Exercises

1. Demonstrate that the eigenvalues of the two-particle permutation operator, $P_{12}$ , are $\pm 1$ .

2. Derive Equations (9.14) and (9.15).

3. Demonstrate that the two-particle permutation operator, $P_{12}$ , is Hermitian.

4. Justify Equation (9.35).

5. Consider the cyclic permutation operator
   $$P_{123} \,\vert k'\rangle\,\vert k''\rangle\,\vert k'''\rangle =\vert k'''\rangle\,\vert k'\rangle\,\vert k''\rangle.$$
   Demonstrate that
   $$P_{123}^{\,2} = P_{123}^{\,\dag },$$
   and
   $$P_{123}\,P_{123}^{\,\dag } =1.$$

6. Consider two identical spin-$1/2$ particles of mass $m$ confined in a cubic box of dimension $L$ . Find the possible energies and wavefunctions of this system in the case of no interaction between the particles.

7. Consider a system of two spin-$1$ particles with no orbital angular momentum (i.e., both particles are in the lowest energy $s$ -state). What are the possible eigenvalues of the total angular momentum of the system, as well as its projection along the $z$ -direction, in the cases in which the particles are non-identical and identical?

8. Consider a hydrogen-like atom consisting of a single electron of charge $-e$ orbiting about a massive nucleus of charge $Z\,e$ (where $Z>0$ ). The eigenstates of the Hamiltonian can be labelled by the conventional quantum numbers $n$ , $l$ , and $m$ .
   1. Show that the energy levels are
      $$E_n = \frac{Z^{\,2}\,E_0}{n^{\,2}},$$
      where $E_0$ is the ground-state energy of a conventional hydrogen atom.
   2. Demonstrate that the first few properly normalized radial wavefunctions, $R_{n\,l}(r)$ , take the form:

      $$R_{1\,0}(r) = \frac{2\,Z^{\,3/2}}{a_0^{\,3/2}}\,\exp\left(-\frac{Z\,r}{a_0}\right),$$

      $$R_{2\,0}(r) = \frac{2\,Z^{\,3/2}}{(2\,a_0)^{3/2}}\left(1-\frac{Z\,r}{2\,a_0}\right)\exp\left(-\frac{Z\,r}{2\,a_0}\right),$$

      $$R_{2\,1}(r) = \frac{Z^{\,3/2}}{\sqrt{3}\,(2\,a_0)^{3/2}}\,\frac{Z\,r}{a_0}\,\exp\left(-\frac{Z\,r}{2\,a_0}\right).$$

      
      
      where $a_0$ is the Bohr radius.

9. Demonstrate that
   $$\int_0^{x_1}dx_2\,x_2^{\,n}\,{\rm e}^{-\beta\,x_2} = \frac{n!}{\b... ...e}^{-\beta\,x_1}}{\beta^{\,n+1}}\sum_{k=0,n}\frac{n!}{k!}\,(\beta\,x_1)^{\,k},$$
   and
   $$\int_{x_1}^\infty dx_2\,x_2^{\,n}\,{\rm e}^{-\beta\,x_2} =\frac{{\rm e}^{-\beta\,x_1}}{\beta^{\,n+1}}\sum_{k=0,n}\frac{n!}{k!}\,(\beta\,x_1)^{\,k}$$
   Here, $n$ is a non-negative integer, and $\beta>0$ .

10. Justify Equation (9.63).

11. Given that the ground-state energy of a helium atom is $-78.98\,{\rm eV}$ , deduce that the ground-state ionization energy (i.e., the minimum energy that must be supplied to remove a single electron from the atom in the ground-state) is $24.56\,{\rm eV}$ .

12. Consider a helium atom in which both electrons are in the $n=2$ , $l=1$ , $m=0$ state (i.e., the $2p$ state). Use the techniques of Section 9.5 to obtain the following estimate for the energy of this state:
    $$E(2p\,2p) = \left(-2+\frac{3\cdot 31}{2^{\,7}}\right)\vert E_0\vert= -17.33\,{\rm eV},$$
    where $E_0$ is the hydrogen ground-state energy. Note that this energy lies well above the energy of the ground-state of singly-ionized helium, which is $E_{{\rm He}+} = 4\,E_0= -54.42\,{\rm eV}$ . This means that a helium atom excited to the $2p\,2p$ state has the option of decaying into a free electron and a singly-ionized helium ion, with the energy of the ejected electron determined by energy conservation. This process is known as autoionization.

