1. No category

Chemistry 460
Dr. Jean M. Standard
Problem Set 5 Solutions
1.
Start with the differential equation
d 2 f (x)
dx
2
−
1
α
f (x) +
f (x) = 0 ,
4
x
where a is a constant.
€
a.) For x on the interval 0 ≤ x ≤ ∞ , obtain the asymptotic solution to the equation in the limit that
x → ∞.
€
To begin, we obtain the asymptotic solution for x → ∞ . As x → ∞ , the last term in the differential
α€
equation,
f (x) , goes to zero. Thus, the asymptotic form of the differential equation is
x
€
d 2 f (x)
dx
€
2
€
=
1
f (x) .
4
The solution can be written in the form of an exponential. Assume a solution of the form
€
f (x) = e± βx ,
where b is a positive constant. The derivatives of the function are
€
f " (x) = ± β e ± βx ,
f " " (x) = β 2 e ± βx .
Substituting the second derivative into the asymptotic equation,
€
β 2 e± βx =
1
4
e± βx .
Therefore, the exponent is given by β 2 = 41 , or β = 12 . The asymptotic solution thus has the form
€
f (x) = e±x / 2 .
€
€
Since the positive exponent goes to infinity as x goes to infinity, the only valid asymptotic solution is with
the negative exponent,
€
f (x) = e−x / 2 .
€
2
1.
Continued
b.) Obtain the differential equation for the remainder.
To get the remainder, we write the full solution as the asymptotic part times the remainder. In this case,
f ( x ) = e−x / 2 g( x ) ,
where the function g(x) is the remainder function.
€
Calculating the derivatives,
f "( x) = −
1
2
(
)
e − x / 2 g( x ) + e − x / 2 g "( x ) = e − x / 2 − 12 g( x ) + g "( x ) ,
[
]
[
f ""( x ) = − 12 e − x / 2 − 12 g ( x ) + g " ( x ) + e − x / 2 − 12 g " ( x ) + g " " ( x )
€
[
]
]
= e − x / 2 g " "( x ) − g "( x ) + 41 g( x ) .
Substituting these expressions into the original differential equation,
€
d 2 f (x)
dx
2
−
1
α
f (x) +
f (x) = 0
4
x
α −x / 2
e − x / 2 g $ $( x ) − g $( x ) + 41 g( x ) − 41 e − x / 2 g( x ) +
e
g( x ) = 0
x
[
]
α
%
(
e − x / 2 ' g $ $ ( x) − g $ ( x) +
g( x) * = 0 .
x
&
)
−x / 2
Dividing both sides of the equation by the factor e
, the equation for the remainder is
€
α
g " "( x) − g "( x) +
g( x) = 0 .
x
€
€
c.) Write the full solution as the asymptotic part times an infinite series in powers of x. Obtain a
recursion relation for the coefficients in the power series.
Next, we assume a power series solution for the remainder,
∞
g( x) =
∑ ak x k .
k=0
€
3
1 c.)
Continued
Evaluating the derivatives,
∞
∑ a k k x k−1 ,
g"( x ) =
k=1
∞
€
g " "( x) =
∑
a k k ( k − 1) x k − 2 .
k=2
Next, the summations are rewritten so that they start at k=0,
€
∞
g"( x ) =
∑ a k k x k−1 ,
k=0
∞
€
g " "( x) =
∑
a k +1 k ( k + 1) x k −1 .
k=0
The derivatives are substituted into the differential equation for the remainder,
€
∞
∑
g " " ( x) − g " ( x) +
a k +1 k ( k + 1) x k −1 −
k=0
∞
∑
α
g( x) = 0
x
a k k x k −1 +
k=0
α
x
∞
∑
ak x k
k=0
Combining the summations and factoring out the powers of x yields
€
∞
∑ { ak+1 k (k + 1) −
a k k + α a k } x k−1 = 0 .
k=0
k−1
In order for a solution to exist, the coefficients of x
must be zero:
€
a k+1 k ( k + 1) − a k k + α a k = 0.
€
Solving for a k+1 leads to the recursion formula:
€
a k+1 =
€
€
k −α
ak .
k ( k + 1)
= 0.
4
1.
Continued
d.) Truncate the series to obtain an expression for α .
If the solution is truncated at the nth term, then a n +1 = 0 .
€
a n +1 = 0 =
n −α
a .
n( n + 1) n
If a n ≠ 0 , then in order for the above equation to be satisfied, we must have
€
n −α
= 0,
n( n + 1)
€
or
€
α = n.
€
2.
The intensity of a pure vibrational spectral transition n → m from a state defined by the wavefunction
ψ n to a state defined by the wavefunction ψ m , is determined by the oscillator strength f nm ,
2
€
fnm = ( En − Em )
€ 3
€
2
ψn (x) x ψ m (x)
,
€
where ψ n is a harmonic oscillator eigenfunction with quantum number n and ψ m is a harmonic
oscillator eigenfunction with quantum number m. The oscillator strength in this equation is valid for
infrared transitions.
€
€
a.) Use the harmonic oscillator eigenfunctions to evaluate ψ n (x) x ψ m (x) . It may be helpful to use the
recursion relation for Hermite polynomials, H v+1 ( z ) = 2 z H v ( z ) − 2 v H v–1 ( z ) . Show that the
integral is nonzero only if m differs from n by one quanta; that is, if m = n ± 1. This is the vibrational
selection rule, which states that a transition occurs only if the vibrational quantum number changes
by 1.
