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7. Quantum Mechanics in Three
Dimensions
Links to specific sections in the text:
7.a. Three dimensional Box and Degenerate States
7.b. Quantization of Angular Momentum
7.c. States of the Hydrogen Atom
7.a Three dimensional Box and Degenerate States
Most problems in the real world have three space dimensions. Thus
far we learned to work with one dimensional quantum mechanics. To
learn about three dimensional quantum mechanics we should first go
back to free particles. Free particles were the starting point of
studying one dimensional quanrum mechanics as well. It is easy to
generalize plane waves to three dimensions. The wave number, k,
becomes a wave vector. So instead of having just kx-wt for the phase of the wave we will allow the
plane wave to move in an arbitrary direction determined by the wave
vector k, the absolute value of which is the wave
number, k. k has the same de Broglie relation with the momentum as
for the one dimensional case, k =2p p/h.
We can imagine that nothing else happened but we rotated the
coordinate system so that the wave vector points into the direction
of k, instead of the x-axis. Then the form of the
plane wave is:
y(r,t) = A exp{ i ( k.r - w t ) } = A exp{ i
2p ( p.r -
E(p) t )/h}.
Again, p is just a parameter of the wavefunction. This describes a
wave propagating in the direction of p. For a
nonrelativistic particle E(p) =
p2 / 2m. This form should
satisfies the time dependent Schrodinger equation with vanishing
potential energy.
We will write down the time dependent Schrodinger equation using
the same rule as we used for the one dimensional case. the
Schrodinger equation should have the form
Eop y(r,t)
= Hop(p,r)
y(r,t).
WE know the form of the energy operator. That has nothing to do
whether the theory is one dimensional or three dimensional. Eop
= i( 2p /h) d/dt. We also learned
the rule for the Hamiltonian operator. To get the quantum expression
just substitute the momentum and coordinate with operators. For a
free particle the Hamiltonian is just the kinetic energy. K =
p2 / 2m. So the question is what
is the momentum operator. We learned about the momentum operator in
the one dimensional case. It is just the operator that brings down a
momentum factor from the one dimensional plane wave wavefunction.
Obviously the requirement is the same for the three dimensional case.
Looking at the three dimensional plane wave it is easy to see that
the components of the momentum are generated if we apply the the
operators, (-ih/2p )d/dx, (-ih/2p )d/dy, and (-ih/2p
)d/dz to the wavefunction. Then we can write for a free particle
Hop(p,r) =
p2 / 2m =
(px2 + py2 +
pz2 )/2m
= ( -h2/4p2d2/dx2 +
-h2/4p2d2/dy2
-h2/4p2d2/dz2)/ 2m =
( -h2/4p2)D/ 2m,
where D = d2/dx2
+ d2/dy2 + d2/dz2 is the
Laplacian, an object thorougly studied in calculus. One can easily
check now that with this definition of the momentum the wavefunction
above, describing plane waves satisfies the Schrodinger equation
i( 2p /h) d/dt y(r,t) = (
-h2/8p2m )Dy(r,t).
Obviously, the one diminsional wavefunction for plane waves also
satisfies this Schrodinger equation, because its partial derivatives
with respect to y and z vanish and the terms not containing the
derivatives with respect to y and z are just the Schrodinger equation
for the free particle in one dimension. Now we can easily generalize
the three dimensional Schrodinger equation to the case when the
potential energy is nonzero. As for the one dimensional case we need
to add a term to the hamiltonian corresponding to the potential
energy. Thus, we obtain
i( 2p /h) d/dt y(r,t) = (
-h2/8p2m )Dy(r,t) + V(r)
y(r,t).
This is the form of the Schrodinger equation used for all single
particle problems. It has had a fantastic success in explaining a
great variety of phenomena.
Just like for the one dimensional system we will mostly
investigate stationary states, states that are eigenstates of the
energy operator, i( 2p /h) d/dt. An
eigenstate of the energy operator satisfies
i( 2p /h) d/dt y(r,t) = E y(r,t),
where E is the eigenvalue of the energy operator, a constant. This
differential equation is easy to solve. It has the form d/dt y = C y, so its
solution is exponential, y(r,t) = exp{ - i 2pE t /h}f(r). Then substituting into
the time dependent Schrodinger equation we obtain the time
independent Schrodinger equation if we substitute this form and
divide by the factor exp{ - i 2pE t /h} which is a common multiplier of every
term. We obtain
E f(r) = (
-h2/8p2m )Df(r) + V(r)
f(r)
The first problem in three dimensional quantum mechanics that we
will investigate will be the states in a three dimensional box, i.e a
potential defined by the condition that it is infinite everywhere
except where 0<x<L, 0<y<L, and 0<z<L. Then the wave
function vanishes outside the box and sinve inside the potential is
zero it satisfies the equation
E f(r) = (
-h2/8p2m )Df(r) = ( -h2/8p2m )(d2/dx2 +
d2/dy2 + d2/dz2 )f(r).
