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Chapter 21: Problem 10

Classify the following functions as symmetric, antisymmetric, or neither in the exchange of electrons 1 and 2 : a. \([1 s(1) 2 s(2)+2 s(1) 1 s(2)] \times[\alpha(1) \beta(2)-\beta(1) \alpha(2)]\) b. \([1 s(1) 2 s(2)+2 s(1) 1 s(2)] \alpha(1) \alpha(2)\) c. \([1 s(1) 2 s(2)+2 s(1) 1 s(2)][\alpha(1) \beta(2)+\beta(1) \alpha(2)]\) d. \([1 s(1) 2 s(2)-2 s(1) 1 s(2)][\alpha(1) \beta(2)+\beta(1) \alpha(2)]\) e. \([1 s(1) 2 s(2)+2 s(1)] s(2)] \times\) \([\alpha(1) \beta(2)-\beta(1) \alpha(2)+\alpha(1) \alpha(2)]\)

Short Answer

Expert verified
In conclusion, functions a, b, and c are symmetric, function d is antisymmetric, and function e is neither symmetric nor antisymmetric.

Step by step solution

01

a. Function a

Exchange electrons 1 and 2 to obtain: \[ [1 s(2) 2 s(1)+2 s(2) 1 s(1)] \times [\alpha(2) \beta(1)-\beta(2) \alpha(1)].\] Compare the original and the exchanged functions, and notice that the sign does not change, which means it is a symmetric function.
02

b. Function b

Exchange electrons 1 and 2 to obtain: \[ [1 s(2) 2 s(1)+2 s(2) 1 s(1)] \alpha(2) \alpha(1).\] Compare the original and the exchanged functions, and notice that the function does not change, which means it is a symmetric function.
03

c. Function c

Exchange electrons 1 and 2 to obtain: \[ [1 s(2) 2 s(1)+2 s(2) 1 s(1)][\alpha(2) \beta(1)+\beta(2) \alpha(1)].\] Compare the original and the exchanged functions, and notice that the function does not change, which means it is a symmetric function.
04

d. Function d

Exchange electrons 1 and 2 to obtain: \[ [1 s(2) 2 s(1)-2 s(2) 1 s(1)][\alpha(2) \beta(1)+\beta(2) \alpha(1)].\] Compare the original and the exchanged functions, and notice that the sign changes, which means it is an antisymmetric function.
05

e. Function e

Exchange electrons 1 and 2 to obtain: \[ [1 s(2) 2 s(1)+2 s(2) 1 s(1)] \times [\alpha(2) \beta(1)-\beta(2) \alpha(1)+\alpha(2) \alpha(1)].\] Compare the original and the exchanged functions, and notice that there is no clear pattern for the sign change; therefore, the function classification is neither symmetric nor antisymmetric. In conclusion, functions a, b, and c are symmetric, function d is antisymmetric, and function e is neither symmetric nor antisymmetric.

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Key Concepts

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

Symmetric and Antisymmetric Wave Functions
In quantum mechanics, wave functions are mathematical descriptions of the quantum state of a system. The symmetry of these wave functions under the exchange of identical particles is a fundamental aspect in the study of quantum systems.

Symmetric wave functions remain unchanged when two particles are exchanged. In contrast, antisymmetric wave functions change their sign upon such an exchange. This property of symmetry or antisymmetry dictates the behavior of different types of particles: bosons have symmetric wave functions, and fermions, such as electrons, have antisymmetric wave functions due to the Pauli exclusion principle.

Understanding the symmetry of wave functions is crucial when determining the possible states for electrons in an atom or molecule. For instance, symmetric functions correspond to bosonic behavior, allowing particles to occupy the same quantum state, while antisymmetric functions describe fermionic behavior, preventing this.
Exchange of Electrons
The exchange of electrons refers to the swapping of positional labels in a wave function that describes a multi-electron system. It's important in quantum mechanics because the properties of identical particles, like electrons, must not depend on the labeling. This concept is illustrated in wave function symmetry problems.

To assess the exchange symmetry, you take a wave function with electrons labeled (1) and (2) and switch these labels. If the overall sign of the wave function is unaffected, it is symmetric. If the exchange inverts the sign, it's antisymmetric. In physical terms, for fermions like electrons, the overall wave function of the system, which includes spatial and spin parts, has to be antisymmetric due to their indistinguishable nature.

