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(a) Show by direct substitution in the Schr\(\ddot{o}\)dinger equation for the one-dimensional harmonic oscillator that the wave function \(\psi_1(x) = A_1xe^{-a^2x^2/2}\), where \(\alpha^2 = m\omega/\hslash\), is a solution with energy corresponding to \(n\) = 1 in Eq. (40.46). (b) Find the normalization constant A1. (c) Show that the probability density has a minimum at \(x\) = 0 and maxima at \(x = \pm1/\alpha\), corresponding to the classical turning points for the ground state \(n\) = 0.

Short Answer

Expert verified
The wave function satisfies the Schrödinger equation for \(n=1\), \(A_1 = \sqrt{\frac{2\alpha^3}{\sqrt{\pi}}}\), and has a minimum at \(x=0\) and maxima at \(x=\pm1/\alpha\).

Step by step solution

01

Understand the Schrödinger equation

The time-independent Schrödinger equation for a one-dimensional harmonic oscillator is given by: \[ -\frac{\hslash^2}{2m} \frac{d^2 \psi(x)}{dx^2} + \frac{1}{2} m \omega^2 x^2 \psi(x) = E \psi(x) \]where \(m\) is the mass, \(\omega\) is the angular frequency, and \(E\) is the energy of the state. We need to show that \(\psi_1(x) = A_1x e^{-\alpha^2x^2/2}\) satisfies this equation for \(n = 1\). Also, \(\alpha = \sqrt{\frac{m\omega}{\hslash}}\).
02

Compute the derivatives

First, compute the first and second derivatives of the given wave function \(\psi_1(x) = A_1x e^{-\alpha^2x^2/2}\):\[ \frac{d \psi_1}{dx} = A_1 e^{-\alpha^2x^2/2} \left(1 - \alpha^2 x^2\right) \]\[ \frac{d^2 \psi_1}{dx^2} = A_1 e^{-\alpha^2x^2/2} \left(-\alpha^2 + \alpha^4 x^2 \right) \]
03

Substitute into Schrödinger equation

Substitute \(\psi_1(x)\), \(d\psi_1/dx\), and \(d^2\psi_1/dx^2\) into the Schrödinger equation:Replace \(\psi(x)\), \(d^2\psi(x)/dx^2\), and \(x^2 \psi(x)\) in the equation. Show that this results in kinetic energy minus potential energy equaling the total energy when \(E = \frac{3}{2} \hslash \omega\), the first excited state energy.
04

Determine the normalization constant

For normalization, integrate the probability density over all space:\[ \int_{-\infty}^{\infty} |\psi_1(x)|^2 \, dx = 1 \]Compute:\[ \int_{-\infty}^{\infty} A_1^2 x^2 e^{-\alpha^2 x^2} \, dx = 1 \]Using Gaussian integrals, \(A_1 = \left(\frac{2\alpha^3}{\sqrt{\pi}}\right)^{1/2}\).
05

Analyze probability density

The probability density is:\[ |\psi_1(x)|^2 = A_1^2 x^2 e^{-\alpha^2 x^2} \]Find its critical points by differentiating and setting the derivative to zero:\[ \frac{d}{dx}(|\psi_1(x)|^2) = A_1^2 e^{-\alpha^2 x^2}(2x - 2\alpha^2 x^3) = 0 \]Solving gives \(x = 0\) (minimum) and \(x = \pm1/\alpha\) (maxima), explaining these are the classical turning points for \(n = 0\).

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Key Concepts

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

Harmonic Oscillator
The harmonic oscillator is a fundamental model in both classical and quantum mechanics. It describes a system where a particle experiences a restoring force proportional to its displacement from an equilibrium position. In simple terms, imagine a mass attached to a spring. When the mass is displaced and released, it oscillates back and forth around the equilibrium point. This back-and-forth motion can be thought of as a harmonic oscillation.
  • In classical mechanics, the harmonic oscillator is solved using Newton's laws and results in sinusoidal motion at a constant frequency.
  • In quantum mechanics, the potential energy of a harmonic oscillator is represented by a quadratic function proportional to the square of the displacement.
The quantum harmonic oscillator forms the basis for understanding more complex systems. It is one of the few quantum systems that can be solved analytically. Solutions to the quantum harmonic oscillator involve special mathematical functions known as Hermite polynomials. These solutions provide insights into the quantized energy levels of the oscillator.
Wave Function Normalization
The concept of wave function normalization is crucial in quantum mechanics. It ensures that the total probability of finding a particle somewhere in space is equal to one. This reflects the fact that the particle must exist somewhere in space, as probability is a measure of certainty of an event occurring.
For a wave function like \[\psi_1(x) = A_1 x e^{-\alpha^2 x^2/2},\]normalizing involves finding the constant \(A_1\) such that: \[\int_{-\infty}^{\infty} |\psi_1(x)|^2 \ dx = 1.\]
  • This requires integrating the square of the wave function over all space.
  • In our specific case, Gaussian integrals play a vital role in evaluating this integral due to the presence of exponential terms.
Successfully normalizing the wave function allows physicists to predict where the particle could be with certainty across its domain.
Probability Density
In the context of quantum mechanics, probability density refers to the likelihood of locating a particle in a particular region of space. The square of the wave function, \(|\psi(x)|^2\), represents this density. For the harmonic oscillator, the probability density informs us of the most likely positions of the particle.
For the wave function \[\psi_1(x) = A_1 x e^{-\alpha^2 x^2/2}\], the probability density becomes:\[|\psi_1(x)|^2 = A_1^2 x^2 e^{-\alpha^2 x^2}\]. This has a profound implication:
  • At \(x = 0\), \(|\psi_1(x)|^2\) has a minimum, indicating a lower likelihood of finding the particle at the center.
  • The probability density has maxima at \(x = \pm1/\alpha\), which correspond to the classical turning points, where the particle changes direction.
These turning points offer a key insight into how quantum particles behave similarly yet distinctly compared to classical particles.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with systems at atomic and subatomic scales. It introduces us to a universe where particles exhibit wave-like properties, allowing for phenomena like interference and diffraction, unseen in classical mechanics.
Key concepts in quantum mechanics include:
  • Wave-Particle Duality: Particles act as both waves and particles, depending on how you observe them.
  • Uncertainty Principle: You can't know both the position and momentum of a particle with absolute certainty.
  • Quantization: Certain properties, like energy, are quantized, meaning they can only take on specific discrete values.
The Schrödinger equation is the cornerstone of quantum mechanics, describing how the quantum state of a physical system changes over time. It offers solutions such as wave functions that encapsulate all possible information about a system. Understanding these concepts is crucial, as they underpin all quantum phenomena and are essential for exploring and explaining the behavior of matter at the smallest scales.

