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Chapter 7: Problem 14

\- Consider a particle in a spherical box of radius \(a\). The Hamiltonian operator for this system is (see Equation 6.43) $$ \hat{H}=-\frac{\hbar^{2}}{2 m r^{2}} \frac{d}{d r}\left(r^{2} \frac{d}{d r}\right)+\frac{\hbar^{2} l(l+1)}{2 m r^{2}} \quad 0

Short Answer

Expert verified
Upper bound energy is \(\frac{\hbar^{2}}{m} a^2 (1 - \ln(a))\).

Step by step solution

01

Understand the Problem

The task is to use the Hamiltonian of a particle in a spherical box with a specific wave function to estimate an upper bound for the ground-state energy. The trial wave function given is \(\phi(r) = a - r\), valid for \(0 < r \leq a\).
02

Hamiltonian for Ground State

The Hamiltonian operator for \(l=0\) simplifies to \(\hat{H} = -\frac{\hbar^{2}}{2 m r^{2}} \frac{d}{d r}\left(r^{2} \frac{d}{d r}\right)\). We begin with this simplified form to evaluate the energy of the given trial wave function.
03

Compute Derivatives of \(\phi(r)\)

Calculate the first derivative: \(\frac{d\phi}{dr} = -1\). Then, compute the second derivative: \(\frac{d^{2}\phi}{dr^{2}} = 0\). These derivatives are needed to apply the Hamiltonian operator.
04

Apply Hamiltonian to \(\phi(r)\)

Plug the derivatives into the Hamiltonian: \(\hat{H} \phi(r) = -\frac{\hbar^{2}}{2 m r^{2}} \frac{d}{dr}\left(r^{2}(-1)\right) = \frac{\hbar^{2}}{2 m r^{2}} \cdot 2r = \frac{\hbar^{2}}{m r}\).
05

Calculate the Energy Expectation Value

The expectation value for energy is given by the integral: \(E = \int_{0}^{a} \phi^*(r) \hat{H} \phi(r) \, dr\). Substitute and compute: \[E = \int_{0}^{a} (a-r) \left(\frac{\hbar^{2}}{m r}\right) (a-r) \, dr = \frac{\hbar^{2}}{m} \int_{0}^{a} \frac{(a-r)^2}{r} \, dr\].
06

Solve the Integral

Evaluate the integral \(\int_{0}^{a} \frac{(a-r)^2}{r} \, dr\). By substitution and algebraic manipulation, solve this integral. The result of this integral is \(a^2(1 - \ln(a) + 0)\).
07

Compute Final Energy Expression

Substitute the integral result back into the energy expression: \[E = \frac{\hbar^{2}}{m} a^2 (1 - \ln(a))\]. This is the estimated upper bound for the ground-state energy.

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Key Concepts

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

Hamiltonian Operator
In quantum mechanics, the Hamiltonian operator, represented as \(\hat{H}\), is a crucial element used to describe the total energy of a system. It acts on the wave function of a quantum system to calculate the energy state of that system.
In this exercise about a spherical box, the Hamiltonian involves both the kinetic and potential energy terms. The significance of the Hamiltonian lies in its ability to give us the energy levels of the system, which are solutions to the Schr枚dinger equation. For a particle inside the spherical box, this Hamiltonian is given by:
  • Kinetic Energy Contribution: \(-\frac{\hbar^2}{2m} abla^2\)
  • Potential Energy: Since the particle is within a potential box, the potential energy inside is zero.
The operator simplifies significantly when applied to the ground state, where the angular momentum quantum number \(l\) is zero. Understanding the role of the Hamiltonian provides insights into how energy states are derived through mathematical operations.
Ground-State Energy
Ground-state energy refers to the lowest possible energy that a quantum mechanical system can have. Unlike classical systems, quantum systems always have energy due to the Heisenberg uncertainty principle. This energy is not zero, even for the ground state.
The importance of calculating an upper bound for the ground state energy, as seen in this exercise, lays the groundwork for verifying how close our estimated energies are to exact solutions. In this particular scenario, the trial wave function \(\phi(r) = a - r\) is used to estimate the energy using different mathematical methods.
Here are a few highlights of the ground-state calculation:
  • The energy is derived from the expectation value \(E = \int_0^a \phi^*(r) \hat{H} \phi(r) \, dr\).
  • An upper bound implies that calculated energies should not underestimate actual values.
  • Verification of calculations can be achieved by comparing with the known exact ground-state energy \(\frac{\pi^2 \hbar^2}{2ma^2}\).
Understanding the methods used to estimate ground-state energies aids significantly in mastering quantum mechanics topics.
Spherical Box Potential
A spherical box potential is a model used in quantum mechanics to describe a system where a particle is confined within a spherical region. This potential is an ideal structure where the boundaries are rigid and defined by a radius \(a\).
Understanding a spherical potential helps in examining how particles behave within confined geometries, similar to electrons within atoms or molecules.
Important features of the spherical box potential include:
  • Boundary conditions: The wave function \(\phi(r)\) must vanish at the boundary \(r=a\), i.e., \(\phi(a) = 0\).
  • The potential outside the sphere is considered to be infinitely large, thus preventing the particle from being found outside \(r > a\).
  • The spherical symmetry simplifies some mathematical aspects and computation of integrals involved in analyzing the system.
Utilizing the spherical box model enhances comprehension of how particles like electrons are influenced by confining potentials.

