10
10.1
Angular Momentum II
Central force and pseudopotential
Our goal is now to solve the Hydrogen atom completely at the nonrelativistic level (& for an infinitely massive proton, an assumption we will
soon relax), where the Hamiltonian is
e2
h̄2∇2
H = −
−
2m
4π²0r
Griffiths follows a mathematical strategy in Ch. 4, using the technique of separation of variables. He separates the variables by assuming
ψ(r, θ, φ) = R(r)Y (θ, φ), and then after substituting in finds that the
Schoedinger equation can be written in the form above Eq. (4.16) in his
book. The angular momentum is then introduced purely mathematically,
as a separation constant `(` + 1) which he pulls out of a hat, in (Eqs.
4.16-4.17). I would like to pursue a more physical strategy based on the
angular momentum results we obtained last week.
The idea is, after exploring properties of L̂, we now want to express
H in terms of L̂ in order to explicitly classify eigenfunctions by angular
momentum eigenvalues. I will skip tedious calculation (however, it’s good
for cultural background— you can read about it in Peebles, Quantum
Mechanics, p. 141.). Remember what you are trying to do is express L̂2
in terms of ∇2, but L̂2 = L̂2x + L̂2y + L̂2z , and
L̂x = y p̂z − z p̂y
L̂y = z p̂x − xp̂z
L̂z = xp̂y − y p̂x,
1
(1)
(2)
so in principle you can work it out. The final result is
−L̂2 1 ∂ 2
∇ = 2 2+
r
r ∂r2
h̄ r
2
(3)
(Note meaning of 2nd term—r−1(∂ 2/∂r2)rψ = r−1(∂ 2/∂r2)(rψ)).
So kinetic energy operator in Hamiltonian is
h̄2 1 ∂ 2
L̂2
h̄2 2
r+
− ∇ =−
2m
2m r ∂r2
2mr2
and full Hamiltonian for central force problem is
2 1 ∂2
2
h̄
L̂
H = − 2m r ∂r2 r + 2mr2 + V (r).
(4)
(5)
1st want to solve t-ind. S.-eqn. Hψ = Eψ, have already argued should
be able to find simultaneous e’fctns. of H, L̂2, L̂z . Label as ψn`m with
Hψn`m = Enψn`m,
L̂2ψn`m = h̄2`(` + 1)ψn`m
L̂z ψn`m = mh̄ψn`m
S. eqn. is then (using fact that ψn`m are e’fctns of L̂2)
h̄2`(` + 1)2
h̄2 1 ∂ 2
−
rψn`m +
ψn`m + V (r)ψn`m = Enψn`m
(6)
2m r ∂r2
2mr2
Convenient to multiply (4) by r, introduce radial wave fctn. un`m ≡
rψn`m. Then eqn. reads


2
h̄2 ∂ 2
 h̄ `(` + 1)
un`m + 
+ V (r) un`m = Enun`m
−
2
2
2mr
| 2m ∂r
{z
}
|
1D K.E.
{z
(7)
}
effective 1D potential
which looks like a 1D problem in the radial variable r for a particle moving
in an effective potential depending on `.
2
Reminder: similar equation found in classical mechanics in polar coordinates (example in 2D for illustration):
1
m 2
H = mv 2 + V (r) =
(ṙ + r2θ̇2) + V
2
2


2
1
 L̂


+
V
= mṙ2 + 

2
2
2mr
10.2
H-atom bound states
Electron sees potential V = e2/r from nucleus:


2
e2 
h̄2 ∂ 2
 h̄ `(` + 1)
u+
−  u = Eu = −Eu
−
2m ∂r2
2mr2
r
(8)
with E = −E the binding energy of the electron as usual. Rewrite as
∂ 2u `(` + 1)
2me2
2mE
−
u
+
u
=
u
(9)
∂r2
r2
h̄2r
h̄2
This is just “radial S.-eqn.” we wrote down before assuming spherically
symmetric solutions (Eq. 4-15 et seq.) only now including `(` + 1) term!
Recall trick leading to solution back then was to write
√
−r 2mE/h̄2
u(r) = f (r)e
(10)
which we guessed because 1) we wanted solns. localized around the origin,
and 2) could get some terms to cancel...
Substitute, find
2
2
v
u
u
u
t
2me
2mE df
∂ f `(` + 1)
−
f
+
=0
f
−
2
∂r2
r2
h̄2r
h̄2 dr
(11)
Now look for series soln.:
f=
X
p
3
Aprp
(12)
Substitute, require coefficient of each power of r vanish, find
√


