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Chapter 9

The Three-Dimensional

Schrödinger Equation
9.1 Introduction
9.2 The Three-Dimensional Schrödinger Equation and Partial Derivatives
9.3 The Two-Dimensional Square Box
9.4 The Two-Dimensional Central-Force Problem
9.5 The Three-Dimensional Central-Force Problem
9.6 Quantization of Angular Momentum
9.7 The Energy Levels of the Hydrogen Atom
9.8 Hydrogenic Wave Functions
9.9 Shells
9.10 Hydrogen-Like Ions
Problems for Chapter 9

9.1 Introduction
In Chapter 8 we studied the one-dimensional Schrödinger equation and saw how it
determines the allowed energies and corresponding wave functions of a particle in one
dimension. If the world in which we lived were one-dimensional, we could now proceed
to apply these ideas to various real systems: atoms, molecules, nuclei, and so on.
However, our world is three-dimensional, and we must first describe how the one-
dimensional equation is generalized to three dimensions.
We shall find that the three-dimensional equation is appreciably more complicated
than its one-dimensional counterpart, involving derivatives with respect to all three
coordinates x, y, and z. Nevertheless, its most important properties will be familiar.
Specifically, in three dimensions, just as in one dimension (and two), the time-
independent Schrödinger equation is a differential equation for the wave function  . For
most systems this equation has acceptable solutions only for certain particular values of
the energy E. Those E for which it has an acceptable solution are the allowed energies of
the system, and the solutions  are the corresponding wave functions.
In this chapter we write down the three-dimensional Schrödinger equation and
describe its solutions for some simple systems, culminating with the hydrogen atom.
Since many of the important features of the three-dimensional equation are already
present in the simpler case of two dimensions, two of our examples will be two-
dimensional.

9.2 The Three-Dimensional Schrödinger Equation and Partial Derivatives
In one dimension the wave function  ( x) for a particle depends on the one coordinate x,
and the time-independent Schrödinger equation has the now familiar form

where U is the particle's potential energy and we temporarily use capital M for the
particle's mass.* In this equation, remember that both  and U are functions of x, whereas
E, although it can take on various values, does not depend on x. In three dimensions we
would naturally expect the wave function  to depend on all three coordinates x, y, and
z; that is,

where

Similarly the potential energy U will normally depend on x, y, and z:

How the differential equation (9.1) generalizes to three dimensions is not so obvious.
Here we shall simply state that the correct generalization of (9.1) is this:

where the three derivatives on the left are the so-called partial derivatives, whose
definition and properties we discuss in a moment. That (9.2) is a possible generalization
of (9.1) is perhaps fairly obvious. That it is the correct generalization is certainly not
obvious. In more advanced texts you will find various arguments that suggest the
correctness of (9.2). However, the ultimate test of any equation is whether its predictions
agree with experiment, and we shall see that the three-dimensional Schrödinger equation
in the form (9.2) has passed this test repeatedly when applied to atomic and subatomic
systems.
The three derivatives that appear in (9.2) are called partial derivatives, and it is
important that you understand what these are. An ordinary derivative, like d / dx , is
defined for a function such as  ( x) that depends on just one variable (for example, the
temperature as a function of position x along a narrow rod). Partial derivatives arise when
one considers functions of two or more variables, such as the temperature as a function of
position (x, y, z) in a three-dimensional room. If  depends on three variables x, y, z, we

*
In the context of the three-dimensional Schrödinger equation, the letter m is traditionally used for
the integer that labels the allowed values of the components of angular momentum. It is to avoid confusion
with this notation that we use M for the mass in the first six sections of this chapter. In Section 9.7 we return
to the hydrogen atom, in which the relevant particle is the electron, whose mass we shall call me.
define the partial derivative  / x as the derivative of  with respect to x, obtained
when we hold y and z fixed. Similarly,  / y is the derivative with respect to y, when x
and z are held constant; and similarly with d / z . Notice that it is customary to use the
symbol  for this new kind of derivative.
The calculation of partial derivatives is very simple in practice, as the following
example shows.
EXAMPLE 9.1 Find the three partial derivatives  / x ,  / y , and d / z for
 ( x, y, z ) = x 2  2 y 3 z  z .

If y and z are held constant, then  has the form

and the rules of ordinary differentiation give

If instead we hold x and z constant, then  has the form

Since the coefficient 2z is constant, the rules of ordinary differentiation give

Finally, if x and y are constant, then

and since 2y3 is constant,

Higher partial derivatives are defined similarly. For example,  2 / x 2 is the
second derivative of  with respect to x, obtained if we hold y and z constant, and so on.
Partial derivatives are really no harder to use than ordinary derivatives, once you
understand their definition. If you have never worked with them before, you might find it
useful to try some of Problems 9.1 to 9.5 right away.
An equation, like the Schrödinger equation (9.2), that involves partial derivatives
is called a partial differential equation. Now that we know what partial derivatives are,
we can discuss methods for solving such equations, and this is what we shall do in the
remainder of this chapter.

9.3 The Two-Dimensional Square Box
Before we consider any three-dimensional systems we consider an example in two
dimensions. This shares several important features of three-dimensional systems, but is
naturally somewhat simpler.
Looking at the Schrödinger equations (9.1) and (9.2) in one and three dimensions,
it is easy to guess, correctly, that the Schrödinger equation for a particle of mass M in two
Here  is a function of the two-dimensional coordinates,  =  (r) =  ( x, y) , and U =
U(r)is the particle's potential energy.
The method for solving the Schrödinger equation depends on the potential-energy
function U(r). In this section we consider a particle confined in a two-dimensional, rigid
square box; that is, a particle for which U(r) is zero inside a square region like that shown
in Fig. 9. 1, but is infinite outside:

A classical example of such a system would be a metal puck sliding on a frictionless,
square air table with perfectly rigid, elastic bumpers at its edges. As a quantum example
we could imagine an electron confined inside a thin square metal sheet.
A classical particle inside a square, rigid box would bounce indefinitely inside the
box. Since U = 0 inside the box, its energy E would be all kinetic and could have any
value in the range

To find the possible energies for the corresponding quantum system we must solve the
Schrödinger equation (9.3) with the potential-energy function (9.4).
Since the particle cannot escape from the box, the wave function  ( x, y) is zero
outside the box, and since  ( x, y) must be continuous, it must also be zero on the
boundary:

Since U(x, y) = 0 inside the box, the Schrödinger equation reduces to

for all x and y inside the box. We must solve this equation, subject to the boundary
conditions (9.5).