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Most popular questions from this chapter

Consider a system subject to the potential $$ V(x)=\frac{k}{2} x^{2}+\frac{\gamma}{6} x^{3}+\frac{\delta}{24} x^{4} $$ Calculate the ground-state energy of this system using a trial function of the form $$ \phi=c_{1} \psi_{0}(x)+c_{2} \psi_{2}(x) $$ where \(\psi_{0}(x)\) and \(\psi_{2}(x)\) are the first two even wave functions of a harmonic oscillator. Why did we not include \(\psi_{1}(x) ?\)

In applying first-order perturbation theory to the helium atom, we must evaluate the integral (Equation 7.50) $$ E^{(1)}=\frac{e^{2}}{4 \pi \varepsilon_{0}} \iint d \mathbf{r}_{1} d \mathbf{r}_{2} \psi_{1 s}^{*}\left(\mathbf{r}_{1}\right) \psi_{1 s}^{*}\left(\mathbf{r}_{2}\right) \frac{1}{r_{12}} \psi_{1 s}\left(\mathbf{r}_{1}\right) \psi_{1 s}\left(\mathbf{r}_{2}\right) $$ where $$ \psi_{1 s}\left(\mathbf{r}_{j}\right)=\left(\frac{Z^{3}}{a_{0}^{3} \pi}\right)^{1 / 2} e^{-Z \mathbf{r}_{j} / a_{0}} $$ and \(Z=2\) for the helium atom. This same integral occurs in a variational treatment of nelium, where in that case the value of \(Z\) is left arbitrary. This problem proves that $$ E^{(1)}=\frac{5 Z}{8}\left(\frac{m_{e} e^{4}}{16 \pi^{2} \varepsilon_{0}^{2} \hbar^{2}}\right) $$ Let \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) be the radius vectors of electron 1 and 2, respectively, and let \(\theta\) be the angle between these two vectors. Now this is generally not the \(\theta\) of spherical coordinates, but if we choose one of the radius vectors, say \(\mathbf{r}_{1}\), to be the \(z\) axis, then the two \(\theta\) 's are the same. Using the law of cosines, $$ r_{12}=\left(r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \theta\right)^{1 / 2} $$ show that \(E^{(1)}\) becomes $$ \begin{aligned} E^{(1)}=& \frac{e^{2}}{4 \pi \varepsilon_{0}} \frac{Z^{6}}{a_{0}^{6} \pi^{2}} \int_{0}^{\infty} d r_{1} e^{-2 Z r_{1} / a_{0} 4 \pi r_{1}^{2} \int_{0}^{\infty} d r_{2} e^{-2 Z r_{2} / a_{0}} r_{2}^{2}} \\ & \times \int_{0}^{2 \pi} d \phi \int_{0}^{\pi} \frac{d \theta \sin \theta}{\left(r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \theta\right)^{1 / 2}} \end{aligned} $$In Problem 7-30 we evaluated the integral that occurs in the first-order perturbation theory treatment of helium (see Equation 7.50). In this problem we will evaluate the integral by another method, one that uses an expansion for \(1 / r_{12}\) that is useful in many applications. We can write \(1 / r_{12}\) as an expansion in terms of spherical harmonics $$ \frac{1}{r_{12}}=\frac{1}{\left|\mathbf{r}_{1}-\mathbf{r}_{2}\right|}=\sum_{l=0}^{\infty} \sum_{m=-l}^{+l} \frac{4 \pi}{2 l+1} \frac{r_{<}^{l}}{r_{>}^{l+1}} Y_{l}^{m}\left(\theta_{1}, \phi_{1}\right) Y_{l}^{m *}\left(\theta_{2}, \phi_{2}\right) $$ where \(\theta_{i}\) and \(\phi_{i}\) are the angles that describe \(\mathbf{r}_{i}\) in a spherical coordinate system and \(r_{<}\)and \(r\) > are, respectively, the smaller and larger values of \(r_{1}\) and \(r_{2} .\) In other words, if \(r_{1}}=r_{2} .\) Substitute \(\psi_{1 s}\left(r_{i}\right)=\left(Z^{3} / a_{0}^{3} \pi\right)^{1 / 2} e^{-Z r_{1} / a_{0}}\), and the above expansion for \(1 / r_{12}\) into Equation \(7.50\), integrate over the angles, and show that all the terms except for the \(l=0, m=0\) term vanish. Show that $$ E^{(1)}=\frac{e^{2}}{4 \pi \varepsilon_{0}} \frac{16 Z^{6}}{a_{0}^{6}} \int_{0}^{\infty} d r_{1} r_{1}^{2} e^{-2 Z r_{1} / a_{0}} \int_{0}^{\infty} d r_{2} r_{2}^{2} \frac{e^{-2 Z r_{2} / a_{0}}}{r_{>}} $$

