1 The Potential Energy Surface in Molecular Quantum Mechanics
17
Let us denote the classical dynamical variables for the electrons collectively as
x, p, and those for the nuclei by X, P and denote the classical Hamiltonian by
H(x, p, X, P). After the customary canonical quantization these variables become
time-independent operators in a Schrödinger representation
x → ˆ
x etc.
In the following it will be important to distinguish between operators and c-numbers,
so in the following we will use the ˆ
x notation for operators, and make no special
choice of representation.
As we have seen, the idea that the kinetic energy of the massive nuclei could be
treated as a perturbation of the electronic motion was first formulated in the framework of the Old Quantum Theory. The idea was to separate the classical Hamiltonian H into two parts to isolate the nuclear momentum variables
H(x, p, X, P) = H o (x, p, X) + κ
4 H 1 (P).
(1.18)
According to Hamilton’s equations for the unperturbed problem
dX
dt
= {X, H o } = 0,
(1.19)
using Poisson-bracket notation, which was interpreted (correctly) as describing the
dynamics of the electrons in the field of stationary nuclei. This was the starting point
of Born and Heisenberg’s calculations [36].
Let us now move to quantum theory and recast (1.18) as an operator relation,
writing the molecular Hamiltonian operator as
ˆ
H(ˆ x, ˆ
p, ˆ
X, ˆ
P) = ˆ
H o (ˆ x, ˆ
p, ˆ
X) + κ
4 ˆ
H 1 ( ˆ
P)
(1.20)
with equation of motion under ˆ
H o
i
d ˆ
X
dt
= [ ˆ
X, ˆ
H o ] = 0
(1.21)
from which we infer the nuclear position operators ˆ
X are constants of the motion
under ˆ
H o . We no longer make the interpretation that follows from (1.19) since specifying precisely the positions {X} for stationary nuclei violates the Uncertainty Principle. Instead (1.21) leads to a completely different conclusion (see below).
We must now take a little bit of care about the definition of the variables, and
dispose of the uninteresting overall motion of the molecule [4]. Since the Coulomb
interaction only depends on interparticle distances it is translation invariant, and
therefore the total momentum operator ˆ
P
ˆ
P =
n
ˆ
p n
17
Let us denote the classical dynamical variables for the electrons collectively as
x, p, and those for the nuclei by X, P and denote the classical Hamiltonian by
H(x, p, X, P). After the customary canonical quantization these variables become
time-independent operators in a Schrödinger representation
x → ˆ
x etc.
In the following it will be important to distinguish between operators and c-numbers,
so in the following we will use the ˆ
x notation for operators, and make no special
choice of representation.
As we have seen, the idea that the kinetic energy of the massive nuclei could be
treated as a perturbation of the electronic motion was first formulated in the framework of the Old Quantum Theory. The idea was to separate the classical Hamiltonian H into two parts to isolate the nuclear momentum variables
H(x, p, X, P) = H o (x, p, X) + κ
4 H 1 (P).
(1.18)
According to Hamilton’s equations for the unperturbed problem
dX
dt
= {X, H o } = 0,
(1.19)
using Poisson-bracket notation, which was interpreted (correctly) as describing the
dynamics of the electrons in the field of stationary nuclei. This was the starting point
of Born and Heisenberg’s calculations [36].
Let us now move to quantum theory and recast (1.18) as an operator relation,
writing the molecular Hamiltonian operator as
ˆ
H(ˆ x, ˆ
p, ˆ
X, ˆ
P) = ˆ
H o (ˆ x, ˆ
p, ˆ
X) + κ
4 ˆ
H 1 ( ˆ
P)
(1.20)
with equation of motion under ˆ
H o
i
d ˆ
X
dt
= [ ˆ
X, ˆ
H o ] = 0
(1.21)
from which we infer the nuclear position operators ˆ
X are constants of the motion
under ˆ
H o . We no longer make the interpretation that follows from (1.19) since specifying precisely the positions {X} for stationary nuclei violates the Uncertainty Principle. Instead (1.21) leads to a completely different conclusion (see below).
We must now take a little bit of care about the definition of the variables, and
dispose of the uninteresting overall motion of the molecule [4]. Since the Coulomb
interaction only depends on interparticle distances it is translation invariant, and
therefore the total momentum operator ˆ
P
ˆ
P =
n
ˆ
p n