Ideas of Quantum Chemistry P100 ppt
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956
F. TRANSLATION vs MOMENTUM and ROTATION vs ANGULAR MOMENTUM
where
ˆ
p =−i
¯
h∇ is the total momentum operator (see Chapter 1). Thus, for trans-
lationswehaveκ ≡T and
ˆ
K ≡
ˆ
p
Rotation and angular momentum operator
Imagine a function f(r) of positions in 3D Cartesian space (think, e.g., about a
probability density distribution centred somewhere in space). Now suppose the
function is to be rotated about the z axis (the unit vector showing its direction is e)
by an angle α, so we have another function, let us denote it by
ˆ
U(α;e)f (r) What is
the relation between f(r) and
ˆ
U(α;e)f (r)? This is what we want to establish. This
relation corresponds to the opposite rotation (i.e. by the angle −α, see Fig. 2.1 and
p. 58) of the coordinate system:
ˆ
U(α;e)f (r) =f
U
−1
r
=f
U(−α;e)r
where U is a 3 ×3 orthogonal matrix. The new coordinates x(α) y(α)z(α) are
expressed by the old coordinates x yz through
1
r
≡
⎛
⎝
x(α)
y(α)
z(α)
⎞
⎠
=
⎛
⎝
cosα sinα 0
−sinα cosα 0
001
⎞
⎠
⎛
⎝
x
y
z
⎞
⎠
Therefore the rotated function
ˆ
U(α;e)f (r) = f (x(α) y(α)z(α)) The func-
tion can be expanded in the Taylor series about α =0:
ˆ
U(α;e)f (r) =f
x(α) y(α)z(α)
=f(xyz)+α
∂f
∂α
α=0
+···
= f(xyz)+α
∂x(α)
∂α
∂f
∂x
+
∂y(α)
∂α
∂f
∂y
+
∂z(α)
∂α
∂f
∂z
α=0
+···
= f(xyz)+α
y
∂
∂x
−x
∂
∂y
f +···
Now instead of the large rotation angle α, let us consider first an infinitesimally
small rotation by angle ε = α/N,whereN is a huge natural number. In such a
situation we retain only the first two terms in the previous equation:
ˆ
U
α
N
;e
f(r) = f(xyz)+
α
N
y
∂
∂x
−x
∂
∂y
f (x yz)
=
1 +
α
N
i
¯
h
i
¯
h
y
∂
∂x
−x
∂
∂y
f =
1 +
α
N
1
i
¯
h
x
ˆ
p
y
−y
ˆ
p
x
f
=
1 −
α
N
i
¯
h
ˆ
J
z
f
1
A positive value of the rotation angle means an anticlockwise motion within the xy plane (x axis
horizontal, y vertical, z axis pointing to us).
2 The Hamiltonian commutes with the total momentum operator
957
If such a rotation is repeated N times, we recover the rotation of the function
bya(possiblylarge)angleα (the limit assures that ε is infinitesimally small):
ˆ
U(α;e)f (r) = lim
N→∞
ˆ
U
α
N
;e
N
f(r) = lim
N→∞
1 −
α
N
i
¯
h
ˆ
J
z
N
f(r)
= exp
−i
α
¯
h
ˆ
J
z
f =exp
−
i
¯
h
αe ·
ˆ
J
f
Thus for rotations
ˆ
U(α;e) =exp(−
i
¯
h
αe ·
ˆ
J), and, therefore, we have κ ≡αe and
ˆ
K ≡
ˆ
J.
