Exercises

1. Consider the two-state system investigated in Section 7.2. Show that the most general expressions for the perturbed energy eigenvalues and eigenstates are

   $$E_1' = E_1 + e_{11}+\frac{\vert e_{12}\vert^{\,2}}{E_1-E_2}+{\cal O}(\epsilon^{\,3}),$$

   $$E_2' = E_2+e_{22}- \frac{\vert e_{12}\vert^{\,2}}{E_1-E_2}+{\cal O}(\epsilon^{\,3}),$$

   
   
   and

   $$\vert 1\rangle' = \vert 1\rangle + \frac{e_{12}^{\,\ast}}{E_1-E_2}\,\vert 2\rangle + {\cal O}(\epsilon^{\,2}),$$

   $$\vert 2\rangle' = \vert 2\rangle -\frac{e_{12}}{E_1-E_2}\,\vert 1\rangle+{\cal O}(\epsilon^{\,2}),$$

   
   
   respectively. Here, $\epsilon = \vert e_{12}\vert/(E_1-E_2)\ll 1$ . You may assume that $\vert e_{11}\vert/(E_1-E_2)$ , $\vert e_{22}\vert/(E_1-E_2)\sim {\cal O}(\epsilon)$ .

2. Consider the two-state system investigated in Section 7.2. Show that if the unperturbed energy eigenstates are also eigenstates of the perturbing Hamiltonian then

   $$E_1' = E_1 + e_{11},$$

   $$E_2' = E_2+e_{22},$$

   
   
   and

   $$\vert 1\rangle' = \vert 1\rangle$$

   $$\vert 2\rangle' = \vert 2\rangle$$

   
   
   to all orders in the perturbation expansion.

3. Consider the two-state system investigated in Section 7.2. Show that if the unperturbed energy eigenstates are degenerate, so that $E_1=E_2=E_{12}$ then the most general expressions for the perturbed energy eigenvalues and eigenstates are
   $$E^\pm = E_{12}+e^\pm,$$
   and
   $$\vert E^\pm\rangle = \langle 1\vert E^\pm\rangle \vert 1\rangle+\langle 2\vert E^\pm\rangle\vert 2\rangle,$$
   respectively, where
   $$e^\pm = \frac{1}{2}\,(e_{11}+e_{22})\pm \frac{1}{2}\left[(e_{11}-e_{22})^2+4\,\vert e_{12}\vert^{\,2}\right]^{1/2},$$
   and
   $$\frac{\langle 1\vert E^\pm\rangle}{\langle 2\vert E^\pm\rangle}=-... ...2}}{e_{11}-e^\pm}\right)=-\left(\frac{e_{22}-e^{\pm}}{e_{12}^{\,\ast}}\right).$$
   Demonstrate that the $\vert E^\pm\rangle$ are the simultaneous eigenkets of the unperturbed Hamiltonian, $H_0$ , and the perturbed Hamiltonian, $H_1$ , and that the $e^\pm$ are the corresponding eigenvalues of $H_1$ .

4. Calculate the lowest-order energy-shift in the ground state of the one-dimensional harmonic oscillator when the perturbation
   $$V = \lambda\,x^{\,4}$$
   is added to
   $$H = \frac{p_x^{\,2}}{2\,m} + \frac{1}{2}\,m\,\omega^{\,2}\,x^{\,2}.$$
   [53]

5. Let $\vert n,l,m\rangle$ denote a properly normalized eigenstate of the hydrogen atom corresponding to the conventional quantum numbers $n$ , $l$ , and $m$ . Show that the only non-zero matrix elements of the operator $z$ between the various $n=3$ states take the values

   $$\langle 3,1,0\vert\,z\,\vert 3,0,0\rangle =\langle 3,0,0\vert\,z\,\vert 3,1,0\rangle = -3\sqrt{6}\,a_0,$$

   $$\langle 3,1,0\vert\,z\,\vert 3,2,0\rangle =\langle 3,2,0\vert\,z\,\vert 3,1,0\rangle = -3\sqrt{3}\,a_0,$$

   $$\langle 3,1,\pm1\vert\,z\,\vert 3,2,\pm 1\rangle = \langle 3,2,\pm 1\vert\,z\,\vert 3,1,\pm 1\rangle = -\frac{9}{2}\,a_0.$$

   
   

6. Calculate the energy-shifts and perturbed eigenstates associated with the linear Stark effect in the $n=3$ state of a hydrogen atom.

