previous home next PDF
The Hydrogen Atom
Michael Fowler 1/12/08
Factoring Out the Center of Mass Motion
The hydrogen atom consists of two particles, the proton and
the electron, interacting via the Coulomb potential 
, where as usual 
. Writing the masses of the two particles as 
Schrödinger’s equation
for the atom is:
But
are not the most
natural position variables for describing this system: since the potential
depends only on the relative position, a better choice is 
defined by:
so 
is the center of mass of the system. It is convenient at the
same time to denote the total mass by 
and the reduced mass by 
Transforming in straightforward fashion to the variables 
Schrödinger’s equation
becomes
Writing the wave function
we can split the equation into two:
and the total system energy is 
Note that the motion
of the center of mass is (of course) just that of a free particle, having a
trivial plane wave solution. From now
on, we shall only be concerned with the relative
motion of the particles. Since the
proton is far heavier than the electron, we will almost always ignore the
difference between the electron mass and the reduced mass, but it should be
noted that the difference is easily detectable spectroscopically: for example,
the lines shift if the proton is replaced by a deuteron (heavy hydrogen).
We’re ready to write Schrödinger’s equation for the hydrogen
atom, dropping the r suffixes in the
second equation above, and writing out 
explicitly in
spherical coordinates:
Factoring Out the Angular Dependence: the Radial Equation
Since the potential is spherically symmetric, the Hamiltonian H commutes with the angular momentum
operators 
so we can construct a
common set of eigenkets of the three operators 
The angular
dependence of these eigenkets is therefore that of the 
’s, so the solutions must be of the form
Now, notice that in the Schrödinger equation above, the
angular part of is exactly the differential operator 
, so operating on 
it will give 
. The spherical
harmonic can then be cancelled from the two sides of the equation
leaving:
it now being apparent that R(r) cannot depend on m.
The radial derivatives simplify if one factors out 1/r from the function R, writing
and temporarily suppressing the E and l to reduce clutter.
The equation becomes:
Rearranging,
Note that this is the same as the Schrödinger equation for a
particle in one dimension, restricted to 
, in a potential (for 
) going to positive infinity at the origin, then negative and
going to zero at large distances, so it always has a minimum for some positive r.
We are interested in bound states of the proton-electron system, so E will be a negative quantity. At large separations, the wave equation simplifies to
having approximate solutions 
The bound states we
are looking for, of course, have exponentially decreasing wave functions at large distances.
Going to a Dimensionless Variable
To further simplify the equation, we introduce the dimensionless variable
giving
where (for reasons which will become apparent shortly) we
have introduced 
defined by
Notice that in transforming from r to the dimensionless variable 
the scaling factor 
depends on energy,
so will be different for different energy bound states!
Consider now the behavior of the wave function near the
origin. The dominant term for
sufficiently small is the centrifugal
one, so
for which the solutions are 
Since the wave
function cannot be singular, we choose the second.
We have established that the wave function decays as 
at large distances, and goes as 
close to the
origin. Factoring out these two
asymptotic behaviors, define 
by
It is straightforward (if tedious) to establish that satisfies the
differential equation:
Putting in a trial series solution 
gives a recurrence
relation between successive coefficients:
For large values of k,
, so 
and therefore 
. This means we have
found the diverging radial wavefunction 
, which is in fact the correct behavior for general values of
the energy.
To find the bound states, we must choose energies such that
the series is not an infinite
one. As long as the series stops
somewhere, the exponential decrease will eventually take over, and yield a
finite (bound state) wave function. Just
as for the simple harmonic oscillator, this can only happen if for some k, 
Inspecting the ratio 
, evidently the condition
for a bound state is that
in which case the
series forterminates at 
From now on, since we know that for the functions we’re interested
in is an integer, we replace 
by n.
To find the energies of these bound states, recall 
and 
, so
(This defines the Rydberg, a popular unit of energy in atomic physics.)
Remarkably, this is the very same series of bound state
energies found by Bohr from his model!
Of course, this had better be the case, since the series of energies
Bohr found correctly accounted for the spectral lines emitted by hot hydrogen
atoms. Notice, though, that there are
some important differences with the Bohr model: the energy here is determined
entirely by n, called the principal quantum number, but, in
contrast to Bohr’s model, n is not the angular momentum. The true
ground state of the hydrogen atom, n
= 1, has zero angular momentum: since
, n = 1 means both l = 0
and k = 0. The ground state wave function is therefore
spherically symmetric, and the function 
is just a constant.
Hence 
and the actual radial
wave function is this divided by r,
and of course suitably normalized.
To write the wave function in terms of r, we need to find . Putting together 
, 
and 
,
where the length
is called the Bohr radius: it is in fact the radius of the lowest orbit in Bohr’s model.
Exercise: check this last statement.
It is worth noting at this point that the energy levels can be written in terms of the Bohr radius a0:
(This is actually obvious: remember that the energies En are identical to those in the Bohr model, in which the radius of the nth orbit is n2a0, so the electrostatic potential energy is -e2/ n2a0, etc.)