This means that, in particular for rotations about the x yz axes (with the cor-
responding unit vectors x y z) we have, respectively
ˆ
U(α;x)
ˆ
J
x
= 0 (F.4)
ˆ
U(α;y)
ˆ
J
y
= 0 (F.5)
ˆ
U(α;z)
ˆ
J
z
= 0 (F.6)
Useful relation
The relation (F.1) means that for any translation or rotation
ˆ
U
ˆ
H
ˆ
U
−1
=
ˆ
H
and taking into account the general form of eq. (F.2) we have for any such trans-
formation a series containing nested commutators (valid for any κ)
ˆ
H =
ˆ
U
ˆ
H
ˆ
U
−1
=exp
−
i
¯
h
κ ·
ˆ
K
ˆ
H exp
i
¯
h
κ ·
ˆ
K
=
1 −
i
¯
h
κ ·
ˆ
K +···
ˆ
H
1 +
i
¯
h
κ ·
ˆ
K +···
=
ˆ
H −
i
¯
h
κ ·
ˆ
K
ˆ
H
−
κ
2
2
¯
h
2
ˆ
K
ˆ
H
ˆ
K
+···
where each term in “+···” contains [
ˆ
K
ˆ
H] This means that to satisfy the equation
we necessarily have
ˆ
K
ˆ
H
=0 (F.7)
2 THE HAMILTONIAN COMMUTES WITH THE TOTAL
MOMENTUM OPERATOR
In particular this means [
ˆ
p
ˆ
H]=0,i.e.
ˆ
p
μ
ˆ
H
=0(F.8)
958
F. TRANSLATION vs MOMENTUM and ROTATION vs ANGULAR MOMENTUM
for μ =x yz. Of course, we also have [
ˆ
p
μ
ˆ
p
ν
]=0forμν =x yz.
Since all these four operators mutually commute, the total wave function is si-
multaneously an eigenfunction of
ˆ
H and
ˆ
p
x
ˆ
p
y
ˆ
p
z
, i.e. the energy and the mo-
mentum of the centre of mass can both be measured (without making any error)
in a space-fixed coordinate system (see Appendix I). From the definition, the mo-
mentum of the centre of mass is identical to the total momentum.
2
3 THE HAMILTONIAN,
ˆ
J
2
AND
ˆ
J
z
DO COMMUTE
Eq. (F.7) for rotations means [
ˆ
J
ˆ
H]=0, i.e. in particular
ˆ
J
x
ˆ
H
= 0 (F.9)
ˆ
J
y
ˆ
H
= 0 (F.10)
ˆ
J
z
ˆ
H
= 0 (F.11)
The components of the angular momentum operators satisfy the following com-
mutation rules:
3
ˆ
J
x
ˆ
J
y
= i
¯
h
ˆ
J
z
ˆ
J
y
ˆ
J
z
= i
¯
h
ˆ
J
x
(F.12)
ˆ
J
z
ˆ
J
x
= i
¯
h
ˆ
J
y
2
Indeed the position vector of the centre of mass is defined as R
CM
=
i
m
i
r
i
i
m
i
, and after differ-
entiation with respect to time (
i
m
i
)
˙
R
CM
=
i
m
i
˙
r
i
=
i
p
i
. The right-hand side represents the
momentum of all the particles (i.e. the total momentum), whereas the left is simply the momentum of
the centre of mass.
3
The commutation relations can be obtained by using the definitions of the operators involved di-
rectly:
ˆ
J
x
=y
ˆ
p
z
−z
ˆ
p
y
etc. For example,
ˆ
J
x
ˆ
J
y
f =
y
ˆ
p
z
−z
ˆ
p
y
z
ˆ
p
x
−x
ˆ
