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Chapter 7: Problem 17

Consider a one-dimensional square well containing two electrons. One electron has \(n=1\) and the other has \(n=2 .\) Plot a two-dimensional contour diagram of the probability distribution of the electrons when their spins are (a) parallel, (b) antiparallel. Devise a measure of the radius of the Fermi hole. Hint. Recall the discussion in Section \(7.11 .\) When the spins are parallel (for example, \(\alpha \alpha\) ) the antisymmetric combination \(\psi_{1}(1) \psi_{2}(2)-\psi_{2}(1) \psi_{1}(2)\) must be used, and when the spins are antiparallel, the symmetric combination must be used. In each case plot \(\psi^{2}\) against axes labelled \(x_{1}\) and \(x_{2}\). Computer graphics may be used to obtain striking diagrams, but a sketch is sufficient.

Short Answer

Expert verified
The state function for parallel spins is antisymmetric and for antiparallel spins is symmetric. The probability distributions are obtained by squaring these state functions. The plots should then show less overlapping for the parallel spins due to the Fermi hole, and more overlapping for the antiparallel spins.

Step by step solution

01

Construction of the State Functions for Parallel Spins

For parallel spins (such as \(\alpha \alpha\)), the antisymmetric combination of wave functions should be used. The wave functions \(\psi_{1}(x)\) and \(\psi_{2}(x)\) correspond to ground state (n=1) and first excited state (n=2) respectively. The state function would therefore be \(\Psi_{\alpha\alpha}(x_{1}, x_{2}) = \psi_{1}(x_{1}) \psi_{2}(x_{2}) - \psi_{2}(x_{1}) \psi_{1}(x_{2})\).
02

Get the Probability Distribution for Parallel Spins

The probability distribution is the square of the absolute value of the state function: \(P_{\alpha\alpha}(x_{1}, x_{2}) = |\Psi_{\alpha\alpha}(x_{1}, x_{2})|^2\). Since the original state function is real, we just square it.
03

Construction of the State Functions for Antiparallel Spins

For antiparallel spins, the symmetric combination of the wave functions should be used. The state function would therefore be \(\Psi_{\alpha\beta}(x_{1}, x_{2}) = \psi_{1}(x_{1}) \psi_{2}(x_{2}) + \psi_{2}(x_{1}) \psi_{1}(x_{2})\).
04

Get the Probability Distribution for Antiparallel Spins

The probability distribution is the square of the absolute value of the state function: \(P_{\alpha\beta}(x_{1}, x_{2}) = |\Psi_{\alpha\beta}(x_{1}, x_{2})|^2\). Since the original state function is real, we just square it.
05

Plotting the probability distributions

Sketch the final probability distributions \(P_{\alpha\alpha}(x_{1}, x_{2})\) and \(P_{\alpha\beta}(x_{1}, x_{2})\) in a two-dimensional contour plot with \(x_{1}\) and \(x_{2}\) as axes. These step requires knowledge of computer graphics, which will enable striking diagrams, but a hand-sketch would suffice.

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Most popular questions from this chapter

Suppose that an electron experiences a shielded Coulomb potential (a Coulomb potential modified by \(\left.\operatorname{afactor} \exp \left(-r / r_{\mathrm{D}}\right), \text { where } r_{\mathrm{D}} \text { is a constant }\right) .\) Evaluate the ratio of spin-orbit coupling constants \(\zeta_{2 \mathrm{p}} / \zeta_{2 \mathrm{p}}^{\circ},\) where \(\zeta_{2 \mathrm{p}}^{\circ}\) is the constant for the unshielded potential. Explore the result of setting \(r_{\mathrm{D}}=k a_{0},\) where \(k\) is a variable parameter.

(a) Calculate the energy difference between the levels with the greatest and smallest values of \(j\) for given \(l\) and \(s\) Each term of a level is \((2 j+1)\) -fold degenerate. (b) Demonstrate that the barycentre (mean energy) of a term is the same as the energy in the absence of spin-orbit coupling. Hint. Weight each level with \(2 j+1\) and sum the energies given in eqn 7.24 from \(j=|l-s|\) to \(j=l+s\) Use the relations $$\begin{array}{l} \sum_{s=0}^{n} s=\frac{1}{2} n(n+1) \quad \sum_{s=0}^{n} s^{2}=\frac{1}{6} n(n+1)(2 n+1) \\ \sum_{s=0}^{n} s^{3}=\frac{1}{4} n^{2}(n+1)^{2} \end{array}$$

\(\mathrm{On}\) the basis of the Thomas-Fermi model of an atom, evaluate the radius within which there is a 50 per cent probability of finding the electron density and evaluate it for the Period 2 elements. Hint. Use the radial distribution function \(r^{2} \rho\).

Demonstrate that for one-electron atoms the selection rules are \(\Delta l=\pm 1, \Delta m_{l}=0,\pm 1,\) and \(\Delta n\) unlimited. Hint. Evaluate the electric-dipole transition moment \(\left\langle n^{\prime}\left|m_{l}^{\prime}\right| \mu | n l m_{l}\right\rangle\) with \(\mu_{x}=-e r \sin \theta \cos \varphi, \mu_{y}=-e r \sin \theta \sin \varphi,\) and \(\mu_{z}=-e r \cos \theta\). The easiest way of evaluating the angular integrals is to recognize that the components just listed are proportional to \(Y_{l m,}\) with \(l=1,\) and to analyse the resulting integral group theoretically.

An excited state of atomic calcium has the electron configuration \(1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{d}^{1} 4 \mathrm{f}^{1}\) (a) Derive all the term symbols (with the appropriate specifications of \(S, L \text { and } J)\) for the electron configuration. (b) Which term symbol corresponds to the lowest energy of this electron configuration? (c) Consider a \(^{3} \mathrm{F}_{2}\) level of calcium derived from a different electron configuration than that shown above. Which of the term symbols determined in part (a) can participate in spectroscopic transitions to this \(^{3} \mathrm{F}_{2}\) level?
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