The oscillator strength integral is
∞
ψn x ψ m = ∫ ψn* ( x ) x ψ m ( x ) dx .
−∞
In order to evaluate this integral, we must develop an expression for the product xψ m . Substituting the
form of the harmonic oscillator wavefunction,
xψ m ( x ) = x N m H m
= N m xH m
The next step is to use the recursion relation, H m+1
solve it for x H m
(
(
( α x) e
( α x) e
−α x 2 /2
−α x 2 /2
.
)
α x = 2 α x H m
(
)
α x − 2 m H m–1
)
αx ,
x Hm
(
)
α x = 1
H m+1
2 α
(
)
α x + m
H m–1
α
(
)
α x .
Substituting this into the expression for xψ m yields
2
! 1
$
m
xψ m ( x ) = N m #
H m+1 α x + H m–1 α x & e−α x /2
"2 α
%
α
2
2
N
mN m
= m H m+1 α x e−α x /2 + H m–1 α x e−α x /2 .
2 α
α
(
(
)
)
(
)
(
)
(
)
α x , and
5
2
a.)
6
continued
Now we can put this expression for x 2ψ v back into the oscillator strength integral and break it up into two
separate integrals,
∞
ψn x ψ m = ∫ ψn* ( x ) x ψ m ( x ) dx
−∞
∞
$ N
= ∫ ψn* ( x ) & m H m+1
%2 α
−∞
ψn x ψ m = Nm ∞ *
∫ ψn ( x ) H m+1
2 α −∞
(
(
2
)
α x e−α x /2 + )
mN m
H m–1
α
2
α x e−α x /2 dx + (
2
'
α x e−α x /2 ) dx
(
)
mN m ∞ *
∫ ψn ( x ) H m–1
α −∞
(
)
2
α x e−α x /2 dx .
The next step is to include the appropriate normalization constants for the m+1 and m–1 wavefunctions by
multiplying and dividing by the requisite factors,
ψn x ψ m = Nm 1 ∞ *
∫ ψn ( x ) $% N m+1H m+1
N m+1 2 α −∞
+ (
2
α x e−α x /2 &' dx )
Nm m ∞ *
∫ ψn ( x ) $% N m−1H m–1
N m−1 α −∞
(
2
α x e−α x /2 &' dx .
)
The terms that appear in bracket in each integral are harmonic oscillator eigenfunctions. Identifying them
yields
ψn x ψ m = Nm 1 ∞ *
N
m ∞ *
∫ ψn ( x ) ψ m+1 ( x ) dx + m
∫ ψn ( x ) ψ m−1 ( x ) dx .
N m+1 2 α −∞
N m−1 α −∞
Because of orthonormality, the first integral on the right is zero unless n=m+1 and the second integral on
the right is zero unless n=m–1. The can be rewritten in a compact form as
ψn x ψ m = Nm 1
N
m
δn,m+1 + m
δn,m−1 ,
N m+1 2 α
N m−1 α
where the factor δ i, j is called a Kronecker delta and is defined as
$0
δ i, j = %
&1
€
i≠ j '
(.
i= j )
Finally, we can use the harmonic oscillator normalization constant N v , which has the form
€
€
1/ 2
$ α '1/ 4 $ 1 '
N v = & ) &&€ v )) .
% π ( % 2 v! (
2
a.)
7
continued
The ratio
Nm
that appears in the first term of the oscillator strength integral is
N m+1
! α $1/4 ! 1 $1/2
# & # m &
Nm
" π % " 2 m! %
= = 2 ( m +1) .
N m+1
$1/2
! α $1/4 !
1
&
# & ## m+1
" π % " 2 ( m +1)! &%
And the ratio
Nm
that appears in the second term of the oscillator strength integral is
N m−1
" α %1/4 " 1 %1/2
$ ' $ m '
Nm
1
# π & # 2 m! &
= = .
1/2
1/4 "
N m−1
2m
%
"α %
1
'
$ ' $$ m−1
# π & # 2 ( m −1)! '&
Substituting, the oscillator strength integral becomes
ψn x ψ m = Nm 1
N
m
δn,m+1 + m
δn,m−1
N m+1 2 α
N m−1 α
= ψn x ψ m = 2 ( m +1)
2 α
δn,m+1 +
m +1
δn,m+1 +
α
1 m
δn,m−1
2m α
m
δn,m−1 .
α
1/2
! mk $
Finally, using the definition α = # 2 & , the oscillator strength integral is
"! %
! ! 2 $1/4
ψn x ψ m = #
& { m +1 δn,m+1 + m δn,m−1 } .
"mk %
2.
8
continued
b.) Using the equation given above, determine the oscillator strengths f nm for m = n – 1 and m = n + 1.
The oscillator strength is
f nm
2
=
( En − Em )
3
€
ψ n (x)
x ψ m (x)
2
.
From part (a), the integral is
€
! ! 2 $1/4
ψn x ψ m = #
& { m +1 δn,m+1 + m δn,m−1 } .
"mk %
Case 1: m = n – 1, or n = m + 1
For this case, the first term in the integral is non-zero and the second term is zero. The integral is
" ! 2 %1/4
" ! 2 %1/4
ψn x ψn−1 = $
m +1 = $
n.