p
z
−
z
ˆ
p
x
−x
ˆ
p
z
y
ˆ
p
z
−z
ˆ
p
y
f
=
y
ˆ
p
z
z
ˆ
p
x
−z
ˆ
p
x
y
ˆ
p
z
−
y
ˆ
p
z
x
ˆ
p
z
−x
ˆ
p
z
y
ˆ
p
z
−
z
ˆ
p
y
z
ˆ
p
x
−z
ˆ
p
x
z
ˆ
p
y
+
z
ˆ
p
y
x
ˆ
p
z
−x
ˆ
p
z
z
ˆ
p
y
f
=
y
ˆ
p
z
z
ˆ
p
x
−z
ˆ
p
x
y
ˆ
p
z
f −
yx
ˆ
p
z
ˆ
p
z
−yx
ˆ
p
z
ˆ
p
z
−
z
2
ˆ
p
y
ˆ
p
x
−z
2
ˆ
p
x
ˆ
p
y
+
xz
ˆ
p
y
ˆ
p
z
−x
ˆ
p
z
z
ˆ
p
y
=
y
ˆ
p
z
z
ˆ
p
x
−yz
ˆ
p
x
ˆ
p
z
f −0 −0 +
xz
ˆ
p
y
ˆ
p
z
−x
ˆ
p
z
z
ˆ
p
y
f
= (−i
¯
h)
2
y
∂f
∂x
−x
∂f
∂y
=i
¯
h
ˆ
J
z
f
3 The Hamiltonian,
ˆ
J
2
and
ˆ
J
z
do commute
959
Eqs. (F.9)–(F.11) are not independent, e.g., eq. (F.11) can be derived from
eqs. (F.9) and (F.10). Indeed,
ˆ
J
z
ˆ
H
=
ˆ
J
z
ˆ
H −
ˆ
H
ˆ
J
z
=
1
i
¯
h
ˆ
J
x
ˆ
J
y
ˆ
H −
1
i
¯
h
ˆ
H
ˆ
J
x
ˆ
J
y
=
1
i
¯
h
ˆ
J
x
ˆ
J
y
ˆ
H −
1
i
¯
h
ˆ
J
x
ˆ
J
y
ˆ
H
= 0
Also, from eqs. (F.9), (F.10) and (F.11) it also follows that
ˆ
J
2
ˆ
H
=0 (F.13)
because from Pythagoras’ theorem
ˆ
J
2
=
ˆ
J
2
x
+
ˆ
J
2
y
+
ˆ
J
2
z
Do
ˆ
J
x
ˆ
J
y
ˆ
J
z
commute with
ˆ
J
2
? Let us check the commutator [
ˆ
J
z
ˆ
J
2
]:
ˆ
J
z
ˆ
J
2
=
ˆ
J
z
ˆ
J
2
x
+
ˆ
J
2
y
+
ˆ
L
2
z
=
ˆ
J
z
ˆ
J
2
x
+
ˆ
J
2
y
=
ˆ
J
z
ˆ
J
2
x
−
ˆ
J
2
x
ˆ
J
z
+
ˆ
J
z
ˆ
J
2
y
−
ˆ
J
2
y
ˆ
J
z
=
i
¯
h
ˆ
J
y
+
ˆ
J
x
ˆ
J
z
ˆ
J
x
−
ˆ
J
x
−i
¯
h
ˆ
J
y
+
ˆ
J
z
ˆ
J
x
+
−i
¯
h
ˆ
J
x
+
ˆ
J
y
ˆ
J
z
ˆ
J
y
−
ˆ
J
y
i
¯
h
ˆ
J
x
+
ˆ
J
z
ˆ
J
y
= 0
Thus,
ˆ
J
z
ˆ
J
2
= 0 (F.14)
and also by the argument of symmetry (the space is isotropic)
ˆ
J
x
ˆ
J
2
= 0 (F.15)
ˆ
J
y
ˆ
J
2
= 0 (F.16)
Now we need to determine the set of the operators that all mutually commute.
Only then can all the physical quantities, to which the operators correspond, have
definite values when measured. Also the wave function can be an eigenfunction
of all of these operators and it can be labelled by quantum numbers, each corre-
sponding to an eigenvalue of the operators in question. We cannot choose, for these
operators, the whole set of
ˆ
H
ˆ
J
x
ˆ
J
y
ˆ
J
z
ˆ
J
2
because, as was shown above,
ˆ
J
x
ˆ
J
y
ˆ
J
z
do not commute among themselves (although they do with
ˆ
H and
ˆ
J
2
).
960
F. TRANSLATION vs MOMENTUM and ROTATION vs ANGULAR MOMENTUM
The only way is to choose as the set of operators either
ˆ
H
ˆ
J
z
ˆ
J
2
or
ˆ
H
ˆ
J
x
ˆ
J
2
or
ˆ
H
ˆ
J
y
ˆ
J
2
Traditionally, we choose
ˆ
H
ˆ
J
z
ˆ
J
2
as the set of mutually com-
muting operators (z is known as the quantization axis).
