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.