/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 For a two-electron system, there... [FREE SOLUTION] | 91Ó°ÊÓ

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For a two-electron system, there are four possible spin functions: 1\. \(\alpha(1) \alpha(2)\) 2\. \(\beta(1) \alpha(2)\) 3\. \(\alpha(1) \beta(2)\) 4\. \(\beta(1) \beta(2)\) The concept of indistinguishability forces us to consider only linear combinations of 2 and 3 \\[ \psi_{\pm}=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2) \pm \beta(1) \alpha(2)] \\] instead of 2 and 3 separately. Show that of the four acceptable spin functions, \(1,4,\) and \(\psi_{\pm}\) three are symmetric and one is antisymmetric. Now for a two-electron system, we can combine spatial wave functions with spin functions. Show that this combination leads to only four allowable combinations: \\[ \begin{array}{c} {[\psi(1) \phi(2)+\psi(2) \phi(1)] \frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\alpha(2) \beta(1)]} \\ {[\psi(1) \phi(2)-\psi(2) \phi(1)][\alpha(1) \alpha(2)]} \\ {[\psi(1) \phi(2)-\psi(2) \phi(1)][\beta(1) \beta(2)]} \end{array} \\] and \\[ [\psi(1) \phi(2)-\psi(2) \phi(1)] \frac{1}{\sqrt{2}}[\alpha(1) \beta(2)+\alpha(2) \beta(1)] \\] where \(\psi\) and \(\phi\) are two spatial wave functions. Show that \(M_{s}=m_{s 1}+m_{n}=0\) for the first of these and that \(M_{s}=1,-1,\) and 0 (in atomic units) for the next three, respectively. Consider the first excited state of a helium atom, in which \(\psi=1 s\) and \(\phi=2 s .\) The first of the four wave functions above, with the symmetric spatial part, will give a higher energy than the remaining three, which form a degenerate set of three. The first state is a singlet state and the degenerate set of three represents a triplet state. Because \(M,\) equals zero and only zero for the singlet state, the singlet state corresponds to \(S=0 .\) The other three, with \(M_{n}=\pm 1,0,\) corresponds to \(S=1 .\) Note that the degeneracy is \(2 S+1\) in each case. Putting all this information into a more mathematical form, given that \(\hat{S}_{\text {bat }}=\hat{s}_{1}+\hat{s}_{2}\) we can show that (Problem \(8-53\) ) \\[ \hat{S}_{\text {vatel }}^{2}[\alpha(1) \beta(2)-\alpha(2) \beta(1)]=0 \\] corresponding to \(S=0,\) and that \\[ \hat{S}_{\text {wal }}^{2}\left[\begin{array}{c} \alpha(1) \alpha(2) \\ \frac{1}{\sqrt{2}}[\alpha(1) \beta(2)+\alpha(2) \beta(1)] \\ \beta(1) \beta(2) \end{array}\right]=2 h^{2}\left[\begin{array}{c} \alpha(1) \alpha(2) \\ \frac{1}{\sqrt{2}}[\alpha(1) \beta(2)+\alpha(2) \beta(1)] \\ \beta(1) \beta(2) \end{array}\right] \\] corresponding to \(S=1\)

Short Answer

Expert verified
Three combinations are symmetric, one is antisymmetric. Only four allowed wave functions exist for two-electron systems.

Step by step solution

01

Identify symmetry properties of spin functions

The spin functions are symmetric if swapping electrons 1 and 2 results in the same function, and antisymmetric if it results in a change of sign. For 1: \( \alpha(1) \alpha(2) \) and 4: \( \beta(1) \beta(2) \), if we swap particles, the function remains the same, so both are symmetric. \( \psi_+ = \frac{1}{\sqrt{2}}[\alpha(1)\beta(2) + \beta(1)\alpha(2)] \) is symmetric because exchange doesn't change the sign, while \( \psi_- = \frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \beta(1)\alpha(2)] \) is antisymmetric, as the sign changes.
02

Determine allowable combinations of wave functions

Upon forming wave functions, we must consider both symmetry in spatial and spin parts. The combination \([\psi(1) \phi(2) + \psi(2) \phi(1)] \frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \beta(1)\alpha(2)]\) is entirely antisymmetric, allowing the first wave function. The other three, \([\psi(1) \phi(2) - \psi(2) \phi(1)]\), combined with symmetric spin parts, form the other allowable combinations.
03

Analyze total spin projections \(M_s\)

