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Show that the atomic determinantal wave function \\[ \psi=\frac{1}{\sqrt{2}}\left|\begin{array}{ll} 1 s \alpha(1) & 1 s \beta(1) \\ 1 s \alpha(2) & 1 s \beta(2) \end{array}\right| \\] is normalized if the 1 s orbitals are normalized.

Short Answer

Expert verified
Since the integral of the square of \(\psi\) equals 1, \(\psi\) is normalized if the 1s orbitals are normalized.

Step by step solution

01

Write the Determinantal Wave Function

The wave function is given in the form of a determinant:\[\psi=\frac{1}{\sqrt{2}}\left|\begin{array}{ll} 1 s \alpha(1) & 1 s \beta(1) \ 1 s \alpha(2) & 1 s \beta(2) \end{array}\right|\] Expanding this determinant, we get:\[\psi = \frac{1}{\sqrt{2}} \times [(1s \alpha(1) \times 1s \beta(2)) - (1s \beta(1) \times 1s \alpha(2))]\]
02

Express Orbitals and Evaluate

Assume the individual 1s orbitals, \(1s\alpha\) and \(1s\beta\), are normalized. Thus:\[\int |1s\alpha(1)|^2 d\tau_1 = \int |1s\beta(1)|^2 d\tau_1 = 1\] and a similar result holds for particle 2 and cross terms vanish because they are orthogonal or result in zero.
03

Define and Use Slater Determinant Properties

For a determinantal wave function like \(\psi\), where the wave function is antisymmetric under particle exchange, normalization follows if the functions involved are orthonormal set. We consider two particles distributed between spin states, i.e., \(\alpha\) and \(\beta\). Orthonormal condition simplifies determinant properties.
04

Compute the Normalization of \(\psi\)

Calculate \(\int |\psi|^2 d\tau_1 d\tau_2\):\[\int |\psi|^2 d\tau_1 d\tau_2 = \frac{1}{2} \int \left( |1s\alpha(1)1s\beta(2)|^2 + |1s\beta(1)1s\alpha(2)|^2 \right) d\tau_1 d\tau_2 - \left( |1s\alpha(1)1s\beta(2)||1s\beta(1)1s\alpha(2)| \right)\]Using normalization and orthogonality, the first two terms add up to 1, and the cross term vanishes, showing total integral equals 1.

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Key Concepts

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

Normalization
Normalization is a fundamental concept in quantum mechanics that ensures the total probability of finding an electron in all space equals one. For a wave function to be normalized, its integral over all space must equal one. In simpler terms, it ensures that the entire probability distribution is accounted for.

In the context of the determinantal wave function, each of the atomic orbitals involved must also be individually normalized. This means, for an orbital such as \(1s\alpha\), the integral of its absolute square across the spatial domain must be one:

\[\int |1s\alpha(x)|^2 \, dx = 1\]This principle ensures that when these orbitals are combined in a wave function, they collectively respect the probability constraints required by the physical scenario. When applied to the determinantal wave function, the normalization of each component guarantees that the entire function is properly anchored to real-world phenomena, such as electron presence around an atom.
Orthogonality
Orthogonality in quantum mechanics refers to the idea that different quantum states are independent of each other. This means that overlapping one state onto another results in zero; essentially, they do not mix or influence each other. Mathematically, this is shown as:

\[\int \phi_i^*(x) \phi_j(x) \, dx = 0 \quad \text{for} \quad i eq j\]Where \(\phi_i\) and \(\phi_j\) are different states. In atomic systems, states with different spins, like \(\alpha\) and \(\beta\), are often orthogonal.

For the determinantal wave function, these orthogonal properties help simplify calculations, ensuring that cross terms vanish when evaluating the integral of the wave function squared. This implies that only combinations where particles remain in the same state contribute any value, providing a clear and computationally effective pathway to solving complex quantum equations.
Slater Determinant
A Slater determinant is a mathematical construct used to describe the wave function of a multi-electron system. It captures the antisymmetry property of fermionic particles, such as electrons, which states that swapping two particles results in a change of sign of the wave function.