'
'
#mk &
#mk &
Substituting, the expression for the oscillator strength becomes
fn,n−1 = 2
( En − En−1 )
3
" ! 2 %1/4
2
= ( En − En−1 ) $
n
'
3
#mk &
" ! 2 %1/2
2
fn,n−1 = ( En − En−1 ) $
' n .
3
#mk &
The harmonic oscillator energy is
(
En = hν 0 n +
1
2
),
so the energy difference is given by
€
En − En−1 = hν 0 ( n + 12 ) − hν 0 ( n −1+ 12 )
= hν 0 .
2
ψn x ψn−1
2
2
9
b.) continued
Substituting,
(* " 2 %1/2 ,*
2
!
fn,n−1 = ( En − En−1 ) ) $
' n 3
*+ # m k &
*.
fn,n−1 = An alternate expression, using ν 0 =
2nhν 0
3
" ! 2 %1/2
$
' .
#mk &
1/ 2
1 $k'
& ) , is given by
2π % m (
1/2
2nhν 0 " ! 2 %
fn,n−1 = $
'
3 #mk &
€
= fn,n−1 = 2n h
6π
1/2
" k %1/2 " ! 2 %
$
'
$ '
#m & #mk &
2n ! 2
.
3m
Case 2: m = n + 1, or n = m – 1
Recall that the integral in the expression for oscillator strength is
! ! 2 $1/4
ψn x ψ m = #
& { m +1 δn,m+1 + m δn,m−1 } .
"mk %
For this case, the first term in the integral is zero and the second term is non-zero. The integral is therefore
ψn x ψn+1
! ! 2 $1/4
! ! 2 $1/4
= #
m = #
n +1 .
&
&
"mk %
"mk %
Substituting, the expression for the oscillator strength becomes
fn,n+1 = 2
( En − En+1 )
3
ψn x ψn+1
2
" ! 2 %1/4
2
= ( En − En+1 ) $
n +1
'
3
#mk &
fn,n+1 = " ! 2 %1/2
2
( En − En+1 ) $ ' (n +1) .
3
#mk &
2
2 b.)
10
continued
The energy difference is
En − En+1 = hν 0 ( n + 12 ) − hν 0 ( n +1+ 12 )
= − hν 0 .
Substituting,
fn,n+1 = Substituting ν 0 =
€
" ! 2 %1/2
2
E
−
E
( n n+1 ) $ ' (n +1)
3
#mk &
fn,n+1 = −
1/2
2 ( n +1) hν 0 " ! 2 %
$
' .
3
#mk &
fn,n+1 = −
2 ( n +1) hν 0
3
1/ 2
1 $k'
& ) ,
2π % m (
= −
fn,n+1 = −
" ! 2 %1/2
$
'
#mk &
1/2
1/2
2 ( n +1) h " k % " ! 2 %
'
$ ' $
6π
#m & #mk &
2 ( n +1) ! 2
.
3m
3.
2
Determine the expectation value of x in the vth state of the harmonic oscillator. [Hint: It might be
helpful to use the recursion relation for Hermite polynomials, H v+1 ( z ) = 2 z H v ( z ) − 2 v H v–1 ( z ) . You
will have to apply this relation twice.]
€
The expectation value is
x 2 = ψv x 2 ψv .
In order to evaluate this integral, we must develop an expression for x 2ψ v . Substituting the form of the
harmonic oscillator wavefunction,
x 2ψv ( x ) = x 2 N v H v
= N v x 2 H v
( α x) e
( α x) e
−α x 2 /2
The next step is to use the recursion relation and solve for x H v
x Hv
Substituting this for x H v
(
αx
(
)
α x = 1
H v+1
2 α
(
−α x 2 /2
(
.
)
αx ,
)
α x + v
H v–1
α
(
)
α x .
) yields
! 1
v
x 2ψv ( x ) = N v x #
H v+1 α x + H v–1 α x
"2 α
α
2
N
vN
= v x H v+1 α x e−α x /2 + v x H v–1
2 α
α
(
(
)
(
)
$
)&% e
(
−α x 2 /2
2
)
α x e−α x /2 .
Now we have to again use the recursion relation for the Hermite polynomials in order to evaluate
x H v+1
(
αx
) and x H (
v–1
)
α x . Thus, we have
and x H v-1
)
1
H v+2
2 α
(
)
1
Hv
2 α
α x + (
α x = (
α x = x H v+1
(
)
α x + )
v+1
Hv
α
v-1
H v-2
α
(
(
αx
)
)
α x .
11
3.
12
continued
Substituting these into the expression for x 2ψ v gives
2
$
Nv ! 1
v+1
#
H v+2 α x + H v α x & e−α x /2 %
2 α "2 α
α
2
$
vN ! 1
v-1
+ v #
H v α x + H v-2 α x & e−α x /2
"
%
α 2 α
α
(N
N
v+1
( ) H α x + N v v H α x + N v v (v-1) H
= ) v H v+2 α x + v
v
v
v-2
2α
2α
α
* 4α
(N
+
2
N ( 2v+1)
N v ( v-1)
x 2ψv ( x ) = ) v H v+2 α x + v
H v α x + v
H v-2 α x , e−α x /2 .
