量子力学
Quantum mechanics
School of Physics and Information Technology
QUANTUM MECHANICS IN
THREE DIMENSIONS
Chapter 4
4.1 Schrödinger Equation in Spherical Coordinates 131
4.2 The Hydrogen atom 145
4.3 Angular Momentum
In the classical theory of central forces, energy and angular momentum are the fundamental conserved quantities, and it plays a significant role in the quantum theory. Now we consider the angular momentum in quantum theory. As we have seen, the stationary states of the hydrogen atom are labeled by three quantum numbers: n, l and m. The n is the principle quantum number which determines the energy of the state, and, l and m, are related to the orbital angular momentum.
Classically, the angular momentum of a particle (with respect to the origin) is given by the formula
which is to say, in components,
,
x z y
L
=
yp
−
zp
.
L
= ×
r
p
.
z y x
L
=
xp
−
yp
,
y x z
L
=
zp
−
xp
In the following section we’ll obtain the eigenvalues of the angular momentum by a purely algebraic technique reminiscent of the one we used in chapter 2 to get the allowed energies of the harmonic oscillator; it is all based on the clever exploitation of commutation relations. After that we will turn to the more
difficult problem of determining the eigenfunctions.
The corresponding quantum operators are obtained by the standard prescription
,
x
i
p
x∂
∂
→
,
y
i
p
y∂
∂
→
.
z
i
p
z∂
∂
→
difficult problem of determining the eigenfunctions.
The operators Lx and Ly do not commute; in fact
4.3.1 Eigenvalues
,
,
x y z y x z
L L
=
yp
−
zp zp
−
xp
[
yp zp
z,
x] [
yp xp
z,
z]
zp zp
y,
xzp xp
y,
z.
=
−
−
+
[ ,
x p
x] [ ,
=
y p
y] [ ,
=
z p
z]
=
i
Of course, we can calculate [Ly,Lz] or [Lz,Lx] as well, but there is no need to From the canonical commutation relations
and the other position and momentum components commute each other. So
,
x y
L L
=
yp
x[
p z
z,
]
+
xp
y[
z p
,
z]
=
i
(
yp
x−
xp
y)
=
i L
z.
y z z x
calculate these separately——we can get them immediately by cyclic permutation of the indices (x y, y z, z x):
These are the fundamental commutation relations for angular momentum; every-thing else follows from them.
,
;
x y z
Notice that Lx, Ly and Lz are incompatible observables. According to the generalized uncertainty principle,
2 2 2
1
[ , ]
2
A B
A B
i
σ σ
≥
2 2 2
1
[
,
]
2
x y
L L
L L
x yi
σ σ
≥
2 2
2 2 2
1
L L
i L
zL
zσ σ
≥
=
Or
It would therefore be futile to look for states that are simultaneously eigenfunctions of Lx and Ly. So does for other components.
2
4
x y
L L
i L
zL
zi
σ σ
≥
=
.
2
zL
Lx
σ
y≥
L
On the other hand, the square of the total angular momentum,
2 2 2 2
[
L L
,
x] [
=
L L
x,
x] [
+
L L
y,
x] [
+
L L
z,
x]
does commutate with the three components of L, for example,
2 2 2 2
,
x y z
L
=
L
+
L
+
L
0
L L L
y[
y,
x] [
L L L
y,
x]
yL L L
z[
z,
x] [
L L L
z,
x]
z= +
+
+
+
Similarly, it follows, of course, that
0
L L L
y[
y,
x] [
L L L
y,
x]
yL L L
z[
z,
x] [
L L L
z,
x]
z= +
+
+
+
(
) (
)
(
) (
)
y z z y z y y z
L
i L
i L L
L i L
i L
L
=
−
+ −
+
+
0
=
2 2 2
So L2 is compatible with each component of L, and we can hope to find
simultaneous eigenstates of L2 and (say) L z:
We will use a “Ladder operator” technique, very similar to the one we applied to the harmonic oscillator back in Section 2.3.1. Let
2
and
z.
L f
=
λ
f
L f
=
µ
f
(2) Introduce ladder operators L+, L-to determine eigenvalues :
so
.
x y
L
±≡
L
±
iL
The commutator with Lz is
[
L L
z,
±] [
=
L L
z,
x]
±
i L L
z,
y=
i L
y± −
i
(
i L
x)
= ±
(
L
x±
iL
y)
And, of course,
If f is an eigenfunction of L2and Lz , so
so L f is an eigenfunction of L2, with the same eigenvalue , and
2
,
0
L L
±=
2
;
z.