4 ROTATION AND TRANSLATION OPERATORS DO NOT
COMMUTE
Now we may think about adding
ˆ
p
x
ˆ
p
y
ˆ
p
z
, to the above set of operators. The
operators
ˆ
H
ˆ
p
x
ˆ
p
y
ˆ
p
z
ˆ
J
2
and
ˆ
J
z
do not represent a set of mutually commuting
operators. The reason for this is that [
ˆ
p
μ
ˆ
J
ν
]=0forμ =ν, which is a consequence
of the fact that, in general, rotation and translation operators do not commute as
shown in Fig. F.1.
5 CONCLUSION
It is, therefore, impossible to make all the operators
ˆ
H
ˆ
p
x
ˆ
p
y
ˆ
p
z
ˆ
J
2
and
ˆ
J
z
com-
mute in a space fixed coordinate system.Whatweareabletodo,though,istowrite
the total wave function
pN
in the space fixed coordinate system as a product of
the plane wave exp(ip
CM
· R
CM
) depending on the centre-of-mass variables and
on the wave function
0N
depending on internal coordinates
4
pN
=
0N
exp(ip
CM
·R
CM
) (F.17)
which is an eigenfunction of the total (i.e. centre-of-mass) momentum operators:
ˆ
p
x
=
ˆ
p
CMx
ˆ
p
y
=
ˆ
p
CMy
ˆ
p
z
=
ˆ
p
CMz
The function
0N
is the total wave function written in the centre-of-mass coor-
dinate system (a special body-fixed coordinate system, see Appendix I), in which
the total angular momentum operators
ˆ
J
2
and
ˆ
J
z
are now defined. The three op-
erators
ˆ
H
ˆ
J
2
and
ˆ
J
z
commute in any space-fixed or body-fixed coordinate system
(including the centre-of-mass coordinate system), and therefore the correspond-
ing physical quantities (energy and angular momentum) have exact values. In this
particular coordinate system:
ˆ
p =
ˆ
p
CM
=0 We may say, therefore, that
in the centre-of-mass coordinate system
ˆ
H
ˆ
p
x
ˆ
p
y
ˆ
p
z
ˆ
J
2
and
ˆ
J
z
all do com-
mute.
4
See Chapter 2 and Appendix I, where the total Hamiltonian is split into the sum of the centre-of-
mass and internal coordinate Hamiltonians; N is the quantum number for the spectroscopic states.
5 Conclusion
961
Fig. F.1. In general, translation
ˆ
U(T) and rotation
ˆ
U(α;e) operators do not commute. The example
shows what happens to a point belonging to the xy plane. (a) A rotation
ˆ
U(α;z) by angle α about
the z axis takes place first and then a translation
ˆ
U(T) by a vector T (restricted to the xy plane)
is carried out. (b) The operations are applied in the reverse order. As we can see the results are
different (two points 1
have different positions in Figs. (a) and (b)), i.e. the two operators do not
commute:
ˆ
U(T)
ˆ
U(α;z) =
ˆ
U(α;z)
ˆ
U(T) This after expanding
ˆ
U(T) = exp[−
i
¯
h
(T
x
ˆ
p
x
+ T
y
ˆ
p
y
)] and
ˆ
U(α;z) =exp(−
i
¯
h
α
ˆ
J
z
) in Taylor series, and taking into account that T
x
T
y
α are arbitrary numbers,
leads to the conclusion that [
ˆ
J
z
ˆ
p
x
] =0and[
ˆ
J
z
ˆ
p
y
] =0. Note, that some translations and rotations
do commute, e.g., [
ˆ
J
z
ˆ
p
z
]=[
ˆ
J
x
ˆ
p
x
]=[
ˆ
J
y
ˆ
p
y
]=0, because we see by inspection (c,d) that any trans-
lation by T =(0 0T
z
) is independent of any rotation about the z axis, etc.