13. Consider an excited state of the helium atom in which one electron is in the ground state, and the other is in the $n, l, m$ state, where $n>1$ .
    1. Demonstrate that the expressions for the direct and exchange integrals given in Section 9.6 reduce to

       $$\frac{I}{\vert E_0\vert} = 2\,a_0\int_0^\infty dr_1\,r_1\,R_{1\,0}(r_1)\,R_{1\,0}(r_1)\left[ \int_0^{r_1}dr_2\,r_2^{\,2}\,R_{n\,l}(r_2)\,R_{n\,l}(r_2)\right.\nonumber$$

       $$\phantom{=}\left.+\int_{r_1}^\infty dr_2\,r_1\,r_2\,R_{n\,l}(r_2)\,R_{n\,l}(r_2)\right],$$

       
       
       and

       $$\frac{J}{\vert E_0\vert} = \frac{2\,a_0}{2\,l+1}\int_0^\infty dr_1\,r_1\,R_{1\,0}(r_1)\,R_... ...\,2}\left(\frac{r_2}{r_1}\right)^l R_{1\,0}(r_2)\,R_{n\,l}(r_2)\right.\nonumber$$

       $$\phantom{=}\left.+\int_{r_1}^\infty dr_2\,r_1\,r_2\left(\frac{r_1}{r_2}\right)^l R_{1\,0}(r_2)\,R_{n\,l}(r_2)\right],$$

       
       
       respectively, where $R_{n\,l}(r)$ is a hydrogen radial wavefunction calculated with a nuclear charge $2\,e$ .

    2. Suppose that the excited electron is in the $n=2$ , $l=1$ , $m=0$ state (i.e., the $2p$ state). Show that

       $$\frac{I}{\vert E_0\vert} = \frac{2^{\,2}\cdot 59}{3^{\,5}}= 0.9712,$$

       $$\frac{J}{\vert E_0\vert} = \frac{2^{\,6}\cdot 7}{3^{\,8}}=0.0683.$$

       
       
       Hence, deduce that the energies of the $1s\,2p$ states of orthohelium and parahelium are

       $$E(1s\,2p)_\pm =(-5+0.9712\pm0.0683)\,\vert E_0\vert = (-4.0288\pm 0.0683)\,\vert E_0\vert$$

       $$=(-54.81\pm0.93)\,{\rm eV},$$

       
       
       where the upper sign corresponds to parahelium.

14. The observed energies of the $1s\,2p$ states of orthohelium and parahelium are

    $$E(1s\,2p)_\pm=(-57.89\pm 0.13)\,{\rm eV},$$

    
    
    where the upper sign corresponds to parahelium [77]. It can be seen that the calculation in the previous exercise has considerably overestimated the size of the exchange integral. The main reason for this is that we neglected to take into account the fact that the $1s$ electron largely shields the $2p$ electron from the nuclear charge (because, on average, the former electron lies much closer to the nucleus that the latter). We can arrive at a better estimate for the exchange integral by using a $R_{1\,0}(r)$ radial function calculated with a nuclear charge $2\,e$ , and a $R_{2\,1}(r)$ radial function calculated with a nuclear charge $e$ . Here, we are assuming that the inner ($1s$ ) electron experiences the full nuclear charge, whereas the outer ($2p$ ) electron only experiences half the nuclear charge, as a consequence of the shielding action of the inner electron. Demonstrate that, in this case, the exchange integral becomes
    $$\frac{J}{\vert E_0\vert} = \frac{2^{\,8}\cdot 7}{5^{\,7}\cdot 3}= 0.00765,$$
    which yields
    $$J = 0.10\,{\rm eV}.$$
    Note that this estimate is much closer to the experimental value ( $0.13\,{\rm eV}$ ) than our previous estimate ( $0.93\,{\rm eV}$ ).

15. Let
    $$J_n(\beta) = \int_{-\infty}^\infty dx\,x^{\,2\,n} \,{\rm e}^{-\beta\,x^{\,2}},$$
    where $n$ is a non-negative integer, and $\beta>0$ . Demonstrate that
    $$J_0(\beta) =\left(\frac{\pi}{\beta}\right)^{1/2},$$
    and
    $$J_{n>0}(\beta) =\frac{1\cdot 3\cdot 5\cdots (2\,n-1)}{2^{\,n}}\left(\frac{\pi}{\beta^{\,2\,n+1}}\right)^{1/2}.$$

16. Consider the general energy eigenvalue problem
    $$H\,\vert\psi\rangle = E\,\vert\psi\rangle.$$
    Suppose that $\vert\psi\rangle$ is a trial solution to the previous equation that is not properly normalized. Prove that
    $$E_0 \leq \frac{\langle \psi \vert\,H\,\vert\psi\rangle}{\langle\psi\vert\psi\rangle},$$
    where $E_0$ is the lowest energy eigenvalue.

