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
Hψ Eψ
(16.3-1)
places
H and E in 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:
p mv
(16.3-2a)
or
p x mv x , p y mv y , and p z mv z
(16.3-2b)
If a particle moves only in the x direction, the kinetic energy is
K
1
2
mv
2
x
p 2
x
2m
(16.3-3)
The expression for the energy as a function of momenta and coordinates is called
Hamilton’s principal function or the classical Hamiltonian, and is denoted by H .
The Hamiltonian is named for Sir
William Rowan Hamilton, 1805–1865,
a great Irish mathematician, physicist,
and astronomer.
We assert that the classical Hamiltonian H is in one-to-one correspondence with
the Hamiltonian operator:
H (x, p x ) K + V
p 2
x
2m
+ V (x)
H −
¯
h
2
2m
d 2
dx 2 + V (x)
(16.3-4)
where K is the kinetic energy and V is the potential energy. In this context the symbol
means “is in one-to-one correspondence with.” The potential energy function V (x)
occurs on both sides of this correspondence in the same way. We assert that the operator
for the potential energy is the multiplication operator V (x):
V V (x)
(16.3-5)
We extend this assertion and postulate that any function of coordinates corresponds to
the operator for multiplication by that function.
The operator that corresponds to the kinetic energy K is now
K
p 2
x
2m
K −
¯
h
2
2m
d 2
dx 2
(16.3-6)
The operator for the square of the x component of the momentum must be
p
2
x − ¯
h
2 d 2
dx 2
(16.3-7)
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
Hψ Eψ
(16.3-1)
places
H and E in 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:
p mv
(16.3-2a)
or
p x mv x , p y mv y , and p z mv z
(16.3-2b)
If a particle moves only in the x direction, the kinetic energy is
K
1
2
mv
2
x
p 2
x
2m
(16.3-3)
The expression for the energy as a function of momenta and coordinates is called
Hamilton’s principal function or the classical Hamiltonian, and is denoted by H .
The Hamiltonian is named for Sir
William Rowan Hamilton, 1805–1865,
a great Irish mathematician, physicist,
and astronomer.
We assert that the classical Hamiltonian H is in one-to-one correspondence with
the Hamiltonian operator:
H (x, p x ) K + V
p 2
x
2m
+ V (x)
H −
¯
h
2
2m
d 2
dx 2 + V (x)
(16.3-4)
where K is the kinetic energy and V is the potential energy. In this context the symbol
means “is in one-to-one correspondence with.” The potential energy function V (x)
occurs on both sides of this correspondence in the same way. We assert that the operator
for the potential energy is the multiplication operator V (x):
V V (x)
(16.3-5)
We extend this assertion and postulate that any function of coordinates corresponds to
the operator for multiplication by that function.
The operator that corresponds to the kinetic energy K is now
K
p 2
x
2m
K −
¯
h
2
2m
d 2
dx 2
(16.3-6)
The operator for the square of the x component of the momentum must be
p
2
x − ¯
h
2 d 2
dx 2
(16.3-7)