4
α
2
α
α
*
-
(
x 2ψv ( x ) = )
(
(
(
)
(
)
)
)
(
(
)
)
(
(
)
)
(
(
+
2
α x , e−α x /2
-
)
)
Now we can put this expression for x 2ψ v back into the expectation value and break the expectation value up
into three separate integrals,
x 2 = ψv x 2 ψv
∞
= ∫ ψv* ( x ) x 2 ψv ( x ) dx
−∞
∞
$N
= ∫ ψv* ( x ) % v H v+2
& 4α
−∞
x 2 = Nv ∞ *
∫ ψv ( x ) H v+2
4α −∞
+ (
(
)
α x + N v ( 2v+1)
Hv
2α
2
)
α x e−α x /2 dx + N v v ( v-1) ∞ *
∫ ψv ( x ) H v-2
α
−∞
(
(
)
α x + N v v ( v-1)
H v-2
α
N v ( 2v+1) ∞ *
∫ ψv ( x ) H v
2α
−∞
(
(
'
2
α x ( e−α x /2 dx
)
)
2
)
α x e−α x /2 dx 2
)
α x e−α x /2 dx .
The next step is to include in the first and third terms the appropriate normalization constants for the v+1 and
v-1 wavefunctions by multiplying and dividing by the requisite factors,
x 2 = Nv 1 ∞ *
∫ ψv ( x ) N v+2 H v+2
N v+2 4α −∞
{
+ (
)
α x e−α x
2
/2
N v v ( v-1) ∞ *
∫ ψv ( x ) N v-2 H v-2
N v-2 α −∞
{
∞
∫ ψ ( x ) { N H (
} dx + 2v+1
2α
−∞
(
)
α x e−α x
2
/2
*
v
v
v
)
α x e−α x
2
/2
} dx } dx .
The terms that appear in bracket in each integral are harmonic oscillator eigenfunctions. Identifying them yields
x 2 = Nv 1 ∞ *
2v+1 ∞ *
N v ( v-1) ∞ *
∫ ψv ( x ) ψv+2 ( x ) dx + ∫ ψv ( x ) ψv ( x ) dx + v
∫ ψv ( x ) ψv-2 ( x ) dx .
N v+2 4α −∞
2α −∞
N v-2 α −∞
3.
13
continued
Using the orthogonality of the harmonic oscillator eigenfunctions, the first and third integrals equal 0. Also,
because the eigenfunctions are normalized, the second integral equals 1. Therefore, the expectation value
becomes
1 !
1$
x 2 = # v + & .
α "
2%
1/2
! mk $
Finally, using the definition α = # 2 & , the expectation value is
"! %
x
2
! ! 2 $1/2 !
1$
= # & # v + & .
"
2%
" mk %
4.
Evaluate the expectation value of px in the vth state of the harmonic oscillator. [Hint: Here it may be
helpful to use the recursion relation for Hermite polynomials as well as another relation for the derivative
of the Hermite polynomials, H v! ( z ) = 2nH v ( z ) .]
€
The expectation value is
px = ψv pˆ x ψv .
In order to evaluate this integral, we must develop an expression for pˆ x ψ v ( x ) . Substituting the form of the
harmonic oscillator wavefunction,
d
N v H v
dx
d
= −i!N v H v
dx
{
pˆ x ψv ( x ) = −i!
( α x) e
{ ( α x) e
−α x 2 /2
−α x 2 /2
}
}.
Using the product and chain rules for the derivative, we have
2
d
H v α x e−α x /2
dx
2
"
d
= −i!N v #e−α x /2 H v α x − α xH v
$
dx
{ (
pˆ x ψv ( x ) = −i!N v
}
)
(
)
2
%
α x e−α x /2 &.
'
(
)
Further use of the chain rule is required to obtain the derivative of the Hermite polynomial that appears in the
first term. For a Hermite polynomial that depends on the argument z, the derivative is
! dz $ dH ( z )
! dz $
d
H v ( z ) = # & v = # & H v' ( z ) .
" dx % dz
" dx %
dx
In this case, z = α x , so that we have
d
Hv
dx
(
)
α x = α H v!
(
)
α x .
Substituting, the expression for pˆ x ψ v ( x ) becomes
{
pˆ x ψv ( x ) = −i!N v e−α x
2
/2
α H v"
(
)
α x − α xH v
(
)
α x e−α x
2
/2
}.
The next step is to use the relation for the derivative of the Hermite polynomial for the first term and the
recursion relation for x H v
(
)
α x in the second term,
H v!
and
x Hv
(
)
(
)
α x = 2vH v-1
α x = 1
H v+1
2 α
(
(
)
α x ,
)
α x + v
H v–1
α
(
)
α x .
14
4.
15
continued
Substituting, we obtain
(
+
2
2
" 1
%
v
pˆ x ψv ( x ) = −i!N v )e−α x /2 α 2vH v-1 α x − α $
H v+1 α x + H v–1 α x ' e−α x /2 ,.
#2 α
&
α
*
2
2
2
.(
.+
α
= −i!N v ) 2v α H v-1 α x e−α x /2 − H v+1 α x e−α x /2 − v α H v–1 α x e−α x /2 ,
2
*.
-.
(
(
.(
pˆ x ψv ( x ) = −i!N v ) v α H v-1
.*
(
)
(
)
)
(
2
)
α x e−α x /2 − α
H v+1
2
(
(
)
)
(
)
2
.+
α x e−α x /2 , .
.-
)
Now we can put this expression for pˆ x ψ v ( x ) back into the expectation value and break the expectation value
up into two separate integrals,
px = ψv pˆ x ψv .
