homework problem set 4 solutions - illinois state university 460/homewor… · homework problem set...
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Chemistry 460 Dr. Jean M. Standard
Homework Problem Set 4 Solutions
1. One of the important properties of Hermitian operators is that their eigenfunctions form a complete set.
This means that any arbitrary function
€
f x( ) may be exactly expressed as a linear combination of eigenfunctions,
€
f x( ) = cn ψn x( )n=1
∞
∑ ,
where
€
cn are the expansion coefficients and
€
ψn x( ) are the eigenfunctions. We showed in class that the expansion coefficients
€
cn may be calculated using the relation
€
cn = ψn* x( ) f x( )
−∞
∞
∫ dx .
Assume that the function you wish to represent is a step function, shown in the figure below and defined by the relation
€
f (x) = 0 C 0
x < 0 0 ≤ x ≤ L
x > L
⎧
⎨ ⎪
⎩ ⎪
⎫
⎬ ⎪
⎭ ⎪
.
Here, C is a constant. Using the eigenfunctions of the particle in an infinite box, which form a complete set, calculate analytically the first six linear expansion coefficients,
€
c1 ,
€
c2 ,
€
c3 ,
€
c4 ,
€
c5 , and
€
c6 , for the step function. Recall that the eigenfunctions of the particle in an infinite box are
€
ψ n x( ) =
2L
sinnπ xL
⎛ ⎝ ⎜
⎞ ⎠ ⎟ , 0 ≤ x ≤ L
0, x < 0, x > L
⎧
⎨ ⎪ ⎪
⎩ ⎪ ⎪
⎫
⎬ ⎪ ⎪
⎭ ⎪ ⎪
,
where n is the quantum number and L is the width of the box. Using the six coefficients you determined, construct a graphical representation of the step potential for the case in which L=2 a.u. and C=1 a.u. Note that since the expansion is truncated after only six terms, it is an approximation rather than an exact result.
x=0 x=L
f(x)=C
2
1. continued The coefficients
€
cn to be calculated are
€
cn = ψn* x( ) f x( )
−∞
∞
∫ dx .
Substituting the form of the function
€
f x( ) and the particle in a box wavefunctions yields
€
cn = C 2L
sin nπxL
⎛
⎝ ⎜
⎞
⎠ ⎟
0
L
∫ dx .
Here, since the particle in a box wavefunctions are 0 outside the range
€
0 ≤ x ≤ L , the limits of integration are reduced to 0 to L. Also, on that range, the function
€
f x( ) equals a constant C, which can be pulled out of the integral. The integral to be evaluated is
€
sin nπxL
⎛
⎝ ⎜
⎞
⎠ ⎟
0
L
∫ dx = − Lnπ
cos nπxL
⎛
⎝ ⎜
⎞
⎠ ⎟
0
L
= − Lnπ
cos nπ( ) − cos 0( )[ ]
sin nπxL
⎛
⎝ ⎜
⎞
⎠ ⎟
0
L
∫ dx = − Lnπ
cos nπ( ) − 1[ ] .
Note that this integral equals 0 for n even, and is non-zero for n odd,
€
sin nπxL
⎛
⎝ ⎜
⎞
⎠ ⎟
0
L
∫ dx = 0, n even
2Lnπ
, n odd
⎧
⎨
⎪ ⎪
⎩
⎪ ⎪
⎫
⎬
⎪ ⎪
⎭
⎪ ⎪
.
Therefore the coefficients are
€
cn = 0, n even
2LCnπ
⋅2L
, n odd
⎧
⎨
⎪ ⎪
⎩
⎪ ⎪
⎫
⎬
⎪ ⎪
⎭
⎪ ⎪
.
3
1. continued The analytical forms of the first six coefficients in the expansion are given below, along with numerical results for the specific case L=2 a.u., C=1 a.u..
n
€
cn (analytic)
€
cn (numerical, a.u.) 1
€
2LCπ
⋅2L
1.2732
2
0
0
3
€
2LC3π
⋅2L
0.4244
4 0
0
5
€
2LC5π
⋅2L
0.2546
6
0
0
Using the expansion coefficients listed above, the step function
€
f x( ) with L=2 and C=1 may be approximately expanded as
€
f x( ) ≈ 1.2732 ⋅sin πx2
⎛
⎝ ⎜
⎞
⎠ ⎟ + 0.4244 ⋅sin 3πx
2⎛
⎝ ⎜
⎞
⎠ ⎟ + 0.2546 ⋅sin 5πx
2⎛
⎝ ⎜
⎞
⎠ ⎟ .
A graph of the six-term expansion of the step function given in the equation above is shown in the figure below.
The dashed line shows the actual step function on the interval between x=0 and 2. The six-term expansion crudely reproduces the step function. Clearly, many more terms are needed in the expansion in order to reproduce the step function more accurately.
4
2. The commutator operator A, B⎡⎣
⎤⎦ for two operators A and B is defined as A, B⎡
⎣⎤⎦ = AB − BA . Using
the definitions of
€
ˆ P and
€
ˆ Q , verify that
€
ˆ P , ˆ Q [ ] = − i .
