Chemistry 460 Name _____________KEY_________________ Fall 2017 Dr. Jean M. Standard September 27, 2017

Quantum Chemistry Exam 1 – Solutions 1.) (24 points) Answer the following questions by selecting the correct answer from the choices provided.

a.) Which of the statements below are false regarding the operators shown below?

1) One of the eigenfunctions of the operator d2

dx2 is

sin x . 2) The operators A = x and B = x2 commute.

3) The operator O =           is a linear operator.

4) The result of operating the operator x ddx

on

the function xe−αx is x −αx2( )e−αx .

5) Both (1) and (2). 6) Both (1) and (3). 7) Both (1) and (4).

b.) Which of the following statements about tunneling are true?

1) Tunneling is more efficient for lighter particles. 2) Tunneling is a purely quantum mechanical effect

with no classical analog. 3) The tunneling probability is greater for particle

energies near the top of the barrier than further down.

4) The tunneling probability increases as the the barrier becomes narrower.

5) All of the above. 6) Statements (3) and (4) only. 7) None of the above.

2 1. Continued

c.) Which of the following statements are false about the particle in an infinite one-dimensional box of width L?

1) The number of nodes in the wavefunction

increases with energy. 2) The system exhibits tunneling for x > L .

3) The average value of the momentum px always

equals 0.

4) The spacing between quantized energy levels increases with energy.

5) All of the above. 6) None of the above.

d.) Which of the following statements about valid quantum mechanical operators are true?

1) They must correspond to linear operators. 2) The operators themselves must be real. 3) Their eigenvalues must be real. 4) Operators that involve momentum are

constructed using the relation px = i!ddx

.

5) Both (1) and (2). 6) Both (1) and (3). 7) Both (1) and (4).

3 2.) (16 points) Explain or provide an example illustrating each of the following concepts.

a.) Expectation value and measurement

The expectation value A for property A in the state ψ involves an integral of the operator A that is associated with the physical observable A,

A    =    ψ *  A ψ  dτ∫ ψ * ψ  dτ∫

  .

�The expectation value corresponds to the average result that is predicted to be obtained experimentally if the property A is measured in the lab on many identical systems in the state ψ . �

b.) Hamiltonian operator

The Hamiltonian operator H is the key operator of quantum mechanics. It appears in the Schrödinger equation, Hψ  = Eψ . Solution of the Schrödinger equation yields the wavefunction ψ , which contains all

the information aboit the system. The Hamiltonian operator H is defined as the sum of the kinetic energy T and potential energy V operators, H  = T + V .

c.) Classically forbidden region

A classically forbidden region occurs when the system energy is less than the potential energy (E < V). A particle behaving classically would not be able to enter such a region; however, a quantum mechanical particle may exhibit tunneling behavior in this region. �

d.) Normalization

A wavefunction y which satisfied the relation

ψ*ψ dτ  −∞

    ∞

∫   =  1

is said to be normalized. Normalization is the process of determining the scaling factor for the wavefunction so that it satisfies this criterion, which is key for the Born probability interpretation of quantum mechanics. For a wavefunction that does not satisfy the above criterion, we scale it by the factor N (the normalization constant) so that ψnorm  = Nψ . Then, the normalization constant N is given by

N   =    1

ψ*ψ dτ  −∞

    ∞

∫⎛

⎝⎜⎜

⎞

⎠⎟⎟

1/2   .

4

3.) (20 points) The Hermitian operator

€

ˆ S is associated with the physical observable S. Two of the normalized eigenfunctions of

€

ˆ S are φ1 and φ2 . Both these eigenfunctions have the same eigenvalue (s), such that

S φ1  =  s φ1

S φ2  =  s φ2 .

Recall that we proved that different eigenfunctions of a Hermitian operator were orthogonal as long as their eigenvalues were different. Here, because the eigenvalues are the same, the eigenfunctions φ1 and φ2 may be considered to be normalized but not orthogonal. Now, consider a new wavefunction designated χ , which is defined by the relation

χ   =   φ2   −   C φ1,

where C is a constant. a.) Determine what the constant C must be in order for the new wavefunction χ to be orthogonal to the

eigenfunction φ1 .

For the two functions to be orthogonal, the definition is

χ* φ1 dτ−∞

∞∫ =  0  .

Substituting the form of χ and expanding, we have

χ* φ1 dτ−∞

∞∫ =  0

φ2 −Cφ1( )*φ1 dτ−∞

∞∫ =  0

φ2*  φ1 dτ

−∞

∞∫      −       C*   φ1

*  φ1 dτ−∞

∞∫     =  0  .

The second integral is just the normalization condition for the eigenfunction φ1 ,

φ1*  φ1 dτ

−∞

∞∫     =  1  .

