Chapter 11, Helium and the periodic table Video Solutions, The Physics of Quantum Mechanics

Problem 1

Show that when the state of a pair of photons is expanded as
$$
|\psi\rangle=\sum_{n n^{\prime}} b_{n n^{\prime}}|n\rangle\left|n^{\prime}\right\rangle
$$
where $\{|n\rangle\}$ is a complete set of single-photon states, the expansion coefficients satisfy $b_{n n^{\prime}}=b_{n^{\prime} n}$.

Nicole Smina

Numerade Educator

08:57

Problem 2

Explain the physical content of writing the wavefunction of a pair of electrons in the form
$$
\left\langle\mathbf{x}, \mathbf{x}^{\prime} \mid \psi\right\rangle=\left(\begin{array}{l}
\psi_{++}\left(\mathbf{x}, \mathbf{x}^{\prime}\right) \\
\psi_{-+}\left(\mathbf{x}, \mathbf{x}^{\prime}\right) \\
\psi_{+-}\left(\mathbf{x}, \mathbf{x}^{\prime}\right) \\
\psi_{--}\left(\mathbf{x}, \mathbf{x}^{\prime}\right)
\end{array}\right)
$$
Which of these functions vanishes when the pair is a spin singlet? What relation holds between the non-zero functions? Suppose $|\psi\rangle$ for a spin singlet can be expanded in terms of products of the single-particle states $|u, \pm\rangle$ and $|v, \pm\rangle$ in which the individual electrons are in the states associated with spatial amplitudes $u(\mathbf{x})$ and $v(\mathbf{x})$ with $S_{z}$ returning $\pm \frac{1}{2} .$ Show that
$$
|\psi\rangle=\frac{1}{2}(|u,-\rangle|v,+\rangle-|v,+\rangle|u,-\rangle-|u,+\rangle|v,-\rangle+|v,-\rangle|u,+\rangle)
$$
and explain why this expansion is consistent with exchange symmetry.
Given the four single-particle states $|u, \pm\rangle$ and $|v, \pm\rangle$, how many linearly independent entangled states of a pair of particles can be constructed if the particles are not identical? How many linearly independent states are possible if the particles are identical fermions? Why are only four of these states accounted for by the states in the first excited level of helium?

Mahnoor Amin

Numerade Educator

07:06

Problem 3

In Chapter 6 we saw that when a state of a composite system has a non-trivial expansion $|\psi\rangle=\sum_{i j} c_{i j}|\mathrm{~A} ; i\rangle|\mathrm{B} ; j\rangle$ in terms of products of states $|\mathrm{A} ; i\rangle$ and $|\mathrm{B} ; j\rangle$ of the individual systems it does not automatically follow that the systems are entangled. By recalling the property that the matrix $c_{i j}$ will have if the systems are not entangled, show that any two electrons are always entangled.

Andrew Eddins

Emory University

06:47

Problem 4

In terms of the position vectors $\mathbf{x}_{\alpha}, \mathbf{x}_{1}$ and $\mathbf{x}_{2}$ of the $\alpha$ particle and two electrons, the centre of mass and relative coordinates of a helium atom are
$$
\mathbf{X} \equiv \frac{m_{\alpha} \mathbf{x}_{\alpha}+m_{\mathrm{e}}\left(\mathbf{x}_{1}+\mathbf{x}_{2}\right)}{m_{t}}, \quad \mathbf{r}_{1} \equiv \mathbf{x}_{1}-\mathbf{X}, \quad \mathbf{r}_{2} \equiv \mathbf{x}_{2}-\mathbf{X}
$$
where $m_{t} \equiv m_{\alpha}+2 m_{\mathrm{e}}$. Write the atom's potential energy operator in terms of the $\mathbf{r}_{i}$
Show that
$$
\begin{gathered}
\frac{\partial}{\partial \mathbf{X}}=\frac{\partial}{\partial \mathbf{x}_{\alpha}}+\frac{\partial}{\partial \mathbf{x}_{1}}+\frac{\partial}{\partial \mathbf{x}_{2}} \\
\frac{\partial}{\partial \mathbf{r}_{1}}=\frac{\partial}{\partial \mathbf{x}_{1}}-\frac{m_{\mathrm{e}}}{m_{\alpha}} \frac{\partial}{\partial \mathbf{x}_{\alpha}} \quad \frac{\partial}{\partial \mathbf{r}_{2}}=\frac{\partial}{\partial \mathbf{x}_{2}}-\frac{m_{\mathrm{e}}}{m_{\alpha}} \frac{\partial}{\partial \mathbf{x}_{\alpha}}
\end{gathered}
$$
and hence that the kinetic energy operator of the helium atom can be written
$$
K=-\frac{\hbar^{2}}{2 m_{t}} \frac{\partial^{2}}{\partial \mathbf{X}^{2}}-\frac{\hbar^{2}}{2 \mu}\left(\frac{\partial^{2}}{\partial \mathbf{r}_{1}^{2}}+\frac{\partial^{2}}{\partial \mathbf{r}_{2}^{2}}\right)-\frac{\hbar^{2}}{2 m_{t}}\left(\frac{\partial}{\partial \mathbf{x}_{1}}-\frac{\partial}{\partial \mathbf{x}_{2}}\right)^{2}
$$
where $\mu \equiv m_{\mathrm{e}}\left(1+2 m_{\mathrm{e}} / m_{\alpha}\right)$. What is the physical interpretation of the third term on the right? Explain why it is reasonable to neglect this term.