L f
=
λ
f
L f
=
µ
f
(
)
( )
( )
(
)
2 2
,
L L f
±=
L
±L f
=
L
±λ
f
=
λ
L f
±so L±f is an eigenfunction of L2, with the same eigenvalue , and
(
)
(
)
z z z z
L L f
±=
L L f
±−
L L f
±+
L L f
±= ±
L f
±+
L
±µ
f
(
)
(
) (
)(
)
z
L L f
±= ±
L f
±+
L
±µ
f
=
µ
±
L f
±Therefore, as
(
) (
)(
)
,
z
L L f
+=
µ
+
L f
+L L f
z(
−) (
=
µ
−
)(
L f
−)
we call L+ the “raising” operator, because it increases the eigenvalue of Lz by , and L- the “lowering” operator, because it lowers the eigenvalue by .
For a given value of , then, we obtain a “ladder” of states, with each “rung” separated from its neighbors by one unit of in the eigenvalue of Lz(see Figure). To ascend the ladder we apply the raising operator, and to descend, the lowering operator. But this process cannot go on forever: Eventually we’re going to reach a state for which the z-component exceeds the total, and that cannot be.
0
t
L f
+=
There must exist a “top rung”, ft, such that
Let l be the eigenvalue of Lz at this top rung (suppose):
,
z t t
L f
=
l
⋅
f
andL f
2 t=
λ
f
t.
µ
µ
+
µ
−
2
µ
+
L f+ f
2
L f+
L f−
l
⋅
l
⋅
t
f
b
Now we calculate the following operators as
(
)(
)
2 2(
)
x y x y x y x y y x
L L
±=
L
±
iL
L
iL
=
L
+
L
i L L
−
L L
(
)
2 2 2 2
z z z z
L
L
i i L
L
L
L
=
−
=
−
±
or, putting it in the other way around,
2 2
z z
L
=
L L
±+
L
L
2
.
L f
=
λ
f
It follows that(
)
(
)
2 2 2 2 2 2
0
(
1) ,
t z z t t t
L f
=
L L
− ++
L
+
L
f
=
+
l
+
l f
=
l l
+
f
and hence
2
.
t t
L f
=
λ
f
2
(
1).
l l
λ
=
+
0
b
L f
−=
Meanwhile, there is also a “bottom rung”, fb, such that
and
L f
2 b=
λ
f
b.
Let be the eigenvalue of l Lz at this bottom rung (suppose):
Then we have
,
b b
z
f
l
f
L
=
⋅
(
)
(
)
2 2 2 2 2 2
0
(
1)
,
b z z b b b
L f
=
L L
+ −+
L
−
L
f
=
+
l
−
l
f
=
l l
−
f
and therefore
2
(
1).
l l
λ
=
−
Comparing with above equation, we see that
λ
=
2l l
(
+
1)
(
1)
(
1)
l
⋅
)
1
(
−
−
l
)
1
(
l
−
) 1 ( −
− l l
f
) 1 (l− l
f
m l
f
l l
f
m
Evidently the eigenvalues of Lz are =m , where m goes from –l to +l in N integer steps. In particular, it follows that l=-l+N, and hence l=N/2, so l must be an integer or a half-integer.
The eigenfunctions are characterized by the number l and m:
;
)
1
(
22 m
l m
l
l
l
f
f
L
=
+
L
zf
lm=
m
f
lm,
where
l
=
0
,
1
/
2
,
1
.
3
/
2
,
;
m
=
−
l
,
−
l
+
1
,
,
l
−
1
,
l
.
l
−
ll
f
−)
1
(
−
−
l
lFor a given value of l, there are 2l+1 different values of m.
Some people like to illustrate this result with the diagram in Figure (l=2). The
arrows are supposed to represent possible angular momenta—in units of they all
have the same length
l
(
l
+
1
)
=
6
=
2
.
45
,
z
L
x
L
y
L
L
1
1
− 0
2
4.3.2 Eigenfunctions
First of all we need to rewrite Lx, Ly and
Lz in spherical coordinates. Now, as
(
−
∇
)
=
−
×
∇
,
×
=
×
=
r
p
r
i
i
r
L
Prove:
and the gradient ▽▽▽▽, in spherical coordinates, is:
m l m l
Y
f
→
meanwhile,;
sin
1
ˆ
1
ˆ
ˆ
φ
θ
φ
θ
θ
∂
∂
+
∂
∂
+
∂
∂
=
∇
r
r
r
r
,
ˆ
r
r
r
=
soBut as
(
r
ˆ
×
r
ˆ
)
=
0
,
( )
r
ˆ
×
θ
ˆ
=
φ
ˆ
,
( )
r
ˆ
×
φ
ˆ
=
−
θ
ˆ
,
and hence
(
)
( )
( )
.
sin
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
∂
∂
×
+
∂
∂
×
+
∂
∂
×
−
=
φ
θ
φ
θ
θ
r
r
r
r
r
r
i
L
.