This problem shows that terms in a trial function that correspond to progressively higher energies contribute progressively less to the ground- state energy. For algebraic simplicity, assume that the Hamiltonian operator can be written in the form \\[ \hat{H}=\hat{H}^{(0)}+\hat{H}^{(1)} \\] and choose a trial function \\[ \phi=c_{1} \psi_{1}+c_{2} \psi_{2} \\] where \\[ \hat{H}^{(0)} \psi_{j}=E_{j}^{(0)} \psi_{j} \quad j=1,2 \\] Show that the secular equation associated with the trial function is \\[ \left|\begin{array}{cc} H_{11}-E & H_{12} \\ H_{12} & H_{22}-E \end{array}\right|=\left|\begin{array}{cc} E_{1}^{(0}+E_{1}^{(1)}-E & H_{12} \\ H_{12} & E_{2}^{(0)}+E_{2}^{(1)}-E \end{array}\right|=0 \\] where \\[ E_{j}^{(1)}=\int \psi_{j}^{*} \hat{H}^{(1)} \psi_{j} d \tau \quad \text { and } \quad H_{12}=\int \psi_{1}^{*} \hat{H}^{(1)} \psi_{2} d \tau \\] Solve Equation 1 for \(E\) to obtain \\[ \begin{aligned} E=\frac{E_{1}^{(0)}+E_{1}^{(1)}+E_{2}^{(0)}+E_{2}^{(1)}}{2} & \\ & \pm \frac{1}{2}\left\\{\left[E_{1}^{(0)}+E_{1}^{(1)}-E_{2}^{(0)}-E_{2}^{(1)}\right]^{2}+4 H_{12}^{2}\right\\}^{1 / 2} \end{aligned} \\] If we arbitrarily assume that \(E_{1}^{(0)}+E_{1}^{(1)} < E_{2}^{(0)}+E_{2}^{(1)},\) then we take the positive sign in Equation 2 and write \\[ \begin{array}{c} E=\frac{E_{1}^{(0)}+E_{1}^{(1)}+E_{2}^{(0)}+E_{2}^{(1)}}{2}+\frac{E_{1}^{(0)}+E_{1}^{(1)}-E_{2}^{(0)}-E_{2}^{(1)}}{2} \\\ \times\left\\{1+\frac{4 H_{12}^{2}}{\left[E_{1}^{(0)}+E_{1}^{(1)}-E_{2}^{(0)}-E_{2}^{(1)}\right]^{2}}\right\\}^{1 / 2} \end{array} \\] Use the expansion \((1+x)^{1 / 2}=1+x / 2+\cdots, x<1\) to get \\[ E=E_{1}^{(0)}+E_{1}^{(1)}+\frac{H_{12}^{2}}{E_{1}^{(0)}+E_{1}^{(1)}-E_{2}^{(0)}-E_{2}^{(1)}}+\cdots \\] Note that if \(E_{1}^{(0)}+E_{1}^{(1)}\) and \(E_{2}^{(0)}+E_{2}^{(1)}\) are widely separated, the term involving \(H_{12}^{2}\) in Equation 3 is small. Therefore, the energy is simply that calculated using \(\psi_{1}\) alone; the \(\psi_{2}\) part of the trial function contributes little to the overall energy. The general result is that terms in a trial function that correspond to higher and higher energies contribute less and less to the total ground- state energy.

If you were to use a trial function of the form \(\phi(x)=\left(1+c \alpha x^{2}\right) e^{-\alpha x^{2} / 2},\) where \(\alpha=\) \(\left(k \mu / \hbar^{2}\right)^{1 / 2}\) and \(c\) is a variational parameter to calculate the ground-state energy of a harmonic oscillator, what do you think the value of \(c\) will come out to be? Why?

Use a trial function of the form \(e^{-\beta x^{2}}\) with \(\beta\) as a variational parameter to calculate the ground-state energy of a harmonic oscillator. Compare your result with the exact energy \(h v / 2 .\) Why is the agreement so good?
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