SEPARATION Of VARIABLES

If we knew nothing at all about partial differential equations, the solution of (9.6)
would be a formidable prospect. Fortunately, there is an extensive mathematical theory of
partial differential equations, which tells us that equations like (9.6) can be solved by a
method called separation of variables. In this method one seeks solutions with the form
where X(x) is a function of x alone and Y(y) a function of y alone. We describe a function
with the form (9.7) as a separated function.* It is certainly not obvious that there will be
any solutions with this separated form. On the other hand, there is nothing to stop us from
seeing if there are such solutions, and we shall find that indeed there are. Furthermore, the
mathematical theory of equations like (9.6) guarantees that any solution of the equation
can be expressed as a sum of separated solutions. This means that once we have found all
of the solutions with the form (9.7), we have, in effect, found all solutions. It is this
mathematical theorem that is the ultimate justification for using separation of variables to
solve the Schrödinger equation for many two- and three-dimensional systems.
To see whether the Schrödinger equation (9.6) does have separated solutions with
the form (9.7), we substitute (9.7) into (9.6). When we do this two simplifications occur.
First, the partial derivatives simplify. Consider, for example,

This derivative is evaluated by treating y as fixed. Therefore, the term Y(y) can be brought
outside, to give

Since X(x) depends only on x, the remaining derivative is an ordinary derivative, which
we can write as

where, as usual, the double prime indicates the second derivative of the function
concerned. Thus

Similarly,

If we substitute (9.8) and (9.9) into the Schrödinger equation (9.6) we find

To separate the terms that depend on x from those that depend on y, we divide by
X(x)Y(y) to give

The right side of this equation is constant (independent of x and y). Thus (9.10) has the
general form

for all x and y (in the box). To see what this implies, we move the function of y over to
the right:

*
Note that by no means can every function of x and y be separated in this way. As simple a
function as x + y cannot be expressed as the product of one function of x and one of y.
This equation asserts that a certain function of x is equal to a quantity that does not
depend on x at all. In other words, this function, which can only depend on x, is in fact
independent of x. This is possible only if the function in question is a constant. We
conclude that the quantity X ( x) / X ( x) in (9.10) is a constant:

If we call this constant  k x2 then (9.11) can be rewritten as

This equation has exactly the form of the Schrödinger equation (8.51) for a particle in a
one-dimensional rigid box,

whose solutions we have already discussed in Chapter 8. In particular, we saw there that
this equation has acceptable solutions only when the constant on the right is negative,
which is why we called the constant in (9.11)  k x2 .
An exactly parallel argument shows that the quantity Y ( x) / Y ( x) in (9.10) has to
be independent of y, that is,

or, if we call this second constant k y2

We see that the method of separation of variables has let us replace the partial differential
equation (9.6), involving the two variables x and y, by two ordinary differential equations
(9.12) and (9.15), one of which involves only the variable x, and the other only y.
Before we seek the acceptable solutions of these two equations, we must return to
the boundary condition that  ( x, y) is zero at the edges of our box (x = 0 or a, and y = 0
or a). Since  ( x, y) = X(x)Y(y) this requires that

and

Now, the differential equation (9.12) and boundary conditions (9.16) for X(x) are exactly
the equation and boundary conditions for a particle in a one-dimensional rigid box. And
we already know the solutions for that problem: The wave function must have the form

where B is a constant. This satisfies the boundary conditions only if kx is an integral
multiple of  / a ,
where nx is any positive integer

Therefore,

The equation and boundary conditions for Y(y) are also the same as those for a one-
dimensional rigid box, and there are acceptable solutions only if

(with ny any positive integer), in which case

Combining (9.19) and (9.21), we find for the complete wave function

where nx and ny are any two positive integers.* In writing (9.23) we have renamed the
constant BC as A; the value of this constant is fixed by the requirement that the integral of
 over the whole box must be 1.
2

Using the form (9.22) we can see the physical significance of the separation
constants kx and ky. If we fix y and move in the x direction, then  varies sinusoidally in
x, with wavelength  = 2 / k x . According to de Broglie, this means that the particle has
momentum in the x direction of magnitude h /  = k x . Since a similar argument can be
applied in the y direction, we conclude that

By analogy with the one-dimensional wave number k, satisfying p = k , we can call kx
and ky by the components of a wave vector. Note, however, that since

the wave function (9.22) is a superposition of states with px   k x , and similarly with
p y   k y . This is the same situation that we encountered in Section 8.4 for the one-
dimensional rigid box and explains the absolute value signs in Eq. (9.24).

ALLOWED ENERGIES

In solving for the wave function (9.22) we have temporarily lost sight of the
energy E. In fact, the last place that E appeared was in (9.10):

*
By labeling these two integers nx and ny we do not wish to imply that there is necessarily a vector
n of which nx and ny are the components, For the moment, nx and ny are simply two integers, one of which
characterizes the function X(x) and the other Y(y).
Now, we know from (9.12) that X  / X is the constant  k x2 , and from (9.18) that kx =
nx  / a . Therefore, X  / X is equal to  nx2  2 / a 2 . Inserting this, and the X7r
corresponding expression for Y  / Y , into (9.25), we obtain

Solving for E, we find that the allowed values of the energy are

where nx and ny are any two positive integers. This energy is the sum of two terms, each
of which has exactly the form of an allowed energy for the one-dimensional box; namely,

where you can think of the subscript 0 as standing for one-dimensional, we can rewrite
the allowed energies for a particle in a two-dimensional square box as

QUANTUM NUMBERS

Just like the one-dimensional box, the two-dimensional box has energy levels that
are quantized. The main difference is that where the one-dimensional energy levels are
characterized by a single integer n, the two-dimensional levels are given by two integers,
nx, and ny. We are going to find many more examples of quantities whose allowed values
are characterized by integers (and sometimes half integers, such as 1 , 1 1 , . . . ). In
2  2

general, any integer or half integer that gives the allowed values of some physical quantity
is called a quantum number. With this terminology we can say that the energy levels of
a particle in a two-dimensional square box are characterized by two quantum numbers, nx
and ny.
The lowest possible energy for the two-dimensional box occurs when both
quantum numbers are equal to 1,

and the corresponding, ground-state, energy is given by (9.29) as

The first excited energy occurs when nx = 1, ny = 2, or vice versa:

In Fig. 9.2 are sketched the lowest four energy levels for the square box.
DEGENERACY

An important new feature of the two-dimensional box is that there can be several
different wave functions for which the particle has the same energy. For example, we saw
that E12 = E21 = 5E0. That is, the state with nx = 1, ny = 2 has the same energy as that with
nx = 2 and ny = 1. The corresponding wave functions are

Since these correspond to different probability densities  , they represent
2

experimentally distinguishable states that happen to have the same energy.
In general, if there are N independent wave functions (N > 1) all with the same
energy E, we say that the energy level E is degenerate and has degeneracy N (or is N-
fold degenerate). If there is only one wave function with energy E, we say that the energy
E is nondegenerate (or has degeneracy 1). Looking at Fig. 9.2, we see that the ground
state and second excited state of the square box are nondegenerate, while the first and
third excited states are both twofold degenerate. In general, most of the levels E11, E22,
E33, . . . are nondegenerate, while most of the levels Enx ,ny with nx  n y are twofold
degenerate, since Enx ,ny = Eny ,nx . A few of the levels have higher degeneracies; for
example, since

it follows that E55 = E17 = E71 and this level is threefold degenerate; since

it follows that E18 = E81 = E47 = E74 and this level is fourfold degenerate.
In Chapter 11 we shall see that degeneracy has an important effect on the structure
and chemical properties of atoms. Therefore, it is important not just to find the energy
levels of a quantum system, but to find the degeneracy of each level.


