13.E: Variational Methods (Exercises)
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 15809

Consider a hydrogenlike 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]

Show that the energy levels are \[E_n = \frac{Z^{\,2}\,E_0}{n^{\,2}},\] where \(E_0\) is the groundstate energy of a conventional hydrogen atom.

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.


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

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 groundstate energy. Note that this energy lies well above the energy of the groundstate of singlyionized 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 singlyionized helium ion, with the energy of the ejected electron determined by energy conservation. This process is known as autoionization.

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]

Consider a particle of mass \(m\) moving in the onedimensional 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 groundstate and firstexcitedstate 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.

Use the variational technique outlined in Section 1.3 to derive the following estimate the groundstate energy of a twoelectron atom with nuclear charge \(Z_0\,e\) in the spinsinglet state: \[E = \frac{(16\,Z_05)^{\,2}}{2^{\,7}}\,E_0,\] where \(E_0\) is the hydrogen groundstate 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 singlyionized 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 \(E_{{\rm Li}_+}= 198.09\,{\rm eV}\) .

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 groundstate energies of twoelectron atoms in the spinsinglet state that are approximately \(1.5\,{\rm eV}\) too high. This is not a particular problem for the helium atom, or the singlyionized lithium ion. However, for the negative hydrogen ion, our estimate for the groundstate energy, \(12.86\,{\rm eV}\), is slightly higher than the groundstate 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]
Table associated with Exercise [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 spinsinglet and spintriplet states, respectively. Given that the Hamiltonian of a twoelectron atom of nuclear charge \(Z\,e\) is \[H = \frac
(click for details){2\,m_e} + \fracCallstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/13:_Variational_Methods/13.E:_Variational_Methods_(Exercises)), /content/body/ol/li[7]/p[2]/span[3]/span[1], line 1, column 1
(click for details){2\,m_e}  \frac{Z\,e^{\,2}}{4\pi\,\epsilon_0\,r_1} \frac{Z\,e^{\,2}}{4\pi\,\epsilon_0\,r_2} + \frac{e^{\,2}}{4\pi\,\epsilon_0\,r_{12}},\] 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 groundstate 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 groundstate energy. Unfortunately, this can only be achieved numerically.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
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 groundstate energy (\(E\)), as well as the corresponding experimentally measured groundstate energy (\(E_{\rm expt}\)) . It can be seen that our new estimate for the groundstate energy of the negative hydrogen ion is now less than the groundstate 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 lowestenergy spintriplet state of a helium atom (i.e., the \(1s\,2s\) spintriplet state).

Repeat the calculation of Section 1.4 using the the trial singleproton 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 groundstate energy, and \[F_+(y)= Z^{\,2} + \frac{2\,Z}{y}\left[ \frac{(1+y)\,{\rm e}^{2\,y} +(12\,y^{\,2}/3)\,{\rm e}^{\,y} + (Z1)\,y\,(1+[1+y]\,{\rm e}^{y})} {1+(1+y+y^{\,2}/3)\,{\rm e}^{y}}\right].\]
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)
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