Notice that the x, y and z dependent terms of the operator acting
on the wavefunction are in separate terms. Such a partial
differential equation is called separabel and can be solved in terms
of the product of wavefunctions each depending only on one variable,
x, y, or z. To prove this let us write f(r) as
f(r) = f1(x)
f2(y) f2(z)
Substituting this trial function into the Schrodinger equation we
obtain
E f1(x) f2(y)
f2(z) = ( -h2/8p2m )[ f1"(x)
f2(y) f2(z) +
f1(x) f2"(y)
f2(z) + f1(x)
f2(y) f2"(z) ],
where the " means second derivative. Then dividing this equation
by the wavefunction, f1(x) f2(y)
f2(z), itself we obtain
E = ( -h2/8p2m )[
f1"(x)/ f1(x)+
f2"(y) /f2(y) +
f2"(z)/f2(z) ].
Now these are terms that depend on different variables and their
sum is equal to a constant. It is possible only if these terms do not
really depend on these variables but are constants themselves. In
other words, calling these constants E1 ,E2 ,
and E3 , we obtain the equations
E1 = ( -h2/8p2m ) f1"(x)/
f1(x), or E1 f1(x) = (
-h2/8p2m ) f1"(x)
E2 = ( -h2/8p2m ) f2"(y)/
f2(y), or E2 f2(y) = (
-h2/8p2m ) f2"(y),
E3 = ( -h2/8p2m ) f3"(z)/
f3(z), or E3 f3(z) = (
-h2/8p2m ) f3"(z)
E = E1 + E2 + E3
These are just one dimensional Schrodinger equations for the three
factors of the wavefunction. In fact these equations and the boundary
conditions are identical to those we solved for the one dimensional
box. Then we know the solutions s well. They have the form of
f1(x) = (2/L)1/2 sin k1 x,a
f2(x) = (2/L)1/2 sin k2 y,
f3(z) = (2/L)1/2 sin k3 z,
where k1 = n1 p /
L, k2 = n2 p /
L, k3 = n3 p /
L, where n1 ,
n2 , and n3 are positive integers. The enrgies
E1 E2 , and E3 are given by
E1 = (h2 n12
/ 8 m L2 ) , E2 = (h2
n22 / 8 m L2 ), and
E3 = (h2 n32 /
8 m L2 ). Then we obtain the energy E as
E = E1 + E2 + E3 = (h2
/ 8 m L2 )( n12 +
n22 + n32 ).
We can easily list now the states of the three dimensional box.
The lowest energy state will have the lowest possible value for
n1 , n2 , and
n3 , n1 =
n2 = n3 = 1. Then the ground state energy of
the particle in the box is
E1,1,1 = (3 h2 / 8 m L2 ).
where the labels of the energy are the three quantum numbers,
n1 , n2 , and
n3 , that the energy eigenvalue depends on. The first
excited state is when say n1 =2,
n2 = n3 = 1. Then the energy is
E2,1,1 = [(4+1+1) h2 / 8 m L2] =
(3 h2 / 4 m L2 ).
Notice however that there are two other states that have the same
energy. These have n2 =2,
n1 = n3 = 1 and n3 =2, n12 = n2 = 1. In other
words, we can say that E2,1,1 = E1,2,1 =
E1,1,2 = (3 h2 / 4 m L2 ). States of
the same system having identical energies are called degenerate.
Notice that the states corresponding to these different quantum
numbers are indeed different, because the wavefunctions have
different dependences on the coordinates. E.g. for the three
degenerate first excited states we have the following three
wavefunctions
f2,1,1(x) = (2/L)3/2 sin ( 2p x / L) sin ( p y / L) sin ( p z / L),
f1,2,1(x) = (2/L)3/2 sin ( p x / L) sin ( 2p y / L) sin ( p z / L),
f1,1,2(x) = (2/L)3/2 sin ( p x / L) sin ( p y / L) sin ( 2p z / L).