17. Consider a particle of mass $m$ moving in the one-dimensional potential
    $$V(x)=\lambda\,x^{\,4},$$
    where $\lambda>0$ . Let

    $$\psi_0(x) = \frac{\alpha^{\,1/2}}{\pi^{\,1/4}}\,{\rm e}^{-\alpha^{\,2}\,x^{\,2}/2},$$

    $$\psi_1(x) = \frac{\sqrt{2}\,\beta^{\,3/2}}{\pi^{\,1/4}}\,x\,{\rm e}^{-\beta^{\,2}\,x^{\,2}/2}.$$

    
    
    Verify that these wavefunctions are properly normalized. Use the variational principle, combined with the plausible trial wavefunctions $\psi_0(x)$ and $\psi_1(x)$ (these wavefunctions are, in fact, the exact ground-state and first-excited-state wavefunctions for a particle moving in the potential $\lambda\,x^{\,2}$ ) to obtain the following estimates for the energies of the ground state, and the first excited state, of the system:

    $$E_0 = \frac{3^{\,4/3}}{4}\left(\frac{\hbar^{\,2}}{2\,m}\right)^{2/3}\lambda^{\,1/3} = 1.082\left(\frac{\hbar^{\,2}}{2\,m}\right)^{2/3}\lambda^{\,1/3},$$

    $$E_1 = \frac{9\cdot 5^{\,1/3}}{4}\left(\frac{\hbar^{\,2}}{2\,m}\right)... ...ambda^{\,1/3}= 3.847\left(\frac{\hbar^{\,2}}{2\,m}\right)^{2/3}\lambda^{\,1/3}.$$

    
    
    The exact numerical factors that should appear in the previous two equations are $1.060$ and $3.800$ , respectively [77]. Hence, it is clear that our approximation to $E_0$ and $E_1$ are fairly accurate.

18. Use the variational technique outlined in Section 9.7 to derive the following estimate the ground-state energy of a two-electron atom with nuclear charge $Z_0\,e$ in the spin-singlet state:
    $$E = \frac{(16\,Z_0-5)^{\,2}}{2^{\,7}}\,E_0,$$
    where $E_0$ is the hydrogen ground-state energy. For the case of a negative hydrogen ion (i.e., $Z_0=1$ ), this formula gives $E_{{\rm H}-}= 0.9453\,E_0=-12.86\,{\rm eV}$ . The experimental value of this energy is $E_{{\rm H}-}= -14.36\,{\rm eV}$ [63]. For the case of a singly-ionized lithium ion (i.e., $Z_0=3$ ), the previous formula gives $E_{{\rm Li}+}= 14.45\,E_0=-196.54\,{\rm eV}$ . The experimental value of this energy is $E_{{\rm Li}_+}= -198.09\,{\rm eV}$ [63].

19. It can be seen from Section 9.7, as well as the previous exercise, that the variational technique described in Section 9.7 yields approximations to the ground-state energies of two-electron atoms in the spin-singlet state that are approximately $1.5\,{\rm eV}$ too high. This is not a particular problem for the helium atom, or the singly-ionized lithium ion. However, for the negative hydrogen ion, our estimate for the ground-state energy, $-12.86\,{\rm eV}$ , is slightly higher than the ground-state energy of a neutral hydrogen atom, $-13.61\,{\rm eV}$ , giving the erroneous impression that it is not energetically favorable for a neutral hydrogen atom to absorb an additional electron to form a negative hydrogen ion (i.e., that the negative hydrogen ion has a negative binding energy).

    
    

    $Z$	$\epsilon$	$Z_1$	$Z_2$	$E \,({\rm eV})$	$E_{\rm expt}\,({\rm eV})$
    $1$	$+1$	$1.04$	$0.28$	$-13.97$	$-14.35$
    $2$	$+1$	$2.18$	$1.19$	$-78.25$	$-78.98$
    $2$	$-1$	$1.97$	$0.32$	$-58.79$	$-59.18$
    $3$	$+1$	$3.29$	$2.08$	$-197.25$	$-198.10$
    $3$	$-1$	$2.93$	$0.60$	$-138.01$	$-139.06$

    
    

    Obviously, we need to perform a more accurate calculation for the case of a negative hydrogen ion. Following Chandrasekhar [21], let us adopt the following trial wavefunction:

    $$\phi({\bf x}_1,{\bf x}_2)=\frac{1}{\sqrt{2}}\left[\psi_1({\bf x}_1)\,\psi_2({\bf x}_2)+\epsilon\,\psi_2({\bf x}_1)\,\psi_1({\bf x}_2)\right],$$
    where

    $$\psi_1({\bf x}) =\frac{1}{\sqrt{\pi}}\left(\frac{Z_1}{a_0}\right)^{3/2}\exp\left(\frac{-Z_1\,r}{a_0}\right),$$

    $$\psi_2({\bf x}) =\frac{1}{\sqrt{\pi}}\left(\frac{Z_2}{a_0}\right)^{3/2}\exp\left(\frac{-Z_2\,r}{a_0}\right).$$