2
CONTOUR MAPS OF

It is often important to know how the probability density  ( x, y) is distributed
2

in space. Because  ( x, y) depends on two variables, it is harder to visualize than in the
2

one-dimensional case. One method that is quite successful is to draw a contour map of
 ( x, y) . Figure 9.3 shows such a contour map for the ground-state density
2

The density  is maximum at the center of the box x = y = a/2. The contours shown are
2

for 
2
equal to 95%, 50%, and 5% of its maximum value. Notice how the contour lines
become more square near the edges of the box. The contour line 
2
= 0 is, of course, the
square boundary of the box itself.
Figure 9.4 shows the same three contour lines for each of three excited states.
Notice how the higher energies correspond to more rapid oscillations of the wave
functions and hence to larger numbers of hills and valleys on the map.
EXAMPLE 9.2 Having solved the Schrödinger equation for a particle in the two-
dimensional square box, one can solve the corresponding three-dimensional problem very
easily (see Problem 9.13). The result is that the allowed energies for a mass M in a rigid
cubical box of side a have the form

where E0  2 2 /(2 M a 2 ) is the same energy introduced in (9.28), and the quantum
numbers nx, ny, nz, are any three positives integers. Use this result to find the lowest five
energy levels and their degeneracies for a mass M in a rigid, cubical box of side a.

Equation (9.32) shows that the energies of a particle in a three-dimensional box
are characterized by three quantum numbers nx, ny, nz. The lowest energy occurs for nx =
ny = nz = 1 and is

The next level corresponds to the three quantum numbers being 2, 1, 1 or 1, 2, 1 or 1, 1,
2:

This level is evidently threefold degenerate. The higher levels are easily calculated, and
the first five levels are found to be as shown in Fig. 9.5.

9.4 The Two-Dimensional Central-Force Problem
Many physical systems involve a particle that moves under the influence of a central
force; that is, a force that always points exactly toward, or away from, a force center O.
The most obvious example is the hydrogen atom, in which the electron is held to the
proton by the central Coulomb force. Other examples where the force is at least
approximately central include the motion of any one electron in a multielectron atom, and
the motion of either atom as it orbits around the other atom in a diatomic molecule.
If the force on a particle is central, it does no work when the particle moves in any
direction perpendicular to the radius vector, as shown in Fig. 9.6. This means that the
particle's potential energy U is constant in any such displacement. Thus U may depend on
the particle's distance, r, from the force center O, but not on its direction, Therefore,
instead of writing the potential energy as U(x, y, z), we can write simply U(r) when the
force is central. This property of central forces will allow us to solve the Schrödinger
equation using separation of variables.
As an introduction to the three-dimensional central-force problem we consider
first a two-dimensional particle moving in a central-force field. We shall not present a
complete solution of the Schrödinger equation for this system, since it is fairly
complicated and we are not really interested in two-dimensional systems anyway.
Nevertheless, we shall carry it far enough to see two important facts: First, like the
energies for a square box, the allowed energies of a two-dimensional particle in a central-
force field are given by two quantum numbers. Second, we shall find that one of the two
quantum numbers is closely connected with the angular momentum of the particle.
Since the potential energy U depends only on r, the distance of the particle from
the force center O, it is natural to adopt r as one of our coordinates. In two dimensions the
simplest way to do this is to use polar coordinates (r ,  ) as defined in Fig. 9.7.* It is a
simple trigonometric exercise to express x and y in terms of r and  as in Fig. 9.7, or
vice versa (Problem 9.14). Note that r is defined as the distance from O to the point of
interest and is therefore always positive: 0  r   . If we increase the angle  by 2
(one complete revolution), we come back to our starting direction. Therefore, you must
remember that   0 . and   0  2 represent exactly the same direction.
The wave function  , which depends on x and y, can just as well be expressed as
a function of r and  :

and the Schrödinger equation can similarly be rewritten in terms of r and  . When one
rewrites the partial derivatives of the Schrödinger equation in terms of r and  , one finds
that

If you have had some experience with handling partial derivatives, you should be able to
verify this rather messy identity (Problem 9.17). Otherwise, it is probably simplest for
now to accept it without proof. In any case, it is helpful to note that all three terms are
dimensonally consistent. Given the identity (9.33), we can rewrite the Schrödinger
equation (9.6) in terms of r and  as

SEPARATION OF VARIABLES

The equation (9.34) can be solved by separation of variables, very much as
described in Section 9.3, except that we now work with the coordinates r and  instead
of x and y. We first seek a solution with the separated form

Substituting (9.35) into (9.34), we find that

*
In two dimensions the angle that we are calling      is more often called    . However, in three
dimensions it is usually called      and our primary interest will be in three dimensions.
where, as before, primes denote differentiation with respect to the argument ( R =
dR / dr ,  = d  / d ). If we now multiply both sides by r 2 /( R ) and regroup terms,
this gives

for all r and  . The equation (9.36) has the form

and this will allow us to separate variables. Notice that the separation in (9.37) occurs
because the potential energy in (9.36) depends only on r (which follows because the force
is central). This explains why we went to the trouble of rewriting the equation in terms of
r and  . It would not have separated if we had used x and y, since U(r) depends on both x
and y (because r =        x 2  y 2 ).
The right side of (9.37) is a function of r but is independent of  . Since the two
sides are equal for all r and  , it follows that the left side is also independent of  , and
hence a constant. By a similar argument the right side is likewise constant, and by (9.37)
these two constants are equal. It is traditional (and, as we shall see, convenient) to call
this constant* - m2. Since each side of (9.36) is equal to the constant - m2, we get two
equations:

and

Once again, separation of variables has reduced a single partial differential equation in
two variables to two separate equations each involving just one variable. The two
equations (9.38) and (9.39) are called the  equation and the radial equation. We discuss
the  equation (9.38) first.
We already know that the general solution of (9.38) is an arbitrary combination of
cos m and sin m , or, equivalently, of eim and eim . We can economize a little in
notation if we use the second pair and if we agree to let m be positive or negative. For
example, both ei and e  i can be written as eim if we let m be ± 1. Thus we can say that
all solutions of (9.38) can be built up from the functions

with m positive or negative.
The  equation (9.38) proved easy to solve, but we must now ask whether there
are any boundary conditions to be met. In fact, there are. For any given r, the points
labeled by  and   2 are the same. Therefore, the wave function  (r ,  ) must satisfy

*
It is to avoid confusion with this m that we are using M for the mass of the particle.
For our separated solutions this means that

that is,  ( ) must be periodic and must repeat itself each time  increases by 2 .
Now it is known from trigonometry that the function cos m is periodic and
repeats itself every 2 if m is an integer, but not otherwise (Fig. 9.8). The same is true of
sin m and hence also of eim = cos m  i sin m . Thus the solution (9.40) of the 
equation is acceptable if and only if m is an integer:

Incidentally, we can now see why it was convenient to use the notation m 2 for the
separation constant in (9.3 8). If we had called it K, for instance, then (9.41) would have
read  K = 0, ± 1, . . .

