Physical Chemistry Third Edition

688 16 The Principles of Quantum Mechanics. II. The Postulates of Quantum Mechanics

16.3 The Operator Corresponding to a Given Variable

We need to find the operator that corresponds to a particular mechanical variable. We

begin by assuming that the Hamiltonian operator is the mathematical operator that is

in one-to-one correspondence with the energy of a system. This is plausible since the

time-independent Schrödinger equation

Hψ̂ Eψ (16.3-1)

placesĤandEin a unique relationship to each other.

Any mechanical variable can be written either as a function of coordinates and

velocities or as a function of coordinates and momenta. It turns out that the variables

need to be expressed in terms of Cartesian coordinates and momentum components.

In Cartesian coordinates the momentum of a particle is its mass times its velocity:

pmv (16.3-2a)

or

pxmvx, pymvy, and pzmvz (16.3-2b)

If a particle moves only in thexdirection, the kinetic energy is

K 

1

2

mv^2 x

p^2 x

2 m

(16.3-3)

The expression for the energy as a function of momenta and coordinates is called
The Hamiltonian is named for Sir Hamilton’s principal functionor theclassical Hamiltonian, and is denoted byH.
William Rowan Hamilton, 1805–1865,
a great Irish mathematician, physicist,
and astronomer.

We assert that the classical HamiltonianH is in one-to-one correspondence with

the Hamiltonian operator:

H(x,px)K +V 

p^2 x

2 m

+V(x)
Ĥ−

h ̄^2

2 m

d^2

dx^2

+V(x) (16.3-4)

whereKis the kinetic energy andVis the potential energy. In this context the symbol


means “is in one-to-one correspondence with.” The potential energy functionV(x)

occurs on both sides of this correspondence in the same way. We assert that the operator

for the potential energy is the multiplication operatorV(x):

V̂
V(x) (16.3-5)

We extend this assertion and postulate thatany function of coordinates corresponds to

the operator for multiplication by that function.

The operator that corresponds to the kinetic energyK is now

K 

p^2 x

2 m

K̂− ̄

h^2

2 m

d^2

dx^2

(16.3-6)

The operator for the square of thexcomponent of the momentum must be

p̂x^2 −h ̄^2

d^2

dx^2

(16.3-7)