∞
= ∫ ψv* ( x ) pˆ x ψv ( x ) dx
−∞
∞
$&
px = −i!N v ∫ ψv* ( x ) % v α H v-1
&'
−∞
∞
px = −i!v α N v ∫ ψv* ( x ) H v-1
−∞
(
(
)
α x − )
α
H v+1
2
2
(
α x e−α x /2 dx + i!
(&
2
α x ) e−α x /2 dx
&*
)
∞
α
N v ∫ ψv* ( x ) H v+1
2
−∞
(
2
)
α x e−α x /2 dx .
The next step is to include in each term the appropriate normalization constants for the v+1 and v-1
wavefunctions by multiplying and dividing by the requisite factors,
px = −i!v α
Nv ∞ *
∫ ψv ( x ) $% N v-1H v-1
N v-1 −∞
(
2
α Nv ∞ *
α x e−α x /2 &' dx + i!
∫ ψv ( x ) $% N v+1H v+1
2 N v+1 −∞
)
(
2
α x e−α x /2 &' dx .
)
The terms that appear in brackets in each integral are the harmonic oscillator eigenfunctions. Identifying them
yields
px = −i!v α
Nv ∞ *
α Nv ∞ *
∫ ψv ( x ) ψv−1 ( x ) dx + i!
∫ ψv ( x ) ψv+1 ( x ) dx .
N v-1 −∞
2 N v+1 −∞
Using the orthogonality of the harmonic oscillator eigenfunctions, both integrals equal 0. Therefore, the
expectation value becomes
px = 0 . 5.
The Hermitian operator Aˆ is associated with the physical observable A. Two of the eigenfunctions of
Aˆ are φ1 and φ 2 . These eigenfunctions are normalized and orthogonal. The eigenvalues of φ1 and φ 2
are 3 and 2, so that
€
€
€
Aˆ φ1 = 3φ1
€
€
€
Aˆ φ 2 = 2 φ 2 .
€
Suppose that a state
ψ
has the form given by
€
ψ =
1
10
φ1 −
3
10
φ2 .
€
Determine the uncertainty ΔA for the state ψ .
€
The uncertainty ΔA is defined as
€
€
$
ΔA = & A 2 − A
%
€
2 '1/ 2
)
(
.
In order to calculate the uncertainty, the expectation values must first be evaluated. The expectation value
€
A 2 is
A2
€
=
ψ Aˆ 2 ψ
ψ ψ
=
ψ Aˆ 2 ψ ,
as long as the wavefunction is normalized. Evaluating the expectation value,
€
2
2
ˆ
A
= ψ A ψ
=
1
φ1 −
10
3
φ2
10
=
1
φ1 Aˆ 2 φ1
10
−
Aˆ 2
1
φ1 −
10
3
φ1 Aˆ 2 φ 2
10
3
φ2
10
3
−
φ1 Aˆ 2 φ 2
10
+
9
φ 2 Aˆ 2 φ 2 .
10
Applying the operator twice,
€
Aˆ 2 φ1 = Aˆ ( 3φ1) = 9φ1 ,
Aˆ 2 φ 2 = Aˆ ( 2φ 2 ) = 4φ 2 .
Substituting,
A
2
A2
€
€
9
12
27
36
=
φ1 φ1 −
φ1 φ 2 −
φ 2 φ1 +
φ2 φ2
10
10
10
10
9
12
27
36
=
⋅1 −
⋅0 −
⋅0 +
⋅1
10
10
10
10
45
=
.
10
16
5.
17
continued
The expectation value A is
A
=
ψ Aˆ ψ
ψ ψ
3
φ2
10
Aˆ
1
φ1 −
10
€
=
ψ Aˆ ψ .
Evaluating the expectation value,
€
A = ψ Aˆ ψ
=
1
φ1 −
10
=
1
φ1 Aˆ φ1
10
−
3
φ1 Aˆ φ 2
10
3
φ2
10
3
−
φ1 Aˆ φ 2
10
+
9
φ 2 Aˆ φ 2 .
10
Using the eigenvalue equations,
€
3
6
9
18
A =
φ1 φ1 −
φ1 φ 2 −
φ 2 φ1 +
φ2 φ2
10
10
10
10
3
6
9
18
=
⋅1 −
⋅0 −
⋅0 +
⋅1
10
10
10
10
21
A =
.
10
Substituting the expectation values into the expression for the uncertainty,
€
2 '1/ 2
$
ΔA = & A 2 − A
)
%
(
1/ 2
$ 45
* 21 -2 '
&
)
=
− , /
+ 10 . )(
&% 10
$ 450 − 441 '1/ 2
= &
)(
%
100
3
ΔA =
.
10
€
6
a.)
18
Evaluate the uncertainties Δx and Δpx for the harmonic oscillator.
The uncertainty Δx is defined as
€
€
[
Δx =
€
The expectation value
2 1/ 2
]
x2 − x
.
x 2 has already been evaluated. From Problem 3,
€
x
! ! 2 $1/2
= #
& ( v + 12 ) .
"mk %
2
The expectation value x is defined as
x = ψv x ψv .
In order to evaluate this integral, we must develop an expression for xψv . Substituting the form of the
harmonic oscillator wavefunction,
xψv ( x ) = x N v H v
(
= N v x H v
)
α x e−α x
(
Substituting this for x H v
(
(
αx
)
α x = 1
H v+1
2 α
(
/2
2
)
α x e−α x /2 .
