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Chapter 7: Problem 8

Consider a three-dimensional, spherically symmetric, isotropic harmonic oscillator with \(V(r)=k r^{2} / 2 .\) Using a trial function \(e^{-\alpha r^{2}}\) with \(\alpha\) as a variational parameter, estimate the ground-state energy. Do the same using \(e^{-\alpha r}\). The Hamiltonian operator is \\[ \hat{H}=-\frac{\hbar^{2}}{2 \mu r^{2}} \frac{d}{d r}\left(r^{2} \frac{d}{d r}\right)+\frac{k}{2} r^{2} \\] Compare these results with the exact ground-state energy, \(E=\frac{3}{2} h v\). Why is one of these so much better than the other?

Short Answer

Expert verified
The Hamiltonian for a three-dimensional, spherically symmetric, isotropic harmonic oscillator is given by:\[ \hat{H}=-\frac{\hbar^{2}}{2 \mu r^{2}} \frac{d}{d r}\left(r^{2} \frac{d}{d r}\right)+\frac{k}{2} r^{2} \] where \( \mu \) is the reduced mass and \( k \) is the force constant of the oscillator.

Step by step solution

01

Define the Hamiltonian

The Hamiltonian for a three-dimensional, spherically symmetric, isotropic harmonic oscillator is given by:\[ \hat{H}=-\frac{\hbar^{2}}{2 \mu r^{2}} \frac{d}{d r}\left(r^{2} \frac{d}{d r}\right)+\frac{k}{2} r^{2} \] where \( \mu \) is the reduced mass and \( k \) is the force constant of the oscillator.
02

Consider the first trial function

First, use a trial function of the form \( \psi(r) = e^{-\alpha r^{2}} \). We need to estimate the expectation value of the Hamiltonian:\[ \langle \hat{H} \rangle = \int \psi^{*}(r) \hat{H} \psi(r) \, d^3r \] This will require expressing the volume element in spherical coordinates: \( d^3r = r^2 \sin(\theta) dr \, d\theta \, d\phi \).

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Key Concepts

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

Variational Method
The variational method is a clever technique used in quantum mechanics to approximate the ground-state energy of a system when an exact solution is not easily attainable. It is particularly useful for complex systems where analytical solutions are challenging to find. The core idea behind the variational method is based on Rayleigh's principle, which states that any trial wave function approximating the true ground-state wave function will yield an expectation value of the Hamiltonian that is greater than or equal to the true ground-state energy. This makes it a very powerful tool in quantum mechanics.

In applying this method, one uses a trial wave function that depends on certain parameters to estimate the ground-state energy. By optimizing these parameters, typically through minimization, the best approximation of the true ground-state energy can be obtained. The steps generally include:
  • Choosing a trial function with adjustable parameters
  • Calculating the expectation value of the Hamiltonian with this trial function
  • Optimizing the parameters to minimize this expectation value
Overall, the variational method provides a systematic approach to obtaining a good approximation of ground-state energies, even in complex quantum systems.
Trial Function
The trial function is a crucial component of the variational method. It serves as an educated guess of what the true wave function of a system might look like. This guesswork begins with selecting a mathematical function, represented usually in terms of one or more variational parameters.

In the context of the quantum harmonic oscillator, the trial functions used are often simple forms like exponential functions. For example, the trial functions \( e^{-\alpha r^2} \) and \( e^{-\alpha r} \) are common choices because they reflect the generally decaying behavior expected for a wave function in a bound state, like that of the harmonic oscillator. Why choose these forms? They are simple to work with mathematically and can be easily scaled or adjusted by altering the parameter \( \alpha \), making them flexible tools for experimentation.

Choosing a good trial function is critical because a well-chosen trial function can come very close to the actual wave function, yielding a more accurate estimate of the ground-state energy. A poor choice might lead us astray. Thus, the skill comes in choosing a function that closely captures the physical behavior of the system while remaining mathematically tractable.
Ground-State Energy Estimation
Estimating the ground-state energy of a quantum system is typically the end goal of performing a variational calculation. The ground-state energy is the lowest energy that a quantum system can achieve, and it has profound implications for understanding the system's behavior.

Using the variational method, the process begins by calculating the expectation value of the Hamiltonian using the selected trial function. For a given trial function \( \psi(r) \), this expectation value is calculated as:\[ \langle \hat{H} \rangle = \int \psi^{*}(r) \hat{H} \psi(r) \, d^3r \]Here, the trial function holds center stage, as it must be normalized and integrated over all space while careful attention is paid to how it transforms under the Hamiltonian operator.

Once calculated, the expectation value is then varied by adjusting the variational parameters - in this case, \( \alpha \) - to minimize \( \langle \hat{H} \rangle \). The value of the parameter(s) that yield the lowest energy expectation is considered the best estimate of the ground-state energy. Comparing this estimated energy with the exact known values, like \( E=\frac{3}{2} h v \), helps to validate the accuracy of the trial function choice.

