Chapter 8, Problem 8B.1AE

Interpretation Introduction

Interpretation:

The wavefunction for the excited state of helium having configuration $1{\text{s}}^{1}2{\text{s}}^1$ has to be stated.

Concept introduction:

For many electron systems, the total wavefunctions are taken as the product of the radial part of the various wavefunctions.  Total wavefunction is the product of spin and radial part.  Therefore, in order to obtain total wavefunction for many electron systems, the possible spins for all the electrons present in the system is taken into consideration.

Answer to Problem 8B.1AE

The wavefunction for the excited state of helium having configuration $1{\text{s}}^{1}2{\text{s}}^1$ is shown below.

    $\begin{array}{c}{\psi }_{{\text{He}}_1}=\frac{1}{\sqrt{2}}\left[\begin{array}{l}\left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)-\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)\\ \left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\end{array}\right]\alpha \left(1\right)\alpha \left(2\right)\\ {\psi }_{{\text{He}}_2}=\frac{1}{\sqrt{2}}\left(\left[\begin{array}{l}\left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)-\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)\\ \left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\end{array}\right]\left[\beta \left(2\right)\alpha \left(1\right)-\beta \left(1\right)\alpha \left(2\right)\right]\right)\\ {\psi }_{{\text{He}}_3}=\frac{1}{\sqrt{2}}\left[\begin{array}{l}\left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)-\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)\\ \left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\end{array}\right]\beta \left(1\right)\beta \left(2\right)\\ {\psi }_{{\text{He}}_4}=\frac{1}{\sqrt{2}}\left(\left[\begin{array}{l}\left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)-\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)\\ \left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\end{array}\right]\left[\beta \left(1\right)\alpha \left(2\right)-\alpha \left(2\right)\beta \left(1\right)\right]\right)\end{array}$

Explanation of Solution

The number of electrons in $\text{He}$ is $2$.  The electronic configuration of excited state is $1{\text{s}}^{1}2{\text{s}}^1$.  Therefore, there are only two rows and two columns and the value of $\text{n}$  in the normalization constant is $2$.  The rows represent electrons $1$ and $2$.  There are two electrons and they can be placed in $1\text{s}$ and $\text{2s}$.  These two electrons can have any spin $\alpha$ or $\beta$.  Therefore, there are four possibilities for the construction of slater determinant.

The total wavefunction is represented as the product of space and spin part of the wavefunction.  The possibilities of arrangement of two electrons are given below.

Possibility 1:

The columns represent the spin orbitals $1\text{s}\alpha$ and $2\text{s}\alpha$.  The slater determinant for the $\text{He}$ is shown below.

    $\begin{array}{c}{\psi }_{\text{He}}=\frac{1}{\sqrt{2}}\begin{array}{cc}1\text{s}\left(r_1\right)\alpha \left(1\right)& \text{1s}\left(r_2\right)\alpha \left(2\right)\\ 2\text{s}\left(r_1\right)\alpha \left(1\right)& 2\text{s}\left(r_2\right)\alpha \left(2\right)\end{array}\\ =\frac{1}{\sqrt{2}}\left[1\text{s}\left(r_1\right)2\text{s}\left(r_2\right)-2\text{s}\left(r_1\right)\text{1s}\left(r_2\right)\right]\alpha \left(1\right)\alpha \left(2\right)\end{array}$

Where,

• $1$ and $2$ represents the electrons.
• $\alpha$ is the spin part of the wavefunction.

Possibility 2:

The columns represent the spin orbitals $1\text{s}\alpha$ and $2\text{s}\beta$.  The slater determinant for the $\text{He}$ is shown below.

    $\begin{array}{c}{\psi }_{\text{He}}=\frac{1}{\sqrt{2}}\begin{array}{cc}1\text{s}\left(r_1\right)\alpha \left(1\right)& \text{1s}\left(r_2\right)\beta \left(1\right)\\ 2\text{s}\left(r_1\right)\alpha \left(2\right)& \text{2s}\left(r_2\right)\beta \left(2\right)\end{array}\\ =\frac{1}{\sqrt{2}}\left(\left[1\text{s}\left(r_1\right)2\text{s}\left(r_2\right)-2\text{s}\left(r_1\right)\text{1s}\left(r_2\right)\right]\left[\beta \left(2\right)\alpha \left(1\right)-\beta \left(1\right)\alpha \left(2\right)\right]\right)\end{array}$

Where,

• $1$ and $2$ represents the electrons.
• $\alpha$ and $\beta$ are the spin part of the wavefunction.