QUANTIZATION OF ANGULAR MOMENTUM

Our conclusion so far is that there are solutions of the Schrödinger equation (9.34)
with the form

provided the quantum number m is an integer. That m must be an integer indicates that
something related to the  dependence of  (r ,  ) is quantized. To decide what this is, let
us fix r so that we can study just the  dependence. If we let  increase, we move around
a circle, as shown in Fig. 9.9, the distance that we travel being s = r  . According to
(9.42), the wave function varies sinusoidally with the distance s:

Now, we saw in Chapter 8 that a wave function

where x is the distance along a line, represents a particle with momentum k along that
line.* Comparison of (9.43) and (9.44) suggests that (9.43) represents a particle with
momentum

in the direction tangential to the circle. If we multiply ptang by r, this gives the angular
momentum

*
Recall that the full time-dependent wave function corresponding to (9.44) is  ( x, t ) =
exp[i(kx  t )] . This wave travels to the right and is sinusoidal in x with wavelength 2 / k , that is, it
has momentum p = h /  = k to the right.
that is, the wave functions (9.42) define states in which the particle has a definite angular
momentum

That m has to be an integer shows that L is quantized in multiples of , as originally
proposed by Bohr. That is, we have justified Bohr's quantization of angular momentum.
You should bear in mind that we have so far discussed the central-force problem
only in two dimensions. For a particle confined to the x-y plane, the angular momentum L
is the same thing as its z component Lz. Thus it is not clear whether the result (9.45) will
apply to L or Lz. when we go on to discuss three-dimensional motion. In fact, we shall see
that in three dimensions it is Lz, that is restricted to integer multiples of :

Of course, there is nothing special about the z axis (in three dimensions), and the general
statement of (9.46) is that any component of the vector L is restricted to integer multiples
of .

THE ENERGY LEVELS

Let us turn now to the radial equation (9.39),

Since this equation contains the energy E, its solution will determine the allowed values
of E. The details of the solution depend on the particular potential-energy function U(r),
which we have not specified. However, we can understand several general features: The
radial equation is an ordinary differential equation, which involves the energy E as a
parameter. Just as with the one-dimensional Schrödinger equation, one can show that
there are acceptable solutions only for certain particular values of E, and these are the
allowed energies of our particle. Notice that the equation (9.47) that determines the
allowed energies depends on the quantum number m. Thus for each value of m we have a
different equation to solve, and will usually get different allowed energies; that is, the
allowed energies of our particle will depend on its angular momentum. This is what one
would expect classically: The more angular momentum the particle has, the more kinetic
energy it will have in its orbital motion; thus, in general, we expect different energies for
different angular momenta.
For each value of m we could imagine finding all the allowed energies. We could
then label them in increasing order by an integer n = 1, 2, 3,. . . , so that the nth level with
angular momentum m would have energy En,m, as shown in Fig. 9. 10. Just as with the
square box, each level is then identified by two quantum numbers. In this case, one of the
quantum numbers, m, identifies the angular momentum, while the other, n, identifies the
energy level for given m.
The radial equation (9.39) involves m only in the term m2/r2. Because this depends
2
on m , we get the same equation, and hence the same energies, whether m is positive or
negative:

In other words, except when m =  there are two states with the same energy, and the
level En,m is twofold degenerate. This is a property that we would also have found in a
classical analysis: Two states that differ only by having Lz =  m are different only
because the particle is orbiting in opposite directions, and we would expect two such
states to have the same energy.
We have not actually found the allowed energies E, of our two-dimensional
particle. These depend on the particular potential-energy function U(r)under
consideration. Whatever the form of U(r), the detailed solution of the radial equation
(9.39) is fairly complicated. Since our real concern is with three-dimensional systems, we
shall not pursue the two-dimensional problem any further here.

9.5 The Three-Dimensional Central-Force Problem
The three-dimensional central-force problem is very similar to the two-dimensional one,
but the additional complexity due to the extra dimension means that we shall have to state
several results without proof
Since the potential energy depends only on r (the particle's distance from the force
center O) we first choose a coordinate system that includes r as one of the coordinates.
For this we use spherical polar coordinates, which are defined in Fig. 9.11. Any point P
is identified by the three coordinates (r , ,  ) where r is the distance from O to P,  is
the angle between the z axis and OP, and  is the angle between the xz plane and the
vertical plane containing OP, as shown. If we imagine P to be a point on the earth's
surface and put the origin O at the earth's center, with the z axis pointing to the north pole,
then  is the colatitude of P (the latitude measured down from the north pole) and  is
its longitude measured in an easterly direction from the xz plane. The angle  lies
between 0, at the north pole, and 2 at the south pole. If  increases from 0 to 2 (with
r and  fixed), then P circles the earth at fixed latitude and returns to its starting point.
The rectangular coordinates (x, y, z) are given in terms of (r , ,  ) by

You should verify these expressions for yourself and derive the corresponding
expressions for (r , ,  ) in terms of (x, y, z) (Problem 9.18).
To write down the Schrödinger equation in terms of spherical coordinates we
must write the derivatives with respect to x, y, z in terms of r,  , and  . When this is
done, one finds that
The proof of this identity is analogous to that of the two-dimensional identity (9.33) (see
Problem 9.17); however, it is appreciably more complicated and is certainly not worth
giving here. We shall simply accept the identity (9.48) and use it to write down the three-
dimensional Schrödinger equation in spherical coordinates:

SEPARATION OF VARIABLES

The three-dimensional Schrödinger equation for a central force, (9.49), can be
solved by separation of variables. We start by seeking a solution with the separated form

If we substitute (9.50) into (9.49), we can rearrange the resulting equation in the form

(For guidance in checking this and the next few steps, see Problem 9.20.) By the now
familiar argument, each side of this equation is equal to the same constant, which we call
m 2 . This gives us two equations:

and a second equation involving r and  . This second equation can next be rearranged in
the form

Once again each side must be equal to a constant, which we shall temporarily call –k.
This gives two final equations with the form (Problem 9.20)

and

We see that separation of variables has reduced the partial differential equation
(9.49) in r,  , and  to three ordinary differential equations, each involving just one
variable. Notice that the  equation (9.52) is exactly the same as Equation (9.38) for the
two-dimensional case. Notice also that neither the  equation (9.52) nor the  equation
(9.53) involves the potential-energy function U(r). This means that the solutions for the
angular functions ( ) and  ( ) will apply to any central-force problem. Finally, note
that both U(r)and E appear only in the radial equation (9.54). Therefore, it is the radial
equation that determines the allowed values of the energy, and these do depend on the
potential-energy function U(r), as one would expect.