Let us list a few higher states of the particle in the three
dimensional box. The next state is when n1 =1, n2 = n3 = 2, because
for this state n2 = n12
+ n22 + n32 = 9.
This corresponding to the energy level
E1,2,2 = [(1+4+4) h2 / 8 m L2] =
(9 h2 / 8 m L2 ).
This level is three fold degenerate as well, because
E2,2,1 = E1,2,2 = E2,1,2 . The next
level,
E3,1,1 = E1,3,1 = E1,1,3 =
[(9+1+1) h2 / 8 m L2] = (11 h2 / 8 m
L2 )
is also three fold degenerate. Then the following level
E2,2,2 is nondegenerate, etc.
In two dimensions the spectrum is similar. The possible energy
levels, as one may guess from the three and one dimensional formulas
E = (h2 / 8 m L2 )( n12
+ n22 + n32 ) and E
= (h2 / 8 m L2 )n2 .
will be
E = (h2 / 8 m L2 )( n12
+ n22 ).
The two dimensional case would correspond to an electron trapped
on the surface of a metal.
Problem: Suppose that a two dimensional box has
different dimensions along the x, y, and z, axes, L1,
L2, and L3 . What are the energy levels of a
particle in such a box?
Solution: Due to the factorization of the wavefunction the
contributions of individual dimension to the total energy are of the
form E = E1 + E2 + E3 . Each term of
this sum is a one dimensional energy. Thus one needs to calculate the
one dimensional energies and sum them to get the total energy. But
one dimensional energies can easily be calculated for an arbitrary
sized box. So one gets for example for a box of size L1,
E1 = (h2 / 8 m L12
)n12
The total energy will be
E = (h2 / 8 m )( n12 /
L12 + n22 /
L22 + n32 /
L32 ).
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7.b. Quantization of Angular Momentum
The next problem we will investigate is the hydrogen atom. Before
we discuss the problem of the hydrogen atom we need to investigate
the type of problem the quantum mechanics of the hydrogen atom would
present. The potential energy appearing in th schrodinger equation is
V(r) = - ke2/r. This potential does not depend on the
coordinates separately, only through the radial distance away from
the center (the hydrogen nucleus, proton), which is r =
(x2 + y 2+ z2 )1/2.
Such a potential is called central. For a central potential the
cartesian coordinates are not the most convenient to use. It is much
better to use polar coordinates. r, q, f.
Remember, these are related to the cartesian coordinates as
r = (x2 + y 2+ z2
)1/2, q =
tan-1(z/r), f =
tan-1(y/x).
The Schrodinger equation contains, besides the potential V(r), the
Laplacian, which is a sum of second derivatives with respect to the
coordinates. The potential is obviously independent of the angles.
The Laplacian can also be obtained if we use the chain rule for
differentiation, i.e. we express d/dx, d/dy, d/dz by the polar
coordinates. Then one obtains
D =
d2/dx2+d2/dy2+d2/dz2
= d2/dr2 + (2/r) d/dr +
(1/r2)Lop2 /
(hbar)2
Now the operator of angular momentum is defined as
Lop =
popxrop.
Substituting the expressions for the momentum components we obtain
Lz = -i d/df
and Lop2 = -
(hbar)2 [ d2/dq2 + cot
q d/dq +
csc2q d2 / df2 ]
Notice that the angular momentum operator is independent of the
radial coordinate, r. Writing the form of the Laplacian into the
Schrodinger equation we obtain
E y(r , q, f) =
- [(hbar)2 / 2m] [
d2/dr2 + (2/r) d/dr] y(r , q, f) + 1/(2 m
r2)Lop2 y(r , q, f) + V(r)
y(r , q, f)
Notice that this equation is almost separable. Just multiply it by
r2 and then the terms either depend on r or the angles.
Then the solution can be written as
y(r , q, f)
= R(r) Y(q, f).
so called radial equation and the spherical equation,
respectively. The factorized solution requires that the function Y
satisfies the equation
Lop2
Ylm(q, f) =
(hbar)2 l(l+1)Ylm(q, f)
are eigenfunctions of the square of the angular momentum operator.
These functions, called spherical harmonics, must be periodic
functions of q and f.
