Chemistry 460 Dr. Jean M. Standard

Homework Problem Set 3 Solutions

1. See Section 2-5 of Lowe and Peterson for more information about this example. When a particle

experiences zero potential energy over its entire coordinate range rather than just in a confined region of space, the particle is called a "free particle".

a.) Show that the general solution to the Schrödinger equation for the free particle in one dimension is

€

ψ(x) = A sin kx + B cos kxor

ψ(x) = C ei kx + De−i kx .

for all values of x from –∞ to +∞ . In these expressions, A, B, C, and D are arbitrary constants (they may be complex) and k is a real, positive constant. You will need to determine how the energy eigenvalue E is related to the constant k for this system in order for the forms shown above to be valid solutions. To show that either one of the forms is a solution, they can be substituted into the Schrödinger equation. Since the potential energy is zero everywhere, the Schrödinger equation for the free particle is

€

−!2

2md 2

dx2 ψ x( ) = Eψ x( ) .

Substituting the form

€

ψ(x) = A sin kx + B cos kx ,

€

−!2

2md 2

dx2 A sin kx + B cos kx{ } = E A sin kx + B cos kx{ } .

Taking the second derivative on the left and simplifying yields

€

−!2

2m−Ak2 sin kx − Bk2 cos kx{ } = E A sin kx + B cos kx{ }

!2k2

2mA sin kx + B cos kx{ } = E A sin kx + B cos kx{ } .

Since the function in the brackets is the same on both sides, we can divide this out to give

€

!2k2

2m = E .

Solving for the constant k,

€

k = 2mE!

.

2

1a.) Continued

So the wavefunction

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ψ x( ) = A sin kx + B cos kx is a solution of the free particle Schrödinger equation if k is related to the energy by the expression given above. Substituting the form

€

ψ(x) = Ceikx + De−ikx ,

€

−!2

2md 2

dx2 Ceikx + De−ikx{ } = E Ceikx + De−ikx{ } .

Taking the second derivative on the left and simplifying yields

€

−!2

2m− k2Ceikx − k2De−ikx{ } = E Ceikx + De−ikx{ }

!2k2

2mCeikx + De−ikx{ } = E Ceikx + De−ikx{ } .

Since the function in the brackets is the same on both sides, we can divide this out to give

€

!2k2

2m = E .

Solving for the constant k,

€

k = 2mE!

.

So the wavefunction

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ψ(x) = Ceikx + De−ikx is also a solution of the free particle Schrödinger equation if k is related to the energy by the expression given above.

b.) Euler’s relation is

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eiα = cosα + i sinα .

Use this relation to show that the two general forms of the free particle solution given in part (a) are equivalent. To show that the two forms are equivalent, we can start with the second form,

€

ψ(x) = Ceikx + De−ikx . From Euler’s relation, the two terms in this wavefunction may be expressed as

€

ei kx = cos kx + i sin kx

e−i kx = cos kx − i sin kx .

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1 b.) Continued

Substituting these forms into the expression for the wavefunction gives

€

ψ(x) = Ceikx + De−ikx

= C cos kx + i sin kx( ) + D cos kx − i sin kx( )ψ(x) = iC − iD( ) sin kx + C +D( ) cos kx .

Since A, B, C, and D are arbitrary constants, we can redefine

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A = iC − iD and

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B = C +D to yield

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ψ(x) = A sin kx + B cos kx . Thus, the two forms of the wavefunction are equivalent.

c.) Special cases of the free particle solution occur when C = 0 or D = 0 in the second form given in part

(a). The two wavefunctions are

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ψ(x) = C ei kx or

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ψ(x) = De−i kx . These are called plane wave solutions and they represent particles moving in the positive and negative x directions, respectively.

To see why the plane wave solutions represent particles traveling in the positive and negative x directions, show that each plane wave solution is also an eigenfunction of the momentum operator. Give the eigenvalue for each of the plane wave solutions.

To show that the plane wave solutions are eigenfunctions of the momentum operator, we must show that

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ˆ p xψ(x) = Pψ(x) , where P is a constant. For

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ψ(x) = C ei kx , application of the momentum operator gives

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ˆ p x Cei kx = − i ! ddx

(Cei kx )

= − i 2!k Cei kx

ˆ p xψ x( ) = !kψ x( ) . So, we see that the plane wave

€

ψ(x) = C ei kx is an eigenfunction of the momentum operator with eigenvalue

€

! k . From postulate 4, when the momentum of the plane wave is measured, the result will be the eigenvalue

€

! k . This means that the momentum of the system is positive since k and

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! are both positive quantities. Thus, we say that a system with wavefunction given by

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ψ(x) = C ei kx represents a particle traveling in the direction of positive x.

4

1 c.) Continued

For

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ψ(x) = De−i kx , application of the momentum operator gives

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ˆ p x De−i kx = − i ! ddx

(De−i kx )

= i 2!k Dei kx

ˆ p xψ x( ) = − !kψ x( ) . So, this time we see that the plane wave

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ψ(x) = De−i kx is an eigenfunction of the momentum operator with eigenvalue

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−! k . From postulate 4, when the momentum of the plane wave is measured, the result will be the eigenvalue

€

−! k . This means that the momentum of the system is negative. Thus, we say that a system with wavefunction given by

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ψ(x) = De−i kx represents a particle traveling in the direction of negative x.

5

2. In class, we considered a particle in a potential of width L and depth

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V0:

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V (x) = ∞

0V0

x < 0

0 ≤ x ≤ L x > L

$

% &

' &

(

) &

* &

.

