent) Schr枚dinger equation for this potential. Hint: First convert it to dimensionless form: $$-y^{\prime \prime}(z)+z y(z)=\epsilon y(z)$$ by letting \(z \equiv a x\) and \(y(z) \equiv(1 / \sqrt{a}) \psi(x)\) (the \(\sqrt{a}\) is just so \(y(z)\) is normalized with respect to \(z\) when \(\psi(x)\) is normalized with respect to \(x\) ) What are the constants \(a\) and \(\varepsilon\) ? Actually, we might as well set \(a \rightarrow 1-\) this amounts to a convenient choice for the unit of length. Find the general solution to this equation (in Mathematica DSolve will do the job). The result is (of course) a linear combination of two (probably unfamiliar) functions. Plot each of them, for \((-15

The gaussian wave packet. 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 \(\left(a x^{2}+b x\right)=y^{2}-\left(b^{2} / 4 a\right) .\) Answer (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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