This requirement makes these eigenvalues are such that l is
always an integer, l=0,1,2,... The eigenfunctions are also
eigenfunctions of the operator Lz, namely
Lz Ylm(q, f)
= hbar m Ylm(q,
f).
where m is another integer. In other words both the angular
momentum and its z component have sharp values in these states. The
wavefunctions are labeled by the quantum numbers l (which is closely
related to the magnitude of the angular momentum) and quantum number
m (which is closely related with the value of the z componenet of the
angular momentum). By the way, at any given integer value of l
spherical functions exist only if m= -l, -l+1,...-1,0,1,...,l-1,l.
The magnitude of m, |m| is always smaller or equal to l. l is called
the orbital quantum number, while m is called the
magnetic quatum number. One can show that the other
components of angular momentum do not have sharp values in these
states. This shows that the angular momentum is quantized in quantum
mechanics, and the magnitude of the angular momentum can only be
|L| = hbar [l(l+1)]1/2, i.e can take
discrete values only. Now this is strange. Is the earth's angular
momentum also quantized?
Problem: Calculate the orbital
quantum number l for the rotation of earth around its poar axis.
Solution: The angular momentum is equal to
L=Iw, where I is the moment of inertia
w and is the angular velocity. The moment
of inertia for a sphere is I= 3MR/5. The mass of earth is M= kg, its
radius, R=6.1¥106m. The angular
velocity w = 2p
/ day = 7.27¥10-5s-1. The result is L
= ¥106kg m s-1.
Dividing this quantity by hbar we obtain L/hbar = . This is a very
large number so we can calculate l by expansion. For large l we have
[l(l+1)]1/2=l [1+1/l]1/2µ l (1 + 1/2l) = l+2. Thus we obtain lµ . This value of l is so large that it
really does not make any difference if we add or take away one unit.
For all practical purposes the angular momentum can be regarded as
continuous for macroscopic processes.
The spherical functions have very simple forms for low values of l
and m. For l=0, m can only be 0 as well. The spherical function is
just a constant.
Ylm(q, f) =
1/(4p)1/2.
The constant value is established so that the wave function would
be normalized. Here we need to pause a little and decide how to
normalize wavefunctions in three dimension.
Problem: Suppose that the
wavefunction of the ground state of the hydrogen atom is
y(r , q, f) =
N e-r/a,
where a is the Bohr radius. N is a yet uknown normalization
constant. Normalize this wavefunction. Also normalize the spherical
function Y00(q, f).
Solution: Normalization means finding factor N
that makes the probability interpretation meaningful. In other words,
since the absolute squared of the wave function is the probability
density, the probability that theparticle is inside the volume dxdy
dz is |y(r , q, f)
|2dxdydz. Then we must require
Ú |y(r , q, f) |2dxdydz = 1 = Ú N2 e-2r/adxdydz.
Now this integral can be much easier done in polar coordinates so
we need to rewrite dxdydz = r2dr sinq dq df. Then we can factorize the integrals
1= Ú |y(r , q, f)
|2dxdydz = N2 Ú e-2r/a r2dr
Ú sinq dq df = N2
[a3/4] [4p ].
This shows that the factor N (the normalization constant) contains
two multipliers: (4/a3)1/2 that normalizes the
radial wavefunction and (4p
)-1/2 that normalizes the spherical function. Then the
wavefunction can be written as
y(r , q, f) =
(4/a3)1/2 e-r/a
Y00(q, f) =
(4/a3)1/2 e-r/a (4p )-1/2 = (p/a3)1/2 e-r/a
The spherical function Y00(q, f) = (4p
)-1/2 is a constant.Then a wavefunction for which the
quantum numbers are l=m=0 is isotropic (it does not depend on the
direction), or we can also say that rotation invariant. Though the
potential energy of the hydrogen atom problem is rotation invariant
the solutions of the Schrodinger equation are not in general
indpendent of the angles. In fact, returning to the Schrodinger
equation in polar coordinates, we can see that
E y(r , q, f) =
- [(hbar)2 / 2m] [
d2/dr2 + (2/r) d/dr] y(r , q, f) + 1/( 2 m
r2)Lop2 y(r , q, f) + V(r)
y(r , q, f)
can be rewritten as (using the separtaion of variables and the
solution of the angular part with spherical functions)
E R(r)Ylm(q, f) =
- [(hbar)2 / 2m] [ d2/dr2 + (2/r) d/dr]
R(r)Ylm(q, f) +
(hbar)2/(2mr2)l(l+1)R(r)Ylm(q, f) + V(r) R(r)Ylm(q, f).
