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Chapter 2: Problem 39

Using \(\mathscr{S}^{2}=\mathscr{S}_{-} \mathscr{S}_{+}+\mathscr{S}_{z}+\mathscr{S}_{z}^{2}\), show that \(\left|{ }^{1} \Psi_{1}^{2}\right\rangle\) is a singlet while \(\left|{ }^{3} \Psi_{1}^{2}\right\rangle,\left|\Psi_{1}^{\overline{2}}\right\rangle\) and \(\left|\Psi_{\overline{1}}^{2}\right\rangle\) are triplets.

Short Answer

Expert verified
\(|{}^{1}\Psi_{1}^{2}\rangle\) is a singlet (\(S = 0\)), while \(|{}^{3}\Psi_{1}^{2}\rangle\), \(|\Psi_{1}^{\overline{2}}\rangle\), and \(|\Psi_{\overline{1}}^{2}\rangle\) are triplets (\(S = 1\)).

Step by step solution

01

- Understanding the Equation

We are given the equation \(\mathscr{S}^{2} = \mathscr{S}_{-}\mathscr{S}_{+} + \mathscr{S}_{z} + \mathscr{S}_{z}^{2}\). This is used to find the eigenvalues of the total spin squared operator \(\mathscr{S}^{2}\). The task is to determine whether certain states are singlets or triplets by substituting each ket into this equation.
02

- Explain Singlet and Triplet States

A singlet state is characterized by having a total spin \(S = 0\), which makes the eigenvalue of \(\mathscr{S}^2\) equal to \(0\). On the other hand, triplet states have a total spin \(S = 1\), resulting in the eigenvalue of \(\mathscr{S}^2\) being \(S(S + 1) = 2\).
03

- Verify Singlet State \(|{}^{1}\Psi_{1}^{2}\rangle\)

To confirm that \(|{}^{1}\Psi_{1}^{2}\rangle\) is a singlet, substitute this state into the equation: \(\mathscr{S}^{2}|{}^{1}\Psi_{1}^{2}\rangle\). Calculate \(\mathscr{S}^2|{}^{1}\Psi_{1}^{2}\rangle = 0\), which confirms the state has \(S = 0\).
04

- Verify Triplet States For Remaining Kets

For each of the remaining states \(|{}^{3}\Psi_{1}^{2}\rangle\), \(|\Psi_{1}^{\overline{2}}\rangle\), and \(|\Psi_{\overline{1}}^{2}\rangle\), substitute them into \(\mathscr{S}^{2}\). Calculate \(\mathscr{S}^2 = 2\) for each, confirming \(S = 1\). These are triplets as they yield the eigenvalue corresponding to \(S(S+1) = 2\).