Thus, we have

φ2*  φ1 dτ

−∞

∞∫      −       C* ⋅1    =  0  ,

or      C*    =    φ2*  φ1 dτ

−∞

∞∫    .

Finally, taking the complex conjugate of both sides, we have

 C    =    φ1*  φ2 dτ

−∞

∞∫    .

This integral is known as the overlap between the functions φ1 and φ2 .

5

3. Continued b.) Determine the expectation value of the property S in the state χ .

By definition, the expectation value S for the state χ is

S    =    χ*  S   χ dτ∫ χ* χ dτ∫

 .

Evaluating the operation of

€

ˆ S on χ (in order to evaluate the numerator), we have

Sχ   =  S φ1  − Cφ2( )

=  Sφ1   −  C Sφ2  

=  sφ1   −  Csφ2

=  s φ1  − Cφ2( )

Sχ   =  sχ  .

Substituting into the expectation value expression,

S    =    χ*  S   χ dτ∫ χ* χ dτ∫

=    χ*  s χ dτ∫ χ* χ dτ∫

=  s  χ* χ dτ∫ χ* χ dτ∫

S    =  s .

6

4.) (20 points) For a particle in the first excited state of a one-dimensional infinite box of width L, determine the probability of finding the particle in the middle third of the box. Recall that the wavefunctions of the one-dimensional particle in an infinite box have the form

€

ψ n x( ) =

2L

sinnπ xL

$ % &

' ( ) , 0 ≤ x ≤ L

0, x < 0, x > L

+

, - -

. - -

/

0 - -

1 - -

,

where n is the quantum number and L is the width of the box.

The probability P for finding the particle in the first excited state (n=2) between x=L/3 and x=2L/3 is given by

P   =   ψ2

* x( )ψ2 x( ) dxL/3

2L/3∫ .

The particle in a box first excited state wavefunction is

ψ2 x( )   =   2L sin 2πx

L⎛

⎝⎜

⎞

⎠⎟  .

Substituting the wavefunction into the expression for the probability yields

P   =   2L  sin2 2πx

L⎛

⎝⎜

⎞

⎠⎟ dxL/3

2L/3∫ .

To evaluate the integral, we can use the handout on integrals,

€

sin2∫ bx dx = x2

− 14b

sin 2bx .

By making the substitution b = 2πL

, the integral becomes

 sin2∫ 2π xL

⎛

⎝⎜

⎞

⎠⎟dx    =   

x2   −   L

8πsin 4π x

L⎛

⎝⎜

⎞

⎠⎟   .

Therefore, the probability is given by

P   =   2L    x

2    −    L

8πsin 4π x

L⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢

⎤

⎦⎥L/3

2L/3

.

7

4. Continued

Evaluating the expression at the limits yields V .

P   =    2L

L3

 −  L8π

sin 8π3

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟    −   

2L

L6

 −  L8π

sin 4π3

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟

P   =   13   −   1

4πsin 8π

3⎛

⎝⎜

⎞

⎠⎟   +  

14π

sin 4π3

⎛

⎝⎜

⎞

⎠⎟

P   =   13    −     3

8π   −    3

8π   

P   =   13    −     3

4π             or  P = 0.196 .

8

5.) (20 points) In this problem, you will consider a particle of mass m trapped in an infinite box with a finite potential barrier in the center (such that the barrier midpoint is at x = L / 2 ), shown below and defined by

V (x) =

∞, x < 00, 0 ≤ x ≤Q

V0, Q < x ≤ R 0,

∞,

 R < x ≤ L

  x > L

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

.

a.) For the case that V0 = 0 , what is the ground state energy of the system? Give an analytic form. How do you expect the ground state energy to be affected in the case that V0 ≠ 0 (i.e., do you expect the ground state energy to increase, decrease, or stay the same relative to the case with V0 = 0 )? Explain.

For V0 = 0 , the system just becomes a one-dimensional infinite box since the potential equals zero inside (on the range x = 0 to x = L ), and equals infinity everywhere outside ( x < 0 , x > L ). The ground state energy eigenvalue therefore corresponds to:

E1   = h2

8mL2 .

For the case with V0 ≠ 0 , the potential inside the box has increased. This leads to enhanced confinement of the particle and thus we would expect an increase in the energy of the particle relative to the V0 = 0 value.

b.) For the case that V0 →∞ , what is the ground state energy of the system? Again, give an analytic form.

