Energy Levels of Hydrogen Atom

Consider a hydrogen atom, for which the potential takes the specific form

$$V(r) = -\frac{e^2}{4\pi\,\epsilon_0\,r}.$$ (396)

The radial eigenfunction $R(r)$ satisfies Equation (395), which can be written

$$\left[\frac{\hbar^2}{2\,\mu} \left(-\frac{1}{r^2} \frac{d}{dr}\,r... ... +\frac{l\,(l+1)}{r^2}\right) -\frac{e^2}{4\pi\,\epsilon_0\,r}- E\right] R = 0.$$ (397)

Here, $\mu = m_e \,m_p/(m_e+ m_p)\simeq m_e$ is the reduced mass, which takes into account the fact that the electron (of mass $m_e$ ) and the proton (of mass $m_p$ ) both rotate about a common centre, which is equivalent to a particle of mass $\mu$ rotating about a fixed point. Let us write the product $r\, R(r)$ as the function $P(r)$ . The above equation transforms to

$$\frac{d^2 P}{d r^2} - \frac{2\,\mu}{\hbar^2}\left[ \frac{l\,(l+1)\,\hbar^2}{2\,\mu \,r^2} - \frac{e^2}{4\pi \,\epsilon_0 \,r}-E\right] P =0,$$ (398)

which is the one-dimensional Schrödinger equation for a particle of mass $\mu$ moving in the effective potential

$$V_{\rm eff}(r) = -\frac{e^2}{4\pi \,\epsilon_0 \,r} + \frac{l\,(l+1)\,\hbar^2}{2\,\mu\, r^2}.$$ (399)

The effective potential has a simple physical interpretation. The first part is the attractive Coulomb potential, and the second part corresponds to the repulsive centrifugal force.

Let

$$a= \sqrt{\frac{-\hbar^2}{2\,\mu \,E}},$$ (400)

and $y=r/a$ , with

$$P(r) = f(y) \exp(-y).$$ (401)

Here, it is assumed that the energy eigenvalue $E$ is negative. Equation (398) transforms to

$$\left(\frac{d^2}{dy^2} -2\,\frac{d}{dy} -\frac{l\,(l+1)}{y^2} + \frac{2\,\mu\, e^2\, a}{4\pi\, \epsilon_0\, \hbar^2\,y}\right) f= 0.$$ (402)

Let us look for a power-law solution of the form

$$f(y) = \sum_{n} c_n\, y^{\,n}.$$ (403)

Substituting this solution into Equation (402), we obtain

$$\sum_n c_n \left[ n\,(n-1)\,y^{\,n-2} - 2\,n\, y^{\,n-1} - l\,(l+... ...+ \frac{2\,\mu\, e^2 \,a}{4\pi\, \epsilon_0 \,\hbar^2}\, y^{\,n-1} \right] = 0.$$ (404)

Equating the coefficients of $y^{\,n-2}$ gives

$$c_n\,[n\,(n-1) - l\,(l+1)] = c_{n-1} \left [2\,(n-1) - \frac{2\,\mu\, e^2\, a}{4\pi\, \epsilon_0\, \hbar^2}\right].$$ (405)

Now, the power law series (403) must terminate at small $n$ , at some positive value of $n$ , otherwise $f(y)$ would behave unphysically as $y\rightarrow 0$ . This is only possible if $[n_{\rm min} (n_{\rm min} -1) - l\,(l+1) ] =0$ , where the first term in the series is $c_{n_{\rm min}}\,y^{\,n_{\rm min}}$ . There are two possibilities: $n_{\rm min} = -l$ or $n_{\rm min} = l+1$ . The former predicts unphysical behavior of the wavefunction at $y=0$ . Thus, we conclude that $n_{\rm min} = l+1$ . Note that for an $l=0$ state there is a finite probability of finding the electron at the nucleus, whereas for an $l>0$ state there is zero probability of finding the electron at the nucleus (i.e., $\vert\psi\vert^{\,2} =0$ at $r=0$ , except when $l=0$ ). Note, also, that it is only possible to obtain sensible behavior of the wavefunction as $r\rightarrow 0$ if $l$ is an integer.

