Phy5646/soham1

(Introduction to Qusntum Mechanics, Griffiths, 2e)Problem 7.14

If the photon has a nonzero mass $(m_{\gamma }\neq 0)$, the Coulomb potential would be rep[laced by the Yukawa potential, $V(r)={\frac {e^{2}}{4\pi \epsilon _{0}}}{\frac {e^{-\mu r}}{r}}$ where $\mu =m_{\gamma }c\hbar$. With a trial wave function of your own devising, estimate the binding energy of a "hydrogen" atom with this potential. Assume $\mu a\ll 1$, and give your answer correct to order $(\mu a)^{2}$

Solution:

The simplest trial function looks exactly like the hydrogen atom ground wavefunction $(\psi ={\frac {1}{\sqrt {\pi a^{3}}}}e^{-r/a}.$ but with $a$ as a variational parameter. The hydrogen atom Hamiltonian is ${\mathcal {H}}={\frac {-\hbar ^{2}}{2m}}-{\frac {e^{2}}{4\pi \epsilon _{0}}}{\frac {1}{r}}=T+V$

For hydrogen atom with standard Coulomb potential (massless photons), we have $\langle T\rangle =\langle V\rangle ,$ with $\langle T\rangle ={\frac {\hbar ^{2}}{2ma^{2}}}$

For the Yukawa potential, $\langle T\rangle$ remains he same, but $\langle V\rangle$ gets modified.

$V={\frac {-e^{2}}{4\pi \epsilon _{0}}}{\frac {1}{\pi a^{3}}}\int _{0}^{\infty }{\frac {e^{-2r/b}e^{-\mu r}}{r}}r^{2}drd\Omega ={\frac {-e^{2}}{4\pi \epsilon _{0}}}{\frac {4}{a^{3}}}\int _{0}^{\infty }e^{-(\mu +2/a)r}rdr={\frac {-e^{2}}{4\pi \epsilon _{0}}}{\frac {4}{a^{3}}}{\frac {1}{(\mu +2/a)^{2}}}=$