Possibility 3:

The columns represent the spin orbitals $1\text{s}\beta$ and $2\text{s}\beta$.  The slater determinant for the $\text{He}$ is shown below.

    $\begin{array}{c}{\psi }_{\text{He}}=\frac{1}{\sqrt{2}}\begin{array}{cc}1\text{s}\left(r_1\right)\beta \left(1\right)& \text{1s}\left(r_2\right)\beta \left(2\right)\\ 2\text{s}\left(r_1\right)\beta \left(1\right)& \text{2s}\left(r_2\right)\beta \left(2\right)\end{array}\\ =\frac{1}{\sqrt{2}}\left[1\text{s}\left(r_1\right)2\text{s}\left(r_2\right)-2\text{s}\left(r_1\right)\text{1s}\left(r_2\right)\right]\beta \left(1\right)\beta \left(2\right)\end{array}$

Where,

• $1$ and $2$ represents the electrons.
• $\beta$ is the spin part of the wavefunction.

Possibility 4:

The columns represent the spin orbitals $1\text{s}\beta$ and $2\text{s}\alpha$.  The slater determinant for the $\text{He}$ is shown below.

    $\begin{array}{c}{\psi }_{\text{He}}=\frac{1}{\sqrt{2}}\begin{array}{cc}1\text{s}\left(r_1\right)\beta \left(1\right)& \text{1s}\left(r_2\right)\alpha \left(2\right)\\ 2\text{s}\left(r_1\right)\beta \left(1\right)& \text{2s}\left(r_2\right)\alpha \left(2\right)\end{array}\\ =\frac{1}{\sqrt{2}}\left(\left[1\text{s}\left(r_1\right)2\text{s}\left(r_2\right)-2\text{s}\left(r_1\right)\text{1s}\left(r_2\right)\right]\left[\beta \left(1\right)\alpha \left(2\right)-\alpha \left(2\right)\beta \left(1\right)\right]\right)\end{array}$

Where,

• $1$ and $2$ represents the electrons.
• $\alpha$ and $\beta$ are the spin part of the wavefunction.

The wavefunctions for $\text{1s}$ and $2\text{s}$ orbitals are given below.

    $\begin{array}{l}{\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }}\left(\frac{Z_{\text{eff}}}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\\ {\psi }_{\text{2s}}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z_{\text{eff}}}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\end{array}$

The values of $Z_{\text{eff}}$ for $\text{1s}$ and $2\text{s}$ orbitals are given as $2$ and $1$ respectively.

Substitute the value of $Z_{\text{eff}}$ in the wavefunctions as shown below.

    $\begin{array}{l}{\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\\ {\psi }_{\text{2s}}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\end{array}$

Substitute the wavefunctions in order to get four possible total wavefunctions for $1{\text{s}}^{1}2{\text{s}}^1$ state of helium as shown below.

    $\begin{array}{c}{\psi }_{{\text{He}}_1}=\frac{1}{\sqrt{2}}\left[\begin{array}{l}\left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)-\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)\\ \left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\end{array}\right]\alpha \left(1\right)\alpha \left(2\right)\\ {\psi }_{{\text{He}}_2}=\frac{1}{\sqrt{2}}\left(\left[\begin{array}{l}\left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)-\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)\\ \left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\end{array}\right]\left[\beta \left(2\right)\alpha \left(1\right)-\beta \left(1\right)\alpha \left(2\right)\right]\right)\\ {\psi }_{{\text{He}}_3}=\frac{1}{\sqrt{2}}\left[\begin{array}{l}\left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)-\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)\\ \left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\end{array}\right]\beta \left(1\right)\beta \left(2\right)\\ {\psi }_{{\text{He}}_4}=\frac{1}{\sqrt{2}}\left(\left[\begin{array}{l}\left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)-\\ \left(\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{2f}{a_0}\right)e^{-2f/3a_0}\right)\\ \left(\frac{1}{\sqrt{\pi }}\left(\frac{2}{a_0}\right)e^{3/2}-\frac{2f}{a_0}\right)\end{array}\right]\left[\beta \left(1\right)\alpha \left(2\right)-\alpha \left(2\right)\beta \left(1\right)\right]\right)\end{array}$