G. VECTOR AND SCALAR
POTENTIALS
Maxwell equations
The electromagnetic field is described by two vector fields: the electric field in-
tensity E and the magnetic field intensity H, both depending on position in space
(Cartesian coordinates x yz)andtimet. The vectors E and H are determined
by the electric charges and their currents. The charges are defined by the charge
charge density
density function ρ(xyzt) such that ρ(xy zt)dV at time t represents the
charge in the infinitesimal volume dV that contains the point (xyz).Theve-
locity of the charge in position x yz measured at time t represents the vec-
tor field v(xyzt),whilethecurrent at point x yz measured at t is equal to
current
i(xyzt)=ρ(x yzt)v(xyzt)
It turns out (as shown by James Maxwell), that HEρand i are interrelated by
the Maxwell equations (c stands for the speed of light)
∇×E +
1
c
∂H
∂t
= 0 (G.1)
∇×H −
1
c
∂E
∂t
=
4π
c
i (G.2)
∇·E =4πρ (G.3)
∇·H = 0 (G.4)
The Maxwell equations have an alternative notation, which involves two new
quantities: the scalar potential φ and the vector potential A that replace E and H:
E =−∇φ −
1
c
∂A
∂t
(G.5)
H =∇×A (G.6)
After inserting E from eq. (G.5) into eq. (G.1), we obtain its automatic satisfac-
tion:
∇×E +
1
c
∂H
∂t
=∇×
−∇φ −
1
c
∂A
∂t
+
1
c
∂H
∂t
=−∇×∇φ −
1
c
∂∇×A
∂t
+
1
c
∂H
∂t
=0
962
G. VECTOR AND SCALAR POTENTIALS
963
because
∇×∇φ =
∂
∂y
∂φ
∂z
−
∂
∂z
∂φ
∂y
∂
∂z
∂φ
∂x
−
∂
∂x
∂φ
∂z
∂
∂x
∂φ
∂y
−
∂
∂y
∂φ
∂x
=[0 0 0]=0 (G.7)
and ∇×A =H.
Eq. (G.4) gives also automatically
∇·(∇×A) =
∂
∂x
∂A
z
∂y
−
∂A
y
∂z
+
∂
∂y
∂A
x
∂z
−
∂A
z
∂x
+
∂
∂z
∂A
y
∂x
−
∂A
x
∂y
=0
Eqs. (G.2) and (G.3) transform into
∇×(∇×A) +
1
c
∂∇φ
∂t
+
1
c
2
∂
2
A
∂t
2
=
4π
c
i
−∇ ·(∇φ) −
1
c
∂∇·A
∂t
= 4πρ
which in view of the identity ∇×(∇×A) =∇(∇·A) −A and ∇·(∇φ) = φ,
gives two additional Maxwell equations (besides eqs. (G.5) and (G.6))
∇
∇·A +
1
c
∂φ
∂t
−A +
1
c
2
∂
2
A
∂t
2
=
4π
c
i (G.8)
φ +
1
c
∇·
∂A
∂t
=−4πρ (G.9)
To characterize the electromagnetic field we may use either E and H or the two
potentials, φ and A.
Arbitrariness of the potentials
φ
and
A
Potentials φ and A are not defined uniquely, i.e. many different potentials lead to the
same intensities of electric and magnetic fields. If we made the following modifica-
tions in φ and A:
φ
= φ −
1
c
∂f
∂t
(G.10)
A
= A +∇f (G.11)
where f is an arbitrary differentiable function (of x yz t), then φ
and A
lead to
the same (see the footnote) E and H:
E
=−∇φ
−
1
c
∂A
∂t
=
−∇φ +
1
c
∇
∂f
∂t
−
1
c
∂A
∂t
+
∂
∂t
(∇f)
=E
H
=∇×A
=∇×A +∇×∇f =H
964
G. VECTOR AND SCALAR POTENTIALS
Choice of potentials
A
and
φ
for a homogeneous magnetic field
From the second Maxwell equation (G.6), we see that, if the magnetic field H
is time-independent, we get the time-independent A. Profiting from the non-
uniqueness of A,wechooseitinsuchawayastosatisfy(whatiscalledthe
Coulombic gauge)
1
Coulombic
gauge
∇·A =0 (G.12)
which diminishes the arbitrariness, but does not remove it.