When solving problems related to wave function symmetry, carefully exchanging the electron labels and checking the resulting sign can provide deep insight into the electronic structure and behavior of quantum systems.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior of matter and energy at the smallest scales of atoms and subatomic particles. This theory provides a mathematical framework for understanding the properties of particles, including their interactions and the probabilities of finding them in various states.

The principles of quantum mechanics are essential for interpreting wave functions and understanding their implications, such as wave function symmetry. Quantum mechanics tells us that particles like electrons don't have precise locations or velocities until they are measured. Instead, they exist in superpositions of all possible states, with their wave functions encoding the probabilities of these states.

Through studying quantum mechanics, students gain the knowledge needed to solve complex problems in advanced chemistry and physics. Quantum mechanics not only explains the behavior of the microscopic world but also underpins much of modern technology, including semiconductors and lasers.
Physical Chemistry Education
Physical chemistry education combines the principles of physics and chemistry to explain how matter behaves on a molecular and atomic level. It plays a pivotal role in preparing students to tackle complex scientific problems through an understanding of quantum mechanics, thermodynamics, kinetics, and spectroscopy.

An effective physical chemistry curriculum ensures students are well-versed with the mathematical tools and concepts necessary to describe and predict the behavior of chemical systems. This includes topics like wave function symmetry and electron exchange, which are crucial for explaining molecular structures and reactions.

Incorporating exercises that involve classifying wave functions can enhance a student's comprehension of molecular orbitals and their importance in chemical bonding and reactivity. An educator's goal in physical chemistry is to break down intricate concepts into more digestible parts, employing step-by-step problem-solving to solidify understanding.

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Most popular questions from this chapter

In this problem, you will use the variational method to find the optimal \(1 s\) wave function for the hydrogen atom starting from the trial function \(\Phi(r)=e^{-\alpha r}\) with \(\alpha\) as the variational parameter. You will minimize \\[ E(\alpha)=\frac{\int \Phi^{*} \hat{H} \Phi d \tau}{\int \Phi^{*} \Phi d \tau} \\] with respect to \(\alpha\) a. Show that \\[ \begin{aligned} \hat{H} \Phi &=-\frac{\hbar^{2}}{2 m_{e}} \frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial \Phi(r)}{\partial r}\right)-\frac{e^{2}}{4 \pi \varepsilon_{0} r} \Phi(r) \\ &=\frac{\alpha \hbar^{2}}{2 m_{e} r^{2}}\left(2 r-\alpha r^{2}\right) e^{-\alpha r}-\frac{e^{2}}{4 \pi \varepsilon_{0} r} e^{-\alpha r} \end{aligned} \\] b. Obtain the result \(\int \Phi^{*} \hat{H} \Phi d \tau=4 \pi \int_{0}^{\infty} r^{2} \Phi^{*} \hat{H} \Phi d r=\) \(\pi \hbar^{2} /\left(2 m_{e} \alpha\right)-e^{2} /\left(4 \varepsilon_{0} \alpha^{2}\right)\) using the standard integrals in the Math Supplement. c. Show that \(\int \Phi^{*} \Phi d \tau=4 \pi \int_{0}^{\infty} r^{2} \Phi^{*} \Phi d r=\pi / \alpha^{3}\) using the standard integrals in the Math Supplement. d. You now have the result \(E(\alpha)=\hbar^{2} \alpha^{2} /\left(2 m_{e}\right)-e^{2} \alpha /\left(4 \pi \varepsilon_{0}\right)\) Minimize this function with respect to \(\alpha\) and obtain the optimal value of \(\alpha\) e. Is \(E\left(\alpha_{\text {optimal }}\right)\) equal to or greater than the true energy? Why?

Calculate the angles that a spin angular momentum vector for an individual electron can make with the \(z\) axis.

Refer to the first ionization energies and electron affinities of the first 11 elements (units of eV) shown in the following table. Why is the magnitude of the electron affinity for a given element smaller than the magnitude of the first ionization energy?

Write the Slater determinant for the ground-state configuration of Be.
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