24 The hydrogen atom
operators can be expressed in polar coordinates as:l+=eiφ(
∂
∂θ+icotθ∂
∂φ)
,
l−=e−iφ(
−
∂
∂θ
+icotθ∂
∂φ)
.
(2.9)
The operatorl+transforms a function with magnetic quantum number
minto another angular momentum eigenfunction that has eigenvalue
m+1. Thusl+is called theraising operator.^6 Thelowering operatorl−(^6) The raising operator contains the fac-
tor eiφ,sothatwhenitactsonaneigen-
function of the formY ∝Θ(θ)eimφ
the resulting function l+Y contains
ei(m+1)φ.Theθ-dependent part of this
function is found below.
changes the magnetic quantum number in the other direction,m→m−
- It is straightforward to prove these statements and other properties
(^7) These properties follow from the com- of these operators;^7 however, the purpose of this section is not to present
mutation relations for angular momen-
tum operators (see Exercise 2.1).
the general theory of angular momentum but simply to outline how to
find the eigenfunctions (of the angular part) of the Schr ̈odinger equation.
Repeated application of the raising operator does not increasem
indefinitely—for each eigenvaluebthere is a maximum value of the mag-
netic quantum number^8 that we shall calll, i.e.mmax=l. The raising
(^8) This statement can be proved rigor-
ously using angular momentum opera-
tors, as shown in quantum mechanics
texts.
operator acting on an eigenfunction withmmaxgives zero since by def-
inition there are no eigenfunctions withm>mmax.Thussolvingthe
equationl+Y= 0 (Exercise 2.11) we find that the eigenfunctions with
mmax=lhave the form
Y∝sinlθeilφ. (2.10)
Substitution back into eqn 2.5 shows that these are eigenfunctionsl^2 with
eigenvalueb=l(l+1), andlis theorbital angular momentum quantum
number. The functionsYl,m(θ, φ) are labelled by their eigenvalues in the
conventional way.^9 Forl=0onlym=0existsandY 0 , 0 is a constant
(^9) The dubious reader can easily check
thatl+Yl,l= 0. It is trivially obvious
thatlzYl,l=lYl,l,wherem=lfor this
function.
with no angular dependence. Forl= 1 we can find the eigenfunctions
by starting from the one withl=1=m(in eqn 2.10) and using the
lowering operator to find the others:
Y 1 , 1 ∝sinθeiφ,
Y 1 , 0 ∝l−Y 1 , 1 ∝cosθ,
Y 1 ,− 1 ∝l−Y 1 , 0 ∝sinθe−iφ.
This gives all three eigenfunctions expected forl=1.^10 Forl=2this
(^10) l−Y 1 ,− 1 =0andm=−1isthelow-
est eigenvalue oflz. Proportional signs
have been used to avoid worrying about
normalisation; this leaves an ambiguity
about the relative phases of the eigen-
functions but we shall choose them in
accordance with usual convention.
procedure gives
Y 2 , 2 ∝ sin^2 θei2φ,
..
.
Y 2 ,− 2 ∝ sin^2 θe−i2φ.
These are the five eigenfunctions withm=2, 1 , 0 ,− 1 ,−2.^11 Normalised
(^11) The relationYl,−m =Y∗
l,mshows
that, ifmmax =l,thenmmin=−l.
Between these two extremes there are
2 l+ 1 possible values of the magnetic
quantum numbermfor eachl.Note
that the orbital angular momentum
quantum numberlis not the same as
the length of the angular momentum
vector (in units of). Quantum me-
chanics tells us only that the expecta-
tion value of the square of the orbital
angular momentum isl(l+ 1), in units
of^2. The length itself does not have a
well-defined value in quantum mechan-
ics and it does not make sense to re-
fer to it. When people say that an
atom has ‘orbital angular momentum
of one, two, etc.’, strictly speaking they
mean that the orbital angular momen-
tum quantum numberlis 1, 2, etc.
angular functions are given in Table 2.1.
Any angular momentum eigenstate can be found from eqn 2.10 by