9.6 Quantization of Angular Momentum
In this section we discuss the two angular equations (9.52) and (9.53) that resulted from
separating the three-dimensional Schrödinger equation. The first of these is exactly the 
equation that arose in the two-dimensional centralforce problem, and it has the same
solutions
Since  ( ) must be periodic with period 2 , it follows, as before, that m must be an
integer

The significance of m is essentially the same as in two dimensions: If we fix r and  and
let  vary, then we move around a circle about the z axis. The radius of this circle is 
= r sin , as shown in Fig. 9.12. In terms of the distance s traveled around this circle, the
angle  is

and we can temporarily rewrite (9.55) as

Comparing this with the familiar one-dimensional wave eikx With momentum           k , we see
that (9.56) represents a state with tangential momentum

If we multiply ptang by the radius  we obtain the z component of angular momentum, Lz
= ptang  . Thus, from (9.5 7) our wave function represents a particle with

as anticipated in Section 9.4.

THE      EQUATION

The second angular equation (9.53), which determines ( ) , is much harder and
we shall have to be satisfied with stating its solutions. The equation is one of the standard
equations of mathematical physics and is called Legendre's equation. It has solutions for
any value of the separation constant k. However, for most values of k, these solutions are
infinite at  = 0 or at  = 2 and are therefore physically unacceptable. It turns out that
the equation has one (and only one) acceptable solution for each k of the form

where l is a positive integer greater than or equal in magnitude to m,

If we denote these acceptable solutions by  lm ( ) , our solutions of the Schrödinger
equation have the form

We shall find the specific form of the function  lm ( ) for a few values of l and m later.
Figure 9.13 shows the function  lm ( ) for the case l = 2 and m = 0, as well as one of the
unacceptable solutions for the case l = 1.75, m = 0.
The physical significance of the quantum number m is, as we have seen, that a
particle with the wave function (9.60) has a definite value of Lz equal to m . In more
advanced texts it is shown that a particle with the wave function (9.60) also has a definite
value for the magnitude of L equal to

That is, the quantum number l identifies the magnitude of L, according to (9.61).
The quantum number m can be any integer (positive or negative) while, for given
m, l can be any integer greater than or equal to the magnitude of m. Turning this around
we can say that l can be any positive integer

while for given 1, m can be any integer less than or equal (in magnitude) to l. That is,

According to (9.61) and (9.62) the possible magnitudes of L are as follows:

quantu             0 1     2     3        4   .
m number, 1:                              ..
magnitu         0 r 4      V      r          .
de,L:       2h 6-h -12h 2_0h ..
When l is large we can approximate l (l  1) by 12 and write

Thus for large l the possible magnitudes of L are close to those of the Bohr model, but for
small l there is an appreciable difference. (See Problems 9.25 and 9.26.)

THE VECTOR MODEL

We have found wave functions for which the magnitude and z componentof L are
quantized. This is, of course, a purely quantum result. Nevertheless, it is sometimes
useful to try to visualize it classically. The magnitude of L is V_1Y+ 1) h, so we imagine
a vector of length L = fl-(l -+I) h. Since L., = m h, this vector must be oriented so that its z
component is m h, and since m can take any of the (21 + 1) different values (9.63) there
are (21 + 1) possible orientations, as shown in Fig. 9.14 for the case l = 2. We can
describe this state of affairs by saying that the spatial orientation of L is quantized, and in
the older literature the quantization of L, was sometimes called "space quantization."
We should emphasize that there is nothing special about the z axis. When we
defined our spherical coordinates we chose to use the z direction as the polar axis (0 = 0),
and when we separated variables the Schrödinger equation led us to wave functions with
a definite value for L_,. If we had chosen the x direction as polar axis, the same procedure
would have produced states with a definite value for L,,, and so on. For the moment it
makes no difference which component of L we choose to focus on, and we shall continue
to work with states that have definite L, However, in more advanced books it is shown
that the Heisenberg uncertainty principle extends to angular momentum and implies that
no two components of L can simultaneously have definite values (except in the special
case that all three components are zero). Therefore, states that have definite L, do not
have definite values of L,, and LY .*
Since our wave functions do not have definite values of LX and Ly. the vectors
shown in Fig. 9.14, with definite components in all directions, are a bit misleading. One
must somehow imagine that the components Lx and LY are random and no longer have
definite values. Since the magnitude and z component are fixed, this means that the vector
L is randomly distributed on a cone as shown in Fig. 9.15. This reflects the quantum
situation, where Lx and LY simply do not have definite values. It is sometimes helpful to
use this classical picture — often called the vector model — as an aid in thinking about
the quantum properties of angular momentum.
The wave functions that we have found for a particle in a central force field have
the form

and represent a particle with definite values for the magnitude and z component of L:

It is sometimes important to know the explicit form of these wave functions. Fortunately,
we shall usually be concerned only with states with small values of /-in chemistry, for
example, all of the electrons most involved in molecular bonding have l = 0 or l -and for
these the angular wave functions are quite simple, as the following example shows.
EXAMPLE 9.3 Writedown the Oequation (9.53) for the cases that I= 0 and that I= 1, m
= 0. Find the angular functions 01,, (O)ei-O explictly for these two cases.

The 0 equation (9.53) is

[Recall that the separation constant k got renamed I(I + I).] If l = 0, the only allowed value
for m is m = 0, and the 0 equation reduces to

By inspection, we see that one solution of this equation is

As stated in connection with (9.58), the 0 equation has only one acceptable solution for
each value of 1 and m. (This is illustrated in Problem 9.27.) Therefore, we need took no

*
The wave functions that have definite LX are different from those with definite L,. However, any
one of the former can be expressed as a sum of the latter, and vice versa. In this sense, it is enough to
consider just those with definite L,. We shall see one example of this in Section 9.8.
further for any other solutions. We see that with 1 = 0 the function 0(0) is independent of
0. Further, with m =.O, eil"'O = l is independent of 0. Thus with l = m = 0 the wave
function (9.64) is actually independent of 0 and 0, and depends only on r:

This means that the probability distribution IVI(r, 0,0)12 for a particle with zero angular
momentum is spherically symmetric. We shall find, for example, that the ground state of
the electron in the hydrogen atom has l = 0 and hence that the hydrogen atom is
spherically symmetric in its ground state.
If l = 1, then m can be m = 1, 0, or - 1. In this example we are asked to consider the case
m = 0, for which the 0 equation (9.65) reads

By inspection we see that the solution of this equation is

Therefore, with 1 = l and m = 0, the complete wave function given by (9.64) is

As is obviously the case whenever m = 0, this wave function is independent of the angle
0. On the other hand, it does depend on 0. Since  0C cos2O, a particle with l = I and
2

m 0 is most likely to be found near the polar axes 0 = 0 and 7r (where cos2 0 1), and has
zero probability of being found in the x-y plane (where 0 = 7r/2 and cos 0 = 0). We shall
find that this distribution of electrons in atoms has important implications for the shape of
many molecules.