7. Suppose that the Hamiltonian, $H$ , for a particular quantum system, is a function of some parameter, $\lambda$ . Let $E_n(\lambda)$ and $\vert n\rangle(\lambda)$ be the eigenvalues and eigenkets of $H(\lambda)$ . Prove the Feynman-Hellmann theorem [45]
   $$\frac{\partial E_n}{\partial\lambda} =\left\langle n(\lambda)\left\vert\,\frac{\partial H}{\partial\lambda}\,\right\vert n(\lambda)\right\rangle.$$
   [61]

8. According to Section 4.6, the Hamiltonian for the radial wavefunction of an energy eigenstate of the hydrogen atom corresponding to the conventional quantum numbers $n$ , $l$ , and $m$ is written
   $$H = -\frac{\hbar^{\,2}}{2\,m_e}\,\frac{1}{r^{\,2}}\,\frac{d}{dr}\... ...^{\,2}}{2\,m_e}\,\frac{l\,(l+1)}{r^{\,2}}-\frac{e^{\,2}}{4\pi\,\epsilon_0\,r}.$$
   Moreover, when expressed as a function of $l$ , the energy eigenvalues are
   $$E_n = \frac{E_0}{(k+l)^{\,2}},$$
   where $E_0$ is the ground-state energy, and $k=n-l$ the number of terms in the power series (4.126). Treating $l$ as a continuous parameter, use the Feynman-Hellmann theorem to prove that
   $$\left\langle \frac{1}{r^{\,2}}\right\rangle= \frac{1}{(l+1/2)\,n^{\,3}\,a_0^{\,2}},$$
   where the expectation value is taken over the energy eigenstate corresponding to the quantum numbers $n$ and $l$ . [61]

9. Demonstrate that
   $$\left\langle \frac{1}{r^{\,3}}\right\rangle =\frac{1}{l\,(l+1/2)\,(l+1)\,n^{\,3}\,a_0^{\,3}},$$
   where the expectation value is taken over the energy eigenstate of the hydrogen atom characterized by the standard quantum numbers $n$ and $l$ . (Hint: Use Kramer's relation--see Exercise 16.)

10. The relativistic definition of the kinetic energy, $K$ , of a particle of total energy $E$ and rest mass $m$ is
    $$K= E - m\,c^{\,2}.$$
    Making use of the standard relativistic result [49]
    $$E = \left[p^{\,2}\,c^{\,2}+m^{\,2}\,c^{\,4}\right]^{1/2},$$
    where $p$ is the particle momentum, demonstrate that, in the non-relativistic limit $p\ll m\,c$ ,
    $$K\simeq \frac{p^{\,2}}{2\,m} - \frac{p^{\,4}}{8\,m^{\,3}\,c^{\,2}}.$$

11. The Hamiltonian of the valence electron in a hydrogen-like atom can be written
    $$H = \frac{p^{\,2}}{2\,m_e} + V(r) + H_R,$$
    where
    $$H_R = - \frac{p^{\,4}}{8\,m_e^{\,3}\,c^{\,2}}$$
    is the first-order correction due to the electron's relativistic mass increase. (See Exercise 10.) Treating $H_R$ as a small perturbation, deduce that it causes an energy-shift in the energy eigenstate, characterized by the standard quantum numbers $n$ , $l$ , $m$ , of
    $${\mit\Delta}E_{nlm} = -\frac{1}{2\,m_e\,c^{\,2}}\left(E_n^{\,2} - 2\,E_n\,\langle V\rangle + \langle V^{\,2}\rangle\right),$$
    where $E_n$ is the unperturbed energy. Finally, show that for the special case of a hydrogen atom, the energy-shift becomes
    $${\mit\Delta}E_{nlm} =\frac{\alpha^{\,2}\,E_n}{n^{\,2}}\left(\frac{n}{l+1/2}-\frac{3}{4}\right),$$
    where $\alpha$ is the fine structure constant.

12. According to Dirac's relativistic electron theory, there is an additional relativistic correction to the Hamiltonian of a valence electron in a hydrogen-like atom that takes the form
    $$H_D = \frac{e^{\,2}\,\hbar^{\,2}}{8\,\epsilon_0\,m_e^{\,2}\,c^{\,2}}\,\delta^{\,3}({\bf x})$$
    [9]. This correction is usually referred to as the Darwin term [24]. Treating $H_D$ as a small perturbation, deduce that, for the special case of a hydrogen atom, it causes an energy-shift in the energy eigenstate, characterized by the standard quantum numbers $n$ , $l$ , $m$ , of
    $${\mit\Delta} E_{nlm} = -\frac{\alpha^{\,2}\,E_n}{n}$$
    for an $l=0$ state, and
    $${\mit\Delta} E_{nlm}=0$$
    for an $l>0$ state. Note that
    $$\vert\psi_{n\,l\,m}({\bf0})\vert = \frac{1}{\sqrt{\pi}\,(n\,a_0)^{3/2}}\,\delta_{l\,0}\,\delta_{m\,0},$$
    where $\psi_{n\,l\,m}({\bf x})$ is the properly normalized wavefunction associated with an energy eigenstate of the hydrogen atom [95].