The next step is to use the recursion relation and solve for x H v
x Hv
2
(
)
α x + )
αx ,
v
H v–1
α
(
)
α x .
) yields
2
! 1
$
v
xψv ( x ) = N v #
H v+1 α x + H v–1 α x & e−α x /2
"2 α
%
α
2
2
Nv
vN
= H v+1 α x e−α x /2 + v H v–1 α x e−α x /2 .
2 α
α
(
(
)
(
)
)
(
)
Now we can put this expression for xψv back into the expectation value and break the expectation value up
into two separate integrals,
x = ψv x ψv
∞
= ∫ ψv* ( x ) x ψv ( x ) dx
−∞
∞
$ N
= ∫ ψv* ( x ) % v H v+1
&2 α
−∞
x = Nv ∞ *
∫ ψv ( x ) H v+1
2 α −∞
(
(
)
α x + )
2
vN v
H v–1
α
α x e−α x /2 dx + (
2
'
α x ( e−α x /2 dx
)
)
vN v ∞ *
∫ ψv ( x ) H v-1
α −∞
(
)
2
α x e−α x /2 dx .
6
a.)
19
continued
The next step is to include in the two terms the appropriate normalization constants for the v+1 and v–1
wavefunctions by multiplying and dividing by the requisite factors,
x = = Nv ∞ *
∫ ψv ( x ) H v+1
2 α −∞
(
2
)
α x e−α x /2 dx + Nv 1 ∞ *
∫ ψv ( x ) $% N v+1H v+1
N v+1 2 α −∞
(
vN v ∞ *
∫ ψv ( x ) H v-1
α −∞
2
N
α x e−α x /2 &' dx + v
N
)
v-1
(
)
2
α x e−α x /2 dx v ∞ *
∫ ψv ( x ) $% N v-1H v-1
α −∞
(
2
α x e−α x /2 &' dx .
)
The terms that appear in bracket in each integral are harmonic oscillator eigenfunctions. Identifying them
yields
x = Nv 1 ∞ *
N v ∞ *
∫ ψv ( x ) ψv+1 ( x ) dx + v
∫ ψv ( x ) ψv-1 ( x ) dx .
N v+1 2 α −∞
N v-1 α −∞
Using the orthogonality of the harmonic oscillator eigenfunctions, both integrals equal 0. Therefore, the
expectation value becomes
x = 0 . Now the uncertainty Δx may be calculated,
€
Δx = #$ x 2 − x
2 %1/2
&
# ' 2 *1/2
%1/2
!
2
1
= - )
, ( v + 2 ) − 0 .
- (mk +
.
$
&
' ! 2 *1/4
1/2
Δx = )
, ( v + 12 ) .
(mk +
~~~~~~~~~~~~~~~~~~~~~~~~~
The uncertainty Δpx is defined as
Δpx =
€
[
px2 − px
2 1/ 2
]
.
2
The expectation value px was shown to be 0 in problem 4; however, the expectation value px must be
€ is
evaluated. The expectation value
€
px2 = ψv pˆ x2 ψv .
€
6
a.)
20
continued
2
In order to evaluate this integral, we must develop an expression for pˆ x ψ v ( x ) . Substituting the form of
the harmonic oscillator wavefunction,
d2
N v H v
dx 2
d2
= −! 2 N v 2 H v
dx
{
pˆ x2 ψv ( x ) = −! 2
(
α x e−α x
)
{ (
α x e−α x
)
2
2
/2
/2
}
}.
Using the product and chain rules for the derivative, we have
2
d2
H v α x e−α x /2
2
dx
2
2
%
d "
d
= −! 2 N v #e−α x /2 H v α x − α xe−α x /2 H v α x &
'
dx $
dx
2
"
2
2
2
d
d
= −! 2 N v # e−α x /2 2 H v α x − α xe−α x /2 H v α x − α e−α x /2 H v
dx
dx
$
{ (
pˆ x2 ψv ( x ) = −! 2 N v
}
)
(
)
(
)
2
+ α 2 x 2 e−α x /2 H v
pˆ x2 ψv ( x ) = −! 2 N v e−α x
2
/2
" d2
# 2 H v
$ dx
(
(
)
(
(
)
α x − α xe−α x
)
α x − 2α x
d
Hv
dx
(
2
/2
)
d
Hv
dx
(
(
αx
)
%
α x & '
)
α x − α H v
)
(
)
α x + α 2 x 2 H v
(
%
α x & .
'
)
Further use of the chain rule is required to obtain the first and second derivatives of the Hermite polynomial
that appears in the first term. For a Hermite polynomial that depends on the argument z, the derivative is
! dz $ dH ( z )
! dz $
d
H v ( z ) = # & v = # & H v' ( z ) .
" dx % dz
" dx %
dx
In this case, z = α x , so that the first derivative is
d
Hv
dx
(
)
α x = α H v!
(
α x .
)
(
)
And the second derivative is
d2
Hv
dx 2
(
)
α x = α H v!! α x .
2
Substituting, the expression for pˆ x ψ v ( x ) becomes
pˆ x2 ψv ( x ) = −! 2 N v e−α x
2
/2
= −! 2 N v e−α x
2
/2
" d2
# 2 H v
$ dx
(
{ α H (((
α x − 2α α xH v(
v
)
α x − 2α x
)
d
Hv
dx
(
(
)
α x − α H v
)
α x − α H v
(
(
)
α x + α 2 x 2 H v
)
α x + α 2 x 2 H v
(
(
%
α x &
'
)
)}
α x .
