Energy Levels of Hydrogen Atom

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

$$V(r) = -\frac{e^{\,2}}{4\pi\,\epsilon_0\,r}.$$ (4.119)

The radial eigenfunction $R(r)$ satisfies Equation (4.118), which can be written

$$\left[\frac{\hbar^{\,2}}{2\,\mu} \left(-\frac{1}{r^{\,2}} \frac{d... ...l\,(l+1)}{r^{\,2}}\right) -\frac{e^{\,2}}{4\pi\,\epsilon_0\,r}- E\right] R = 0.$$ (4.120)

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 orbit about a common centre of mass, which is equivalent to a particle of mass $\mu$ orbiting about a fixed point [50]. However, given that $m_e\ll m_p$ , we can safely replace $\mu$ by $m_e$ . (The correction involved in using $\mu$ , rather than $m_e$ , in the analysis is actually less than that involved in neglecting the electron's relativistic mass increase.) Let us write the product $r\, R(r)$ as the function $P(r)$ . The previous equation transforms to

$$\frac{d^{\,2} P}{d r^{\,2}} - \frac{2\,m_e}{\hbar^{\,2}}\left[ \f... ...{\,2}}{2\,m_e \,r^{\,2}} - \frac{e^{\,2}}{4\pi \,\epsilon_0 \,r}-E\right] P =0,$$ (4.121)

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

$$V_{\rm eff}(r) = -\frac{e^{\,2}}{4\pi \,\epsilon_0 \,r} + \frac{l\,(l+1)\,\hbar^{\,2}}{2\,m_e\, r^{\,2}}.$$ (4.122)

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

$$a= \sqrt{\frac{-\hbar^{\,2}}{2\,m_e \,E}},$$ (4.123)

and $y=r/a$ , with

$$P(r) = f(y) \exp(-y).$$ (4.124)

Here, it is assumed that the energy eigenvalue $E$ is negative. Equation (4.121) transforms to

$$\left(\frac{d^{\,2}}{dy^{\,2}} -2\,\frac{d}{dy} -\frac{l\,(l+1)}{... ... + \frac{2\,m_e\, e^{\,2}\, a}{4\pi\, \epsilon_0\, \hbar^{\,2}\,y}\right) f= 0.$$ (4.125)

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

$$f(y) = \sum_{k} c_k\, y^{\,k}.$$ (4.126)

Substituting this solution into Equation (4.125), we obtain

$$\sum_k c_k \left[ k\,(k-1)\,y^{\,k-2} - 2\,k\, y^{\,k-1} - l\,(l+... ...2\,m_e\, e^{\,2} \,a}{4\pi\, \epsilon_0 \,\hbar^{\,2}}\, y^{\,k-1} \right] = 0.$$ (4.127)

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

$$c_k\,[k\,(k-1) - l\,(l+1)] = c_{k-1} \left [2\,(k-1) - \frac{2\,m_e\, e^{\,2}\, a}{4\pi\, \epsilon_0\, \hbar^{\,2}}\right].$$ (4.128)

Now, the power-law series (4.126) must terminate at small $k$ , at some positive value of $k$ , otherwise $f(y)$ would diverge unphysically as $y\rightarrow 0$ . This is only possible if $[k_{\rm min} \,(k_{\rm min} -1) - l\,(l+1) ] =0$ , where the first term in the series is $c_{k_{\rm min}}\,y^{\,k_{\rm min}}$ . There are two possibilities: $k_{\rm min} = -l$ or $k_{\rm min} = l+1$ . The former predicts unphysical divergence of the wavefunction at $y=0$ . Thus, we conclude that $k_{\rm min} = l+1$ , and that the first term in the series is $c_{l+1}\,y^{\,l+1}$ . 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 $l=0$ ). Furthermore, 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 (4.126) is

$$\frac{c_k \,y}{c_{k-1}} = \frac{2\, y}{k},$$ (4.129)

according to Equation (4.128). This is the same as the ratio of successive terms in the series

$$\sum_k \frac{(2\,y)^{\,k}}{k!},$$ (4.130)

which converges to $\exp(2\,y)$ . We conclude that $f(y)\rightarrow \exp(2\,y)$ as $y\rightarrow \infty$ . It follows from Equation (4.124) that $R(r) \rightarrow \exp(r/a) /r$ as $r\rightarrow \infty$ . This does not correspond to physically acceptable behavior of the wavefunction, because $\int d^{\,3}{\bf x}\, \vert\psi\vert^{\,2}$ must be finite. The only way in which we can avoid this unphysical behavior is if the series (4.126) terminates at some maximum value of $k$ . According to the recursion relation (4.128), this is only possible if

$$\frac{m_e\, e^{\,2} \,a}{4\pi \,\epsilon_0\, \hbar^{\,2}} = n,$$ (4.131)

where the last term in the series is $c_n\, y^{\,n}$ . It follows from Equation (4.123) that the energy eigenvalues are quantized, and can only take the values

$$E = \frac{E_0}{n^{\,2}},$$ (4.132)

where

$$E_0 = - \frac{m_e\, e^{\,4}}{32\pi^2\,\epsilon_0^{\,2}\, \hbar^{\,2}} = - 13.6\,{\rm eV}$$ (4.133)

is the energy of the ground state (i.e., the lowest energy state). Here, $n$ is a positive integer that must exceed the quantum number $l$ , otherwise there would be no terms in the series (4.126). Expression (4.132) is known as the Bohr formula, because it was derived heuristically by Niels Bohr in 1913 [10].

In summary, the properly normalized wavefunction of a hydrogen atom is written

$$\psi(r, \theta, \varphi) = R_{n\,l}(r)\, Y_{l\,m} (\theta, \varphi),$$ (4.134)

where

$$R_{n\,l}(r) = \frac{u_{n\,l}(r/a)}{r/a},$$ (4.135)

and

$$a = n\,a_0.$$ (4.136)

Here,

$$a_0 =\frac{4\pi\, \epsilon_0\,\hbar^{\,2}}{m_e \,e^{\,2}} = 5.3\times 10^{-11}\,\,\,{\rm meters}$$ (4.137)

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

$$\left[\frac{d^{\,2}}{dx^{\,2}}-\frac{l\,(l+1)}{x^{\,2}} + \frac{2\,n}{x} - 1\right] u_{n\,l} = 0$$ (4.138)

that satisfies the normalization constraint

$$a^{\,3}\int_0^\infty dx\,[u_{n\,l}(x)]^{\,2}= \int_0^\infty dr\,r^{\,2}\left[R_{nl}(r)\right]^{\,2} = 1.$$ (4.139)

Finally, the $Y_{l\,m}(\theta,\varphi)$ 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. Incidentally, the quantum numbers $n$ , $l$ , and $m$ are conventionally referred to as the principle quantum number, the azimuthal quantum number, and the magnetic quantum number, respectively.

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 (4.132) are independent of the quantum number $l$ , despite the fact that $l$ appears in the radial eigenfunction equation (4.138). 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.