1
ˆ
ˆ
∂
∂
−
∂
∂
−
=
φ
θ
θ
θ
φ
i
,
ˆ
ˆ
ˆ
L
j
L
k
i
L
L
=
x+
y+
zAs in cartesian coordinates
.
sin
∂
−
∂
−
=
φ
θ
θ
θ
φ
i
we use the relations
Thus
∂
∂
−
+
−
∂
∂
+
−
=
φ
θ
θ
φ
θ
φ
θ
θ
φ
φ
sin
1
)
ˆ
sin
ˆ
sin
cos
ˆ
cos
(cos
)
ˆ
cos
ˆ
sin
(
i
j
i
j
k
i
L
i
i
ˆ
sin
cos
cos
sin
∂
∂
−
∂
∂
−
=
φ
θ
θ
φ
θ
φ
j
i
ˆ
sin
cos
sin
cos
∂
∂
−
∂
∂
+
φ
θ
θ
φ
θ
φ
k
i
ˆ
φ
∂
∂
+
k
L
j
L
i
Lx
ˆ
+
yˆ
+
zˆ
We can also determine the raising and lowering operators:
y x
iL
L
L
±=
±
±
∂
∂
−
∂
∂
−
=
φ
φ
θ
θ
φ
cot
cos
sin
i
cos
cot
sin
∂
,
∂
−
∂
∂
φ
φ
θ
θ
φ
i
i
i
∂
∂
±
−
∂
∂
±
−
=
φ
φ
φ
θ
θ
φ
φ
cos
)
cot
(cos
sin
)
sin
(
i
i
,
2 2 z zL
L
L
L
L
±=
−
±
2 2 2 2
,
x y z
L
=
L
+
L
+
L
By using or we have
.
sin
1
sin
sin
1
2 2 2 2 2∂
∂
+
∂
∂
∂
∂
−
=
φ
θ
θ
θ
θ
θ
L
We are now to determine
f
lm(
θ
,
φ
),
which is an eigenfunction of ,L
22
∂
∂
∂
.
)
1
(
sin
1
sin
sin
1
2 2 2 2 2 2 m l m l ml
f
l
l
f
f
L
=
+
∂
∂
+
∂
∂
∂
∂
−
=
φ
θ
θ
θ
θ
θ
But this is precisely the “angular equation”. And it’s also an eigenfunction of Lz, with the eigenvalue of m :
,
m l m l m lz
f
mf
but this is equivalent to the azimuthal equation (Eq.4.21).
)
,
(
)
,
(
θ
φ
mθ
φ
m
Y
f
=
).
1
(
sin
1
sin
sin
1
2 22
∂
=
−
+
∂
+
∂
∂
∂
∂
l
l
Y
Y
Y
Y
θ
θ
θ
θ
θ
φ
) , (θ φ m l Y
We have already solved this system of equations: The result is the spherical harmonic, .
Conclusion: Spherical harmonics are eigenfunctions of L2 and L
z.
Recalling what we have done in Section 4.1 that we solved the Schrodinger equation by separation of variables, we have inadvertently constructed
simultaneous eigenfunctions of the three commuting operators H, L2 and L
z:
,
ψ
ψ
E
H
=
)
,
(
)
,
(
θ
φ
lθ
φ
l
Y
f
=
)
,
(
)
(
)
,
,
(
θ
φ
θ
φ
ψ
r
=
R
r
Y
That is m l m l m
l
L
f
l
l
f
Y
L
2=
2=
2(
+
1
)
.
)
1
(
)
(
)
1
(
)
(
)
(
2 2 22
2
ψ
=
=
=
+
=
+
ψ
l
l
Y
r
R
l
l
Y
L
r
R
Y
r
R
L
L
lm lm lmm l m l m l
z
f
mf
i
f
L
=
∂
∂
=
φ
m l m lz
Y
mY
L
=
L
zψ
=
m
ψ
Then at last we have
H
ψ
=
E
ψ
,
L
2ψ
=
2l
(
l
+
1
)
ψ
,
L
zψ
=
m
ψ
.
As.
sin
1
sin
sin
1
2 2 2 2 2∂
∂
+
∂
∂
∂
∂
−
=
φ
θ
θ
θ
θ
θ
Y
L
ψ
ψ
φ
ψ
θ
θ
ψ
θ
θ
θ
ψ
E
V
r
r
r
r
r
r
m
∂
+
=
We can rewrite the Schrodinger equation more compactly:
ψ
ψ
ψ
E
V
L
r
r
r
mr
∂
+
+
=
∂
∂
∂
−
2 2 22
2
1
Question:
There is only one inconsistency between the functions
By algebraic method the angular momentum permits l to take on half-integer values, whereas separation of variables yielded
eigenfunctions only for integer values. What is?
The l can be half-integer that is turned out to be important in the following sections. Spin!
