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Chapter 40: Problem 56

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

Short Answer

Expert verified
After direct substitution, the wave function \(\psi_{1}(x)=A_{1} x e^{-\alpha^{2} x^{2} / 2},\) is validated as a solution to Schr枚dinger鈥檚 equation. Upon solving the normalization condition integral, we obtain the value of normalization constant \(A_{1}\). Derivative of probability density provides us with positions of minima and maxima. The maximum values occur at \(x=\pm 1 / \alpha\), and minimum value is at \(x=0\).

Step by step solution

01

Direct Substitution in Schr枚dinger Equation

First, it's necessary to show that \(\psi_{1}(x)=A_{1} x e^{-\alpha^{2} x^{2} / 2}\) solves the Schr枚dinger equation for the harmonic oscillator. The Schr枚dinger equation for the harmonic oscillator is given by \[ (-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} + \frac{1}{2}m\omega^{2}x^{2})\psi = E\psi \] Substitute the given \(\psi\) into this equation and simplify.
02

Find the Normalization Constant \(A_{1}\)

After confirming the provided wave function solves the Schr枚dinger equation, you need to find the normalization constant \(A_{1}\). This will ensure the probability density integrates to 1 over all space. To find \(A_{1}\), set up and solve the integral \[ \int_{-\infty}^{\infty}|\psi_{1}(x)|^{2}dx = 1 \]
03

Analyze Probability Density

Lastly, demonstrate that the probability density has a minimum at \(x=0\) and maxima at \(x=\pm 1 / \alpha\). To do this, compute the derivative of the probability density \(\rho(x) = |\psi_{1}(x)|^{2}\), and then identify the points where this derivative is 0, as these correspond to the minima and maxima of \(\rho(x)\).

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Key Concepts

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

Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It is notably different from classical mechanics, as it introduces unique concepts such as quantization of energy, wave-particle duality, and the uncertainty principle.

One of the crucial aspects of quantum mechanics is the Schr枚dinger equation, which is a key mathematical equation that describes how the quantum state of a physical system changes over time. It forms the basis of wave mechanics and allows physicists to calculate the probability of finding a particle in a particular region of space, rather than predicting its exact location and momentum, which is not possible due to the Heisenberg Uncertainty Principle.
Harmonic Oscillator Wave Functions
In quantum mechanics, the harmonic oscillator is a system that experiences a restoring force proportional to the displacement from its equilibrium position. It is analogous to a mass attached to a spring in classical mechanics but with quantized energy levels.

The wave functions for the quantum harmonic oscillator are known as eigenfunctions and represent the state of the system with specific quantized energy levels. These functions, typically denoted as \(\psi_n(x)\), where \(n\) is the principal quantum number, are solutions to the Schr枚dinger equation for the harmonic oscillator potential. Each wave function corresponds to a particular energy state, labeled by \(n\), which determines the allowable energy levels for the oscillator.
Probability Density in Quantum Systems
Probability density \(\rho(x)\) in quantum mechanics is a measure of the likelihood of finding a particle in a particular position in space. The probability density is the square of the modulus of the wave function, \(\rho(x) = |\psi(x)|^2\).

For the harmonic oscillator, the probability density reflects the particle's position probability distribution and varies with different energy states. In the first excited state (\(n=1\)), the probability density has a characteristic shape with a minimum at the equilibrium position (\(x=0\)) and maxima at certain positions, which correlates to the energy levels of the system. These positions often align with the classical turning points, reinforcing the correspondence principle, which states that the behavior of systems described by quantum mechanics replicates classical physics in the limit of large quantum numbers.
Normalization of Wave Functions
Normalization of wave functions is a fundamental step in quantum mechanics, ensuring that the wave functions represent valid probability densities. A normalized wave function allows the total probability of finding a particle in all space to equal one, which is a physical requirement since the particle must exist somewhere in space.

To normalize a wave function \(\psi(x)\), the integral of the probability density must be set to one across all space: \[ \int_{-\infty}^{\infty}|\psi(x)|^{2}dx = 1 \] Normalizing the wave function for a harmonic oscillator entails finding the normalization constant \(A_n\) that will make this integral be unity. This constant is then used in calculations of measurable quantities like the expectation value of position or momentum.

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Most popular questions from this chapter

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 \(3097 \mathrm{MeV} ;\) the \(\psi(2 \mathrm{~S})\) particle is an excited state of this same quark-antiquark combination, with a rest energy of \(3686 \mathrm{MeV}\). A simplified representation of the potential energy of interaction between a quark and an antiquark is \(U(x)=A|x|,\) 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 $$ \int_{a}^{b} \sqrt{2 m[E-U(x)]} d x=\frac{n h}{2} \quad(n=1,2,3, \ldots) $$ 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 \(U(x)=A|x|\) 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 $$ E_{n}=\frac{1}{2 m}\left(\frac{3 m A h}{4}\right)^{2 / 3} n^{2 / 3} \quad(n=1,2,3, \ldots) $$ (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?

The penetration distance \(\eta\) in a finite potential well is the distance at which the wave function has decreased to \(1 / e\) of the (b) wave function at the classical turning point: $$ \psi(x=L+\eta)=\frac{1}{e} \psi(L) $$ The penetration distance can be shown to be $$\eta=\frac{\hbar}{\sqrt{2 m\left(U_{0}-E\right)}}$$ The probability of finding the particle beyond the penetration distance is nearly zero. (a) Find \(\eta\) for an electron having a kinetic energy of \(13 \mathrm{eV}\) in a potential well with \(U_{0}=20 \mathrm{eV} .\) (b) Find \(\eta\) for a \(20.0 \mathrm{MeV}\) proton trapped in a 30.0 -MeV-deep potential well.

An electron is bound in a square well that has a depth equal to six times the ground-level energy \(E_{1-\mathrm{IDW}}\) of an infinite well of the same width. The longest-wavelength photon that is absorbed by this electron has a wavelength of \(582 \mathrm{nm}\). Determine the width of the well.

In your research on new solid-state devices, you are studying a solid-state structure that can be modeled accurately as an electron in a one-dimensional infinite potential well (box) of width \(L\). In one of your experiments, electromagnetic radiation is absorbed in transitions in which the initial state is the \(n=1\) ground state. You measure that light of frequency \(f=9.0 \times 10^{14} \mathrm{~Hz}\) is absorbed and that the next higher absorbed frequency is \(16.9 \times 10^{14} \mathrm{~Hz}\). (a) What is quantum number \(n\) for the final state in each of the transitions that leads to the absorption of photons of these frequencies? (b) What is the width \(L\) of the potential well? (c) What is the longest wavelength in air of light that can be absorbed by an electron if it is initially in the \(n=1\) state?

An electron is bound in a square well of width \(1.50 \mathrm{nm}\) and 40.25 depth \(U_{0}=6 E_{1-\mathrm{IDW}}\). If the electron is initially in the ground level and absorbs a photon, what maximum wavelength can the photon have and still liberate the electron from the well?
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