Nom all the angular dependence is gone from the equation, except
for the multiplier Ylm(q,
f) that appears in every term and can be cancelled. Then we
obtain the radial equation
E Rl(r)= -
[(hbar)2 / 2m] [
d2/dr2 + (2/r) d/dr] Rl(r) +
(hbar)2/(2mr2)l(l+1)Rl(r) + V(r) Rl(r).
where we explicitly display the dependence of the radial
wavefunction on angular momentum quantum number l. Note that the
radial wavefunction does not depend on the quantum number m
(calledusually the magnetic quantum number, for reasons we will see
later). Also note that the term (1/2mr2)l(l+1) appears
just like an addition to the potential energy. It is called the
centrifugal barrier and appears as a repulsive potential. The
effective potential appearing in the Schrodinger equation is
Veff(r) = (hbar)2/(2mr2)
l(l+1) + V(r).
Unless the potential energy, V(r) is as singular as
1/r2 near the origin (what never happens in practical
physical problems), the centrifugal barrier dominates the effective
potential at small r (provided l is not 0). Due to this strong
repulsion the particle cannot approach the origin. Consequently for
all but nonvanishing values of the angular momentum the wavefunction
must vanish at the origin. The higher the angular momentum is the
stronger is this repulsion and the faster the wavefunction vanishes
at the origin. In fact, one can show that Rl(r) µ rl near r=0.
Let us examine now the allowed values of angular momentum more
closely.
Example: Draw the possible angular momentum vectors if
the orbital quantum number, l=3.
Solution: If l=3, then |L| = (3x4)1/2
hbar= 3.46 hbar, and Lz = m hbar, where the possible
values of m are m=-3,-2,-1,0,1,2,3. The components Lx and
Ly are not sharp and cannot be determined. Then for a
given m the angular momentum vector has a fixed magnitude and z
component, but undetermined in the x and y directions. Thus the end
of the angular momentum vector can be on a circle of radius r = hbar [l(l+1)-m2]1/2=
hbar [12 - m2]1/2. Projecting these to the xz
plane these circles look like lines. Then we have the following
picture:
The x and y components of the angular momentum are completely
undetermined when the z component is determined. In fact, on each
circle represented on the picture above as a horizontal line, every
angular momentum vector is equally probable. So when one attempts to
measure the angular momentum in such a state then one obtains all
possible choices with equal probability.
Notice that this picture and anything we said about the angular
momentum, its possible values, its eigenfunctions have nothing to do
with the physical problem at hand. It follows from the nature of
quantum mechanics and of angular momentum. It is somehow a properety
of space itself. Therefore it is sometimes called "space
quantization."
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7.c. States of the Hydrogen Atom
The radial Schrodinger equation for the hydrogen atom is written
as
E Rl(r)= -
[(hbar)2 / 2m] [
d2/dr2 + (2/r) d/dr] Rl(r) +
(1/2mr2)l(l+1)Rl(r)
-( Ze2k/ r)Rl(r).
Here we use a generalized hydrogen-like potential (introducing the
factor Z) rather than a straight forward potential for the Coulomb
force between the proton and the electron, because then we will be
able to discuss ionized atoms of higher Z as well. The exact
solutions of this equation can be given for every orbital quantum
number, l. Infinitely many states exist for every l. These states can
be labeled by an iteger quantum number n, called the
principal quantum number, and written as
Rn,l(r). By convention we choose n>l. In other
words,say for l=2 the possible values of n are n=3,4,5,... One can
turn it around and say that for any given n the allowed values of l
are l=0,1,2,....,n-1. In particular for the lowest n, n=1 the only l
allowed is l=0. The eigenvalues of the hamiltonian, E, turn out to be
independent of l. On the language we learned earlier, they are
usually degenerate. The energy values are:
En = - (ke2/2a0)
(Z2/n2),
where as before, a0 is the Bohr radius. This expression
coincides with the Bohr energy levels, only it was derived from a
fundamental theory, rather then using ad-hoc assumptions. The n
quantum number we introduce is exactly Bohr's quantum number. Now we
can look at states with increasing values of n. n=1 is the ground
state. if n=1, then l=0. If l=0, then m=0 as well (remember
|m|<=l). Thus, there is only one wavefunction corresponding to the
n=1 level. The level is not degenerate. the wavefunction is Y(r , q,f) =
R1,0(r) Y00(q,f). Remember that
Y00(q,f) is really
independent of the angles, it is a constant. The next level is n=2.
