If the photon had a nonzero mass $\left({\mathbf{m}}_{\gamma }\ne \mathbf{0}\right)$, the Coulomb potential would be replaced by the Yukawa potential,

$V\left(r\right)=-\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{e^{-\mu r}}{r}$ (8.73).

Where$\mu =m_{\gamma }c/\hslash$ . 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$(\mu a{)}^2$ .

The trial wave function of the form:

$\psi \left(x\right)=\frac{1}{\sqrt{\pi b^3}}{e}^{-r/b}$

Same hydrogen but replaced a with b.

For hydrogen,

So for trial wave function, .

Let $\psi =\frac{1}{\sqrt{\pi b^3}}{e}^{-r/b}$(same as hydrogen, but with a→b adjustable).

we have $⟨T⟩=-E_1=\frac{{\hslash }^2}{2m{a}^2}$for hydrogen, so in this case $⟨T⟩=\frac{{\hslash }^2}{2m{b}^2}.$

$⟨T⟩=-E_n;⟨V⟩=2E_n$ (4.218).

$\begin{array}{rcl}⟨V⟩& =& -\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{4\pi }{\pi b^3}{\int }_0^{\infty }{e}^{-2r/b}\frac{e^{-\mu r}}{r}{r}^{2}dr=-\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{4}{b^3}{\int }_0^{\infty }{e}^{-(\mu +2/b)r}rdr\\ & =& -\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{4}{b^3}\frac{1}{{(\mu +2/b)}^2}\\ & =& -\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{1}{b{\left(1+\frac{\mu b}{2}\right)}^2}\end{array}$

Now, calculate ,

$⟨H⟩=\frac{{\hslash }^2}{2m{b}^2}-\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{1}{b{\left(1+\frac{\mu b}{2}\right)}^2}.$

$\begin{array}{rcl}\frac{\partial ⟨H⟩}{\partial b}& =& -\frac{{\hslash }^2}{m{b}^3}+\frac{e^2}{4\mathrm{\pi }{\in }_0}\left[\frac{1}{b^2{(1+\mu b/2)}^2}+\frac{\mu }{b{(1+\mu b/2)}^3}\right]\\ & =& -\frac{{\hslash }^2}{m{b}^3}+\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{(1+3\mu b/2)}{b^2{(1+\mu b/2)}^3}\\ & =& 0\end{array}$

$$\begin{gathered} \frac{{\hslash }^2}{m}\left(\frac{4\mathrm{\pi }{\in }_0}{e^2}\right)=b\frac{(1+3\mu b/2)}{{(1+\mu b/2)}^3}, \\ \text{or}b\frac{(1+3\mu b/2)}{{(1+\mu b/2)}^3}=a. \end{gathered}$$

This determines b, but unfortunately it’s a cubic equation. So we use the fact that μ is small to obtain a suitable approximate solution. If μ = 0 , then b = a (of course), so$\mu a\ll 1;\mu b\ll 1$

. We’ll expand in powers of $\mu b$:

$$\begin{gathered} a\approx b\left(1+\frac{3\mu b}{2}\right)\left[1-\frac{3\mu b}{2}+6{\left(\frac{\mu b}{2}\right)}^2\right] \\ \approx b\left[1-\frac{9}{4}{\left(\mu b\right)}^2+\frac{6}{4}{\left(\mu b\right)}^2\right] \\ =b\left[1-\frac{3}{4}{\left(\mu b\right)}^2\right]. \end{gathered}$$

Since the $\frac{3}{4}(\mu b{)}^2$ term is already a second-order correction, we can replace b by a:

$$\begin{gathered} b\approx \frac{a}{\left[1-\frac{3}{4}{\left(\mu b\right)}^2\right]} \\ \approx a\left[1+\frac{3}{4}{\left(\mu a\right)}^2\right] \end{gathered}$$

$⟨H{⟩}_{\min }=\frac{{\hslash }^2}{2m{a}^2{\left[1+\frac{3}{4}{\left(\mu a\right)}^2\right]}^2}-\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{1}{a\left[1+\frac{3}{4}{\left(\mu a\right)}^2\right]{\left[1+\frac{1}{2}\left(\mu a\right)\right]}^2}$

$$\begin{gathered} \approx \frac{{\hslash }^2}{2m{a}^2}\left[1-2\frac{3}{4}{\left(\mu a\right)}^2\right]-\frac{e^2}{4\mathrm{\pi }{\in }_0}\frac{1}{a}\left[1-\frac{3}{4}{\left(\mu a\right)}^2\right]\left[1-2\frac{\mu a}{2}+3{\left(\frac{\mu a}{2}\right)}^2\right]. \\ =-E_1\left[1-\frac{3}{2}{\left(\mu a\right)}^2\right]+2E_1\left[1-\mu a+\frac{3}{4}{\left(\mu a\right)}^2-\frac{3}{4}{\left(\mu a\right)}^2\right] \\ =E_1\left[1-2\left(\mu a\right)+\frac{3}{2}{\left(\mu a\right)}^2\right]. \end{gathered}$$