8
1 Operations
1.3 Operations and Operators
Besides functions, we must also consider the action of operations on operators. In
quantum chemistry, operators, such as the Hamiltonian, H, are usually spatial functions and, as such, are transformed in the same way as ordinary functions, e.g.,
H ′ (P 1 ) = H( ˆ
R −1 P 1 ). So why devote a special section to this? Well, operators are
different from functions in the sense that they also operate on a subsequent argument, which is itself usually a function. Hence, when symmetry is applied to an
operator, it will also affect whatever follows the operator. Symmetry operations act
on the entire expression at once. This can be stated for a general operator O as
follows:
ˆ
ROf = O
′ ˆ
Rf
(1.17)
From this we can identify the transformed operator O ′ by smuggling ˆ
R −1 ˆ
R (= ˆ
E)
into the left-hand side of the equation:
ˆ
RO ˆ
R
−1 ˆ
Rf = O
′ ˆ
Rf
(1.18)
This equality is true for any function f and thus implies 4 that the operators preceding ˆ
Rf on both sides must be the same:
O
′ = ˆ
RO ˆ
R
−1
(1.19)
This equation provides the algebraic formalism for the transformation of an operator. In the case where O ′ = O, we say that the operator is invariant under the
symmetry operation. Equation (1.19) then reduces to
ˆ
RO − O ˆ
R =[ ˆ
R,O]=0
(1.20)
This bracket is know as the commutator of ˆ
R and O. If the commutator vanishes, we
say that ˆ
R and O commute. This is typically the case for the Hamiltonian. As an application, we shall study the functional transformations of the differential operators
∂
∂x ,
∂
∂y under a rotation around the positive z-axis. To find the transformed operators, we have to work out expressions such as
∂
∂x ′ where x ′ = x( ˆ
R −1 P 1 ). Hence, we
are confronted with a functional form, viz., the derivative operator, of a transformed
argument, x ′ , but this is precisely where classical analysis comes to our rescue because it provides the chain rule needed to work out the coordinate change. We have:
∂
∂x ′ =
∂x
∂x ′
∂
∂x
+
∂y
∂x ′
∂
∂y
(1.21)
In order to evaluate this equation, we have to determine the partial derivatives in the
transformed coordinates. Using the result in Sect. 1.1 but keeping in mind that the
4 The fact that two operators transform a given function in the same way does not automatically
imply that those operators are the same. Operators will be the same if this relationship extends over
the entire Hilbert space.
1 Operations
1.3 Operations and Operators
Besides functions, we must also consider the action of operations on operators. In
quantum chemistry, operators, such as the Hamiltonian, H, are usually spatial functions and, as such, are transformed in the same way as ordinary functions, e.g.,
H ′ (P 1 ) = H( ˆ
R −1 P 1 ). So why devote a special section to this? Well, operators are
different from functions in the sense that they also operate on a subsequent argument, which is itself usually a function. Hence, when symmetry is applied to an
operator, it will also affect whatever follows the operator. Symmetry operations act
on the entire expression at once. This can be stated for a general operator O as
follows:
ˆ
ROf = O
′ ˆ
Rf
(1.17)
From this we can identify the transformed operator O ′ by smuggling ˆ
R −1 ˆ
R (= ˆ
E)
into the left-hand side of the equation:
ˆ
RO ˆ
R
−1 ˆ
Rf = O
′ ˆ
Rf
(1.18)
This equality is true for any function f and thus implies 4 that the operators preceding ˆ
Rf on both sides must be the same:
O
′ = ˆ
RO ˆ
R
−1
(1.19)
This equation provides the algebraic formalism for the transformation of an operator. In the case where O ′ = O, we say that the operator is invariant under the
symmetry operation. Equation (1.19) then reduces to
ˆ
RO − O ˆ
R =[ ˆ
R,O]=0
(1.20)
This bracket is know as the commutator of ˆ
R and O. If the commutator vanishes, we
say that ˆ
R and O commute. This is typically the case for the Hamiltonian. As an application, we shall study the functional transformations of the differential operators
∂
∂x ,
∂
∂y under a rotation around the positive z-axis. To find the transformed operators, we have to work out expressions such as
∂
∂x ′ where x ′ = x( ˆ
R −1 P 1 ). Hence, we
are confronted with a functional form, viz., the derivative operator, of a transformed
argument, x ′ , but this is precisely where classical analysis comes to our rescue because it provides the chain rule needed to work out the coordinate change. We have:
∂
∂x ′ =
∂x
∂x ′
∂
∂x
+
∂y
∂x ′
∂
∂y
(1.21)
In order to evaluate this equation, we have to determine the partial derivatives in the
transformed coordinates. Using the result in Sect. 1.1 but keeping in mind that the
4 The fact that two operators transform a given function in the same way does not automatically
imply that those operators are the same. Operators will be the same if this relationship extends over
the entire Hilbert space.