Moving on to the excited states: for n = 2, we have a choice: either
the radial function can have one term, as before, but now the angular momentum l = 1 (since n = k + l + 1);
or can have two terms (so k = 1), and l = 0. Both options give the same energy, -0.25 Ry, since n is the same, and the energy only depends on n. In fact, there are four states at this energy, since 
has states with m = 1, m = 0 and m = -1,
and l = 0 has the one state m = 0.
(For the moment, we are not counting the extra factor of 2 from the two
possible spin orientations of the
electron.)
For n = 3, there are 9 states altogether: l = 0 gives one, l = 1 gives 3 and l = 2 gives 5 different m values. In fact, for principal quantum number n there are n2 degenerate states. (n2 being the sum of the first n odd integers.)
The states can be mapped out, energy vertically, angular momentum horizontally:
The energy 
the levels are
labeled nl, n being
the principal quantum number and the traditional notation for angular momentum l is given at the bottom of the diagram.
The two red diagonally downward arrows are the first two transitions in the
spectroscopic Balmer series, four lines of which gave Bohr the clue that led to
his model. (A photon has spin one, so
when the atom emits a single photon its own angular momentum must change by
one.) The corresponding series of transitions
to the 1s ground state are in the
ultraviolet, they are called the Lyman series.
Wave Functions for some Low-n States
From now on, we label the wave functions with the quantum
numbers, 
, so the ground state is the spherically symmetric 
.
For this state 
, where 
, with w0
a constant, and 
.
So, as a function of r,
with N an easily evaluated normalization
constant:
For n = 2, l = 1 the function is still a single
term, a constant, but now 
, and, for n
= 2, 
, remembering the
energy-dependence of k.
Therefore 
. Again, evaluating
the normalization constant is routine, yielding
.
The wave functions for the other m-values, 
have the cosq in 
replaced by 
respectively (from the
earlier discussion of the ’s).
The other n = 2 state has l = 0, so from n = k + l + 1, we have k = 1 and the series for w has two terms, k = 0 and k = 1, the ratio being
for the relevant values: k
= 0, l = 0, n = 2. So 
. For n = 2, 
, the normalized wave function is
Note that the zero angular momentum wave functions are nonzero and have nonzero slope at the origin. This means that the full three dimensional wave functions have a slope discontinuity there! But this is fine—the potential is infinite at the origin. (Actually, the proton is not a point charge, so really the kink will be smoothed out over a volume of the size of the proton—a very tiny effect.)
General Solution of the Radial Equation
In practice, the first few radial functions can be constructed
fairly easily using the method presented above, but it should be noted that the
differential equation for
is in fact
where k, p are integers, and 
is a Laguerre polynomial
(Messiah, page 482).
The two equations are the same if z = 2r, and the solution to the radial equation is therefore
Quoting Messiah, the Laguerre polynomials 
, and the associated
Laguerre polynomials are given by:
(These representations can be found neatly by solving
Laplace’s equation using – surprise – a
But what do they look like?
The function 
is zero at the origin
(apart from the trivial case 
) and zero at infinity, always positive and having nonzero
slope except at its maximum value, 
. The p derivatives bring in p separated zeroes, easily checked by
sketching the curves generated by successive differentiation. Therefore, , a polynomial of degree p,
has p real positive zeroes, and value
at the origin
, since the only nonzero term at z = 0 is that generated by all
p differential operators
acting on 
.
The associated Laguerre polynomial is generated by differentiating 
k times. Now has p + k real positive zeroes, differentiating it
gives a polynomial one degree lower, with zeroes which must be one in each
interval between the zeroes of . This argument
remains valid for successive differentiations, so must have p real separate zeroes.
Putting all this together, and translating back from to r, the radial solutions are:
with N the
normalization constant. 
, since the normalization is 
, 
gives a better idea of
at what distance from the proton the electron is most likely to be found.
Here are the three n = 3 radial wave functions:
The number of nodes, the radial quantum number, is 3 - l - 1. (Note: The relative normalizations are correct here, but not the overall normalization.)
For higher n values,
the wave functions become reminiscent of classical mechanics. for example, for n = 10, the highest angular momentum state probability distribution
peaks at 
, the Bohr orbit radius:
whereas for n = 10, l = 0, we find:
Notice this peaks just below twice the Bohr radius. This can be understood from classical mechanics: for an inverse square force law, elliptical orbits with the same semimajor axis have the same energy. The l = n - 1 orbit is a circle, the l = 0 orbit is a long thin ellipse, so it extends almost twice as far from the origin as the circle. Furthermore, as the size of the peaks indicates, the orbiting electron spends more time at the far distance, since it will be moving very slowly. The “local wavelength” gives an idea of how the kinetic energy varies with distance from the proton.
(Note: the normalizations in the above graphs are only approximate.)