The given determinantal wave function can be represented as:

\[\psi = \frac{1}{\sqrt{2}} \left| \begin{array}{cc} 1s\alpha(1) & 1s\beta(1) \ 1s\alpha(2) & 1s\beta(2) \end{array} \right|\]This format ensures antisymmetry inherently. If two particles swap, the determinant changes sign, showing adherence to Pauli's Exclusion Principle. This principle prevents two electrons with the same quantum numbers from occupying the same space.
Wave Function
The wave function is a foundational element in quantum mechanics, representing the quantum state of a system. It contains all the information about a system's particles.

In mathematical terms, a wave function \(\psi(x)\) allows us to predict probabilities about a particle's position and momentum. For electrons in atoms, the atomic orbitals, like those in the Slater determinant, act as specific wave functions.

The square modulus of a wave function, \(|\psi(x)|^2\), gives the probability density of finding a particle at position \(x\). Thus, the role of a wave function is to map quantum probabilities into forms that can correspond to real-world measurements and observations. When calculated within the framework of quantum mechanics principles, these functions serve as the basis for understanding and predicting how particles behave on the atomic and subatomic levels.

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Most popular questions from this chapter

Consider a helium atom in an excited state in which one of its \(1 s\) electrons is raised to the \(2 s\) level, so that its electron configuration is \(1 s 2 s\). Argue that because the two orbitals are different, there are four possible determinantal wave functions for this system: $$ \begin{aligned} &\phi_{1}=\frac{1}{\sqrt{2}}\left|\begin{array}{ll} 1 s \alpha(1) & 2 s \alpha(1) \\ 1 s \alpha(2) & 2 s \alpha(2) \end{array}\right| \\ &\phi_{2}=\frac{1}{\sqrt{2}}\left|\begin{array}{ll} 1 s \beta(1) & 2 s \beta(1) \\ 1 s \beta(2) & 2 s \beta(2) \end{array}\right| \\ &\phi_{3}=\frac{1}{\sqrt{2}}\left|\begin{array}{ll} 1 s \alpha(1) & 2 s \beta(1) \\ 1 s \alpha(2) & 2 s \beta(2) \end{array}\right| \\ &\phi_{4}=\frac{1}{\sqrt{2}}\left|\begin{array}{ll} 1 s \beta(1) & 2 s \alpha(1) \\ 1 s \beta(2) & 2 s \alpha(2) \end{array}\right| \end{aligned} $$ To calculate the energy of the \(1 s 2 s\) configuration, assume the variational function $$ \psi=c_{1} \phi_{1}+c_{2} \phi_{2}+c_{3} \phi_{3}+c_{4} \phi_{4} $$ Show that the secular equation associated with this linear combination trial function is (this is the only lengthy part of this problem and at least you have the answer in front of you; be sure to remember that the \(1 s\) and \(2 s\) orbitals here are eigenfunctions of the hydrogenlike Hamiltonian operator) $$ \left|\begin{array}{cccc} E_{0}+J-K-E & 0 & 0 & 0 \\ 0 & E_{0}+J-K-E & 0 & 0 \\ 0 & 0 & E_{0}+J-E & -K \\ 0 & 0 & -K & E_{0}+J-E \end{array}\right|=0 $$ where$$ \begin{aligned} &J=\iint d \tau_{1} d \tau_{2} 1 s(1) 1 s(1)\left(\frac{1}{r_{12}}\right) 2 s(2) 2 s(2) \\ &K=\iint d \tau_{1} d \tau_{2} 1 s(1) 2 s(1)\left(\frac{1}{r_{12}}\right) 1 s(2) 2 s(2) \end{aligned} $$ and \(E_{0}\) is the energy without the \(1 / r_{12}\) term in the helium atom Hamiltonian operator. Show that $$ E_{0}=-\frac{5}{2} E_{\mathrm{h}} $$ Explain why \(J\) is called an atomic Coulombic integral and \(K\) is called an atomic exchange integral.Even though the above secular determinant is \(4 \times 4\) and appears to give a fourth-degree polynomial in \(E\), note that it really consists of two \(1 \times 1\) blocks and a \(2 \times 2\) block. Show that this symmetry in the determinant reduces the determinantal equation to $$ \left(E_{0}+J-K-E\right)^{2}\left|\begin{array}{cc} E_{0}+J-E & -K \\ -K & E_{0}+J-E \end{array}\right|=0 $$ and that this equation gives the four roots $$ \begin{aligned} E &=E_{0}+J-K \\ &=E_{0}+J \pm K \end{aligned} $$ Show that the wave function corresponding to the positive sign in \(E\) in the \(E_{0}+J \pm K\) is $$ \psi_{3}=\frac{1}{\sqrt{2}}\left(\phi_{3}-\phi_{4}\right) $$ and that corresponding to the negative sign in \(E_{0}+J \pm K\) is $$ \psi_{4}=\frac{1}{\sqrt{2}}\left(\phi_{3}+\phi_{4}\right) $$ Now show that both \(\psi_{3}\) and \(\psi_{4}\) can be factored into a spatial part and a spin part, even though \(\phi_{3}\) and \(\phi_{4}\) separately cannot. Furthermore, let $$ \psi_{1}=\phi_{1} \quad \text { and } \quad \psi_{2}=\phi_{2} $$ and show that both of these can be factored also. Using the argument given in Problem 8-24, group these four wave functions \(\left(\psi_{1}, \psi_{2}, \psi_{3}, \psi_{4}\right)\) into a singlet state and a triplet state. Now calculate the energy of the singlet and triplet states in terms of \(E_{0}, J\), and \(K\). Argue that \(J>0\). Given that \(K>0\) also, does the singlet state or the triplet state have the lower energy? The values of \(J\) and \(K\) when hydrogenlike wave functions with \(Z=2\) are used are \(J=10 / 27 E_{\mathrm{b}}\) and \(K=32 /(27)^{2} E_{\mathrm{h}}\). Using the ground-state wave function $$ \phi=\frac{1}{\sqrt{2}}\left|\begin{array}{ll} 1 s \alpha(1) & 1 s \beta(1) \\ 1 s \alpha(2) & 1 s \beta(2) \end{array}\right| $$ show that the first-order perturbation theory result is \(E=-11 / 4 E_{\mathrm{b}}\) if hydrogenlike wave functions with \(Z=2\) are used. Use this value of \(E\) to calculate the energy difference between the ground state and the first excited singlet state and the first triplet state of helium. The experimental values of these energy differences are \(159700 \mathrm{~cm}^{-1}\) and 166 \(200 \mathrm{~cm}^{-1}\), respectively (cf. Figure \(\left.8.5\right)\).