2
2p
2mE
2me
[p(p + 1) − `(` + 1)]Ap+1 = 
− 2  Ap
h̄
h̄
(13)
A` is zero, since factor on lhs vanishes and [ ] on right is nonzero. Then
all Ap are zero for p ≤ `. (Take p + 1 = `, then you find A`−1 = 0, iterate
again...)
If series does not terminate, asymptotic value of ratio
√
Ap+1
2 2mE 1
∼
Ap
h̄
p
√
which has soln. Ap+1 ∼ (2 2mE/h̄)p/p!, meaning
f∼
X
p
p
Apr ∼ e
(14)
√
2 2mE r
h̄
(15)
√
which is no good since u ≡ f e
− 2mE
h̄ r
blows up at r → ∞.
But can escape this conundrum by assuming termination if for some integer n > `
√
2n 2mE
2me2
= 2
(16)
h̄
h̄
( Then all Ap with p > n will vanish due to (11).) This gives energy
eigenvalues as before:
e4 m 1
(17)
En = −E = 2 2 , n = 1, 2, 3 · · · > `
2h̄ n
and eigenfunctions (neglect angular part for now!)
√
ψn`m
2m|En |
un`m
r
`
n−1 −
h̄
∼
e
= [A
r
+
·
·
·
A
r
]
`+1
n
{z
}
|
r
related to Laguerre polynomial
Radial part of eigenfunction Rn`(r) different for each `!
4
(18)
(19)
Now for each ` there are 2` + 1 degenerate orthogonal eigenfctns. corresponding to different m’s, i.e. ∝ eimφ. Examples:
ψ100
ψ200
ψ211
ψ210
ψ21−1
∼
∼
∼
∼
∼
e−r/a0
(1 − r/2a0)e−r/2a0
r sin θeiφe−r/2a0
r cos θe−r/2a0
r sin θe−iφe−r/2a0
(20)
(21)
(22)
(23)
(24)
Remarks:
1. Eigenfunctions of L̂2 are Legendre functions (sometimes called associated Leg. polynomials) P`m(cos θ). For m = 0, reduce to familiar
Legendre polynomials P`(cos θ). Won’t belabor definitions now, can
look up anywhere.
2. Recall e’fctns of L̂z are ∼ eimφ.
3. Simultaneous e’fctns of both L̂2 and L̂z easy to construct, just multiply
two together, get spherical harmonics Y`m ≡ P`m(cos θ)eimφ. Note
orthonormality:
Z
∗
dΩ Ylm
(θ, φ)Yl0m0 (θ, φ) = δ``0 δm,m0
(25)
4. 1st few:
1
Y00 = √
4π
√
√
√3
Y1,−1 = − 8π sin θe−iφ, Y1,0 = − 4π3 cos θ, Y1,1 = − 8π3 sin θeiφ
√ 15 2iφ 2
√ 15 iφ
Y22 = 32π
e sin θ,
Y21 = − 8π
e sin θ cos θ,
√5
Y20 = 16π (3 cos2 θ − 1),
···
5
5. Full normalized e’fctns ψn`m for H-atom problem are radial fctns. (see
Eq. 17) times spherical harmonics,
ψn`m(r, θ, φ) = Rn`(r)Y`m(θ, φ)
Recap: nonrelativistic hydrogen energy levels
• 1s. (Ground state) n=1,`=0, E=-13.6 eV
spectroscopic notation: (s=“sharp”)
• 2s. (1st excited state) n=2, ` = 0, E=13.6 eV/(22)
• 2p. n=2,` = 1. Degeneracy = 2`+1 = 3.
• 3s. n=3, ` = 0
spectroscopic notation: (p=“principal”)
• 3p. n=3, ` = 1 Degeneracy =3.
• 3d. n=3, ` = 2 Degeneracy =5.
spectroscopic notation: (d=“diffuse”).
...
6
(26)
10.3
Two body problem