Mahipal Kumawat

Numerade Educator

20:00

Problem 5

Show that the exchange integral
$$
\int \mathrm{d}^{3} \mathbf{x} \mathrm{d}^{3} \mathbf{x}^{\prime} \frac{\Psi_{1}^{*}(\mathbf{x}) \Psi_{2}(\mathbf{x}) \Psi_{2}^{*}\left(\mathbf{x}^{\prime}\right) \Psi_{1}\left(\mathbf{x}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|}
$$
is real for any single-particle wavefunctions $\Psi_{1}$ and $\Psi_{2}$.

Mirza Aslam Beig

Numerade Educator

02:03

Problem 6

The $\mathrm{H}^{-}$ion consists of two electrons bound to a proton. Estimate its ground-state energy by adapting the calculation of helium's groundstate energy that uses the variational principle. Show that using singleparticle wavefunctions $u(\mathbf{x}) \propto \mathrm{e}^{-r / a}$ the expectation of the Hamiltonian is
$$
\langle H\rangle_{a}=\mathcal{R}\left(2 x^{2}-\frac{11}{4} x\right) \quad \text { where } \quad x \equiv \frac{a_{0}}{a}
$$
Hence find that the binding energy of $\mathrm{H}^{-}$is $\sim 0.945 \mathcal{R}$. Will $\mathrm{H}^{-}$be a stable ion?

Hafiz Shahzaib

Numerade Educator

06:53

Problem 7

Assume that a LiH molecule comprises a $\mathrm{Li}^{+}$ion electrostatically bound to an $\mathrm{H}^{-}$ion, and that in the molecule's ground state the kinetic energies of the ions can be neglected. Let the centres of the two ions be separated by a distance $b$ and calculate the resulting electrostatic binding energy under the assumption that they attract like point charges. Given that the ionisation energy of $\mathrm{Li}$ is $0.40 \mathcal{R}$ and using the result of Problem $11.6$, show that the molecule has less energy than that of well separated hydrogen and lithium atoms for $b<4.4 a_{0} .$ Does this calculation suggest that $\mathrm{LiH}$ is a stable molecule? Is it safe to neglect the kinetic energies of the ions within the molecule?

Prachita Kush

Numerade Educator

02:19

Problem 8

Two spin-one gyros are in a box. Express the states $|j, m\rangle$ in which the box has definite angular momentum as linear combinations of the states $|1, m\rangle\left|1, m^{\prime}\right\rangle$ in which the individual gyros have definite angular momentum. Hence show that
$$
|0,0\rangle=\frac{1}{\sqrt{3}}(|1,-1\rangle|1,1\rangle-|1,0\rangle|1,0\rangle+|1,1\rangle|1,-1\rangle)
$$
By considering the symmetries of your expressions, explain why the ground state of carbon has $l=1$ rather than $l=2$ or 0 . What is the total spin angular momentum of a $\mathrm{C}$ atom?

Dominador Tan

Numerade Educator

03:38

Problem 9

Suppose we have three spin-one gyros in a box. Express the state $|0,0\rangle$ of the box in which it has no angular momentum as a linear combination of the states $|1, m\rangle\left|1, m^{\prime}\right\rangle\left|1, m^{\prime \prime}\right\rangle$ in which the individual gyros have well-defined angular momenta. Hint: start with just two gyros in the box, giving states $|j, m\rangle$ of the box, and argue that only for a single value of $j$ will it be possible to get $|0,0\rangle$ by adding the third gyro; use results from Problem $11.8$
Explain the relevance of your result to the fact that the ground state of nitrogen has $l=0$. Deduce the value of the total electron spin of an $\mathrm{N}$ atom.

Penny Riley

Numerade Educator

02:27

Problem 10

Consider a system made of three spin-half particles with individual spin states $|\pm\rangle$. Write down a linear combination of states such as $|+\rangle|+\rangle|-\rangle$ (with two spins up and one down) that is symmetric under any exchange of spin eigenvalues $\pm$. Write down three other totally symmetric states and say what total spin your states correspond to.
Show that it is not possible to construct a linear combination of products of $|\pm\rangle$ which is totally antisymmetric.
What consequences do these results have for the structure of atoms such as nitrogen that have three valence electrons?

Zachary Warner

Numerade Educator