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CHGN 351
FALL 1999
PRACTICE MIDTERM 1 - SOLUTIONS
Principles of Quantum Mechanics: True or False
1. In a photoelectric experiment, the work function is the sum of the incident photon
energy and the electron kinetic energy. F
2. The Bohr model predicts an ionization energy for hydrogen atom that is finite. T
3. Any wave on a classical string can be expressed as a linear combination of the normal
modes. T
4. The state function  can never be negative. F
5. The probability density for a particle-in-a-box can never be negative. T
6. The state function (x) for a particle-in-a-box (size a) must be of the form
A sin(nx/a), where n is a positive integer, and A a constant. F
7. The state function for a particle-in-a-box must be a real function. F
8. The system’s state function  must be an eigenfunction of the system’s Hamiltonian.
F
9. The smaller the box a particle is in, the higher the kinetic energy in the ground state.
T
10. The ground state of a particle in a two-dimensional box is degenerate. F
11. A measurement of the momentum for a particle-in-a-box ground state can only give
two possible values. T
12. The time-dependent state function is always equal to a function of time multiplied by
a function of the coordinates. F
13. The momentum and Hamiltonian operators commute for a free particle. T
14. If  is an eigenfunction of the linear operator Aˆ , then c is an eigenfunction of Aˆ ,
where c is an arbitrary constant. T
15. If we measure the property A when the system’s state function is not an eigenfunction
of Aˆ , then we can get a result that is not an eigenvalue of Aˆ . F
Phenomena and significance of the quantum theory
1. Write a couple of sentences explaining what deBroglie’s hypothesis is, some
experimental verifications of the hypothesis, and some significant consequences.
DeBrogie hypothesized that if light can have particle-like properties (as well as wave-like
properties), then matter can have wave-like properties (as well as particle-like ones).
Specifically, matter can have a wavelength that obeys the same relation that light obeys,
namely  = h / p . The wavelike nature of matter was confirmed in the interference
patterns seen in the scattering of electrons and atoms off of crystals, for example. This
revolutionary hypothesis about matter inspired Schrodinger to develop his equation which
governs the dynamics of matter on the atomic and molecular scale, providing the basis for
calculating molecular properties from first principles.
2. List a few ways in which an electron behaves like a macroscopic particle, and in which
it behaves like a macroscopic wave. Similarly, list a few ways in which an electron
doesn't behave like a particle, and doesn't behave like a wave.
An electron is like a particle in that it comes in a discrete packet, and carries a discrete
amount of charge, and it is like a wave in that it can exhibit interference and exist in
stationary states (like a standing wave). An electron doesn’t behave like a macroscopic
particle in that its position may not be known to arbitrary precision, and doesn’t behave
like a macroscopic wave in that an attempt to detect the electron finds it at one particular
position, and not with intensity spread out over a spatial region.
Calculating expectation values: the harmonic oscillator ground state
We will discuss the harmonic oscillator in class, but here we give without derivation a
stationary state solution to the harmonic oscillator Schrodinger equation. For this
problem, calculate expectation values based on the wavefunction of the ground state
0(x) = ()1/4 exp(-x2/2)
where  = (k  / hbar2)1/2 , k is the spring constant, and  is the reduced mass. The
parameter x describes the coordinate of the particle in a harmonic potential V(x) = kx2/2,
and can take any value from - to .
a) Show this satisfies the harmonic oscillator Schrodinger’s Equation, where the potential
energy term in the Hamiltonian is given by V(x) = kx2/2.
First we identify the Hamiltonian to be the sum of the kinetic and potential energy
operators:
2
d2 1 2
Hˆ  
 kx
2m dx 2 2
We then verify that the given wavefunction satisfies Schrodinger’s equation:
Hˆ  0 (x)  E 0 (x)
d 2 0 (x) 1 2

 kx  0 (x)  E  0 (x)
2m dx 2
2
2
We use the results

d 2 0 (x) d 2  1/ 4
 2 
exp( x 2 / 2)
2


dx
dx  


d
  
 x exp( x 2 / 2)
  dx 
1/ 4
 
2 2
2
2
    x exp( x / 2)   exp( x / 2)

1/ 4
 
2 2
2
   ( x   )exp( x / 2)

1/ 4
d  0 (x)
 ( 2 x 2   ) 0 (x)
2
dx
2
to find for the Schrodinger’s equation:
d  0 (x) 1 2

 kx  0 (x)  E  0 (x)
2m dx 2
2
2
2
2
1
( 2 x 2   ) 0 (x)  kx 2 0 (x)  E 0 (x)
2m
2
2
1 2
2 2

( x   )  kx  E
2m
2
2
km
1
E
( 2 x 2   )  kx2
2m
2
2

k
E

2m 2

Thus we see that 0(x) is indeed an eigenfunctions with eigenvalue given by the last
equation.
b) Determine <x> and <p> for the ground state.

 x 

0
(x)* x  0 (x)dx  0

The last integral can either be evaluated explicitly, or by recognizing the fact that the
integrand is odd, since it is the product of two even functions (0(x)), and one odd
function, x.
Similarly,

d 
 p    0 (x)* i
 (x)dx
 dx  0


 i

0
(x)* x exp(x 2 / 2)dx  0

for the same reason.
c) Determine the average potential energy, kx2/2, and the average kinetic energy, p2/(2).
-------------------------Some integrals you may find helpful:

e

ax 2
dx 

a
;

2 ax
 x e dx 
2


2
a 3/ 2

ax
 xe dx is a very simple integral that requires no algebra (remember symmetry!)
2



1 2
1
kx    0 (x)* kx 2 0 (x)dx
2
2





*
 1/ 4
 1 2  1/ 4

2
2
  exp( x / 2) 2 kx   exp( x / 2)dx




1/ 2 
1 
 k  
2  
x
2
exp( x 2 )dx

1  
 3/ 2
k

2   2
1
k
1
 kx 2 
 E
2
4 2

1/ 2
Thus the average potential energy is half of the average energy for the harmonic
oscillator. (This is a general feature of the harmonic oscillator.)

2
p2
1
d 2 
* 


  0 (x)  2m dx2  0 (x)dx
2m
2m 
This last integral could also be evaluated explicitly, but it is simpler to recognize that the
above is the average kinetic energy, which is the average energy minus the average
potential energy, namely:
p2
1 2
1
k

 E    kx   E 
2m
2
2
4