.
and
lmm
l
Y
f
In classical mechanics, a rigid object admits two kinds of angular momentum:
4.4 Spin
,
p
r
L
=
×
orbital angular momentum
associated with the motion of the center of mass, and spin
o
r
p
,
ω
I
S
=
center of masswhere I is the moment of inertia, associated with the motion about the center of mass.
But, in quantum mechanics, an analogous thing happens, and there is a
absolutely fundamental distinction. In addition to orbital angular momentum, associated ( in the case of hydrogen) with motion of the electron around the nucleus (and described by the spherical harmonics), the electron also carries
another form of angular momentum, which is nothing to do with motion in space (and which is not, therefore, described by any function of the position variables r, , ) but which is somewhat analogous to classical spin.
,
ω
I
However, the electron (as far as we know) is a structureless point particle, and its spin angular momentum cannot be decomposed into orbital angular momenta of constituent parts. Suffice it to say that elementary particles carry intrinsic angular momentum (S) in addition to their “extrinsic” angular momentum (L).
The algebraic theory of spin is a carbon copy of the theory of orbital angular momentum, beginning with the fundamental commutation relations:
.
]
,
[
,
]
,
[
,
]
,
[
S
xS
y=
i
S
zS
yS
z=
i
S
xS
zS
x=
i
S
yIt follows (as before) that the eigenvectors of S2 and S
z satisfy
and for
;
,
)
1
(
,
22 2
m
s
s
s
m
s
S
f
S
sm=
=
+
;
,
,
m
m
s
m
s
S
f
S
z sm=
z=
;
)
1
(
,
)
1
(
)
1
(
,
=
+
−
±
±
±
s
m
s
s
m
m
s
m
S
y x
iS
S
But this time the eigenvectors are not spherical harmonics ( they are not
functions of r, , at all), and there is no a priori reason to exclude the half-integer values of s and m:
1. It so happens, that every elementary particle has a specific and immutable
.
,
1
,
,
1
,
;
,
2
3
,
1
,
2
1
,
0
m
s
s
s
s
s
=
=
−
−
+
−
Notes:
1. It so happens, that every elementary particle has a specific and immutable value of s, which we call the spin of that particular species: pi mesons have spin 0; electrons have spin 1/2; photons have spin 1; deltas have spin 3/2; gravitons have spin 2; and so on.
By far the most important case is s=1/2, for this is the spin of the particles that make up ordinary matter (protons, neutrons, and electrons), as well as all quarks and all leptons. Moreover, once you understand spin 1/2, it is a simpler matter to work out the formalism for any higher spin. These are just two eigenstates:
2
1
=
s
m
s
,
2
1
=
m
Spin up4.4.1 Spin 1/2
m
,
1
2
1
,
2
1
+
Using above two states as a basis vectors, the general state of a spin-1/2 particle can be expressed as a two-element column matrix (or spinor):
2
m
s
,
↓
+
↑
=
+
=
+
=
=
a
b
a
+b
−a
b
b
a
χ
χ
χ
1
0
0
1
m
,
2
1
2
2
2
1
,
2
1
−
2
1
−
=
with +
=↑=
representing spin up, and for spin down.0
1
χ
−=↓=
1
0
χ
Meanwhile, the spin operators become 2×2 matrices, which we can work out by noting their effect on + and -.
,
,
)
1
(
,
2 2m
s
s
s
m
s
S
=
+
,
3
)
1
1
(
1
1
,
1
2 2 22
=
χ
=
+
χ
=
χ
S
S
By using,
4
3
)
1
2
1
(
2
1
2
1
,
2
1
2 2 22
+ +
+
=
+
=
=
S
χ
χ
χ
S
,
4
3
)
1
2
1
(
2
1
2
1
,
2
1
2 2 22
− −
−
=
+
=
=
−
S
χ
χ
χ
S
If we write S2 as a matrix with undetermined elements,
Then the first equation says
so
.
0
=
e
=
0
1
4
3
0
1
2f
e
d
c
,
0
4
3
2=
e
c
,
4
3
2=
c
The second equation says
=
1
0
4
3
1
0
2f
e
d
c
,
4
3
0
2
=
f
d
so
,
0
=
d
.
4
3
2=
Finally, we have
.
1
0
0
1
4
3
22
=
S
Similarly, we write Sz as a matrix with undetermined elements,
=
,
f
e
d
c
S
zand use
,
2
1
+
+
=
χ
χ
z
S
,
2
−−
=
−
χ
χ
z
S
We have
,
0
1
2
1
0
1
=
f
e
d
c
−
=
1
0
2
1
1
0
f
e
d
c
,
1
0
0
1
2
1
−
=
z
Mean well for “ladder operators”
S
±=
S
x±
iS
yso
,
,
0
+ − ++
+
χ
=
S
χ
=
χ
S
,
0
,
=
=
− − − +−
χ
χ
S
χ
S
,
0
0
1
0
=
+S
.
