المحاضرة 1 الفيزياء الذرية

Atomic model Atomic model
E. Rutherford E. Rutherford
(1871-1937)
? The size of the nucleus is very small
compared to the orbit radius
? Electrons orbit nuclei as planets orbit
around the Sun – the Coulomb force
? Problem: rotating electrons (due to
centripetal acceleration) emit EM waves –
bound to loose energy and fall on the
nucleus – no stable atoms…
N. Bohr (1885 N. Bohr (1885-1962)
nucleus (protons + neutrons) and a “cloud” of
negatively charged electrons that surround
the nucleus.
? The electrons do not emit radiation while
on the orbit Atoms are made of a positively charged

The Bohr model of the atom
• Both energy and angular momentum are
quantized
• Electrons move in specific “orbits”. Only
certain orbits are allowed. Orbits are
characterized by their energy a and angular
momentum
• The orbits do not have to be spherical
• Electrons can jump from one orbit to another
by absorbing or emitting energy.

Chapter 1
Angular Momentum
Understanding the quantum mechanics of angular momentum is fundamental in
theoretical studies of atomic structure and atomic transitions. Atomic energy
levels are classified according to angular momentum and selection rules for radiative
transitions between levels are governed by angular-momentum addition
rules. Therefore, in this first chapter, we review angular-momentum commutation
relations, angular-momentum eigenstates, and the rules for combining
two angular-momentum eigenstates to find a third. We make use of angularmomentum
diagrams as useful mnemonic aids in practical atomic structure calculations.
A more detailed version of much of the material in this chapter can
be found in Edmonds (1974).
1.1 Orbital Angular Momentum - Spherical
Harmonics
Classically, the angular momentum of a particle is the cross product of its position
vector r = (x, y, z) and its momentum vector p = (px, py, pz):
L = r × p.
The quantum mechanical orbital angular momentum operator is defined in the
same way with p replaced by the momentum operator p ? ?i¯h?. Thus, the
Cartesian components of L are
Lx =
h¯
i
³
y
?
?z ? z
?
?y
´
, Ly =
h¯
i
³
z
?
?x ? x
?
?z
´
, Lz =
h¯
i
³
x
?
?y ? y
?
?x
´
. (1.1)
With the aid of the commutation relations between p and r:
[px, x] = ?i¯h, [py, y] = ?i¯h, [pz, z] = ?i¯h, (1.2)
1
2 CHAPTER 1. ANGULAR MOMENTUM
one easily establishes the following commutation relations for the Cartesian
components of the quantum mechanical angular momentum operator:
LxLy ? LyLx = i¯hLz, LyLz ? LzLy = i¯hLx, LzLx ? LxLz = i¯hLy.
(1.3)
Since the components of L do not commute with each other, it is not possible to
find simultaneous eigenstates of any two of these three operators. The operator
L
2 = L
2
x + L
2
y + L
2
z
, however, commutes with each component of L. It is, therefore,
possible to find a simultaneous eigenstate of L
2 and any one component of
L. It is conventional to seek eigenstates of L
2 and Lz.
1.1.1 Quantum Mechanics of Angular Momentum
Many of the important quantum mechanical properties of the angular momentum
operator are consequences of the commutation relations (1.3) alone. To
study these properties, we introduce three abstract operators Jx, Jy, and Jz
satisfying the commutation relations,
JxJy ? JyJx = iJz , JyJz ? JzJy = iJx , JzJx ? JxJz = iJy . (1.4)
The unit of angular momentum in Eq.(1.4) is chosen to be ¯h, so the factor of
¯h on the right-hand side of Eq.(1.3) does not appear in Eq.(1.4). The sum of
the squares of the three operators J
2 = J
2
x + J
2
y + J
2
z
can be shown to commute
with each of the three components. In particular,
[J
2
, Jz] = 0 . (1.5)
The operators J+ = Jx +iJy and J? = Jx ?iJy also commute with the angular
momentum squared:
[J
2
, J±] = 0 . (1.6)
Moreover, J+ and J? satisfy the following commutation relations with Jz:
[Jz, J±] = ±J± . (1.7)
One can express J
2
in terms of J+, J? and Jz through the relations
J
2 = J+J? + J
2
z ? Jz , (1.8)
J
2 = J?J+ + J
2
z + Jz . (1.9)
We introduce simultaneous eigenstates |?, mi of the two commuting operators
J
2 and Jz:
J
2
|?, mi = ? |?, mi, (1.10)
Jz|?, mi = m |?, mi, (1.11)
and we note that the states J±|?, mi are also eigenstates of J
2 with eigenvalue
?. Moreover, with the aid of Eq.(1.7), one can establish that J+|?, mi and
J?|?, mi are eigenstates of Jz with eigenvalues m ± 1, respectively:
JzJ+|?, mi = (m + 1) J+|?, mi, (1.12)
JzJ?|?, mi = (m ? 1) J?|?, mi. (1.13)
1.1. ORBITAL ANGULAR MOMENTUM - SPHERICAL HARMONICS 3