9.7 The Energy Levels of the Hydrogen Atom
Of the three equations that resulted from separating the Schrödinger equation, we
have now discussed the two angular ones. It remains to consider the radial equation
(9.54), which we rewrite as

[Remember that the separation constant kin (9.54) got renamed 1(1 + 1).]This is the
equation that determines the allowed values of the energy E, which win, of course,
depend on the precise form of the potential-energy function U(r). Since the equation
involves the angular-momentum quantum number 1, the allowed values of E win
generally depend on l as well. Notice, however, that (9.69) does not involve the quantum
number m. Thus the allowed values of E will not depend on m; that is, for a given
magnitude of L equal to Nfl-(l+ 1) h, we shall find the same allowed energies for all (21 +
1) different orientations given by m = 1, 1 - 1, . . . , - 1. This is just what we would expect
classically: Since the force field is spherically symmetric, the energy of the particle
cannot depend on the orientation of its orbit. Quantum mechanically it means that in any
centralforce problem, a level with L = VI(I + 1) h will always be at least (21 + I)-fold
degenerate. As we shall see shortly, it can sometimes be more degenerate, since two
states with different l may happen to have the same energy.
The detailed solution of the differential equation (9.69) depends on the potential-
energy function U(r). As a first and very important example, we consider the electron
bound to a proton in a hydrogen atom, for which

If we substitute (9.70) into (9.69), we obtain the differential equation

where we have replaced m by m,, the mass of the electron. This equation has been studied
extensively by mathematical physicists. Here we must be content with simply stating the
facts about its solutions (but see Problems 9.33 and 9.34 for some simple special cases):
The equation (9.71) has acceptable solutions only if E has the form

where n is any integer greater than L That is, the energy is quantized, and its allowed
values are given by (9.72). You may recognize the first factor in (9.72) as the Rydberg
energy, originally defined in (6.22),

Thus, solution of the three-dimensional Schrödinger equation for a hydrogen atom has
brought us back to exactly the energy levels

predicted by the Bohr model. Since these levels are known to be correct, this is a most
satisfactory result.
The possible values of the quantum number l are the integers I 0, 1, 2, . . . , and for
each value of 1, we have stated, the radial equation has a solution only if n is an integer
greater than h

Turning these statements around, we can say that the possible values of n are the positive
integers,

and that, for each value of n, l can be any integer less than n,

that is,

For the ground state, n = 1, the only possible value of l is l = 0, and the ground
state of hydrogen therefore has zero angular momentum. With l = 0, the only possible
value of m is m = 0 and the ground state is characterized by the unique set of quantum
numbers
Notice that although the Schrödinger equation and the Bohr model give the same energy
for the ground state, there is an important difference: Whereas the Bohr model assumed a
magnitude L = l h in the ground state, the Schrödinger equation predicts that L = 0, a
prediction that is borne out by experiment.
For the first excited level, n = 2, there are two possible values of 1, namely 0 or 1.
If l = 0, then m can only be 0; but with l = I there are three possible orientations of L,
given by m = 1, 0, or - 1. Thus there are four independent wave functions for the first
excited level, with quantum numbers:

This means that the first excited level is fourfold degenerate.
For the nth level, there are n possible values of L, given by l = 0, 1, (n - 1). To
display this graphically, it is convenient to draw energy-level diagrams in which the ener .
'Sy is p otted upward as usual, but with the different angular momenta L = V1(l + 1) h
shown separately by plotting l horizontally. The first four levels of the hydrogen atom are
plotted in this way in Fig. 9.16.
In Figure 9.16 we have introduced the code letters s, p, d, f . . . , which are
traditionally used to identify the magnitude of the angular momentum. These are as
follows:

These code letters are a survival from early attempts to classify spectral lines; in
particular, s, p, d, and f stood for sharp, principal, diffuse, and fundamental After f the
letters continue alphabetically, although code letters are seldom used for values of l
greater than 6. When specifying the values of n and l it is traditional to give the number n
followed by the code letter for L Thus the ground state of hydrogen is called I s; the first
excited level can be 2s or 2p; and so on. Lower case letters, s, p, d, . . . , are generally
used when discussing a single electron, and capitals, S, P, D, . when discussing the total
angular momentum of a multielectron atom.
Even when n and l are specified, there are still (21 + 1) distinct states
corresponding to the (21 + 1) orientations m = 1, 1 - 1, . . . , - L For s states (1 = 0) there
isjust one orientation, forp states (I = 1) there are (2 X 1) + l = 3, for d states (I = 2) there
are (2 X 2) + l = 5, and so on. These numbers are shown in parentheses on the right of
each horizontal bar in Fig. 9.16. The total degeneracy of any level can be found by adding
all of these numbers for the level in question. For example, the n = l level is
nondegenerate; the n = 2 level has degeneracy 4; the n = 3 level 9. The nth level has 1 =
0, 1, . (n - 1) and hence has degeneracy* (Problem 9.29)

*
Actually, the total degeneracy is twice this answer. This is because the electron has another
degree of freedom, called spin, which can be thought of as the angular momentum due to its spinning on its
own axis (much as the earth spins on its north-south axis). This spin can have two possible orientations, and
for each of the states described here, there are really two states, one for each orientation of the spin. This
will be discussed in Chapter 10.
In conclusion, the stationary states of hydrogen can be identified by three quantum
numbers, n, 1, and m. The numbers l and m characterize the magnitude and z component
of the angular momentum L. The number n determines the energy as E,, = - EA In 2 and,
for this reason, is often called the principal quantum number. It is a peculiarity of the
hydrogen atom that the energy depends only on n and is independent of L We shall see
that in other atoms the energy of an electron is determined mainly by n, but does,
nonetheless, depend on l as well.

9.8 Hydrogenic Wave Functions
In many applications of atomic physics it is important to know at least the qualitative
behavior of the electron wave functions. In this section we discuss the wave functions for
the lowest two levels in the hydrogen atom.