LetustaketheexampleofanatominahomogeneousmagneticfieldH.Let
us locate the origin of the coordinate system on the nucleus, the choice being
quite natural for an atom, and let us construct the vector potential at position
r =(xyz) as
A(r) =
1
2
[H ×r] (G.13)
As has been shown above, this is not a unique choice, there are an infinite num-
ber of them. All the choices are equivalent from the mathematical and physical
point of view, they differ however by a peanut, the economy of computations. It
appears that this choice of A is at least a logical one. The choice is also consistent
with the Coulombic gauge (eq. (G.12)), because
∇·A =
1
2
∇·[H ×r]=
1
2
∇·[H ×r]
=
1
2
∇·[H
y
z −yH
z
H
z
x −zH
x
H
x
y −xH
y
]
=
1
2
∂
∂x
(H
y
z −yH
z
) +
∂
∂y
(H
z
x −zH
x
) +
∂
∂z
(H
x
y −xH
y
)
=0
and also with the Maxwell equations (p. 962), because
∇×A
=
1
2
∇×[H ×r]=
1
2
∇·[H ×r]
=
1
2
∇×[H
y
z −yH
z
H
z
x −zH
x
H
x
y −xH
y
]
=
1
2
∂
∂y
(H
x
y −xH
y
) −
∂
∂z
(H
z
x −zH
x
)
∂
∂z
(H
y
z −yH
z
) −
∂
∂x
(H
x
y −xH
y
)
∂
∂x
(H
z
x −zH
x
) −
∂
∂y
(H
y
z −yH
z
)
=H
Thus, this is the correct choice.
1
The Coulombic gauge, even if only one of the possibilities, is almost exclusively used in molecular
physics. The word “gauge” comes from the railway terminology referring to the different distances
between the rails.
G. VECTOR AND SCALAR POTENTIALS
965
Practical importance of this choice
An example of possible choices of A is given in Fig. G.1.
If we shifted the vector potential origin far from the physical system under con-
sideration (Fig. G.1.b), the values of |A| on all the particles of the system would be
gigantic. A would be practically homogeneous within the atom or molecule. If we
calculated ∇×A =H on a particle of the system, then however horrifying it might
look, we would obtain almost 0, because ∇×A means the differentiation of A,and
for a homogeneous field this yields zero. Thus we are going to study the system in
a magnetic field, but the field disappeared! Very high accuracy would be needed
to calculate ∇×A correctly as differences between two large numbers, which is
always a risky business numerically due to the cancellation of accuracies. It is there-
fore seen that the numerical results do depend critically on the choice of the origin of
A (arbitrary from the point of view of mathematics and physics). It is always better
to have the origin inside the system.
Vector potential causes the wave function to change phase
The Schrödinger equation for a particle of mass m and charge q is
−
¯
h
2
2m
(r) +V=E(r)
where V =qφ with φ standing for the scalar electric potential.
The probability density of finding the particle at a given position depends on
|| rather than itself. This means that the wave function could be harmlessly
multiplied by a phase factor
(r) =(r) exp[−
iq
¯
hc
χ(r)],whereχ(r) could be any
(smooth
2
) function of the particle’s position r.Thenwehave||=|
| at any r.
If
(r) is as good as is, it would be nice if it kindly satisfied the Schrödinger
equation like does, of course with the same eigenvalue
−
¯
h
2
2m
(r) +V
(r) =E
(r)
Let us see what profound consequences this has. The left-hand side of the last
equation can be transformed as follows
−
¯
h
2
2m
(r) +V
(r)
=−
¯
h
2
2m
exp
−
iq
¯
hc
χ
+exp
−
iq
¯
hc
χ
+2(∇)
∇exp
−
iq
¯
hc
χ)
+V exp
−
iq
¯
hc
χ
=−
¯
h
2
2m
exp
−
iq
¯
hc
χ
+∇
−
iq
¯
hc
exp
−
iq
¯
hc
χ
∇χ
2
See Fig. 2.5.