We showed in class that requiring the wavefunction to be continuous across the boundaries between the regions led to an equation for energy quantization:

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z cot z + α2 − z2 = 0 , where z = kL and

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α = 2mV0L2

!2

#

$ % %

&

' ( (

12

.

a.) Consider the case where α = 6.0. Numerically determine the allowed values of z that solve the

equation above (i.e., find the roots of the equation). You should be able to obtain reasonable starting guesses by looking at the graphical solution presented in class.

We need to solve the equation

€

z cot z + α2 − z2 = 0

for z with

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α = 6.0. Since

€

3π2 ≤ α ≤ 5π

2 , there will be two solutions. From the graphical solution given in

class, reasonable starting estimates of the solutions are

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z ≈ 3 and

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z ≈ 5.5.

We can use Solver in Microsoft Excel and a guess of

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z ≈ 3 to find a root of the above equation. Solver returns the result z = 2.679. Next, using Solver with a guess of

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z ≈ 5.5 returns the result z = 5.226.

x=0 x=L

V=V0

I II III

6

2. Continued

b.) Given L = 3.0 bohr and

€

V0 = 2.0 hartrees, use the values of z that you obtained above to determine the allowed energies (in a.u.) for the electron in the box. Compare your results to the energies of an electron in an infinite box of the same width L.

Next, to get the energies, we use the equation (given in the handout),

€

E = !2z2

2mL2 .

With L=3.0 bohr, m=1 a.u., and

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! = 1 a.u., the energies are:

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E1 = 12 z12

2 ⋅1 ⋅3.02 = 2.679( )2

2 ⋅ 3.02 , or E1 = 0.399 hartrees .

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E2 = 12 z22

2 ⋅1 ⋅3.02 = 5.226( )2

2 ⋅3.02 , or E2 = 1.517 hartrees .

7

3. Consider a particle in a finite potential well of width L and depth

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V0:

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V (x) = V0

0V0

x < 0 0 ≤ x ≤ L

x > L

#

$ %

& %

'

( %

) %

.

In this problem, you will construct the analytic solution for

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E < V0 , the bound states of the system.

a.) Write down the general solutions for the wavefunction in regions I, II, and III. The general solutions for the wavefunctions in regions I-III for the case

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E < V0 are

ψI x( )   =   Ceλx  + De−λx

ψII x( )   =   Asinkx  + Bcoskx

ψIII x( )   =   Feλx  + Ge−λx  ,

where k for region II and

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λ for regions I and III are defined as

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k = 2mE!

,

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λ = 2m V0 − E( )!

.

b.) Test the general solutions in regions where x→±∞ and eliminate any terms that cause the wavefuntion to blow up.

Regions I and III must be tested because x→−∞ in region I, and x→+∞ in region III. In region I, the term De−λx goes to infinity as x→−∞ ; therefore, it must be eliminated from the general solution. The solution in region I becomes:

ψI x( )   =   Ceλx  . In region III, the term Feλx goes to infinity as x→+∞ ; therefore, it must be eliminated from the general solution. The solution in region III then becomes:

ψIII x( )   =   Ge−λx  .

x=0 x=L

V=V0

I II III

8

3. Continued c.) Apply the boundary conditions to match the wavefunction and its first derivative at x = 0. Use the

equations to simplify the general solutions by elimination of some of the arbitrary constants.

The boundary conditions at x = 0 are

ψI 0( ) = ψII 0( )     and      ʹψI 0( ) =  ʹψII 0( ) . Substituting the forms of the wavefunctions into the matching equation leads to

Ce0   =  Asin0  +  Bcos0 . Since

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sin 0 = 0 and

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cos 0 = 1, the equation becomes

C   =  B .

Substituting the forms of the first derivatives into the matching equation leads to

 λCe0   =   Ak cos0  −  Bk sin0 . Again, with

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sin 0 = 0 and

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cos 0 = 1, the equation becomes

λC   =   Ak  ,

or      A  =  λCk

.

Therefore, the solution for the wavefunction in region I remains

ψI x( )   =   Ceλx  . The solution for the wavefunction in region II becomes

ψII x( )   =   λCk

sinkx   +  C coskx  .

Here, we have eliminated two of the arbitrary constansts, A and B, by using the matching conditions to relate them to the constant C.

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3. Continued

d.) Apply the boundary conditions to match the wavefunction and its first derivative at x = L. Use the ratio of the equations for matching the first derivative and the wavefunction and at x = L to obtain an equation containing no arbitrary constants. This is the equation which leads to energy quantization for the particle in a finite box.

The boundary conditions at x = L are

ψII L( ) = ψIII L( )    and    ʹψII L( ) =  ʹψIII L( ) . Substituting the forms of the wavefunctions and their first derivatives leads to the equations

ψII L( ) = ψIII L( )   ⇒    λCk

sinkL   +  C coskL    =   Ge−λL

ʹψII L( ) =  ʹψIII L( )    ⇒     λC coskL   −  kC sinkL   =   −λGe−λL .

Taking the ratio of these equations yields

ʹψ II L( )ψII L( )

  =   ʹψ III L( )

ψIII L( ) .

Substituting,

λC coskL   −  kC sinkLλCk

sinkL   +  C coskL    =    −λGe

−λL

Ge−λL,

or     λ coskL   −  k sinkLλk

sinkL   +  coskL    =   −λ .