THE GROUND STATE

The ground state is the I s state with n = I and l = 0. Since 1 = 0, m has to be zero
and, as discussed below (9.66), the wave function is spherically symmetric (that is, is
independent of 0 and 0 and depends only on r):

The radial function r 1, (r) is determined by the radial equation (9.71), which we can
rewrite (for the particular case that 1 = 0) as

If we recall that h 'Am,ke2) is the Bohr radius a,6 and that ER = ke2l2aB, we can rewrite
this equation more simply as

For the case n = 1, it is easy to verify that the solution of this equation is (Problem
9.33)

This wave function is plotted in Fig. 9.17. Since 
2
is the probability density for the
electron, it is clear from this picture that the probability density is maximum at the origin.
In fact, it is characteristic of all s states (states with zero angular momentum) that  is
2

nonzero at the origin; whereas for any state with 1:0 0, IV12 is zero at the origin. This
situation is easy to understand classically: A classical particle can be found at r = 0 only if
its angular momentum is zero. This difference between states with l = 0 and those with l
> 0 has important consequences in multielectron atoms, as we discuss in Chapter 11. It
also means that the exact energy of s states is slightly dependent on the spatial extent of
the nucleus. In fact, careful measurements of energies of atomic electrons in s states have
been used to measure nuclear radii.
Since the electron's potential energy depends only on its distance from the
nucleus, it is often more important to know the probability of finding the electron at any
particular distance from the nucleus than to know the probability ofits being at any
specific position. More precisely, we seek the probability of finding it anywhere between
the distances r and r + dr from 0, that is, anywhere in a spherical shell between the radii r
and r + A This can be evaluated if we recall that the probability of finding the electron in
a small volume dV is IV12 dV. The volume of this spherical shell is the area of the
sphere, 47rr2, times its thickness, dr

For the ground state of hydrogen the wave function depends on r only and is the same at
all points in this thin shell. Therefore, the required probability is just

P(between r and r + dr) = 
2
dV = JR(r)J247rr2 dr.

We can rewrite this as

We have dropped the subscripts Is in these important relations, since they are in fact true
for all wave functions.
An important feature of the function (9.82) is the factor of r, which comes from
the factor 4;rr 2 in the volume of the spherical shell (9.80). It means that when we discuss
the probability of different distances r (as opposed to different positions), large distances
are more heavily weighted, just because larger r corresponds to larger spherical shells,
with more volume than those with small r.
For the ground state of hydrogen, with wave function (9.79), the radial probability
density is

This is plotted in Fig. 9.18.* Perhaps its most striking property is that its maximum is at r
= a,,. That is, the most probable distance between the electron and proton in the Is state is
the Bohr radius aB. Thus although quantum mechanics gives a very different picture of
the hydrogen atom (with the electron's probability density spread continuously through

*
Note that the radial density in Fig. 9.18 is zero at r= 0 even though the wave function itself is not,
This is due to the factor r2 in (9.82).
space), it agrees exactly with the Bohr model as to the electron's most probable radius in
the ground state.
Armed with the radial density P,, (r) one can calculate several important
properties of the atom. Problems 9.31 and 9.35 to 9.37 contain some examples, and here
is another.
EXAMPLE 9.4 Find the constant A in the Is wave function r is = Ae-rlaB and the
average value of the potential energy for the ground state of hydrogen.

The constant A is determined by the requirement that the total probability of
finding the electron at any radius must be 1:

Substituting (9.83) we find that

This integral can be evaluated with two integrations by parts to give a 3
B14 (Problem 9.35). Therefore,

and

To find the average value of the potential energy U(r)we multiply U(r)by the
probability P(r) dr that the electron be found at distance r, and integrate over all r:

If we substitute U(r)= - kelfr and replace P(r) by (9.8 3) this gives

The integral can be evaluated by parts as a 2 /4, and we find that

Note that this quantum value for the mean potential energy agrees exactly with the
potential energy of the Bohr model.

THE 2s WAVE FUNCTION

In the n = 2 level, with E ER /4, we have seen that there are four independent
wave functions to consider. Of these, the 2s wave function depends only on r
where R2,(r) is determined by the radial equation (9.77) to be (Problem 9.38)*

The probability of finding the electron between distances r and r + dr from the
origin is again given by P(r) dr with

This function is plotted in Fig. 9.19. As we would expect, it is peaked at a much larger
radius than the I s function. Specifically, the most probable radius for the 2s state is r -
5.2aB, in approximate (though not exact) agreement with the second Bohr radius, r =
4a,,. An important feature of the 2s distribution is the small secondary maximum much
closer to r = 0 (at r - 0.76aB). This means that there is a small (but not negligible)
probability of finding the 2s electron close to the nucleus.

THE 2p, WAVE FUNCTION

There are three 2p wave functions, corresponding to the three possible
orientations of an I I state. In Section 9.6 (Example 9.3) we saw that the angular part of
the m 0 wave function is 0(0) = cos 0, and the complete wave function is therefore

For reasons that we shall see in a moment, this is often called the 2p., wave function. The
radial function R2,(r) is found by solving the radial equation for E = - E,,~ 14 and 1 = 1.
This gives (Problem 9.39)

Notice that R2P (r) is zero at r = 0. Thus the probability density  is zero at the origin, a
2

result that applies (as already mentioned) to any state with nonzero angular momentum.
Substituting (9.90) into (9.89), we find for the complete wave function of the 2p
state with m = 0:

Since this depends on r and 0 it is harder to visualize than the l = 0 wave functions (which
depend on r only). One way to show its main features is to draw a contour map of the
probability density  in the x-z plane, as shown in Fig. 9.20(a). Since  is
2                                                       2

independent of 0, one would find the same picture in any other plane containing the z
axis, and one obtains the full three-dimensional distribution simply by rotating Fig.
9.20(a) about the z axis. Figure 9.20(b) shows a perspective view of the 75% contour
obtained in this way.

*
As usual, A denotes a constant, which is determined by the normalization condition (9.84). For
simplicity we use the same symbol, A, for all such constants, but we do not wish to imply that they have the
same value for all wave functions.
The probability density 
2
for (9.91) is largest on the z axis (where cos 0=±I) at
the points z=±2aB, and is zero in the x-y plane (where cos 0 = 0). The region in which the
electron is most likely to be found consists of two approximately spherical volumes
centered on the z axis, one above and the other below the x-y plane, as shown in Fig.
9.20(b). It is because the electron is concentrated near the z axis that the 2p state with m =
0 is called the 2p, state.

THE 2p, AND 2p y WAVE FUNCTIONS

There are still two more 2p states to be discussed. An easy way to write these
down is to note that the 2p, wave function (9.91) can be rewritten as

since r cos 0 = z. Now, the Schrödinger equation, from which this was derived, involves
each of the coordinates x, y, z in exactly the same way. Thus if (9.92) is a solution, so
must be the two functions obtained from (9.92) by replacing z with x or with y:

and

The properties of these two wave functions are very similar to those of v2p., except that
where  V/2,, is concentrated near the z axis, V/2,,,, is concentrated near the x axis and
V2PY near the y axis. Figure 9.21 shows perspective views of all three wave functions.
We shall see in Chapter 16 that the concentration of the electron near one of the axes in
each of these states has important implications for the shape of some molecules.
As with s waves, it is often important to know the probability of finding the
electron at a certain distance from the origin (as opposed to that for finding it at one
particular position). Because the 2p wave functions depend on 0 and 0 as well as r, the
probability of finding the electron between r and r + dr must be calculated by integrating
over the angles 0 and 0. However, the result is exactly the same as (9.82) for s waves