In the exercise, comparing different trial functions, \( e^{-\alpha r^2} \) vs. \( e^{-\alpha r} \), illustrates how different selections may yield different estimates, demonstrating the importance of choosing a trial function that mirrors the actual physical properties of the system.

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Most popular questions from this chapter

Calculate the first-order correction to the energy of a particle constrained to move within the region \(0 \leq x \leq a\) in the potential $$ \begin{aligned} V(x) &=V_{0} x & & 0 \leq x \leq \frac{a}{2} \\ &=V_{0}(a-x) & & \frac{a}{2} \leq x \leq a \end{aligned} $$ where \(V_{0}\) is a constant.

This problem involves the proof of the variational principle, Equation 7.4. Let \(\hat{H} \psi_{n}=\) \(E_{n} \psi_{n}\) be the problem of interest, and let \(\phi\) be our approximation to \(\psi_{e^{-}}\)Even though we do not know the \(\psi_{n}\), we can express \(\phi\) formally as $$ \phi=\sum_{n} c_{n} \psi_{n} $$ where the \(c_{n}\) are constants. Using the fact that the \(\psi_{n}\) are orthonormal, show that $$ c_{n}=\int \psi_{n}^{*} \phi d \tau $$ We do not know the \(\psi_{n}\), however, so Equation 1 is what we call a formal expansion. Now substitute Equation 1 into $$ E_{\phi}=\frac{\int \phi^{*} \hat{H} \phi d \tau}{\int \phi^{*} \phi d \tau} $$ to obtain $$ E_{\phi}=\frac{\sum_{n} c_{n}^{*} c_{n} E_{n}}{\sum_{n} c_{n}^{*} c_{\kappa}} $$ Subtract \(E_{0}\) from the left side of the above equation and \(E_{0} \sum_{n} c_{n}^{*} c_{n} / \sum_{n} c_{n}^{*} c_{n}\) from the right side to obtain, $$ E_{\theta}-E_{0}=\frac{\sum_{n} c_{n}^{*} c_{n}\left(E_{n}-E_{0}\right)}{\sum_{n} c_{n}^{*} c_{n}} $$ Now explain why every term on the right side is positive, proving that \(E_{\phi} \geq E_{0^{\prime}}\)

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 helium, 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} \\] Letting \(x=\cos \theta,\) show that the integrand over \(\theta\) is \\[ \begin{aligned} \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}} &=\int_{-1}^{1} \frac{d x}{\left(r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} x\right)^{1 / 2}} \\ &=\frac{2}{r_{1}} \quad r_{1} > r_{2} \\ &=\frac{2}{r_{2}} \quad r_{1} < r_{2} \end{aligned} \\] Substituting this result into \(E^{(1)},\) show that \\[ \begin{aligned} E^{(1)}=\frac{e^{2}}{4 \pi \varepsilon_{0}} \frac{16 Z^{6}}{a_{0}^{6}} \int_{0}^{\infty} d r_{1} e^{-2 Z_{1} / a_{0}} r_{1}^{2}\left(\frac{1}{r_{1}} \int_{0}^{r_{1}} d r_{2} e^{-2 Z_{2} / a_{0}} r_{2}^{2}\right.\\\ &\left.+\int_{r_{1}}^{\infty} d r_{2} e^{-2 Z r_{2} / a_{0}} r_{2}\right) \\ =& \frac{e^{2}}{4 \pi \varepsilon_{0}} \frac{4 Z^{3}}{a_{0}^{3}} \int_{0}^{\infty} d r_{1} e^{-22 r_{1} / a_{0}} r_{1}^{2}\left[\frac{1}{r_{1}}-e^{-2 Z_{1} / a_{0}}\left(\frac{Z}{a_{0}}+\frac{1}{r_{1}}\right)\right] \\ =& \frac{5}{8} Z\left(\frac{e^{2}}{4 \pi \varepsilon_{0} a_{0}}\right)=\frac{5}{8} Z\left(\frac{m_{\mathrm{e}} e^{4}}{16 \pi^{2} \varepsilon_{0}^{2} \hbar^{2}}\right) \end{aligned} \\] Show that the energy through first order is \\[ \begin{aligned} E^{(0)}+E^{(1)} &=\left(-Z^{2}+\frac{5}{8} Z\right)\left(\frac{m_{\mathrm{e}} e^{4}}{16 \pi^{2} \varepsilon_{0}^{2} \hbar^{2}}\right)=-\frac{11}{4}\left(\frac{m_{\mathrm{e}} e^{4}}{16 \pi^{2} \varepsilon_{0}^{2} \hbar^{2}}\right) \\ &=-2.75\left(\frac{m_{\mathrm{e}} e^{4}}{16 \pi^{2} \varepsilon_{0}^{2} \hbar^{2}}\right) \end{aligned} \\] compared with the exact result, \(E_{\text {exact }}=-2.9037\left(m_{\mathrm{e}} e^{4} / 16 \pi^{2} \varepsilon_{0}^{2} \hbar^{2}\right)\)

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 \(\psi_{1}(x) ?\)

\- 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
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