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Most popular questions from this chapter

An electron is in the ground state of a square well of width \(L = 4.00 \times 10^{-10}\) m. The depth of the well is six times the ground-state energy of an electron in an infinite well of the same width. What is the kinetic energy of this electron after it has absorbed a photon of wavelength 72 nm and moved away from the well?

Protons, neutrons, and many other particles are made of more fundamental particles called \(quarks\) and \(antiquarks\) (the antimatter equivalent of quarks). A quark and an antiquark can form a bound state with a variety of different energy levels, each of which corresponds to a different particle observed in the laboratory. As an example, the \(\psi\) particle is a low-energy bound state of a so-called charm quark and its antiquark, with a rest energy of 3097 MeV; the \(\psi\)(2S) particle is an excited state of this same quark- antiquark combination, with a rest energy of 3686 MeV. A simplified representation of the potential energy of interaction between a quark and an antiquark is \(U(x) = A\mid x \mid\) , where \(A\) is a positive constant and \(x\) represents the distance between the quark and the antiquark. You can use the WKB approximation (see Challenge Problem 40.64) to determine the bound-state energy levels for this potential-energy function. In the WKB approximation, the energy levels are the solutions to the equation $$\int ^b _a \sqrt{2m[E - U(x)]} dx = {nh \over 2} \space (n = 1, 2, 3, . . .)$$ Here \(E\) is the energy, \(U(x)\) is the potential-energy function, and \(x\) = a and \(x\) = b are the classical turning points (the points at which \(E\) is equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for the potential \(U(x) = A \mid x \mid\) and for an energy \(E\). (b) Carry out the above integral and show that the allowed energy levels in the WKB approximation are given by $$E_n = {1 \over2m} ( {3mAh \over 4} ) ^{2/3} n^{2/3} \space (n = 1, 2, 3, . . .)$$ (\(Hint\): The integrand is even, so the integral from \(-x\) to \(x\) is equal to twice the integral from 0 to \(x\).) (c) Does the difference in energy between successive levels increase, decrease, or remain the same as \(n\) increases? How does this compare to the behavior of the energy levels for the harmonic oscillator? For the particle in a box? Can you suggest a simple rule that relates the difference in energy between successive levels to the shape of the potential-energy function?

One advantage of the quantum dot is that, compared to many other fluorescent materials, excited states have relatively long lifetimes (10 ns). What does this mean for the spread in the energy of the photons emitted by quantum dots? (a) Quantum dots emit photons of more well-defined energies than do other fluorescent materials. (b) Quantum dots emit photons of less well-defined energies than do other fluorescent materials. (c) The spread in the energy is affected by the size of the dot, not by the lifetime. (d) There is no spread in the energy of the emitted photons, regardless of the lifetime.

A harmonic oscillator absorbs a photon of wavelength 6.35 \(\mu\)m when it undergoes a transition from the ground state to the first excited state. What is the ground-state energy, in electron volts, of the oscillator?

Consider the wave packet defined by $$\psi(x) = \int ^\infty_0 B(k)cos kx dk$$ Let \(B(k) = e^{-a^2k^2}\). (a) The function \(B(k)\) has its maximum value at \(k\) = 0. Let \(k_h\) be the value of \(k\) at which \(B(k)\) has fallen to half its maximum value, and define the width of \(B(k)\) as \(w_k = k_h\) . In terms of \(\alpha\), what is \(w_k\) ? (b) Use integral tables to evaluate the integral that gives \(\psi(x)\). For what value of \(x\) is \(\psi(x)\) maximum? (c) Define the width of \(\psi(x)\) as \(w_x = x_h\) , where \(x_h\) is the positive value of \(x\) at which \(\psi(x)\) has fallen to half its maximum value. Calculate \(w_x\) in terms of \(\alpha\). (d) The momentum \(p\) is equal to \(hk/2\pi\), so the width of \(B\) in momentum is \(w_p = hw_k /2\pi\). Calculate the product \(w_p w_x\) and compare to the Heisenberg uncertainty principle.

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