where, for any of the 2p states,

This function is plotted in Fig. 9.22, where we see that P2P(r) is maximum at r = 4aB
(Problem 9.44); that is, the most probable radius for the 2p states agrees exactly with the
radius of the second circular Bohr orbit.
Before leaving the 2p wave functions, we should mention a final complication. In
our general discussion of the central-force problem, we saw that for any p state (I = 1)
there must be three possible orientations given by m = 1, 0, or - 1. In the case of the 2p
states we found, explictly, three independent wave functions 2p,, 2p., and2p, It turns out
that these latter three wave functions are not exactly the same as the former. Specifically,
the 2p, state is precisely the m = 0 state. (This was how we derived it.) On the other hand,
the 2p, state is not the m = 1, nor the m = - 1, state. Instead, the 2p., wave function is the
sum of the wave functions for m 1, while the 2p y function is their difference (Problem
9.41).
The important property of the 2p states is this: Any 2p wave function can be
written as a linear combination of the three wave functions with m = 1, 0, and - 1, or as a
combination of the wave functions 2p, 2p., and 2p, Which set of three functions we
choose to focus on is largely a matter of convenience, and for our purposes the three
functions 2p, 2p y, and 2p, are usually more suitable.
This situation is very similar to what we saw in Sections 8.6 and 8.7 when solving
the differential equation V" = - k 2 V. Any solution of that equation could be expressed as
a linear combination,

of the two solutions sin kx and cos kx, or as a combination,

of the two solutions e:':ikx. When seeking the energy levels of a rigid box we found it
convenient to use the pair sin kx and cos kx to apply the boundary conditions V/(O) =
VI(a) = 0. On the other hand, to interpret the solutions in terms of momentum, it was
convenient to re-express them in terms of the pair e-1:1kx, as in Eq. (8.75).

9.9 Shells
We have seen that the most probable radius for the Is state of hydrogen is r = aB, while
those for the 2s and 2p states are r - 5.2 aB and r = 4 aB. For the 3s, 3p, and 3d states the
most probable radii are 13.IaB, 12aB, and 9aB1 respectively. These results are illustrated
in Fig. 9.23, which shows the radial densities and most probable radii for all of the states
concerned.
Figure 9.23 suggests what is found to be true for all of the states with which we
shall be concerned: For all the different states with a given value of n, the most probable
radii are quite close to one another, and are reasonably well separated from those with any
other value of n. This important property is illustrated in a different way in Fig. 9.24,
which shows how the most probable radii for the states with quantum number n tend to
bunch together in concentric spherical shells with radii close to the Bohr values of n 2 aa.
For this reason the word shell is often used for the set of all states with a given value of n.
In the hydrogen atom one can characterize a shell in two different ways that are
exactly equivalent. As we have just seen, for all states in the nth shell, the most probable
distances of the electron from the nucleus are clustered close to the Bohr value n 2 aB.
Alternatively, since all states of a given shell have the same value of n, they all have the
same energy. Thus the word "shell" can refer either to a clustering in space (what we
could call a spatial shell) or to a clustering in energy (an energy shell).
The notion of shells is very important in atoms with more than one electron, as we
shall see in Chapter 11. We shall find that the possible states of any one electron in a
multielectron atom can be identified by the same three quantum numbers, n, 1, m, that
label the states of hydrogen. Furthermore, just as with hydrogen, 0 states with a given n
is well separated from the most probable radius for any other value of n. Thus we can
speak of spatial shells, as a characteristic clustering of the radial distributions for given n,
in just the same sense as in hydrogen.
On the other hand, the allowed energies of any one electron in a multielectron
atom are more complicated than those of hydrogen. In particular, we shall find that states
with the same principal quantum number, n, do not necessarily have the same energy.
Nevertheless, the states can be grouped into energy shells, such that all levels within one
shell are closer to one another than to any level in a neighboring shell. However, these
energy shells do not correspond to unique values of n: States with the same n may belong
to different shells, and one shell may contain states with different values of n. For
example, in many atoms the 3s and 3p levels are close to one another but are quite well
separated from the 3d level, which is closer to the 4s and 4p levels. In this case the 3s and
3p levels form one energy shell, and the 3d, 4s, and 4p another.
Unfortunately, the word "shell" is commonly used (without qualification) to
denote both what we have called a spatial shell and what we have called an energy shell.
We shall discuss all this in more detail in Chapter 11. We mention it here only to
emphasize that the simple situation in hydrogen (for which the grouping of states
according to energy is exactly the same as the grouping according to distance from the
nucleus) is unique to hydrogen.

9.10 Hydrogen-Like Ions
In Chapter 6 we saw that Bohr's model of the hydrogen atom could be easily generalized
to any hydrogen-like ion (that is, a single electron bound to a nucleus of charge Ze). The
modern, Schrödinger theory of hydrogen can be generalized in exactly the same way. The
potential energy of the electron in hydrogen is U = - ke2lr, that of the electron in a
hydrogen-like ion is U = - Zke2lr. Thus the Schrödinger equation for the latter case differs
from that of the former only in that ke2 is replaced by Zke2 in the potential-energy
function.* Therefore, we can convert our hydrogen solutions into solutions for the
hydrogen-like ion simply by substituting Zke whenever the term ke2 appears. This lets us
draw three important conclusions with almost no additional labor.
First, the properties of the angular wave functions and the allowed values of
angular momentum do not involve the potential energy U at all. Therefore, these angular
properties are exactly the same for any hydrogen-like ion as for hydrogen itself
Second, the Schrödinger equation for hydrogen has acceptable solutions only for
the allowed energies,

*
Throughout this chapter we are ignoring motion of the nucleus. If we were to include this, there
would be a second difference between the hydrogen atom and the hydrogen-like ion, because of the
different masses of the nuclei. This small effect can be allowed for by introducing a reduced mass, as
described briefly in Section 6.8, but we shall ignore it here.
Replacing ke 2 by Zke2' We find for the allowed energies of a hydrogen-like ion:

Third, the spatial extent of the hydrogen wave functions is determined by the Bohr radius,

thus the corresponding parameter for a hydrogen-like ion is

For example, the ground-state wave function of hydrogen is V/1, = A exp[-rlaB];
therefore, that for a hydrogen-like ion, with aB replaced by aBIZ, is

Since all wave functions are modified in the same way, each state of a hydrogen-like ion
is pulled inward by a factor I IZ, compared to the corresponding state in hydrogen.
The relationship between the quantum properties of the hydrogen atom and the
hydrogen-like ion is closely analogous to the corresponding relationship for the Bohr
model. This provides the ultimate justification for the several properties of hydrogen-like
ions described in Chapter 6 in connection with the Bohr model.
When we discuss multielectron atoms in Chapter 11, we shall make extensive use
of the two results (9.96) and (9.97). These are so important, let us close by reiterating
them in words: When an electron moves around a total charge Ze, its allowed energies are
Z2 times the allowed energies of a hydrogen atom; and its spatial distribution is scaled
inward by a factor of 1 1Z compared to hydrogen.

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