0
1
0
0
=
−S
,
↑
↓=
+S
.
↓
↑=
−S
Now y xiS
S
S
±=
±
, so)
(
2
1
− ++
=
S
S
S
x)
(
2
1
− +−
These are the famous Pauli spin matrices. Notice that S , S S , and S2 are all
σ
2
=
S
Since Sx, Sy and Sz all carry a factor of /2, it is tidier to define
where
,
0
1
1
0
=
x
σ
,
0
0
−
=
i
i
y
σ
.
1
0
0
1
−
=
z
σ
These are the famous Pauli spin matrices. Notice that Sx, Sy Sz, and S2 are all
hermitian (as they should be, since they represent observables). On the other hand, S+ and S- are not hermitian—evidently they are not observable.
The eigenspinors of Sz are (of course)
=
+
0
1
χ
−=
1
0
χ
(eigenvalue );
− +
=
S
x2
1
,
1
↓
↑=
i
S
y=↓
=↑
−+
,
+ −
=
S
x2
1
− +
=
i
S
y2
1
,
2
1
↑
↓=
x
S
,
2
1
↓
↑=
x
S
,
2
↓
↑=
i
S
y+ −
=
−
i
S
y2
1
,
↑
↓=
+
S
.
↓
↑=
−
S
− +
=
i
S
y2
,
2
1
↑
−
↓=
i
S
y,
+ −
+
χ
=
χ
S
− +
−
χ
=
χ
Since these are the only possibilities, that is (i.e. the spinor must be normalized)
.
1
|
|
|
|
a
2+
b
2=
If you measure Sz on a particle in the general state
,
1
0
0
1
− +
+
=
+
=
=
χ
χ
χ
a
b
a
b
b
a
you could get spin up with probability , or spin down with probability .
2
2
|
|
a
2
|
|
b
2
−
.
1
|
|
|
|
a
2+
b
2=
But what if, instead, you chose to measure Sx? What are the possible results, and what are their respective probabilities? According to the generalized statistical interpretation, we need to know the eigenvalues and eigenspinors of Sx. The characteristic equation of Sx is
=
0
2
/
2
/
0
x
S
0
2
/
2
/
=
−
−
λ
λ
2
±
=
Evidently the normalized eigenspinors of Sx are
Not surprisingly, the possible value for Sx are the same as those for Sz . The eigenspinors are obtained in the usual way:
±
=
±
=
β
α
α
β
β
α
β
α
2
0
1
1
0
2
α
β
=
±
=
+
1
2
1
) (x
χ
(eigenvalue );2
(eigenvalue ).
2
−
−
=
−
2
1
1
) (x
χ
As the eigenvectors of a hermitian matrix, they span the space; the generic spinor can be expressed as a linear combination of them:
2
1
2
−
2
2
1
.
2
2
) ( )
(x
a
b
xb
a
b
a
b
a
− +
−
+
+
=
=
χ
χ
χ
Now, if you measure Sx, the probability of getting /2 is 1/2|a+b|2, and the - /2
Discussion: See book on page 176.
=
+
0
1
χ
z-component Sz ?
A spinning charged particle constitutes a magnetic dipole. Its magnetic dipole moment, , is proportional to its spin angular momentum, S:
4.4.2 Electron in Magnetic Field
;
S
=
B
The proportional constant, , is called the gyromagnetic ratio and
m
e
⋅
−
=
2
so the Hamiltonian of a spinning charged particle, at rest in a magnetic field B, is
B
H
=
−
⋅
When a magnetic dipole is placed in a magnetic field B, it experiences a
torque , which is to line it up parallel to the field (just like a compass needle). The energy associated with this torque is
B
×
µ
B
S
Imagine a particle of spin 1/2 at rest in a uniform magnetic field, which points in the z-direction:
1. Larmor precession:
k
B
B
=
0( )
.
1
0
0
1
2
0 0
0
−
−
=
−
=
⋅
−
=
B
k
S
B
S
B
H
γ
zThe Hamiltonian, in matrix form, is
The eigenstates of H are the same as those of Sz
1
0
2
−
with the energy
,
+
χ
,
−
χ
with the energy,
2
0
B
E
+=
−
,
2
0
B
E
−=
+
Evidently the energy is lowest when the dipole moment is parallel to the field——just as it would be classically.
B
,
χ
χ
H
t
i
=
∂
∂
Since the Hamiltonian is dependent, the general solution to the time-dependent Schrodinger equation,
− +
+
=
χ
χ
χ
(
0
)
a
b
can be expressed in terms of the stationary states:General initial state t=0:
γ
Clearly, the constant a and b are determined by the initial conditions:
.