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Key Concepts

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

Spin Operators
In quantum chemistry, spin operators are crucial for understanding the behavior of particles like electrons that possess spin. These operators are mathematical constructs used to analyze spin angular momentum. The total spin operator, denoted by \( \mathscr{S} \), combines the individual spin components such as \( \mathscr{S}_x \), \( \mathscr{S}_y \), and \( \mathscr{S}_z \). The square of this operator, represented by \( \mathscr{S}^2 \), provides valuable information about the system's total spin. The equation \( \mathscr{S}^2 = \mathscr{S}_{-}\mathscr{S}_{+} + \mathscr{S}_{z} + \mathscr{S}_{z}^{2} \) expresses this relationship. - \( \mathscr{S}_z \) refers to the z-component of the spin.- \( \mathscr{S}_{-} \) and \( \mathscr{S}_{+} \) are the lowering and raising operators, respectively, which alter the magnetic quantum number associated with the spin.These operators follow specific commutation relationships that aid in determining a system's quantum behavior. By applying these operators to quantum states, one can find eigenvalues and eigenfunctions, revealing insights into the systemic spin. This is particularly important when evaluating whether states are singlets or triplets, as these have distinct physical and chemical implications.
Singlet and Triplet States
Singlet and triplet states are terms used in quantum mechanics to describe the combined spin states of two electrons. They arise from the quantum property that electrons have spin, usually considered as either 'up' (+1/2) or 'down' (-1/2). - **Singlet State**: This state results when two electrons are paired such that their spins cancel each other out, leading to a total spin \( S = 0 \). The corresponding eigenvalue for \( \mathscr{S}^2 \) is therefore \( 0 \). The wavefunction for the singlet state is symmetric with respect to electron exchange, typically written as \( |{}^{1}\Psi\rangle \).- **Triplet States**: Triplets occur when the additive spin of the two electrons results in a total spin of \( S = 1 \). The possible configurations mean the electrons' spins do not entirely cancel each other out. Thus, the \( \mathscr{S}^2 \) eigenvalue becomes \( S(S + 1) = 2 \). Triplet wavefunctions are anti-symmetric upon swapping of electron positions. These states are fundamental in processes like chemical bonding and spectroscopy, where they influence the resulting molecule's physical properties. Understanding the nature of these spin states helps predict the behavior of molecules in an excited state.
Eigenvalues in Quantum Mechanics
In quantum mechanics, eigenvalues are essential to understanding the behavior of quantum systems. They arise from the application of operators, such as the Hamiltonian or spin operators, to quantum states. When these operators act on a state, the result is typically expressed as an eigenvalue multiplied by the original state.- **Eigenvalue Equation**: Applied to a spin operator, this equation takes the form \( \mathscr{S}^2 | \Psi \rangle = \lambda | \Psi \rangle \). Here, \( \lambda \) is the eigenvalue that indicates how the system's spin is quantified. - **Total Spin Eigenvalues**: When using the total spin squared operator \( \mathscr{S}^2 \), the eigenvalues can reveal whether a state is a singlet or triplet, providing insight into its magnetic properties. A singlet state has an eigenvalue of zero, while a triplet state yields a value of two.Eigenvalues are not just numbers; they relate to measurable properties such as energy and angular momentum. This concept allows quantum chemists to predict potential outcomes and behaviors of particles, enhancing the understanding of material properties and reaction mechanisms.
Electronic Structure Theory
Electronic structure theory is the bedrock of understanding how atoms and molecules form bonds and what happens when they react. It deeply involves quantum mechanical principles to explain the electronic arrangements and their energy levels within an atom or molecule. - **Wavefunctions**: At the heart of this theory are wavefunctions, which describe the quantum states of electrons. They contain all the information about a particle's position and spin. - **Hartree-Fock and Post-Hartree-Fock Methods**: To solve for electron distribution, various computational methods are employed. Hartree-Fock theory considers electron motions to be independent, simplifying calculations. More advanced methods, like configuration interaction and coupled-cluster, offer greater accuracy by considering electron correlation. By employing such methods, chemists can predict molecular geometries, reaction pathways, and properties. Understanding electronic structure is critical for visualizing chemical bonding and reactivity, which is essential for fields like material science, pharmacology, and nanotechnology. Accurate electronic structure computations enable innovations in designing new molecules and materials with desired properties.

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Most popular questions from this chapter

Consider an operator \(\mathscr{A}\) that commutes with the Hamiltonian. Suppose \(|\Phi\rangle\) is an eigenfunction of \(\mathscr{H}\) with eigenvalue \(E\). Show that \(\mathscr{A}|\Phi\rangle\) is also an eigenfunction of \(\mathscr{H}\) with eigenvalue \(E .\) Thus if \(|\Phi\rangle\) is (energetically) nondegenerate, then \(\mathscr{A}|\Phi\rangle\) is at most a constant multiple of \(|\Phi\rangle(\) i.e., \(\mathscr{A}|\Phi\rangle=a|\Phi\rangle)\) and hence \(|\Phi\rangle\) is an eigenfunction of \(\mathscr{A} .\) In case of degeneracies, we can always construct appropriate linear combinations of the degenerate eigenfunctions of \(\mathscr{H}\) that are also eigenfunctions of \(\mathscr{A}\).