13. Consider an energy eigenstate of the hydrogen atom characterized by the standard quantum numbers $n$ , $l$ , and $m$ . Show that the energy-shift due to spin-orbit coupling takes the form
    $${\mit\Delta}E_{nlm} = -\frac{\alpha^{\,2}\,E_n}{n^{\,2}}\left[\frac{n}{2\,(l+1/2)\,(l+1)}\right]$$
    for $j=l+1/2$ , and
    $${\mit\Delta}E_{nlm} = \frac{\alpha^{\,2}\,E_n}{n^{\,2}}\left[\frac{n}{2\,(l+1/2)\,l}\right]$$
    for $j=l-1/2$ , and
    $${\mit\Delta} E_{nlm}=0$$
    for the special case of an $l=0$ state. Here, $j$ is the standard quantum number associated with the magnitude of the sum of the electron's orbital and spin angular momenta. (See Section 7.7.)

14. Demonstrate that if the energy-shifts due to the electron's relativistic mass increase, the Darwin term, and spin-orbit coupling, calculated in the previous three exercises for an energy eigenstate of the hydrogen atom, characterized by the standard quantum numbers $n$ , $l$ , and $m$ , are added together then the net fine structure energy-shift can be written
    $${\mit\Delta} E_{nlm} = \frac{\alpha^{\,2}\,E_n}{n^{\,2}}\left(\frac{n}{j+1/2}-\frac{3}{4}\right).$$
    Here, $E_n$ is the unperturbed energy, $\alpha$ the fine structure constant, and $j$ the standard quantum number associated with the magnitude of the sum of the electron's orbital and spin angular momenta.

    Show that fine structure causes the energy of the $(2p)_{3/2}$ states of a hydrogen atom to exceed those of the $(2p)_{1/2}$ and $(2s)_{1/2}$ states by $4.5\times 10^{-5}\,{\rm eV}$ .

15. The linear Stark effect exhibited by the $n=2$ states of a hydrogen atom depends crucially on the supposed degeneracy of these states. However, this degeneracy is lifted by fine structure. Consequently, the expressions for the Stark energy shifts derived in Section 7.6 are only valid when the energy splitting predicted by the linear Stark effect greatly exceeds that caused by fine structure. Deduce that this is the case when
    $$\vert{\bf E}\vert\gg 1\times 10^5\,{\rm V\,m}^{\,-1}.$$

16. Demonstrate that
    $$Y_{l\,m_l}\,\,\chi_\pm = \left(\frac{l\pm m_l+1}{2\,l+1}\right)^{... ...left(\frac{l\mp m_l}{2\,l+1}\right)^{1/2}{\cal Y}_{l\,\,m_l\pm 1/2}^{\,l-1/2},$$
    where the $Y_{l\,m_l}$ are spherical harmonics, the $\chi_\pm$ are standard Pauli two-component spinors, and the ${\cal Y}^{\,j}_{l\,m_j}$ are spin-angular functions. In particular, show that

    $$Y_{0\,0}\,\,\chi_\pm = {\cal Y}_{0\,\,\pm1/2}^{\,1/2},$$

    $$Y_{1\,0}\,\,\chi_\pm = \sqrt{\frac{2}{3}}\,{\cal Y}_{1\,\,\pm 1/2}^{\,3/2} \mp\sqrt{\frac{1}{3}}\,{\cal Y}_{1\,\,\pm 1/2}^{\,1/2},$$

    $$Y_{1\,\mp 1}\,\,\chi_\pm = \sqrt{\frac{1}{3}}\,{\cal Y}_{1\,\,\mp 1/2}^{\,{3/2}}\mp \sqrt{\frac{2}{3}}\,{\cal Y}_{1\,\,\mp 1/2}^{\,1/2},$$

    $$Y_{1\,\pm 1}\,\,\chi_\pm = {\cal Y}_{1\,\,\pm 3/2}^{\,3/2}.$$

    
    