The first combination involves a spin configuration \( \psi_- \), which has a total spin projection \(M_s = 0\). The other combinations paired with \(\alpha(1) \alpha(2)\), \(\beta(1) \beta(2)\), and \(\psi_+\) each result in projections of \(M_s = 1\), \(M_s = -1\), and \(M_s = 0\), respectively.
04

Discuss implications for helium atom states

For a helium system with \(\psi = 1s\) and \(\phi = 2s\), a symmetric spatial function corresponds to higher energy and a singlet state \(S=0\), due to degeneracy \(2S + 1 = 1\). In contrast, the antisymmetric spatial forms correspond to triplet states \(S=1\), with a degeneracy of 3, confirming the lowering of energy.

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Key Concepts

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

Spin Symmetry
In quantum mechanics, spin symmetry refers to the behavior of a wave function when the particles it describes are swapped. Two-electron systems can exhibit different types of spin symmetry: symmetric or antisymmetric. When the wave function remains unchanged upon the exchange of electrons, it is said to be symmetric. Conversely, if the wave function's sign changes, it is antisymmetric.

For instance, in a two-electron system:
  • The functions \[\alpha(1) \alpha(2)\quad \text{and}\quad \beta(1) \beta(2)\]are symmetric because swapping electrons 1 and 2 does not alter these functions.
  • The function \[\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) + \beta(1)\alpha(2)]\]is also symmetric as exchanging particles keeps the combination intact.
  • The function \[\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \beta(1)\alpha(2)]\]is antisymmetric, where a swap flips the sign.
Spin symmetry is crucial as it lays the foundation for determining whether the spatial wave function needs to be symmetric or antisymmetric to follow the Pauli exclusion principle.
Spatial Wave Function
The spatial wave function describes the spatial part of a quantum system, constituting the probability amplitude for the positions of the particles. For a two-electron system, spatial symmetry is dictated by the interchange of the spatial coordinates of the electrons.

Two-electron spatial wave functions can be:
  • Symmetric: \[[\psi(1) \phi(2) + \psi(2) \phi(1)]\]This expression remains unchanged under exchange of spatial coordinates of electrons, indicating symmetry.
  • Antisymmetric: \[[\psi(1) \phi(2) - \psi(2) \phi(1)]\]This changes sign under electron coordinate exchange, portraying antisymmetry.
The combination of the spatial wave function with the spin function ensures the overall wave function complies with the exclusion principle. Therefore, it must be either symmetrical in spin and antisymmetrical in spatial parts or vice versa.
Singlet and Triplet States
In quantum systems, particularly in two-electron systems, the terms singlet and triplet describe particular states with different total spin characteristics. These terms arise from combining the two electron spins in distinct ways:

  • Singlet State:
    This state has a total spin (\[S=0\]), leading to only one configuration, from its name "singlet." It is formed with an antisymmetric spin part;\[\frac{1}{\sqrt{2}}[\alpha(1)\beta(2) - \beta(1)\alpha(2)]\], which pairs with symmetric spatial parts to satisfy quantum requirements.
  • Triplet State:
    This state features total spin \[S=1\], resulting in three possible configurations, corresponding to projections of \[M_s = 1, 0, -1\]. The spin configurations are symmetric forms such as\[\alpha(1) \alpha(2), \beta(1) \beta(2), \text{and} \frac{1}{\sqrt{2}}[\alpha(1)\beta(2) + \beta(1)\alpha(2)]\]. The triplet states are associated with an antisymmetrical spatial configuration.
Both states play a crucial role in distinguishing the different energy levels and degeneracies of quantum systems, such as in the helium atom's excited state.
Total Spin Projection
The total spin projection, denoted as \[M_s\], pertains to the sum of the spin quantum numbers of all electrons in a system, giving the component of total spin along a specified direction (usually the z-axis). For a two-electron system, the various combinations of spins yield distinct projections.

Consider these projections for both singlet and triplet states:
  • Singlet State:
    With one configuration,\[M_s = 0\], as it's the only viable sum from \[S=0\].The singlet contributes a solitary degenerate level.
  • Triplet States:
    For \[S=1\], the three configurations render projections of \[M_s = 1, 0, -1\].Each corresponds to different angular momentum contributions, contributing to a triply degenerate energy level.
In systems like helium's first excited state, these projections illustrate how different configurations lead to varying energy levels, bounding electron states into distinct spectral lines.

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Most popular questions from this chapter

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 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?

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?

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 \(\left.8.5\right)\).

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