Energy levels of the hydrogen atom

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

$$V(r) = -\frac{e^2}{4\pi\epsilon_0 r}.$$ (403)

The radial eigenfunction $R(r)$ satisfies Eq. (402), which can be written

$$\left[\frac{\hbar^2}{2 \mu} \left(-\frac{1}{r^2} \frac{d}{d... ...)}{r^2}\right) -\frac{e^2}{4\pi\epsilon_0 r}- E\right] R = 0.$$ (404)

Here, $\mu = m_e m_p/(m_e+ m_p)$ 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

$$\frac{d^2 P}{d r^2} - \frac{2 \mu}{\hbar^2}\left( \frac{l ... ...2 \mu r^2} - \frac{e^2}{4\pi \epsilon_0 r}-E\right) P =0,$$ (405)

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

$$V_{\rm eff}(r) = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{l (l+1) \hbar^2}{2 \mu r^2}.$$ (406)

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

$$a= \sqrt{\frac{-\hbar^2}{2 \mu E}},$$ (407)

and $y=r/a$, with

$$P(r) = f(y) \exp(-y).$$ (408)

Here, it is assumed that the energy eigenvalue $E$ is negative. Equation (405) transforms to

$$\left(\frac{d^2}{dy^2} -2 \frac{d}{dy} -\frac{l (l+1)}{y^2... ...c{2 \mu e^2 a}{4\pi \epsilon_0 \hbar^2 y}\right) f= 0.$$ (409)

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

$$f(y) = \sum_{n} c_n y^n.$$ (410)

Substituting this solution into Eq. (409), we obtain

$$\sum_n c_n \left\{ n (n-1) y^{n-2} - 2 n y^{n-1} - l (... ... e^2 a}{4\pi \epsilon_0 \hbar^2} y^{n-1} \right\} = 0.$$ (411)

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

$$c_n[ n (n-1) - l (l+1) ] = c_{n-1} \left [2 (n-1) - \frac{2 \mu e^2 a}{4\pi \epsilon_0 \hbar^2}\right].$$ (412)

Now, the power law series (410) must terminate at small $n$, at some positive value of $n$, otherwise $f(y)$ behaves 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 behaviour of the wave-function 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 behaviour of the wave-function as $r\rightarrow 0$ if $l$ is an integer.

For large values of $y$, the ratio of successive terms in the series (410) is

$$\frac{c_n y}{c_{n-1}} = \frac{2 y}{n},$$ (413)

according to Eq. (412). This is the same as the ratio of successive terms in the series

$$\sum_n \frac{(2 y)^n}{n!},$$ (414)

which converges to $\exp(2 y)$. We conclude that $f(y)\rightarrow \exp(2 y)$ as $y\rightarrow \infty$. It follows from Eq. (408) that $R(r) \rightarrow \exp(r/a) /r$ as $r\rightarrow \infty$. This does not correspond to physically acceptable behaviour of the wave-function, since $\int \vert\psi\vert^2 dV$ must be finite. The only way in which we can avoid this unphysical behaviour is if the series (410) terminates at some maximum value of $n$. According to the recursion relation (412), this is only possible if

$$\frac{\mu e^2 a}{4\pi \epsilon_0 \hbar^2} = n,$$ (415)

where the last term in the series is $c_n y^n$. It follows from Eq. (407) that the energy eigenvalues are quantized, and can only take the values

$$E = - \frac{\mu e^4}{32\pi^2\epsilon_0^{ 2} \hbar^2 n^2}.$$ (416)

Here, $n$ is a positive integer which must exceed the quantum number $l$, otherwise there would be no terms in the series (410).

It is clear that the wave-function for a hydrogen atom can be written

$$\psi(r, \theta, \varphi) = R(r/a) Y_l^m (\theta, \varphi),$$ (417)

where

$$a =\frac{n 4\pi \epsilon_0 \hbar^2}{\mu e^2} = 5.3\times 10^{-11} n {\rm meters},$$ (418)

and $R(x)$ is a well-behaved solution of the differential equation

$$\left(\frac{1}{x^2} \frac{d}{dx} x^2 \frac{d}{dx}-\frac{l (l+1)}{x^2} + \frac{2 n}{x} - 1\right) R = 0.$$ (419)

Finally, the $Y_l^m$ are spherical harmonics. The restrictions on the quantum numbers are $\vert m\vert \leq l< n$. Here, $n$ is a positive integer, $l$ is a non-negative integer, and $m$ is 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 energy of the ground state is

$$E_0 = - \frac{\mu e^4}{32\pi^2 \epsilon_0^{ 2} \hbar^2} = -13.6 {\rm electron volts}.$$ (420)

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 Eq. (416) are independent of the quantum number $l$, despite the fact that $l$ appears in the radial eigenfunction equation (419). This is a special property of a $1/r$ Coulomb potential.

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