For large values of $y$ , the ratio of successive terms in the series (403) is

$$\frac{c_n \,y}{c_{n-1}} = \frac{2\, y}{n},$$ (406)

according to Equation (405). This is the same as the ratio of successive terms in the series

$$\sum_n \frac{(2\,y)^{\,n}}{n!},$$ (407)

which converges to $\exp(2\,y)$ . We conclude that $f(y)\rightarrow \exp(2\,y)$ as $y\rightarrow \infty$ . It follows from Equation (401) that $R(r) \rightarrow \exp(r/a) /r$ as $r\rightarrow \infty$ . This does not correspond to physically acceptable behavior of the wavefunction, since $\int d^3x'\, \vert\psi\vert^{\,2}$ must be finite. The only way in which we can avoid this unphysical behavior is if the series (403) terminates at some maximum value of $n$ . According to the recursion relation (405), this is only possible if

$$\frac{\mu\, e^2 \,a}{4\pi \,\epsilon_0\, \hbar^2} = n,$$ (408)

where the last term in the series is $c_n\, y^{\,n}$ . It follows from Equation (400) that the energy eigenvalues are quantized, and can only take the values

$$E = \frac{E_0}{n^2},$$ (409)

where

$$E_0 = - \frac{\mu\, e^4}{32\pi^2\,\epsilon_0^{\,2}\, \hbar^2} = - 13.6\,{\rm eV}$$ (410)

is the ground state energy. Here, $n$ is a positive integer which must exceed the quantum number $l$ , otherwise there would be no terms in the series (403).

The properly normalized wavefunction of a hydrogen atom is written

$$\psi(r, \theta, \varphi) = R_{n\,l}(r)\, Y_{l\,m} (\theta, \varphi),$$ (411)

where

$$R_{n\,l}(r) = {\cal R}_{n\,l}(r/a),$$ (412)

and

$$a = n\,a_0.$$ (413)

Here,

$$a_0 =\frac{4\pi\, \epsilon_0\,\hbar^2}{\mu \,e^2} = 5.3\times 10^{-11}\,\,\,{\rm meters}$$ (414)

is the Bohr radius, and ${\cal R}_{n\,l}(x)$ is a well-behaved solution of the differential equation

$$\left[\frac{1}{x^2} \frac{d}{dx}\, x^2 \,\frac{d}{dx}-\frac{l\,(l+1)}{x^2} + \frac{2\,n}{x} - 1\right] {\cal R}_{n\,l} = 0$$ (415)

that is consistent with the normalization constraint

$$\int_0^\infty dr\,r^2\,[R_{n\,l}(r)]^{\,2}= 1.$$ (416)

Finally, the $Y_{l\,m}$ are spherical harmonics. The restrictions on the quantum numbers are $\vert m\vert \leq l< n$ , where $n$ is a positive integer, $l$ a non-negative integer, and $m$ an integer.

The ground state of hydrogen corresponds to $n=1$ . The only permissible values of the other quantum numbers are $l=0$ and $m=0$ . Thus, the ground state is a spherically symmetric, zero angular momentum state. The next energy level corresponds to $n=2$ . The other quantum numbers are allowed to take the values $l=0$ , $m=0$ or $l=1$ , $m=-1, 0, 1$ . Thus, there are $n=2$ states with non-zero angular momentum. Note that the energy levels given in Equation (409) are independent of the quantum number $l$ , despite the fact that $l$ appears in the radial eigenfunction equation (415). This is a special property of a $1/r$ Coulomb potential.

In addition to the quantized negative energy states of the hydrogen atom, which we have just found, there is also a continuum of unbound positive energy states.