6\. Determine the term symbols associated with an \(n p^{1}\) electron configuration. Show that these term symbols are the same as for an \(n p^{5}\) electron configuration. Which term symbol represents the ground state?

The spin operators, \(\hat{S}_{x}, \hat{S}_{y},\) and \(\hat{s},\) like all angular momentum operators, obey the commutation relations (Problem \(6-13\) ) \\[ \left[\hat{S}_{x}, \hat{S}_{y}\right]=i \hbar \hat{S}_{z} \quad\left[\hat{S}_{y}, \hat{S}_{z}\right]=i \hbar \hat{S}_{x} \quad\left[\hat{S}_{z}, \hat{S}_{x}\right]=i \hbar \hat{S}_{y} \\] Define the (non-Hermitian) operators \\[ \hat{S}_{+}=\hat{S}_{x}+i \hat{s}_{y} \quad \hat{S}_{-}=\hat{S}_{x}-i \hat{S}_{y} \\] and show that \\[ \left[\hat{S}_{z}, \hat{S}_{+}\right]=\hbar \hat{S} \\] and \\[ \left[\hat{S}_{z}, \hat{S}_{-}\right]=-\hbar \hat{S}_{-} \\] Now show that \\[ \hat{S}_{+} \hat{S}_{-}=\hat{S}^{2}-\hat{S}_{z}^{2}+\hbar \hat{S}_{z} \\] and that \\[ \hat{S}_{-} \hat{S}_{+}=\hat{S}^{2}-\hat{S}_{z}^{2}-\hbar \hat{S}_{z} \\] where \\[ \hat{S}^{2}=\hat{S}_{x}^{2}+\hat{S}_{y}^{2}+\hat{S}_{z}^{2} \\]

3\. Why is the radial dependence of multielectron atomic wave functions in the Hartree-Fock approximation different from the radial dependence of hydrogen atomic wave functions?

The term symbols for an \(n p^{3}\) electron configuration are \(^{2} P,^{2} D,\) and \(^{4} S\). Calculate the values of \(J\) associated with each of these term symbols. Which term symbol represents the ground state?

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