17. Taking electron spin into account, the $n=2$ energy eigenstates of the hydrogen atom in the presence of an external electric field are

    $$\vert 1\pm\rangle = \sqrt{\frac{1}{2}}\,\vert 2,0,0\rangle\,\chi_\pm -\sqrt{\frac{1}{2}}\,\vert 2,1,0\rangle\,\chi_\pm,$$

    $$\vert 2\pm\rangle = \sqrt{\frac{1}{2}}\,\vert 2,0,0\rangle\,\chi_\pm +\sqrt{\frac{1}{2}}\,\vert 2,1,0\rangle\,\chi_\pm,$$

    $$\vert 3\pm\rangle = \vert 2,1,1\rangle\,\chi_\pm,$$

    $$\vert 4\pm\rangle = \vert 2,1,-1\rangle\,\chi_\pm.$$

    
    
    (See Section 7.6.) Here, $\vert n,l,m\rangle$ is an unperturbed energy eigenket corresponding to the standard quantum numbers $n$ , $l$ , and $m$ . Moreover, the energies of these states are

    $$E_{1\pm} = E_2 +3\,e\,a_0\,\vert{\bf E}\vert,$$

    $$E_{2\pm} = E_2-3\,e\,a_0\,\vert{\bf E}\vert,$$

    $$E_{3\pm } = E_2,$$

    $$E_{4\pm} = E_2,$$

    
    
    where ${\bf E}$ is the external electric field-strength, and $E_2$ the unperturbed energy eigenvalue corresponding to $n=2$ . The fine structure Hamiltonian can be written
    $$H_{FS} = H_{LS} + H_R + H_D,$$
    where $H_{LS}$ is the spin-orbit Hamiltonian (see Section 7.7), and $H_R$ and $H_D$ are defined in Exercises 11 and 12, respectively. Treating $H_{FS}$ as a small perturbation, deduce that fine structure modifies the energies of the previously defined eigenstates such that

    $$E_{1\pm} = E_2\left(1+\frac{11}{48}\,\alpha^{\,2}\right)+3\,e\,a_0\,\vert{\bf E}\vert,$$

    $$E_{2\pm} = E_2\left(1+\frac{11}{48}\,\alpha^{\,2}\right)-3\,e\,a_0\,\vert{\bf E}\vert,$$

    $$E_{3-} =E_2\left(1+\frac{11}{48}\,\alpha^{\,2}\right),$$

    $$E_{4+} =E_2\left(1+\frac{11}{48}\,\alpha^{\,2}\right),$$

    $$E_{3+} = E_2\left(1+\frac{11}{48}\,\alpha^{\,2}\right)-\frac{1}{6}\,\alpha^{\,2}\,E_2,$$

    $$E_{4-} = E_2\left(1+\frac{11}{48}\,\alpha^{\,2}\right) -\frac{1}{6}\,\alpha^{\,2}\,E_2.$$

    
    

18. Consider the $n=2$ energy eigenstates of the hydrogen atom in the Paschen-Back limit. (See Section 7.8.) These states are conveniently labeled using the standard quantum numbers $n$ , $l$ , $m_l$ , and $m_s$ . Treating the fine structure Hamiltonian, $H_{FS}$ , defined in the previous exercise, as a small perturbation, show that the perturbed energies of the various states are

    $$E_{2,1,1,1/2} = E_2\left(1+\frac{5}{16}\,\alpha^{\,2}\right) +2\,\mu_B\,B - \frac{1}{3}\,\alpha^{\,2}\,E_2,$$

    $$E_{2,1,0,1/2} = E_2\left(1+\frac{5}{16}\,\alpha^{\,2}\right) +\mu_B\,B - \frac{1}{6}\,\alpha^{\,2}\,E_2,$$

    $$E_{2,0,0,1/2} = E_2\left(1+\frac{5}{16}\,\alpha^{\,2}\right) +\mu_B\,B,$$

    $$E_{2,1,1,-1/2} = E_2\left(1+\frac{5}{16}\,\alpha^{\,2}\right),$$

    $$E_{2,1,-1,1/2} = E_2\left(1+\frac{5}{16}\,\alpha^{\,2}\right),$$

    $$E_{2,0,0,-1/2} = E_2\left(1+\frac{5}{16}\,\alpha^{\,2}\right) -\mu_B\,B,$$

    $$E_{2,1,0,-1/2} = E_2\left(1+\frac{5}{16}\,\alpha^{\,2}\right) -\mu_B\,B - \frac{1}{6}\,\alpha^{\,2}\,E_2,$$

    $$E_{2,1,-1,-1/2} = E_2\left(1+\frac{5}{16}\,\alpha^{\,2}\right) -2\,\mu_B\,B- \frac{1}{3}\,\alpha^{\,2}\,E_2.$$

    
    
    Here, $\mu_B$ is the Bohr magnetron, and $B$ the external magnetic field-strength.

19. Justify Equation (7.143).