)
(
/22 / /
/
0 0
=
+
=
− −− −
+ + − i B t
t B i t
iE t
iE
be
ae
e
b
e
a
t
γγ
χ
χ
χ
,
)
0
(
=
++
−=
b
a
b
a
χ
χ
,
2
cos
=
α
a
α
where is a fixed angle whose physical significance will appear in a moment. Thus
As |a|2+|b|2=1, without essential loss of generality, we will write
,
2
sin
=
α
b
.
)
2
/
sin(
)
2
/
cos(
)
(
2 / 2 / 2 / 2 / 0 0 0 0=
=
− − t B i t B i t B i t B ie
e
be
ae
t
γ γ γ γα
α
χ
To get a feel for what is happening here, let’s calculate the expectation value of S, as a function of time:
=
=
=
(
t
)
S
(
t
)
+(
t
)
S
(
t
)
S
xχ
xχ
χ
xχ
− /2 2 / 0 0
)
2
/
sin(
)
2
/
cos(
t B i t B ie
e
γ γα
α
0
1
1
0
2
)
)
2
/
sin(
)
2
/
(cos(
α
e
−iγB0t/2α
e
+iγB0t/2).
cos(
sin
2
α
γ
B
0t
Similarly,
=
=
+(
t
)
S
(
t
)
S
yχ
yχ
sin
sin(
),
2
α
γ
B
0t
−
=
=
+(
t
)
S
(
t
)
S
zχ
zχ
cos
.
2
α
Finally, we have
30
=
α
determine the initial state.
=
x
S
sin
cos(
).
2
α
ω
t
=
y
S
sin
sin(
),
2
α
ω
t
−
=
z
S
cos
.
2
α
30
=
α
0
B
=
30
=
α
S
),
sin(
sin
2
α
t
−
.
t
)
cos(
sin
2
α
0
2
0
.
B
=
=
Evidently <S> is tilted at a constant angle to the z-axis, and precesses about the field at the Larmor frequency
α
just as it would classically.
,
0
B
=
ω
2. The Stern-Gerlach experiment:
Experimental conditions:
(1) A heavy neutral atomic beam — Ag atom, for example.
How to determine the atomic spin: all the inner electrons of the atom are paired, in such a way that their spin and orbital momenta cancel and the net spin is simply that of the outermost—unpaired—electron. if we use silver Using neutral atom is to avoid the large-scale deflection that would otherwise result form the Lorentz force, and heavy so we can construct localized wave packets and treat the motion in terms of classical particle trajectories.
(2) Inhomogeneous magnetic field
spin is simply that of the outermost—unpaired—electron. if we use silver atoms, for example, there is only one outmost unpaired electron there, so in this case s=1/2, and hence the beam splits in two (2s+1=2).
In a inhomogeneous magnetic field, there is not only a torque, but also a force, on a magnetic dipole:
).
(
)
(
B
S
B
H
F
=
−∇
=
∇
⋅
=
∇
⋅
This force can be used to separate out particles with a particular spin orientation, as follows.
Imagine a beam of relatively heavy neutral atoms, traveling in the y direction, 1). Classical picture theory:
Imagine a beam of relatively heavy neutral atoms, traveling in the y direction, which passes through a region of inhomogeneous magnetic field—say,
,
)
(
B
0z
k
i
x
B
=
−
+
+
where B0 is a strong uniform field and the constant describes a small deviation from homogeneity. Only the z-component of B is important, while x-component of B here is for
.
0
=
⋅
∇
B
0
.
)
k
S
i
S
(
F
=
−
x+
zBut because of the Larmor precession about B0, Sx oscillates rapidly, and
averages to zero; the net force is in the z direction:
[4.170]
,
z
z
S
F
=
Then the force on these atoms isand the beam is deflected up and down, in proportion to the z component of the spin angular momentum. Classically, we’d expect a smear, but in fact the beam splits into 2s+1 separate streams, beautifully demonstrating the
quantization of angular momentum.
>
≤
≤
+
−
<
=
.
for
,
0
,
0
for
)
(
,
0
for
,
0
0
T
t
T
t
S
z
B
t
H
zWe examine the process from the perspective of a reference frame that moves along with the beam. In this frame the Hamiltonian starts out zero, turns on for a time T, and then turns off again:
2). Quantum picture theory:
Suppose the atom has spin 1/2, and starts out in the state
,
)
(
=
χ
++
χ
−χ
t
a
b
fort
≤
0
.