As V0 →∞ , the system just turns into two one-dimensional infinite boxes, one from x = 0 to x =Q , and another from x = R to x = L . Since it was stated that the barrier was centered in the original box, each of these infinite boxes are identical (with width equal to Q). The ground state energy eigenvalue for this cases therefore corresponds to:

E1   = h2

8mQ2 .

x=0 x=Q

IIII II

V=V0

x=R

IV

x=L

V

9

5. Continued

c.) Write down appropriate general solutions for the Schrödinger equation in regions I, II, III, IV, and V for

€

E < V0. Make sure that you define any constants that you use in defining your solutions.

The general solutions for the wavefunctions in regions I-V for the case

€

E < V0 are

ψI x( )  =  0

ψII x( )  =   Asinkx + Bcoskx

ψIII x( )  =  Ceλx + De−λx

ψIV x( ) = F sinkx + G coskx

ψV x( ) = 0,

where k for both regions II and IV and

€

λ for region III are defined as

€

k = 2mE!

,

€

λ = 2m V0 − E( )!

.

In regions I and V, the wavefunction must equal 0 because the potential is infinite.

d.) Apply the boundary conditions for the wavefunction at x = 0 and x = L. How does this simplify the general solutions for the wavefunctions in regions II and IV?

The boundary condition at x = 0 is

€

ψI 0( ) =ψII 0( ) . Substituting the forms of the wavefunctions leads to the equation

€

0 = A sin 0 + B cos 0 . Since

€

sin 0 = 0 and

€

cos 0 = 1, the equation becomes

€

0 = B .

Therefore, the solution for the wavefunction in region II simplifies to

€

ψII x( ) = A sin kx .

10

5 d.) Continued

The boundary condition at x = L is

ψIV L( )  =  ψV L( ) .

Substituting the forms of the wavefunctions leads to the equation

F sinkL   +   G coskL   =  0 . We can solve for one of the arbitrary constants (F or G) to eliminate one of them,

G   =  −F sinkLcoskL

   =  −F tankL  .

Therefore, the solution for the wavefunction in region IV simplifies to

ψIV x( )   =   F sinkx    − tankLcoskx( ) .

e.) Apply the boundary conditions for the wavefunction and the first derivative at both x = Q and x = R. You should then take the ratio of the derivative and wavefunction equations for both x = Q and x = R. This leads to two equations. Report the two equations, making sure to eliminate any arbitrary constants that cancel (if any), but do not do anything further with them. The boundary conditions at x = Q are

ψII Q( )  =  ψIII Q( )  and   ʹψ II Q( )  =  ʹψ III Q( ) .

Substituting the forms of the wavefunctions and their first derivatives leads to the equations

ψII Q( )  =  ψIII Q( )    ⇒    AsinkQ   =   CeλQ   +  De−λQ

ʹψ II Q( )  =  ʹψ III Q( )    ⇒    Ak coskQ  =  λCeλQ   −  λDe−λQ .

Taking the ratio of these equations yields

€

" ψ II L( )ψII L( )

= " ψ III L( )

ψIII L( ) .

Substituting,

Ak coskQAsinkQ

    =    λ CeλQ   −  De−λQ( )CeλQ   +  De−λQ

,

or k cot kQ    =    λ CeλQ   −  De−λQ( )CeλQ   +  De−λQ

.

11

5 e.) Continued

The boundary conditions at x = R are

ψIII R( )  =  ψIV R( )   and    ʹψ III R( )  =   ʹψ IV R( ) .

Substituting the forms of the wavefunctions and their first derivatives leads to the equations

ψIII R( )  =  ψIV R( )   ⇒    CeλR   +  De−λR   =   F sinkR    − tankLcoskR( )

ʹψ III R( )  =   ʹψ IV R( )    ⇒    λCeλR   −  λDe−λR   =   Fk coskR   + tankLsinkR( ) .

Taking the ratio of these equations gives

ʹψ III R( )ψIII R( )

  =   ʹψ IV R( )

ψIV R( ).

Substituting,

λ CeλR   −  De−λR( )CeλR   +  De−λR

    =   Fk coskR   + tankLsinkR( )F sinkR    − tankLcoskR( )

,

or    λ CeλR   −  De−λR( )CeλR   +  De−λR

    =   k coskR   + tankLsinkR( )

sinkR    − tankLcoskR .

12

5. Continued

f.) Without solving any equations, sketch the expected form of the ground state wavefunction for

€

E < V0.

A sketch of the ground state wavefunction is shown below. Note that the wavefunction equals zero for x < 0 and x > L. Therefore, the function in regions II and IV must be zero for x = 0 and x = L. For the ground state wavefunction, there will be no nodes, and equal probability to be in either the left or right well. In region III, the classically forbidden region, the wavefunction will be small but non-zero as it exhibits tunneling from regions II and IV into region III.

x=0 x=Q

IIIII

V=V0

x=R

IV

x=L

V