Chemistry 362
Dr. Jean M. Standard
Problem Set 2 Solutions
1.
Determine the outcomes of operating the following operators on the functions listed. (In these functions,
a is a constant.)
a.) operator: d/dx ; function: x e −ax
d
x e −ax
dx
(
)
= e −ax − a x e −ax
€
2
; function: 3x + 5
b.) operator:
€
2
The result is 3x + 5 . No further simplification can be done.
€
€
2
2
c.) operator: d / dx ; function: x sin 3x
€
d2
d
x sin 3x ) =
(sin 3x + 3x cos 3x)
2 (
dx
dx
€
€
= 3cos 3x + 3cos 3x − 9x sin 3x
= 6 cos 3x − 9x sin 3x .
2.
€
Consider two operators, Aˆ = x and Bˆ = d / dx . For the function f (x) = x e −ax , where a is a constant,
calculate the following quantities.
a.) Aˆ Bˆ f€( x )
€
€
€
d
Aˆ Bˆ f ( x ) = x
x e −ax
dx
(
)
(
= x e −ax − a x e −ax
)
Aˆ Bˆ f ( x ) = x e −ax − a x 2 e −ax .
€
b.) Bˆ Aˆ f ( x )
d
x x e −ax
dx
d
=
x 2 e −ax
dx
= 2x e −ax − a x 2 e −ax .
Bˆ Aˆ f ( x ) =
€
(
Bˆ Aˆ f ( x )
€
(
)
)
c.) Are the results from parts (a) and (b) the same? If the results are not the same, it is said that
the two operators do not commute.
No, in this case, the results are not the same. That is, Aˆ Bˆ f ( x ) ≠ Bˆ Aˆ f ( x ) . Therefore, the
operators Aˆ and Bˆ do not commute.
€
€
€
2
2
3.
ikx
−ax
. In these
Which of the following functions are eigenfunctions of d/dx? e , cos kx, kx, e
functions, k and a are constants. For those functions that are eigenfunctions, give the eigenvalues.
a.)
d ikx
e
= ike ikx
dx
€
ikx
The function e is an eigenfunction of d/dx with eigenvalue ik.
€
b.)
d
€
cos kx = − k sin kx
dx
The function cos kx is not an eigenfunction of d/dx.
€
c.)
d
€ = k
kx
dx
The function kx is not an eigenfunction of d/dx.
€
d.)
2
d −ax 2
e€
= − 2ax e −ax
dx
2
−ax
The function e
is not an eigenfunction of d/dx.
€
€
2
4.
−ax
2
2
Demonstrate whether or not the function f (x) = e
is an eigenfunction of the operator d / dx . If the
function is an eigenfunction, give the eigenvalue.
The first derivative is
2
d −ax 2
e
=€− 2ax e −ax .
dx
€
2
d 2 −ax 2
The second derivative is
e
= 4a 2 x 2 − 2a e −ax .
2
dx
€
(
)
2
−ax
Since this result does not equal a constant times the original function, f (x) = e
is NOT an eigenfunction of
2
2
d
/
dx
the operator €
.
€
€
3
5.
For a particle moving in three dimensions, the classical expression for the kinetic energy T is given by:
T=
1
px2 + py2 + pz2 .
2m
(
)
Give the corresponding expression for the quantum mechanical kinetic energy operator in three
dimensions, Tˆ .
€
To convert a classical component of momentum, pα , where α = x, y,or z , into a quantum mechanical operator
pˆα €
, we use the relation,
€
pˆα €= − i!
€
d
.
dα
For the momentum squared (as appears in the kinetic energy expression), we therefore have
€
pˆα2 = − ! 2
d2
.
dα 2
Substituting these momentum operators, the quantum mechanical kinetic energy operator Tˆ becomes
€
ˆ
T =
2
2 &
1 # 2 d2
2 d
2 d
−
!
−
!
%−!
(,
2m $
dx 2
dy 2
dz 2 ' €
and factoring out the −! 2 terms gives
€
!2 # d 2
d2
d2 &
Tˆ = −
+
% 2 +
( .
2m $ dx
dy 2
dz 2 '
€
Since Tˆ involves three variables rather than just one, it is more appropriate to use partial derivatives,
€
!2 $ ∂ 2
∂2
∂2 '
Tˆ = −
+
& 2 +
) .
2m % ∂x
∂y 2
∂z 2 (
€
€
6.
Indicate which functions are acceptable wavefunctions:
x 2 - not acceptable (goes to
e−x
€
€
€
€
∞ as x → ∞)
- not acceptable (goes to ∞ as x → − ∞ )
€
€
e x - not acceptable (goes to ∞ as x → ∞)
€
€
e−ax
2
- acceptable
€
€
4
7.
$ nπ x '
Normalize the function Ψ(x) = sin&
) , where n is a positive integer and a is a constant. The range of x
% a (
is 0 to a. (You will have to look up the integral in a table.)
For a normalized wavefunction Ψnorm , the integral over all space must be equal to 1.
€
Let Ψnorm = NΨ , where N is the normalization constant. For this problem, the range of the coordinate x is 0 to
a. Therefore, for normalization we require
€
∫
€
a
0
2
Ψnorm
(x) dx = 1 .
Substituting Ψnorm = NΨ leads to
€
N2
∫
∫
a
€
a
0
Ψ 2 (x) dx = 1 .
$ nπ x '
Substituting Ψ(x) = sin&
) , we have
% a ( €
N2
€
0
# nπx &
sin 2 %
( dx = 1 .
$ a '
To evaluate the integral, we can look it up in a table. From the CRC,
€
∫ sin bx dx
2
=
x
sin2bx
−
.
2
4b
nπ
By making the substitution b =
, we obtain
a
€
∫
€
a
0
a
+x
$ nπx '
$ 2nπx '.
a
sin 2 &
sin&
) dx = - −
)0 .
% a (
4nπ % a (/0
,2
Evaluating the limits yields
€
∫
a
0
$ nπx '
$a
' $
'
a
a
sin 2 &
sin( 2nπ ) ) − & 0 −
sin( 0) ) .
) dx = & −
% a (
%2
( %
(
4nπ
4nπ
Note that since sin2nπ = 0 and sin0 = 0 , the first and last terms drop out, and we are left with
€
€
∫
€
a
0
$ nπx '
a
sin 2 &
.
) dx =
% a (
2
2
Finally, substituting this result into the normalization condition, N
€
"a%
N 2$ ' = 1 ,
#2&
€
or N =
Therefore, the normalized wavefunction Ψnorm is given by
€
Ψnorm =
€
€
$ nπx '
2
sin&
) .
% a (
a
∫
a
0
# nπx &
sin 2 %
( dx = 1 , we have
$ a '
2
.
a
5
8.
Is the function in problem 7 an eigenfunction of the linear momentum operator, pˆ x ?
Since pˆ x = −i!
d
, the eigenvalue equation would be
dx
€
d
Ψ(x)
dx
% nπx (
d
= −i! sin'
*
dx & a )
% nπ ( % nπx (
pˆ x Ψ(x) = −i!' * cos'
*
& a ) & a )
pˆ x Ψ(x) = −i!
€
Since the right side is not equal to a constant times the original function, the function Ψ(x) from problem 7 is
NOT an eigenfunction of the operator pˆ x .
€
€
€