While the Hamiltonian acts, evolves in the usual way:
χ
(t),
)
(
t
a
e
iE+t/b
e
−iE−t/− −
+
+
=
χ
χ
χ
for0
≤
t
≤
T
,
where we know
,
2
)
(
B
0z
E
±=
+
z
S
z
B
S
and hence it emerges in the sense
(
)
( )(
)
( ),
)
(
t
ae
i B0T /2e
i T /2 zbe
i B0T /2e
−i T /2 z− −
+
+
=
γ γχ
fort
≥
T
.
corresponds to a momentum in z-direction of p, the two terms above carry momentum in the z direction; the spin up component has momentum
As the eigenfunction of momentum
f
pz
e
i(p/ )z2
1
)
(
π
=
momentum in the z direction; the spin up component has momentum
and it moves in the plus-z direction; the spin-down component has the opposite momentum, and it moves in the minus-z direction. Thus the beam splits in two.
,
2
T
F
T
p
z=
=
z,
z
z
S
F
=
Comparison: in classical point of view,
,
2
=
z
S
,
2
The significance of Stern-Gerlach experiment:
(1) The experiment demonstrated the spatial quantization of the quantum theory. However, this experiment was performed before the notion of spin
was proposed; and later it turned out that the two split lines are due to the
spin of the outermost electron of silver.
(2) Measurement of the state——spin-up state or spin-down state.
Suppose now that we have two spin-1/2 particle, the electron and the proton in the ground state of hydrogen. Each can have spin up or spin down, so there are four possibilities in all:
.
,
,
,
↑↓
↓↑
↓↓
↑↑
The first arrow refers to the electron and
4.4.3 Addition of Angular Momenta
o
r
) 1 (
S
) 2 (
S
proton1. Simplest example:
Question: What is the total angular momentum of the atom? The first arrow refers to the electron and
the second to the proton.
.
) 2 ( )
1 (
S
S
S
≡
+
electron
Let total angular momenta of the system is
r
Above composite states of the system can be represented by 1 2
.
,
) 2 ( )
1 (
z z
z
S
S
S
=
+
Applying
Note that S(1) acts only on
1, and S(2) acts only on 2. That is
(
)
1 2) 2 ( 2 1 ) 1 ( 2 1 ) 2 ( ) 1 ( 2
1
S
S
S
S
S
z=
z+
z=
z+
z(
)
(
2)
1 1 2 2 1 2) 2 ( 1 2 1 ) 1 (
m
m
S
S
z+
z=
+
=
(
m
1+
m
2)
1 2=
=
2 1
S
z(
m
1+
m
2)
1 2So m (the quantum number for the composite system) is just m1+m2:
=
2 1
S
z(
m
1+
m
2)
1 21
2
1
2
1
:
=
+
=
↑↑
m
0
2
1
2
1
:
=
−
+
=
↓↑
m
0
2
1
2
1
:
=
−
=
↑↓
m
1
2
1
2
1
:
=
−
−
=
−
↑↓
m
But we get two states with m=0 ? According to general theory of angular momentum, the state can be generated by
So if we apply the lowering operator,
.
)
1
(
,
)
1
(
)
1
(
,
=
+
−
±
±
±
s
m
s
s
m
m
s
m
S
) 2 ( )
1 (
− −
−
=
S
+
S
S
to the state↑↑
( )
↑↑
=
(
−↑
)
↑
+
↑
(
−↑
)
=
( )
↓
↑
+
↑
( )
↓
−
) 2 ( )
1 (
S
S
S
(
↓↑
+
↑↓
)
;
=
if we apply the lowering operator,
S
−=
S
−(1)+
S
−(2) to the state(
↓↑
+
↑↓
)
=
−
S
(
S
−(1)↓
)
↑
+
↓
(
S
−(2)↑
) (
+
S
−(1)↑
)
↓
+
↑
(
S
−(2)↓
)
( ) ( )
↓
+
↓
↓
+
=
↓↓
↓
+
=
0
0
2
Evidently the three states are in the same set with s=1, which are:
=↓↓
−
↑↓
+
↑↓
=
=↑↑
=
1
,
1
)
(
2
1
1,0
,
1
,
1
,
1
m
The total s=1 and m=-1,0,1 and there have three states as
,
1
1,
,
1,0
,
1
,
1
−
m
s
,
This is called the triplet combination.
Meanwhile, the other orthogonal state with m=0 carries s=0:
)
(
2
1
0,0
=
↑↓
−
↑↓
This is called the singlet combination.
0
)
(
2
1
=
↑↓
−
↑↓
−
S
0
)
(
2
1
=
↑↓
−
↑↓
+
Conclusion: the combination of two spin-1/2 particles can carry a total spin of 1 or 0, depending on whether they occupy the triplet or the singlet
configuration. Now we prove it by getting the eigenvalue of S2.
