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13.E: Variational Methods (Exercises)

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    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]

      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 spin-singlet and spin-triplet states, respectively. Given that the Hamiltonian of a two-electron atom of nuclear charge \(Z\,e\) is \[H = \frac

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      {2\,m_e} + \frac
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      {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]


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

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