Skip to main content
Physics LibreTexts

7.P: Exercises

  1. Calculate the energy-shift in the ground state of the one-dimensional harmonic oscillator when the perturbation


    $\displaystyle V = \lambda\,x^4


    is added to


    $\displaystyle H = \frac{p_x^{\,2}}{2\,m} + \frac{1}{2}\,m\,\omega^2\,x^2.


    The properly normalized ground-state wavefunction is


    $\displaystyle \psi(x) = \left(\frac{m\,\omega}{\pi\,\hbar}\right)^{1/4}\,\exp\left(-\frac{m\,\omega^2\,x^2}{2\,\hbar}\right).



  2. Calculate the energy-shifts due to the first-order Stark effect in the $ n=3$ state of a hydrogen atom. You do not need to perform all of the integrals, but you should construct the correct linear combinations of states.


  3. The Hamiltonian of the valence electron in a hydrogen-like atom can be written


    $\displaystyle H = \frac{p^2}{2\,m_e} + V(r) - \frac{p^4}{8\,m_e^{\,3}\,c^2}.


    Here, the final term on the right-hand side is the first-order correction due to the electron's relativistic mass increase. Treating this term 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


    $\displaystyle {\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, and $ \alpha$ the fine structure constant.


  4. Consider an energy eigenstate of the hydrogen atom characterized by the standard quantum numbers $ n$ , $ l$ , and $ m$ . Show that if the energy-shift due to spin-orbit coupling (see Section 7.7) is added to that due to the electron's relativistic mass increase (see previous exercise) then the net fine structure energy-shift can be written


    $\displaystyle {\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=l\pm 1/2$ the quantum number associated with the magnitude of the sum of the electron's orbital and spin angular momenta. You will need to use the following standard results for a hydrogen atom:


    $\displaystyle \left\langle \frac{a_0}{r}\right\rangle$ $\displaystyle = \frac{1}{n^2},$    
    $\displaystyle \left\langle \frac{a_0^{\,2}}{r^2}\right\rangle$ $\displaystyle = \frac{1}{(l+1/2)\,n^3},$    
    $\displaystyle \left\langle \frac{a_0^{\,3}}{r^3}\right\rangle$ $\displaystyle = \frac{1}{l\,(l+1/2)\,(l+1)\,n^3}.$    


    Here, $ a_0$ is the Bohr radius. Assuming that the above formula for the energy shift is valid for $ l=0$ (which it is), 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}$ .