As
Applying it on the triplet, we have
(
(1) (2)) (
(1) (2))
2
S
S
S
S
S
S
S
≡
⋅
=
+
⋅
+
( ) ( )
(1) 2 (2) 2 (1) (2)2
S
S
S
S
+
+
⋅
=
Applying it on the triplet, we have
( )
↑↑
⋅
=
⋅
(2) (1) (2) )1 (
1
,
1
(1)
S
S
S
S
(
↑
)(
↑
) (
+
↑
)(
↑
) (
+
↑
)(
↑
)
=
(1) (2) (1) (2) (1) (2)z z
y y
x
x
S
S
S
S
S
S
↑
↑
+
↓
↓
+
↓
↓
=
2
2
2
2
2
( )
↓↓
−
( )
↓↓
+
( )
↑↑
=
( )
↑↑
=
4
4
4
4
2 2
2 2
( )
↑↑
⋅
=
⋅
(2) (1) (2) )1 (
1
,
1
S
S
S
S
1
,
1
4
2
=
( ) ( )
↑↑
=
1
1
+
1
↑
↑=
3
( )
↑↑
2 2
2 ) 1 (
S
( ) ( )
↑↑
=
+
↑
↑=
( )
↑↑
4
1
2
2
2 )
1 (
S
( )
( )
2
1
,
1
4
2
4
3
4
3
1
,
1
22 2
2 2
2
=
↑↑
=
+
+
↑↑
=
S
S
1
2
)
1
(
s
+
=
s
=
( )
↑↓
⋅
(2)) 1 (
(2)
S
S
(
(1)↑
)(
(2)↓
) (
+
(1)↑
)(
(2)↓
) (
+
(1)↑
)(
(2)↓
)
z z
y y
x
x
S
S
S
S
S
S
↓
−
↑
+
↑
−
↓
+
↑
↓
=
2
2
2
2
2
2
i
i
( )
↑↓
=
⋅
(2) )1 (
S
S
(
↓↑
−
↑↓
)
=
2
2
(
↓↑
−
↑↓
)
=
2
4
( )
↓↑
=
⋅
(2) )1 (
S
S
(
2
↑↓
−
↓↑
)
4
2
Similarly,
Then
(
↑↓
+
↓↑
)
⋅
=
⋅
2
1
0
,
1
(1) (2)) 2 ( )
1 (
S
S
(
)
(
)
1
,
0
4
2
1
4
2
2
2
1
4
2 2 2=
↑↓
+
↓↑
=
↓↑
−
↑↓
+
↑↓
−
↓↑
=
0
,
1
4
0
,
1
2 ) 2 ( ) 1 (⋅
=
S
S
( )
2
1
,
1
4
2
4
3
4
3
0
,
1
2 2 2 22
=
+
+
↑↑
=
S
( )
4
4
4
Similarly,
( )
2
1
,
1
4
2
4
3
4
3
1
,
1
2 2 2 22
−
=
+
+
↓↓
=
−
S
The eigenvalue of S2on triplet is
2
2=
1
(
1
+
1
)
2s
=
1
.
1
2
)
1
(
s
+
=
s
=
Some examples: book
2. General theory of addition of angular momenta:
If you combine spin
s
1 with spins
2, what total spinss
can you get ? The answer is that you get every spin from (s
1+
s
2) down to (s
1- s
2)——or (s
2-
s
1), ifs
2>s
1——in integer steps:.
|
|
,
),
2
(
),
1
(
),
(
s
1s
2s
1s
2s
1s
2s
1s
2s
=
+
+
−
+
−
−
s
s
The spin states for
s
1 is: |s1 m1>;The combined state for the total spin
s
of the system is |s m>.The spin states for
s
2 is: |s2 m2>. The direct product states for composite state is: |s1 m1> |s2 m2>.Then the combined state |s m> with total spin
s
and z-component m will be some linear combination of composite states |s1 m1> |s2 m2> :2 2 1
1
2 1
2 1
2 1
,
m
C
s
m
s
m
s
m m m
s s
m m m
= +
The constants are called 1 2 Clebsch-Gordan coefficients. 2 1 s s m m m
C
m
s
C
m
s
m
s
s s s s m m m,
2 1 2 1 2 2 1 1=
Or reverselySee the table:
2 2 1 1 2 1 2 1 2 1
,
m
C
s
m
s
m
s
m m m s s m m m = +=
2
1
,
2
1
2
1
,
2
1
2
1
2
1
,
2
1
2
1
,
2
1
2
1
0
,
1
=
−
+
−
(
↑↓
+
↓↑
)
=
1
,
1
0
,
2
2
1
0
,
1
1
,
2
6
1
1
,
1
2
,
2
2
1
1
,
2
=
−
+
−
1
,
1
1
,
2
5
1
0
,
1
0
,
2
5
3
1
,
1
1
,
2
5
1
0
,
m
s
C
m
s
m
s
s
s s s
m m
m
,
2 1
2 1
2 2 1
1
=
2
1
,
2
1
3
1
2
1
,
3
2
15
1
2
1
,
2
5
5
3
0
,
1
2
1
,
2
3
−
+
Group theory