Now both l=0 and l=1 are allowed. If l=1, then m could take values
m=-1,0,1. Altogether there are 4 different wavefunctions
corresponding to the same energy, E2. These wavefunctions
are
R2,0(r)Y00(q,f),
R2,1(r)Y1-1(q,f),
R2,1(r)Y10(q,f),
R2,1(r)Y11(q,f)
all distinct. Then this energy level is fourfold degenerate.
Problem: What is the degeneracy of level n , for
general n?
Solution: the possible values of l are l=0,1,2,...,n-1. For each l
the possible states have m values from -l to l. Alltogether 2l+1, for
each l. These numbers must be added up to n-1 to get the total
degeneracy. The sum of (2l+1) is twice the sum of l from 0 to n-1 and
the sum of 1 from zero to n-1. The sum of l is a geometric series.
The result is n(n-1)/2. Multiplied by 2 one obtains n(n-1). Then the
sum of 1 from 0 to n-1 is just the number of n-s, which is n. Then
the total degeneracy is N=n(n-1)+N=n2. The states
belonging to a single n form a shell. The 2l+1
states belonging to a given value of n and given value of l are
called a subshell.
For orbital quantum number, l=0 (called s-wavefunctions) do not
vanish at the origin. In fact, we already described the n=1
wavefunction
y1,0,0(r , q, f) = (p/a3)1/2 e-r/a
The normalized radial wavefunction is R1,0(r) =
(4/a3)1/2 e-r/a
Problem: Determine the probability, P, that the
electon is insidee the Bohr radius in the hydrogen atom.
Solution: One would think at first sight that the probability
density is |R1,0(r) |2. This is false however.
The wavefunction is normalized as
Ú |R1,0(r) |2
r2 dr =1
Then the probability to find the electron between r andr+dr is
P(r) dr = |R1,0(r) |2 r2 dr
The physical significance of the multiplier r2 is
clear. The density |R1,0(r) |2 is proposrtional
to the probability to find the electron in unit volume near r. But
then the smaller r gets the less chance to find suvh a volume. One
could say that r is approximately constant in a shell of radius r.
The volume of such a shell is proportional to r2 (It is
4pr2dr ). Having the
probability distribution we just have to integrate it over the
relevant volume:
P = Ú | (4/a3)1/2
e-r/a |2 r2 dr = (4/a3)
Ú e-2r/a r2 dr,
Where the limits of the integral are 0 and a, since we want to
find the probability to find th eelectron at r<a. We need to
integrate by parts repeatedly:
P = - (4/a3) (a/2) e-2r/a r2
|0a + (4/a3) (a/2)2 Ú e-2r/a r dr = -
2 e-2 -
(4/a3) (a/2)22 e-2r/a r
|0a + (4/a3) (a/2)22
Ú e-2r/a dr
= -4 e-2
- (4/a3) (a/2)32 e-2r/a
|0a = 1 - 5
e-2
µ 0.333.
Comments concerning the energy levels and states of the hydrogen
atom:
- If the nucleus is not proton tbut has charge Z, then in the
Schrodinger equation e2 should be substituted by
e2Z. Since that is the only appearence of e in the
eequation, we get all results modified if in wave function and in
energies we substitute e2 by e2Z. That means
a should be substituted by a/Z (a is invereley proportional to
e2). In particular we already said earlier that
En = - (ke2/2a)
(Z2/n2) and we get that R1,0(r) =
(4Z3/a3)1/2 e-rZ/a
- The set of sharp observables in the eigenstates of the
hydrogen atom are the energy , En, the magnitude of the
angular momdetum, |L|, and the z component of the angular
momentum, Lz. They are determined by the three quantum
numbers n, l, and m.
- As we indicated above, the independence of En from
m is a consequence of rotation symmetry: the radial Schrodinger
equation itself is indpeendent of m, so the solution and
eigenvalue cannot dpend on it either. The energy levels are also
independent of l, however. This is kind of an "accidental"
degeneracy and it is not valid for potentials other than the
Coulomb potential. Energy levels for other rotation symmetric
(central) potentials do depend on l.
The ground state wavefunction can be best represented by the
figure below, which is again a projection of the wavefunction to a
plane (it is otherwise spherically symmetric, rotation invariant).
