Phy5646/soham1

(Introduction to Quantum 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 $\mu a\ll 1$, where $a={\frac {4\pi \epsilon _{0}\hbar ^{2}}{me_{2}}}$, and give your answer correct to order $(\mu a)^{2}$

Solution:

The simplest trial function looks exactly like the hydrogen atom ground wavefunction, but with $a$ changed to $b$. $(\psi ={\frac {1}{\sqrt {\pi b^{3}}}}e^{-r/b}.$. $b$ acts 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 $\langle T\rangle =-E_{1}{\frac {\hbar ^{2}}{2ma^{2}}}$

For the Yukawa potential, $\langle T\rangle ={\frac {\hbar ^{2}}{2mb^{2}}}$ and

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

Or,${\mathcal {H}}=\langle T\rangle +\langle V\rangle ={\frac {\hbar ^{2}}{2mb^{2}}}-{\frac {-e^{2}}{4\pi \epsilon _{0}}}{\frac {1}{b(1+\mu b/2)^{2}}}$

${\frac {\partial {\mathcal {H}}}{\partial b}}=0={\frac {\hbar ^{2}}{mb^{3}}}+{\frac {-e^{2}}{4\pi \epsilon _{0}}}\left[{\frac {1}{b(1+\mu b/2)^{2}}}+{\frac {\mu }{b(1+\mu b/2)^{3}}}\right]=frac{\hbar ^{2}}{mb^{3}}+{\frac {-e^{2}}{4\pi \epsilon _{0}b^{2}}}\left[{\frac {1+3\mu b/2)}{(1+\mu b/2)^{3}}}\right]$

Or, $frac{4\pi \epsilon _{0}\hbar ^{2}}{me_{2}}=b{\frac {1+3\mu b/2)}{(1+\mu b/2)^{3}}}$ Or, $b{\frac {1+3\mu b/2)}{(1+\mu b/2)^{3}}}=a$ Now since $\mu a\ll 1,<math>\mu b\ll 1$</math>. So

$b(1+3\mu b/2)\left[1-{\frac {3\mu b}{2}}+6{\frac {\mu ^{2}b^{2}}{4}}\right]\approx a$ Or, $a\approx =b\left[1-3(\mu b)^{2}/4\right]$

In the second order term, we can replace $b$ by $a$ So

$b\approx a\left[1+3(\mu a)^{2}/4\right]$

With this optimized value of $b$, we can find out $\langle {\mathcal {H}}\rangle$

$\langle {\mathcal {H}}\rangle ={\frac {\hbar ^{2}}{2ma^{2}\left[1+3(\mu a)^{2}/4\right]}}-{\frac {e^{2}}{4\pi \epsilon _{0}}}{\frac {1}{a\left[1+3(\mu a)^{2}/4\right]\left[1+\mu a/2\right]^{2}}}$ where we have approximated $b$ by $a$ in the last bracketed term in the denominator.

Or,$\langle {\mathcal {H}}\rangle ={\frac {\hbar ^{2}}{2ma^{2}}}\left[1-6(\mu a)^{2}/4\right]-{\frac {-e^{2}}{4\pi \epsilon _{0}a}}\left[1-3(\mu a)^{2}/4\right]\left[1-2(\mu a)/2+3(\mu a/2)^{2}\right]=-E_{1}\left[1-3(\mu a)^{2}/2\right]+2E_{1}\left[1-\mu a-{\frac {3}{4}}(\mu a)^{2}+{\frac {3}{4}}(\mu a)^{2}\right]$ Or, $\langle {\mathcal {H}}\rangle =E_{1}\left[1-2\mu a+{\frac {3}{2}}(\mu a)^{2}\right]$