Given a state $$ |K\rangle=\left|\chi_{1} \chi_{2} \cdots \chi_{N}\right\rangle=a_{1}^{\dagger} a_{2}^{\dagger} \cdots a_{N}^{\dagger}|\rangle $$ show that \(\left\langle K\left|a_{i}^{\dagger} a_{j}\right| K\right\rangle=1\) if \(i=j\) and \(i \in\\{1,2, \ldots, N\\}\), but is zero otherwise.

Consider the Slater determinants $$ \begin{aligned} &|K\rangle=\left|\chi_{i} \chi_{j}\right\rangle \\ &|L\rangle=\left|\chi_{k} \chi_{l}\right\rangle \end{aligned} $$ Show that $$ \langle K \mid L\rangle=\delta_{i k} \delta_{j l}-\delta_{i l} \delta_{j k} $$ Note that the overlap is zero unless: 1) \(k=i\) and \(l=j\), in which case \(|L\rangle=|K\rangle\) and the overlap is unity and 2) \(k=j\) and \(l=i\) in which case \(|L\rangle=\left|\chi_{j} \chi_{i}\right\rangle=\) \(-|K\rangle\) and the overlap is minus one.

The Hartree-Fock ground state energy for an \(N\)-electron system is \({ }^{N} E_{0}=\left\langle{ }^{N} \Psi_{0}|\mathscr{H}|^{N} \Psi_{0}\right\rangle .\) Consider a state of the ionized system (in which an electron has been removed from spin orbital \(\chi_{a}\) ) with energy \({ }^{N-1} E_{a}=\left\langle^{N-1} \Psi_{a}|\mathscr{H}|^{\mid N-1} \Psi_{a}\right\rangle\), where \(\left.\left.\right|^{N-1} \Psi_{a}\right\rangle\) is a single determinant with all spin orbitals but \(\chi_{a}\) occupied, $$ \left.\left.\right|^{N-1} \Psi_{a}\right\rangle=\left|\chi_{1} \chi_{2} \cdots \chi_{a-1} \chi_{a+1} \cdots \chi_{N}\right\rangle $$ Show, using the rules in the tables, that the energy required for this ionization process is $$ { }^{N} E_{0}-{ }^{N-1} E_{a}=\langle a|h| a\rangle+\sum_{b}^{N}\langle a b|| a b\rangle $$ To show the power and simplicity of the mnemonic device introduced in this subsection, let us derive the above result without doing any algebra. Consider the representation of \(\left|{ }^{N} \Psi_{0}\right\rangle\) in Fig. 2.4. If we remove an electron from \(\chi_{a}\), we lose the "one-electron energy" contribution \(\langle a|h| a\rangle\) to \({ }^{N} E_{0}\). Moreover, we lose the pair-wise contributions arising from the "interaction" of the electron in \(\chi_{a}\) with the remaining electrons (i.e., \(\left.\sum_{b \neq a}^{N}\langle a b|| a b\rangle\right)\) Because \(\langle a a|| a a\rangle=0\), the above result follows immediately.

Given two nondegenerate eigenfunctions of a hermitian operator \(\mathscr{L}\) that commutes with \(\mathscr{H}\), i.e., \(, \mathcal{L}\left|\Psi_{1}\right\rangle=a_{1}\left|\Psi_{1}\right\rangle, \mathscr{A}\left|\Psi_{2}\right\rangle=\) \(a_{2}\left|\Psi_{2}\right\rangle, a_{1} \neq a_{2}\), show that \(\left\langle\Psi_{1}|\mathscr{H}| \Psi_{2}\right\rangle=0\). Thus the matrix element of the Hamiltonian between, say, singlet and triplet spin-adapted configurations is zero.
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