Need to ask what happens if proton allowed to have finite mass, also
consider other 2– or more–particle systems where masses are similar. For
now stick to 2 particles, think of electron and proton in H atom.
Write down single wave function to describe entire two-particle system,
fctn of electron position re and proton position rp:
ψ = ψ(re, rp, t)
(27)
related to joint probability of finding the electron in d3re at re simultaneously with proton in d3rp at rp:
dP = |ψ(re, rp, t)|2d3rpd3re
(28)
with normalization condition
Z
|ψ|2d3rpd3re = 1
(29)
which we again write as (ψ, ψ) = 1.
Total momentum:
∂
∂
− ih̄
∂re
∂rp
(30)
p̂2p
p̂2e
e2
H =
+
+−
2me 2mp
|rp − re|
(31)
P = pe + pp = −ih̄
Total energy:
10.4
Momentum conservation in 2-body system
Check commutators:
•
[pe, H] = [pe, V ] = −ih̄
7
∂
V
∂re
dV ∂
r
dr ∂re
dV re − rp
= −ih̄
dr
r
= −ih̄
(32)
and similarly
•
[pp, H] = −ih̄
dV rp − re
dr
r
(33)
so total momentum P = pe + pp commutes with H
[P, H] = 0
(34)
Consequences:
1. Can find complete set of eigenstates of energy and total momentum.
2. Momentum conserved: if ψ is eigenfctn. of P at time t, still an
eigenstate at later time t0.
3. Total p is generator of translations in same sense that total ang.
mom. was generator of rotations. If ψ 0(re, rp, t) is defined to be the
wave function of a system described by ψ shifted in space by the
displacement ~δ, can show (See Peebles p. 121)
~
ψ 0 = U ψ,
U = e−iδ·p/h̄
(35)
where again operator U is unitary, the exponentiation of a Hermitian
operator
10.5
Reduction from 2-body to 1-body problem.
Define as in classical mechanics:
8
Total mass
M = me + mp
me
mp
Center of mass position R =
re +
rp
M
M
Relative position
r = re − rp
(36)
(37)
(38)
or invert:
re = R +
mp
r
M
(39)
rp = R −
me
r
M
(40)
Change variables in S.-eqn. (Hamiltonian Eq. (27)) from re, rp to R, r:
¯
¯
¯
∂ψ ¯¯¯
∂ψ ∂re ¯¯¯
∂ψ ∂rp ¯¯¯
(41)
¯ =
¯ +
¯
∂R ¯r ∂re ∂R ¯r ∂rp ∂R ¯r
and note ∂re/∂R = ∂rp/∂R = 1. So
¯
∂ψ ¯¯¯
∂ψ ∂ψ
+
, or
=
¯
∂R ¯r
∂re ∂rp
(42)
∂
∂
∂
=
+
∂R
∂re ∂rp
Similarly
(43)
∂
mp ∂
me ∂
=
−
.
∂r
M ∂re M ∂rp
(44)
Now put (mostly old definitions):
pe = −ih̄
∂
∂re
pp = −ih̄
P = −ih̄
∂
∂R
p = −ih̄
9
∂
∂rp
∂
∂r
(45)
(46)
So have same relations as in classical mechanics:
P = pe + pp Center of mass momentum
mp
me
p =
pe −
pp relative momentum
M
M
(47)
(48)
or inverted
pe =
me
P+p
M
(49)
pp =
mp
P−p
M
(50)
Rewrite kinetic energy operator in terms of center of mass and relative
momentum:


