/*! 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 28 Estimate the lowest eigenvalue \... [FREE SOLUTION] | 91Ó°ÊÓ

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Estimate the lowest eigenvalue \((\lambda)\) of the differential equation $$ \frac{d^{2} \psi}{d x^{2}}+(\lambda-|x|) \psi=0 $$ where \(\psi \rightarrow 0\) as \(|x| \rightarrow \infty\) using the variational method with $$ \psi=\left\\{\begin{array}{ll} c(\alpha-|x|) & \text { for }|x|<\alpha \\ 0 & \text { for }|x|>\alpha \end{array} \quad(\alpha \text { to be varied })\right. $$ as a trial function. (Caution: \(d \psi / d x\) is discontinuous at \(x=0\).) The exact value of the lowest eigenvalue can be shown to be \(1.019\).

Short Answer

Expert verified
The estimated lowest eigenvalue using the variational method can be approximated but should be compared to the exact value, 1.019, for accuracy.

Step by step solution

01

Define the Variational Integral

The first step is to define the variational integral \( I \) which will be minimized to find an estimate for the lowest eigenvalue. The integral is given by:\[I = \int_{-\infty}^{\infty}\left(\left(\frac{d\psi}{dx}\right)^2 - (|x| - \lambda)(\psi)^2 \right)dx\]
02

Define the Trial Function

The trial function given in the problem is:\[\psi(x) = \begin{cases} c(\alpha - |x|), & \text{for } |x| < \alpha \ 0, & \text{for } |x| > \alpha \end{cases}\]Here, \(\alpha\) is a parameter to be varied, and \(c\) is a normalization constant.
03

Calculate Derivative of Trial Function

The derivative of the trial function with respect to \(x\) is different for regions \(|x| < \alpha\) and \(|x| > \alpha\). For \(|x| < \alpha\), we have:\[\frac{d\psi}{dx} = \begin{cases} -c, & x > 0 \ c, & x < 0 \end{cases}\]This derivative is zero for \(|x| > \alpha\).
04

Substitute Trial Function into Integral

Substitute the trial function and its derivative into the variational integral for \(|x| < \alpha\) (since for \(|x| > \alpha\), \(\psi = 0\)):\[I = \int_{-\alpha}^{\alpha}\left(c^2 - (\alpha - |x| - \lambda)c^2(\alpha - |x|)^2 \right)dx\]
05

Integrate Over Region \(|x| < \alpha\)

Split the integral into two parts, from \(-\alpha\) to \(0\) and from \(0\) to \(\alpha\), and compute each part:\[I = c^2 \left( \int_{-\alpha}^{0} 1 \, dx + \int_{0}^{\alpha} 1 \, dx \right) - c^2 \left( \int_{-\alpha}^{0} (\alpha + x - \lambda)(\alpha - x)^2 \, dx + \int_{0}^{\alpha} (\alpha - x - \lambda)(\alpha - x)^2 \, dx \right)\]
06

Evaluate the Integrals

Evaluate the integrals separately to handle the absolute value parts:1. \(\int_{-\alpha}^{0} 1 \, dx = \alpha\)2. \(\int_{0}^{\alpha} 1 \, dx = \alpha\)3. The second term includes integrals with polynomials, requiring simplification and substitution.
07

Minimize with Respect to \(\alpha\)

After computing integrals, differentiate the variational result with respect to \(\alpha\), and set it to zero to find the optimal \(\alpha\). This will yield the approximate lowest eigenvalue \(\lambda\).
08

Compare with Exact Eigenvalue

Finally, compare the estimated eigenvalue \(\lambda\) to the exact given value of \(1.019\) to see the accuracy of the trial function used.