The lines represent constant values of the vavefunction, the heavier
the line is the larger the constant value is:
problem: Calculate the most probable
value of of the distance of the electron from the nucleus in the
ground state of the hydrogen like atom.
Solution: The normalized radial ground state
wavefunction is
R1,0(r) = (4Z3/a3)1/2
e-rZ/a
Then the probability distribution is given by
P(r) =|R1,0(r)|2 r2,
where the last factor is the space factor needed to be included
due to the size of shells of thickness dr at radius r. The integral
of Ú P(r) dr =1, is normalized. The most
probable radius is obtained at points where P(r) is maximum. The
derivative of P(r) must vanish at points where it is maximum. Thus
P'(r) µ d(r2 e-2rZ/a)/dr = 2r
e-2rZ/a - (2Z/a)r2 e-2rZ/a = 2r
e-2rZ/a ( 1 - Z r / a ) = 0.
This happens at either r=0, shich is obviously not a maximum (at
r=0 P(r)=0, a minimum), and at r= a/Z, which is a true maximum.
Problem: Calculate the expectation value of the
distance of the electron from the nucleus.
Solution: The expectation value is calculated as
<r> = Ú r P(r) dr =
(4Z3/a3) Ú r3
e-2rZ/a dr
It is the simplest to calculate this integral by substituting
first x = 2rZ/a. Then we obtain
<r> = (4Z3/a3) (a / 2Z)4
Ú x3 e-x dx = 3!
a/4Z= 3a / 2Z,
because Ú xn e-x
dx = n!.
Let us examine the first excited states of
hydrogen like atoms. They have an important role in atomic structure
and chemical bonds. The first excited states are characterized by
principal quantum number n=2. As we learned n=2 allows orbital
quantum numbers l=0 and 1 only. Thus there are four possible states
Y2,0,0(r, q,f) =
R2,0(r)Y00(q,f)
= (Z/2a0)3/2 (2 - Zr/a) e-rZ/2a (1/4p)1/2,
Y2,1,-1(r, q,f) =
R2,1(r)Y1-1(q,f) = (Z/2a0)3/2
(Zr/31/2a)
e-rZ/2a [ - 1/2(3/2p)1/2 sin q
e-if ],
Y2,1,0(r, q,f) =
R2,1(r)Y10(q,f) = (Z/2a0)3/2
(Zr/31/2a) e-rZ/2a
1/2(3/p)1/2 cos q ,
Y2,1,1(r, q,f) =
R2,1(r)Y11(q,f) = (Z/2a0)3/2
(Zr/31/2a) e-rZ/2a
[ 1/2(3/2p)1/2 sin q
e+if ].
The last factors are the appropriate spherical harmonics
everywhere. The general structure of the the radial wavefunctions,
Rn,l(r) is that they have an exponential factor
e-rZ/na and that is multiplied by an (n-1)st order
polynomial, which has a form of rl and another (n-l-1)st
order polynomial. Knowing this we can predict the form of the wave
function R2,1(r). It must have the form of
e-rZ/2a multiplied by rl=r, and a n-l-1=0 order
polynomial, 1. Thus, up to normalization
R2,1(r) µ r
e-rZ/2a
Rememeber that all these states are degenerate, they have the same
energy. It is interesting to plot these wavefunctions in a way
similar to how we did in for the ground state wavefunction.
instead of the wavefunctions Y2,1,-1(r, q,f)
and Y2,1,1(r, q,f) we may use their normalized linear
combinations. One can always do that with degenerate states. We will
define
Yz2,1(r, q,f) = Y2,1,0(r, q,f) = C z e-rZ/2a
Yx2,1(r, q,f) = [Y2,1,1(r, q,f) + Y2,1,-1(r, q,f)] / 21/2 =
C x e-rZ/2a
Yy2,1(r, q,f) = [Y2,1,1(r, q,f) - Y2,1,-1(r, f)] / i 21/2 =
C y e-rZ/2a
where C = (Z/2a0)3/2
(Z/31/2a0)
1/2(3/p)1/2
These equation follow from the identities
z = r cos q,
x = r sin q [ e+if +
e-if
]/ 2,
y = r sin q [ e+if -
e-if
]/ 2i.
This shows that the three wavefunctions above are have exactly the
same form except they are oriented along the three axes. Their
orientation is orthogonal to each other. By convention we say that
the electron described by these wave functions belong to the 2p
shell, 2 meaning n=2 and p meaning l=1. The 2s shell wave function
has n=2 and l=0.
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