Using
€
ˆ P = − i ddQ
,
€
ˆ P , ˆ Q [ ] f Q( ) = ˆ P ˆ Q − ˆ Q ˆ P ( ) f Q( )
= − i ddQ
Q + i Q ddQ
⎛
⎝ ⎜
⎞
⎠ ⎟ f Q( )
= − i ddQ
Q f Q( )( ) + i Q ʹ f Q( )
= − i f Q( ) + Q ʹ f Q( )( ) + i Q ʹ f Q( )= − i f Q( ) .
Therefore,
€
ˆ P , ˆ Q [ ] = − i .
5
3. Start with the differential equation
€
d 2 f (x)dx2 − 1
4f (x) + α
xf (x) = 0 ,
where α 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
equation,
€
αxf (x) , goes to zero. Thus, the asymptotic form of the differential equation is
€
d 2 f (x)dx2 = 1
4f (x) .
The solution can be written in the form of an exponential. Assume a solution of the form
€
f (x) = e±βx , where β is a positive constant. The derivatives of the function are
€
ʹ f (x) = ± β e±βx , ʹ f (x) = β 2 e±βx . Substituting the second derivative into the asymptotic equation,
€
β 2 e±βx = 14 e±βx .
Therefore, the exponent is given by
€
β 2 = 14 , 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 .
6
3. 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,
€
ʹ f x( ) = − 12 e−x / 2 g x( ) + e−x / 2 ʹ g x( ) = e−x / 2 − 1
2 g x( ) + ʹ g x( )( ) ,
€
ʹ f x( ) = − 12 e−x / 2 − 1
2 g x( ) + ʹ g x( )[ ] + e−x / 2 − 12 ʹ g x( ) + ʹ g x( )[ ]
= e−x / 2 ʹ g x( ) − ʹ g x( ) + 14 g x( )[ ].
Substituting these expressions into the original differential equation,
€
d 2 f (x)dx2 −
14
f (x) + αx
f (x) = 0
e−x / 2 ʹ g x( ) − ʹ g x( ) + 14 g x( )[ ] − 1
4 e−x / 2g x( ) + αx
e−x / 2g x( ) = 0
e−x / 2 ʹ g x( ) − ʹ g x( ) + αx
g x( )⎡
⎣ ⎢ ⎤
⎦ ⎥ = 0 .
Dividing both sides of the equation by the factor
€
e−x / 2 , the equation for the remainder is
€
ʹ g x( ) − ʹ g x( ) + αx
g x( ) = 0 .
7
3. continued
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 xk
k=0
∞
∑ .
Evaluating the derivatives,
€
ʹ g x( ) = ak k xk−1
k=1
∞
∑ ,
€
ʹ g x( ) = ak k k −1( ) xk−2
k= 2
∞
∑ .
Next, the summations are rewritten so that they start at k=0,
€
ʹ g x( ) = ak k xk−1
k=0
∞
∑ ,
€
ʹ g x( ) = ak+1 k k + 1( ) xk−1
k= 0
∞
∑ .
The derivatives are substituted into the differential equation for the remainder,
€
ʹ g x( ) − ʹ g x( ) + αx
g x( ) = 0
ak+1 k k + 1( ) xk−1
k= 0
∞
∑ − ak k xk−1
k= 0
∞
∑ + αx
ak xk
k= 0
∞
∑ = 0 .
Combining the summations and factoring out the powers of x yields
€
ak+1 k k +1( ) − ak k + α ak{ } xk−1
k=0
∞
∑ = 0 .
In order for a solution to exist, the coefficients of
€
xk−1 must be zero:
€
ak+1 k k +1( ) − ak k + α ak = 0 . Solving for
€
ak+1 leads to the recursion formula:
€
ak+1 = k −αk k +1( )
ak .
8
3. continued
d.) Truncate the series to obtain an expression for α .
If the solution is truncated at the nth term, then
€
an+1 = 0.
€
an+1 = 0 = n −αn n + 1( )
an .
If
€
an ≠ 0 , then in order for the above equation to be satisfied, we must have
€
n −αn n +1( )
= 0 ,
or
€
α = n .
9
4. Given the recursion formula for the Hermite polynomials,
€
an+2 = 2n + 1 − 2εn +1( ) n + 2( )
an ,
and
€
H4 (Q) = 16Q4 − 48Q2 +12 , verify that the coefficients of the v = 4 polynomial are correct starting with a4 = 2
4 =16 .
Since
€
ε = v + 12 , we can write
€
an+2 = 2n + 1 − 2 v + 1
2( )n +1( ) n + 2( )
an .
For v=4, the recursion formula is
€
an+2 = 2n − 8n +1( ) n + 2( )
an .
Substituting n=2 into the recursion formula,
€
a4 = −43 ⋅4
a2
= − 13a2 ,
or a2 = − 3a4 .
Substituting
€
a4
= 24 = 16, we have
€
a2 = − 3a4
= − 3 ⋅16a2 = − 48 .
Substituting n=0 into the recursion formula,
€
a2 = −82
a0
= − 4a0 ,
or a0 = − 14
a2 .
Substituting
€
a2
= −48, we have
€
a0 = − 14a2
= − 14⋅ −48( )
a0 = 12 .
Since
€
H4 (Q) = a4 Q4 + a2 Q
2 + a0 = 16Q4 − 48Q2 +12 , we see that the coefficients of the v=4 Hermite polynomial given are verified.