6-2
Stationary States
Recall that the general solution to the time-dependent Schrödinger equation is the
superposition-of-states wave packet defined by
!
x
,
y
,
z
,
t
( ) =
c
n
"
n
n
=
1
#
$
x
,
y
,
z
( )
e
%
iE
n
t
/
h
(6-1)
Let us consider what is required for the probability density at any position in space
not
to
be varying with time. That is, the system is “stationary” or in a “stationary state”. The
probability density at a particular position is
!
*
x
,
y
,
z
,
t
( )
!
x
,
y
,
z
,
t
( )
=
c
m
*
"
m
*
m
=
1
#
$
x
,
y
,
z
( )
e
iE
m
t
/
h
c
n
"
n
n
=
1
#
$
x
,
y
,
z
( )
e
%
iE
n
t
/
h
=
c
m
*
c
n
"
m
*
x
,
y
,
z
( )
"
n
x
,
y
,
z
( )
e
iE
m
t
/
h
e
%
iE
n
t
/
h
m
,
n
=
1
#
$
=
c
i
2
"
i
*
x
,
y
,
z
( )
"
i
x
,
y
,
z
( )
i
=
1
#
$
+
c
m
*
c
n
"
m
*
x
,
y
,
z
( )
"
n
x
,
y
,
z
( )
e
i E
m
%
E
n
( )
t
/
h
m
&
n
#
$
(6-2)
Notice that we are not integrating over all space here, we are only asking about a
particular position, so the various terms in the second sum of the bottom equality are
not
necessarily zero. It is these terms that include a time dependence, so for a system to have
a stationary probability density, we must find a way to make every term in the second
sum equal to zero. There is only one way to do this (unless all states are degenerate):
every value of {
c
} must be zero except for a single one. That is
!
x
,
y
,
z
,
t
( ) =
c
j
"
j
x
,
y
,
z
( )
e
#
iE
j
t
/
h
(6-3)
Since both the time-dependent and time-independent wave functions are normalized over
all space, note that
!
x
,
y
,
z
,
t
( )
!
x
,
y
,
z
,
t
( )
=
j
"
j
x
,
y
,
z
( )
e
#
iE
j
t
/
h
c
j
"
j
x
,
y
,
z
( )
e
#
iE
j
t
/
h
=
c
j
*
e
iE
j
t
/
h
c
j
e
#
iE
j
t
/
h
"
j
x
,
y
,
z
( )
"
j
x
,
y
,
z
( )
=
c
j
2
(6-4)
Thus, the square modulus of
c
j
must be one (since
Ψ
is normalized).