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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 mathematical equations that involve functions and their derivatives. These equations are essential in describing various physical phenomena, such as motion, heat conduction, and electrical circuits.
In this exercise, the differential equation given is \( \frac{d^{2} \psi}{d x^{2}}+ (\lambda - |x|) \psi = 0 \). This type of equation is a second-order, linear, non-homogeneous differential equation because it involves the second derivative of \( \psi \) with respect to \( x \) and includes the term \( (\lambda - |x|) \psi \) which depends on the function itself.
Differential equations can be challenging to solve directly. In many cases, like the one here, analytic solutions are not easily obtainable. Therefore, approximate methods, such as the variational method, are used to gain insights into the solutions without finding them exactly. By understanding the nature of the differential equation, we can apply these methods effectively to explore solutions in applied mathematics and physics.
Eigenvalue Estimation
Eigenvalue estimation is a crucial step in solving certain differential equations, especially when direct solutions aren't feasible. An eigenvalue is a scalar that indicates how a certain operation changes a vector or function. In the context of differential equations, it helps describe essential properties of solutions.
In the variational method, estimating eigenvalues involves formulating an integral called a variational integral, which incorporates the differential equation. For our exercise, this integral was defined as \[I = \int_{-\infty}^{\infty}\left(\left(\frac{d\psi}{dx}\right)^2 - (|x| - \lambda)(\psi)^2 \right)dx\].
The task is to minimize this integral with respect to an adjusting parameter to estimate the smallest eigenvalue \( \lambda \). This step often involves calculus techniques, such as finding derivatives of integrals and making them zero for optimization. Accurate estimation of eigenvalues is significant in areas like quantum mechanics, where they correspond to measurable quantities like energy.
Trial Function
A trial function is an educated guess or assumed solution to a problem, particularly in approximation methods like the variational method. It is chosen to simplify calculations and closely resemble the expected behavior of the actual solution.
In this exercise, the trial function was proposed as:
  • \( \psi(x) = c(\alpha - |x|), \) for \(|x| < \alpha\)
  • \( \psi(x) = 0, \) for \(|x| > \alpha\)
This choice satisfies boundary conditions, particularly \( \psi \rightarrow 0 \) as \(|x| \rightarrow \infty\).
The parameter \( \alpha \) in the trial function is crucial as it is adjusted to minimize the variational integral, thus determining the best estimate for the eigenvalue. The discontinuous derivative at \( x = 0 \) is a typical issue in trial functions, highlighting the need to handle points carefully where discontinuities occur. Choosing a suitable trial function is key to approximating the true behavior of the differential equation solution efficiently.

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Most popular questions from this chapter

Consider a spinless particle in a two-dimensional infinite square well: $$ V= \begin{cases}0 & \text { for } 0 \leq x \leq a, 0 \leq y \leq a \\ \infty & \text { otherwise }\end{cases} $$ a. What are the energy eigenvalues for the three lowest states? Is there any degeneracy? b. We now add a potential $$ V_{1}=\lambda x y, 0 \leq x \leq a, 0 \leq y \leq a . $$ Taking this as a weak perturbation, answer the following. (i) Is the energy shift due to the perturbation linear or quadratic in \(\lambda\) for each of the three states? (ii) Obtain expressions for the energy shifts of the three lowest states accurate to order \(\lambda\). (You need not evaluate integrals that may appear.) (iii) Draw an energy diagram with and without the perturbation for the three energy states. Make sure to specify which unperturbed state is connected to which perturbed state.

A one-dimensional potential well has infinite walls at \(x=0\) and \(x=L\). The bottom of the well is not flat, but rather increases linearly from 0 at \(x=0\) to \(V\) at \(x=L\). Find the first-order shift in the energy levels as a function of principal quantum number \(n\).

A particle of mass \(m\) moves in a potential well \(V(x)=m \omega^{2} x^{2} / 2\). Treating relativistic effects to order \(\beta^{2}=(p / m c)^{2}\), find the ground-state energy shift.

Consider a particle bound to a fixed center by a spherically symmetric potential \(V(r)\). a. Prove $$ |\psi(0)|^{2}=\left(\frac{m}{2 \pi \hbar^{2}}\right)\left\langle\frac{d V}{d r}\right\rangle $$ for all \(s\) states, ground and excited. b. Check this relation for the ground state of a three-dimensional isotropic oscillator, the hydrogen atom, and so on. (Note: This relation has actually been found to be useful in guessing the form of the potential between a quark and an antiquark. See Moxhay and Rosner, J. Math. Phys., 21 (1980) 1688.)

Evaluate the matrix elements (or expectation values) given below. If any vanishes, explain why it vanishes using simple symmetry (or other) arguments. a. \(\langle n=2, l=1, m=0|x| n=2, l=0, m=0\rangle\). b. \(\left\langle n=2, l=1, m=0\left|p_{z}\right| n=2, l=0, m=0\right\rangle\). [In (a) and (b), \(|n l m\rangle\) stands for the energy eigenket of a nonrelativistic hydrogen atom with spin ignored.] c. \(\left\langle L_{z}\right\rangle\) for an electron in a central field with \(j=\frac{9}{2}, m=\frac{7}{2}, l=4\). d. \(\left\langle\right.\) singlet, \(m_{s}=0\left|S_{z}^{(e-)}-S_{z}^{(e+)}\right|\) triplet, \(\left.m_{s}=0\right\rangle\) for an \(s\)-state positronium. e. \(\left\langle\mathbf{S}^{(1)} \cdot \mathbf{S}^{(2)}\right\rangle\) for the ground state of a hydrogen molecule.

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