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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle T\rangle = \langle V\rangle,} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle T\rangle = \frac{\hbar^{2}}{2ma^2}}

For the Yukawa potential, $\langle T\rangle$ remains he same, but Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle V \rangle} gets modified.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}dr d\Omega = \frac{-e^2}{4\pi \epsilon_{0}}\frac{4}{a^{3}} \int_{0}^{\infty}e^{-(\mu + 2/a)r} r dr = \frac{-e^2}{4\pi \epsilon_{0}}\frac{4}{a^{3}}\frac{1}{(\mu+2/a)^2} =}