$$\require{cancel}$$

13.E: Variational Methods (Exercises)

1. 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 $$m$$.[ex9.8]

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: \begin{aligned} R_{1\,0}(r) &= \frac{2\,Z^{\,3/2}}{a_0^{\,3/2}}\,\exp\left(-\frac{Z\,r}{a_0}\right),\\[0.5ex] 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),\\[0.5ex] 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).\end{aligned} where $$a_0$$ is the Bohr radius.

2. 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 $$24.56\,{\rm eV}$$.

3. 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 1.3 to obtain the following estimate for the energy of this state: $E(2p\,2p) = \left(-2+\frac{3\cdot 31}{2^{\,7}}\right)|E_0|= -17.33\,{\rm eV},$ 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.

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

5. Consider a particle of mass $$m$$ moving in the one-dimensional potential $V(x)=\lambda\,x^{\,4},$ where $$\lambda>0$$. Let \begin{aligned} \psi_0(x)&= \frac{\alpha^{\,1/2}}{\pi^{\,1/4}}\,{\rm e}^{-\alpha^{\,2}\,x^{\,2}/2},\\[0.5ex] \psi_1(x) &= \frac{\sqrt{2}\,\beta^{\,3/2}}{\pi^{\,1/4}}\,x\,{\rm e}^{-\beta^{\,2}\,x^{\,2}/2}.\end{aligned} 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: \begin{aligned} 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},\\[0.5ex] E_1&= \frac{9\cdot 5^{\,1/3}}{4}\left(\frac{\hbar^{\,2}}{2\,m}\right)^{2/3} \lambda^{\,1/3}= 3.847\left(\frac{\hbar^{\,2}}{2\,m}\right)^{2/3}\lambda^{\,1/3}.\end{aligned} The exact numerical factors that should appear in the previous two equations are $$1.060$$ and $$3.800$$, respectively . Hence, it is clear that our approximation to $$E_0$$ and $$E_1$$ are fairly accurate.

6. Use the variational technique outlined in Section 1.3 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 $$E_{{\rm H}-}= -14.36\,{\rm eV}$$ . 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 $$E_{{\rm Li}_+}= -198.09\,{\rm eV}$$ .

7. It can be seen from Section 1.3, as well as the previous exercise, that the variational technique described in Section 1.3 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). [ex13.7]

 $$Z$$ $$\epsilon$$ $$Z_1$$ $$Z_2$$ $$E \,({\rm eV})$$ $$E_{\rm expt}\,({\rm eV})$$ [0.5ex] $$1$$ $$+1$$ $$1.04$$ $$0.28$$ $$-13.97$$ $$-14.35$$ [0.5ex] $$2$$ $$+1$$ $$2.18$$ $$1.19$$ $$-78.25$$ $$-78.98$$ [0.5ex] $$2$$ $$-1$$ $$1.97$$ $$0.32$$ $$-58.79$$ $$-59.18$$ [0.5ex] $$3$$ $$+1$$ $$3.29$$ $$2.08$$ $$-197.25$$ $$-198.10$$ [0.5ex] $$3$$ $$-1$$ $$2.93$$ $$0.60$$ $$-138.01$$ $$-139.06$$

Obviously, we need to perform a more accurate calculation for the case of a negative hydrogen ion. Following Chandrasekhar , let us adopt the following trial wavefunction: $\phi({\bf r}_1,{\bf r}_2)=\frac{1}{\sqrt{2}}\left[\psi_1({\bf r}_1)\,\psi_2({\bf r}_2)+\epsilon\,\psi_2({\bf r}_1)\,\psi_1({\bf r}_2)\right],$ where \begin{aligned} \psi_1({\bf r}) &=\frac{1}{\sqrt{\pi}}\left(\frac{Z_1}{a_0}\right)^{3/2}\exp\left(\frac{-Z_1\,r}{a_0}\right),\\[0.5ex] \psi_2({\bf r}) &=\frac{1}{\sqrt{\pi}}\left(\frac{Z_2}{a_0}\right)^{3/2}\exp\left(\frac{-Z_2\,r}{a_0}\right).\end{aligned} Here, $$r=|{\bf r}|$$, $$a_0$$ 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 $H = \frac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/13:_Variational_Methods/13.E:_Variational_Methods_(Exercises)), /content/body/ol/li[7]/p[2]/span[3]/span[1], line 1, column 1  {2\,m_e} + \frac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/13:_Variational_Methods/13.E:_Variational_Methods_(Exercises)), /content/body/ol/li[7]/p[2]/span[3]/span[2], line 1, column 1  {2\,m_e} - \frac{Z\,e^{\,2}}{4\pi\,\epsilon_0\,r_1} -\frac{Z\,e^{\,2}}{4\pi\,\epsilon_0\,r_2} + \frac{e^{\,2}}{4\pi\,\epsilon_0\,r_{12}},$ show that the expectation value of $$H$$ (i.e., $$\langle H\rangle = \langle\phi|\,H\,|\phi\rangle/\langle \phi|\phi\rangle$$) is \begin{aligned} \frac{\langle H\rangle}{|E_0|}& = \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.\\[0.5ex]&\phantom{=}\left.\left.-\epsilon\left(2\,Z-\frac{5}{8}\right)x\,y^{\,6}+\frac{1}{2}\,\epsilon\,y^{\,8}\right]\right/\left(x^{\,6}+\epsilon\,y^{\,6}\right),\end{aligned} 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.

Table [table1] 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 ($$E_{\rm expt}$$) . 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).

8. Repeat the calculation of Section 1.4 using the the trial single-proton wavefunction $\psi_0({\bf r}) = \frac{1}{\sqrt{\pi}\,a^{\,3/2}}\,{\rm e}^{-r/a},$ where $$a=a_0/Z$$, $$r= |{\bf r}|$$, $$a_0$$ is the Bohr radius, and $$Z$$ an adjustable parameter . 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 $F_+(y)= -Z^{\,2} + \frac{2\,Z}{y}\left[ \frac{(1+y)\,{\rm e}^{-2\,y} +(1-2\,y^{\,2}/3)\,{\rm e}^{\,-y} + (Z-1)\,y\,(1+[1+y]\,{\rm e}^{-y})} {1+(1+y+y^{\,2}/3)\,{\rm e}^{-y}}\right].$

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 $$E_{\rm bind} = 0.173\,|E_0|= 2.35\,{\rm eV}$$, respectively. (See Figure [fh2pa].) These values are far closer to the experimentally determined values, $$R_0=1.06\times 10^{-10}\,{\rm m}$$ and $$E_{\rm bind} = 2.8\,{\rm eV}$$ , than those derived in Section 1.4. [ex9.23]

Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)