6
a.)
21
continued
For the second derivative term involving the Hermite polynomial, we may use the differential equation for
the Hermite polynomials,
H v!! (Q ) − 2Q H v! (Q ) + ( 2ε −1) H v (Q ) = 0.
Here, Q = α x and ε = v +1 / 2 . Solving the differential equation for the second derivative,
H v!!
(
)
α x = 2 α x H v!
(
)
α x − 2v H v
(
)
αx .
2
Substituting, the terms involving the first derivatives cancel, and the expression for pˆ x ψ v ( x ) becomes
2
{ α H ""( α x) − 2α α xH " ( α x) − α H ( α x) + α x H ( α x) }
= −! N e
{2α α x H " ( α x) − 2v α H ( α x) − 2α α xH " ( α x) − α H ( α x)
+ α x H ( α x )}
= −! N e
{ −2v α H ( α x) − α H ( α x) + α x H ( α x) }
( x ) = −! N e
{ −α (2v+1) H ( α x) + α x H ( α x) }.
pˆ x2 ψv ( x ) = −! 2 N v e−α x
/2
2 2
v
v
v
v
2
2
−α x /2
v
v
v
v
v
2 2
v
2
2
−α x /2
v
pˆ x2 ψv
2 2
v
v
2
2
−α x /2
v
2 2
v
v
v
The last term in this expression, involving x 2 H v , was worked out in problem 3. The result is
x2Hv
(
)
α x = 1
H v+2
4α
(
)
α x + (2v+1) H
2α
v
(
)
α x + v ( v-1)
H v-2
α
(
)
αx .
Substitution gives
pˆ x2 ψv ( x ) = −! 2 N v e−α x
2
/2
{ −α (2v+1) H (
v
)
α x + α 2 x 2 H v
(
)}
αx (
" 1
(2v+1) H α x + v (v-1) H
αx
) −α ( 2v+1) H v α x + α 2 $ H v+2 α x + v
v-2
4
α
2α
α
*+
#
(
,
2
α ( 2v+1)
α
= −! 2 N v e−α x /2 )−α ( 2v+1) H v α x + H v+2 α x + H v α x + α v ( v-1) H v-2 α x 4
2
+
.
2
(
,
α
α
pˆ x2 ψv ( x ) = −! 2 N v e−α x /2 )− ( 2v+1) H v α x + H v+2 α x + α v ( v-1) H v-2 α x - .
+ 2
.
4
= −! 2 N v e−α x
2
/2 *
(
)
(
(
)
(
)
(
)
(
)
)
(
(
)
)
(
(
(
)
)
% ,*
)'& -*.
6
a.)
22
continued
2
Now we can put this expression for pˆ x ψ v ( x ) back into the expectation value and break the expectation
value up into three separate integrals,
px2 = ψv pˆ x2 ψv
∞
= ∫ ψv* ( x ) pˆ x2 ψv ( x ) dx
−∞
∞
$ α
= −! 2 N v ∫ ψv* ( x ) %− ( 2v+1) H v
& 2
−∞
px2 = ! 2 N v
∞
α
(2v+1) ∫ ψv* ( x ) H v
2
−∞
(
α
α x + H v+2
4
(
)
(
2
)
α x e−α x /2 dx − ! 2 N v
∞
−! 2 N vα v ( v-1) ∫ ψv* ( x ) H v-2
−∞
(
)
α x + α v ( v-1) H v-2
α ∞ *
∫ ψv ( x ) H v+2
4 −∞
(
(
2
'
α x ( e−α x /2 dx
)
)
2
)
α x e−α x /2 dx 2
)
α x e−α x /2 dx .
The next step is to include in the second and third terms the appropriate normalization constants for the v+2
and v–2 wavefunctions by multiplying and dividing by the requisite factors,
px2 = ∞
! 2α
(2v+1) ∫ ψv* ( x ) $% N v H v
2
−∞
−! 2α v ( v-1)
(
2
! 2α N v ∞ *
α x e−α x /2 &' dx − ∫ ψv ( x ) $% N v+2 H v+2
4 N v+2 −∞
)
Nv ∞ *
∫ ψv ( x ) $% N v-2 H v-2
N v-2 −∞
(
(
2
α x e−α x /2 &' dx )
2
α x e−α x /2 &' dx .
)
The terms that appear in bracket in each integral are harmonic oscillator eigenfunctions. Identifying them
yields
px2 = ∞
! 2α
! 2α N v ∞ *
∫ ψv ( x ) ψv+2 ( x ) dx (2v+1) ∫ ψv* ( x ) ψv ( x ) dx − 2
4 N v+2 −∞
−∞
− ! 2α v ( v-1)
Nv ∞ *
∫ ψv ( x ) ψv-2 ( x ) dx .
N v-2 −∞
Using the orthogonality of the harmonic oscillator eigenfunctions, the second and third integrals equal 0.
The first integral yields 1 due to normalization of the harmonic oscillator eigenfunctions. Therefore, the
expectation value becomes
! 2α
(2v+1) 2
!
1$
= ! 2α # v + & .
"
2%
px2 = 1/2
! mk $
Finally, using the definition α = # 2 & , the expectation value is
"! %
(
px2 = mk! 2
1/2 !
)
1$
# v + & .
"
2%
6
a.)