Since J+ raises the eigenvalue m by one unit, and J? lowers it by one unit,
these operators are referred to as raising and lowering operators, respectively.
Furthermore, since J
2
x + J
2
y
is a positive definite hermitian operator, it follows
that
? ? m2
.
By repeated application of J? to eigenstates of Jz, one can obtain states of arbitrarily
small eigenvalue m, violating this bound, unless for some state |?, m1i,
J?|?, m1i = 0.
Similarly, repeated application of J+ leads to arbitrarily large values of m, unless
for some state |?, m2i
J+|?, m2i = 0.
Since m2
is bounded, we infer the existence of the two states |?, m1i and |?, m2i.
Starting from the state |?, m1i and applying the operator J+ repeatedly, one
must eventually reach the state |?, m2i; otherwise the value of m would increase
indefinitely. It follows that
m2 ? m1 = k, (1.14)
where k ? 0 is the number of times that J+ must be applied to the state |?, m1i
in order to reach the state |?, m2i. One finds from Eqs.(1.8,1.9) that
?|?, m1i = (m2
1 ? m1)|?, m1i,
?|?, m2i = (m2
2 + m2)|?, m2i,
leading to the identities
? = m2
1 ? m1 = m2
2 + m2, (1.15)
which can be rewritten
(m2 ? m1 + 1)(m2 + m1) = 0. (1.16)
Since the first term on the left of Eq.(1.16) is positive definite, it follows that
m1 = ?m2. The upper bound m2 can be rewritten in terms of the integer k in
Eq.(1.14) as
m2 = k/2 = j.
The value of j is either integer or half integer, depending on whether k is even
or odd:
j = 0,
1
2
, 1,
3
2
, · · · .
It follows from Eq.(1.15) that the eigenvalue of J
2
is
? = j(j + 1). (1.17)
The number of possible m eigenvalues for a given value of j is k + 1 = 2j + 1.
The possible values of m are
m = j, j ? 1, j ? 2, · · · , ?j.
4 CHAPTER 1. ANGULAR MOMENTUM
x
y
z
? r
?
Figure 1.1: Transformation from rectangular to spherical coordinates.
Since J? = J
†
+, it follows that
J+|?, mi = ?|?, m + 1i, h?, m|J? = ?
?
h?, m + 1|.
Evaluating the expectation of J
2 = J?J+ +J
2
z +Jz in the state |?, mi, one finds
|?|
2 = j(j + 1) ? m(m + 1).
Choosing the phase of ? to be real and positive, leads to the relations
J+|?, mi =
p
(j + m + 1)(j ? m)|?, m + 1i, (1.18)
J?|?, mi =
p
(j ? m + 1)(j + m)|?, m ? 1i. (1.19)
1.1.2 Spherical Coordinates - Spherical Harmonics
Let us apply the general results derived in Section 1.1.1 to the orbital angular
momentum operator L. For this purpose, it is most convenient to transform
Eqs.(1.1) to spherical coordinates (Fig. 1.1):
x = r sin ? cos ?, y = r sin ? sin ?, z = r cos ?,
r =
p
x
2 + y
2 + z
2, ? = arccos z/r, ? = arctan y/x.
In spherical coordinates, the components of L are
Lx = i¯h
µ
sin ?
?
??
+ cos ? cot ?
?
??
¶
, (1.20)
Ly = i¯h
µ
? cos ?
?
??
+ sin ? cot ?
?
??
¶
, (1.21)
Lz = ?i¯h
?
??
, (1.22)
1.1. ORBITAL ANGULAR MOMENTUM - SPHERICAL HARMONICS 5
and the square of the angular momentum is
L
2 = ?¯h
2
µ
1
sin ?
?
??
sin ?
?
??
+
1
sin2
?
?
2
??2
¶
. (1.23)
Combining the equations for Lx and Ly, we obtain the following expressions for
the orbital angular momentum raising and lowering operators:
L± = ¯he±i?µ
±
?
??
+ i cot ?
?
??
¶
. (1.24)
The simultaneous eigenfunctions of L
2 and Lz are called spherical harmonics.
They are designated by Ylm(?, ?). We decompose Ylm(?, ?) into a product of a
function of ? and a function of ?:
Ylm(?, ?) = ?l,m(?)?m(?).
The eigenvalue equation LzYl,m(?, ?) = ¯hmYl,m(?, ?) leads to the equation
?i
d?m(?)
d? = m?m(?), (1.25)
for ?m(?). The single valued solution to this equation, normalized so that
Z 2?
0
|?m(?)|
2
d? = 1 , (1.26)
is
?m(?) =
1
?
2?
e
im?
, (1.27)
where m is an integer. The eigenvalue equation L
2Yl,m(?, ?) = ¯h
2
l(l +
1)Yl,m(?, ?) leads to the differential equation
µ
1
sin ?
d
d? sin ?
d
d? ?
m2
sin2
?
+ l(l + 1)¶
?l,m(?) = 0 , (1.28)
for the function ?l,m(?). The orbital angular momentum quantum number l
must be an integer since m is an integer.
One can generate solutions to Eq.(1.28) by recurrence, starting with the
solution for m = ?l and stepping forward in m using the raising operator L+,
or starting with the solution for m = l and stepping backward using the lowering
operator L?. The function ?l,?l(?) satisfies the differential equation
L+?l,?l(?)??l(?) = ¯h??l+1(?)
µ
?
d
d? + l cot ?
¶
?l,?l(?) = 0 ,
which can be easily solved to give ?l,?l(?) = c sinl
?, where c is an arbitrary
constant. Normalizing this solution so that
Z ?
0
|?l,?l(?)|
2
sin ?d? = 1, (1.29)
6 CHAPTER 1. ANGULAR MOMENTUM
one obtains
?l,?l(?) =
1
2
l
l!
r
(2l + 1)!
2
sinl
? . (1.30)