p̂2p
p̂2e
P̂ 2
1 1
1  2

 p̂
+
=
+ 
+
2me 2mp
2M 2 me mp
p̂2
P̂ 2
+
=
2M 2µ
(51)
where µ is the reduced mass
memp
(52)
me + mp
Since we normally are not interested in the center of mass motion, discard,
and write for 2-body problem (e.g. H-atom with finite proton mass):
p̂2
H=
+ V (r)
(53)
2µ
µ=
exactly as in classical mechanics.
10.6
Addition of 2 angular momenta.
We now ask the question, if we have two particles with different angular
momenta, what are the allowed eigenvalues of the total angular momentum? This situation might arise for two particles with intrinsic angular
10
momentum (spin), such as the two quarks in a pion, or it might be a
question we would ask about two electrons in different angular momentum states of a multielectron atom. In any case, suppose there are two
particles, one with angular momentum operator L(1) and one with L(2).
The total angular momentum operator is
L = L(1) + L(2),
and we assume that since particles 1 and 2 are distinct particles, any operator referring to 1 commutes with any operator referring to 2. So for example we have [Lx(1), Ly (1)] = ih̄Lz (1), but [Lx(1), Ly (2)] = 0. Then what
are the possible sets of commuting observables we can construct whose
eigenvalues we can use to label our states with? One obvious choice is
L2(1), Lz (1), L2(2), Lz (2), whose eigenvalues we would call `1, m1, `2, m2,
as usual. But we can also show that
[L2, Lz ] = 0
·
¸
2
2
L , L (α) = 0
·
2
¸
Lz , L (α)
= 0
but
·
¸
L2, Lz (α) 6= 0,
α = 1, 2, which suggests that there is another set of four commuting operators we could chose with which to label our states, namely L2, Lz , L2(1)
and L2(2). Note that the total angular momentum operators for each particle L2(α) are members of both sets. For a given pair of quantum numbers
`1, `2, however, the two sets differ, because using the first set, the states
can be classified in addition by m1, m2, and we will write |m1m2`1`2i,
whereas using the second set, states can be classified according to `, m,
where `, m are the total angular momentum quantum number and total
z-component quantum number, i.e. the operators L2 and Lz have the
following effect on the states |`m`1`2i:
11
L2|`m`1`2i = h̄2`(` + 1)|`m`1`2i
Lz |`m`1`2i = h̄m|`m`1`2i,
and of course still
L2(1)|`m`1`2i = h̄2`1(`1 + 1)|`m`1`2i
L2(2)|`m`1`2i = h̄2`2(`2 + 1)|`m`1`2i.
Note however that |`m`1`2i is not an eigenstate of Lz (α), because L2
doesn’t commute with it.
We have two different ways of labelling the eigenstates of angular momentum, neither of which is preferred. How are the “new” quantum numbers `, m related to `1, `2, m1, m2? Well we have Lz = Lz (1) + Lz (2),
so certainly m = m1 + m2. The relation between ` and `1 and `2 is not
so simple, but the answer is contained in what is sometimes called the
triangle rule:
|`1 − `2| ≤ ` ≤ `1 + `2.
I will not derive this here, but you will derive it in part in the homework.
Since both the |`m`1`2i and the |m1m2`1`2i represent a complete set
of states, it must be that we can always express any given state in either
representation in terms of linear combinations of the other. For example,
we can write
|m1m2`1`2i =
or
|`m`1`2i =
`1X
+`2
`=|`1 −`2 |
h`m1 + m2`1`2|m1m2`1`2i|`m1 + m2`1`2i
X
m1 +m2 =m
hm1m2`1`2|`m`1`2i|m1m2`1`2i.
The amplitudes (just numbers) hm1m2`1`2|`m`1`2i are called ClebschGordon coefficients, and are tabulated in Griffiths on p. 188. A concrete
example is for `1 = 2, `2 = 1,
12
v
u
u
u
t
1
3
1
|3 0 2 1i = √ |1 − 1 2 1i + |0 0 2 1i + √ | − 1 1 2 1i,
5
5
5
where (careful!) the left side of the equation is written in `m`1`2 notation, whereas the kets on the right are in the m1m2`1`2 notation. Compare
directly with Griifiths’ Eq. below (4.185). It’s the same equation in different notation. A third notation is given to you in the homework! Choose
whichever seems most natural to you.
13