23
continued
Now the uncertainty Δpx may be calculated,
Δpx = #$ px2 − px
#
= - mk! 2
$
(
(
Δpx = mk! 2
2 %1/2
&
1/2
1*
2%
) v + , − 0 .
(
2+
&
1/2 '
)
1/2
1*
) v + , .
(
2+
1/4 '
)
b.) Show that the product Δx Δpx for the ground state of the harmonic oscillator has precisely the
minimum value given by the Uncertainty Principle. In other words, show that for the harmonic
oscillator ground state, the equality in the Uncertainty Principle holds.
€
We can substitute the uncertainty in position and momentum from part (a) to determine the uncertainty
product,
(" 2 %1/4
+
1/2 (
!
Δx Δpx = *$
' ( v + 12 ) -* m k ! 2
*# m k &
-)
)
,
(
1/4
1 1/2 +
2
-,
) (v + )
Δx Δpx = ! ( v + 12 ) .
For the ground state, v=0, so the uncertainty product is
Δx Δpx =
!
.
2
!
The uncertainty principle states that Δx Δpx ≥
. Thus, the harmonic oscillator ground state yields the
2
€
minimum uncertainty product. That is, for the ground state of the harmonic oscillator, the equality holds.
€
c.) Does the equality in the Uncertainty Principle hold for the first excited state of the harmonic
oscillator? Demonstrate whether or not it does.
From part (b), the uncertainty product is
Δx Δpx = ! ( v + 12 ) .
For the first excited state, v=1, and the uncertainty product becomes
Δx Δpx =
3!
.
2
Since this is larger than ! / 2 , the equality in the uncertainty principle does not hold for the first excited
state of the harmonic oscillator. Note,
€ though, that the uncertainty principle is satisfied since the result for
the uncertainty product is larger than ! / 2 .
€
€
7.
Consider a system in a free particle state with positive momentum, ψ (x) = e ikx . Calculate the
uncertainty in the momentum, Δpx , explicitly. Discuss your result. What would Δx be for this state?
The uncertainty in momentum is defined as
€
€
€
Δpx =
[p
2
x
− px
2 1/ 2
]
The expectation value of momentum is
€
px
=
ψ pˆ x ψ
.
ψ ψ
Evaluating the operation of the momentum operator,
€
d ikx
e
dx
pˆ x ψ = !k ψ .
pˆ x ψ = − i!
Substituting,
€
px
px
=
ψ pˆ x ψ
ψ ψ
=
!k ψ ψ
ψ ψ
= !k .
The expectation value of the square of momentum is
€
px2
=
ψ pˆ x2 ψ
.
ψ ψ
Evaluating the effect of the operator,
€
pˆ x2 ψ = − ! 2
d2
dx 2
2
e ikx
= − ! 2 (ik ) e ikx
pˆ x2 ψ = ! 2 k 2 ψ .
€
.
24
7.
25
continued
Substituting,
px2
px2
=
ψ pˆ x2 ψ
ψ ψ
=
! 2k 2 ψ ψ
ψ ψ
= ! 2k 2 .
The uncertainty in momentum is therefore given by
€
[p
= [! k
2
x
Δpx =
2 2
2 1/ 2
]
− (!k ) ]
− px
2 1/ 2
Δpx = 0 .
This result could have been predicted by noting that the wavefunction ψ (x) = e ikx is an eigenfunction of the
momentum operator. When the €
momentum is measured, the result is therefore the eigenvalue exactly; thus, the
uncertainty in the momentum is zero.
€
From the uncertainty principle,
Δx Δpx ≥
Solving for the uncertainty in the position,
€
Δx ≥
!
.
2
!
.
2 Δpx
Since the uncertainty in the momentum is zero, this becomes
€
!
!
Δx ≥
=
,
2 Δpx
2⋅ 0
Δx = ∞.
Because the momentum is known exactly, we see as a consequence of the uncertainty principle that the position
is totally uncertain.
€
8.
26
Show that two commuting operators can have the same set of eigenfunctions.
Let us refer to the two commuting operators as Aˆ and Bˆ . Since the operators commute, we know that
€
[ Aˆ , Bˆ ]
€
= 0.
Assume that the eigenvalue equation for the operator Aˆ is given by
€
Aˆ ψ i = a i ψ i ,
€
where the functions ψ are eigenfunctions of the operator Aˆ .
i
€
Next, operate on both sides of this equation with the operator Bˆ .
€
€
ˆ
B Aˆ ψ i = a i Bˆ ψ i .
€
Since the operators Aˆ and Bˆ commute, we can substitute Bˆ Aˆ = Aˆ Bˆ on the left,
€
Bˆ Aˆ ψ i = a i Bˆ ψ i
€
€
Aˆ Bˆ ψ€ = a Bˆ ψ ,
i
i
or Aˆ Bˆ ψ i
(
)
i
= a i Bˆ ψ i .
( )
Notice that this equation is an eigenvalue equation for the operator Aˆ with eigenfunction Bˆ ψ i . But we started
€
Aˆ being the functions ψ i . Therefore, the eigenfunction Bˆ ψ i must be
with the eigenfunctions of the operator
proportional to ψ i , or
€
€
ˆ
€
B ψ€ = c ψ ,
€
i
i
€
where c is a constant. This equation is an eigenvalue equation for the operator Bˆ with eigenfunction ψ i .
€ the same set of eigenfunctions, ψ i .
Therefore, the operators Aˆ and Bˆ have
€
€
€
€
€