    
    
    Here, $r=\vert{\bf x}\vert$ , $a_0$ is the Bohr radius, and $Z_1$ , $Z_2$ are adjustable parameters. Moreover, $\epsilon$ takes the values $+1$ and $-1$ for the spin-singlet and spin-triplet states, respectively. Given that the Hamiltonian of a two-electron atom of nuclear charge $Z\,e$ is
    $$H = \frac{{\bf p}_1^{\,2}}{2\,m_e} + \frac{{\bf p}_2^{\,2}}{2\,m_... ...{Z\,e^{\,2}}{4\pi\,\epsilon_0\,r_2} + \frac{e^{\,2}}{4\pi\,\epsilon_0\,r_{12}},$$
    show that the expectation value of $H$ (i.e., $\langle H\rangle = \langle\phi\vert\,H\,\vert\phi\rangle/\langle \phi\vert\phi\rangle$ ) is

    $$\frac{\langle H\rangle}{\vert E_0\vert} = \left[x^{\,8}-2\,Z\,x^{\,7}- \frac{1}{2}\,x^{\,6}\,y^{\,2} + \frac{1}{2}\,x^{\,5}\,y^{\,2}+\frac{1}{8}\,x^{\,3}\,y^{\,4}\right.$$

    $$\phantom{=}\left.\left.-\epsilon\left(2\,Z-\frac{5}{8}\right)x\,y... ...c{1}{2}\,\epsilon\,y^{\,8}\right]\right/\left(x^{\,6}+\epsilon\,y^{\,6}\right),$$

    
    
    where $E_0$ is the hydrogen ground-state energy, $x= Z_1+Z_2$ , and $y=2\sqrt{Z_1\,Z_2}$ . We now need to minimize $\langle H\rangle$ with respect to variations in $Z_1$ and $Z_2$ to obtain an estimate for the ground-state energy. Unfortunately, this can only be achieved numerically.

    The previous table shows the numerically determined values of $Z_1$ and $Z_2$ that minimize $\langle H\rangle$ for various choices of $Z$ and $\epsilon$ . The table also shows the estimate for the ground-state energy ($E$ ), as well as the corresponding experimentally measured ground-state energy ( $E_{\rm expt}$ ) [63,66]. It can be seen that our new estimate for the ground-state energy of the negative hydrogen ion is now less than the ground-state energy of a neutral hydrogen atom, which demonstrates that the negative hydrogen ion has a positive (albeit, small) binding energy. Incidentally, the case $Z=2$ , $\epsilon=-1$ yields a good estimate for the energy of the lowest-energy spin-triplet state of a helium atom (i.e., the $1s\,2s$ spin-triplet state).

20. Justify Equation (9.108).

21. Justify Equations (9.116) and (9.117).

22. Justify Equations (9.119) and (9.120).

23. Repeat the calculation of Section 9.8 using the the trial single-proton wavefunction
    $$\psi_0({\bf x}) = \frac{1}{\sqrt{\pi}\,a^{\,3/2}}\,{\rm e}^{-r/a},$$
    where $a=a_0/Z$ , $r=\vert{\bf x}\vert$ , $a_0$ is the Bohr radius, and $Z$ an adjustable parameter [48]. Show that the energy of the hydrogen molecule ion, assuming a molecular wavefunction that is even under exchange of proton positions, can be written
    $$E_{\rm total} = - F_+(R/a)\,E_0,$$
    where $R$ is the proton separation, $E_0$ the hydrogen ground-state energy, and
    $$F_+(y)= -Z^{\,2} + \frac{2\,Z}{y}\left[ \frac{(1+y)\,{\rm e}^{-2\... ...+ (Z-1)\,y\,(1+[1+y]\,{\rm e}^{-y})} {1+(1+y+y^{\,2}/3)\,{\rm e}^{-y}}\right].$$

    It can be shown, numerically, that the previous function attains its minimum value, $F_+ = -1.173$ , when $Z= 1.238$ and $y=2.480$ . This leads to predictions for the equilibrium separation between the two protons, and the binding energy of the molecule, of $R_0= (2.480/1.238)\,a_0= 2.003\,a_0= 1.06\times 10^{\,-10}\,{\rm m}$ and $E_{\rm bind} = 0.173\,\vert E_0\vert= 2.35\,{\rm eV}$ , respectively. (See Figure 9.1.) These values are far closer to the experimentally determined values, $R_0=1.06\times 10^{-10}\,{\rm m}$ and $E_{\rm bind} = 2.8\,{\rm eV}$ [53], than those derived in Section 9.8.