Download Physics Tutorial 19 Solutions

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Physics Tutorial 19 Solutions
Quantum Physics II
Self-check Questions:
S1. State the position-momentum and energy-time uncertainty principles.
h
h
x p 
, E t 
4
4
S2. Do the uncertainty principles arise because of imprecision in measuring
instruments?
No. Heisenberg was careful to point out that the uncertainty is not due to the measuring
instruments but due to the fundamental, intrinsic nature of matter.
S3. Does the position-momentum uncertainty principle imply that a precise measurement
of a particle’s position is impossible?
No. It is just saying that the more precisely the position of a particle is determined, the less
precisely the momentum is known in this instant, and vice versa.
S4. How can one get the wave function of a body?
In quantum mechanics, the wave function can be gotten by solving the Schrodinger’s equation
for the specified conditions.
S5. Does the wave function have any physical meaning?
No. It is just a mathematical representation that scientists use to describe the quantum particle.
As the wave function does not have any physical meaning, it can have real and complex parts.
2
However, the absolute square of the wave function  is always real and has a real physical
meaning.
S6. How do we calculate the probability of finding the particle within any spatial region
from the wave function?
2
The absolute square of the wave function  is probability density function, and gives the
probability of finding the body at a particular location x. If the body is moving only in 1-D along
the x-axis, the integral 
x2
x1
2
 dx gives the probability of finding the body between x2 and x1.
S7. What is the meaning of normalizing a wave function?
A wave function that obeys



2
 dx  1 is said to be normalized.
S8. What is a potential barrier?
A potential barrier refers to a potential energy distribution of which energy height is higher than
the incident particle’s energy.
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S9. How is quantum tunnelling different from classical physics?
Classically, the particle would not be able to move past the region of the potential barrier, but
quantum mechanically, there is a probability that it can tunnel through.
S10.Can you recall the equations for the transmission coefficient and reflection
coefficient?
2m(U  E )
T  exp( 2kL ), k 
,
R T 1
Data:
Planck constant
Elementary charge
Rest mass of electron
Rest mass of proton
h = 6.63 x 1034 Js
e = 1.60 x 1019 C
me = 9.11 x 1031 kg
mp = 1.67 x 1027 kg
Heisenberg position-momentum uncertainty principle
1. An electron moves in a straight line with a constant speed v = 1.00 x 106 ms-1 which has
been measured to a precision of 0.20%. What is the maximum precision with which its
position could be simultaneously measured?
[Ans: 2.90 108 m]
Using the position-momentum uncertainty principle,
h
x p 
4
h
6.63  1034
x 

 2.90  108 m
(4 )(m )v (4 )(9.11 1031 )(0.0020  1.00  106 )
2. What is the minimum uncertainty in position as imposed by the uncertainty principle on a
100g baseball thrown at 42  1 m/s ?
[Ans: 5.28 1034 m]
Minimum uncertainty in the position of the baseball is
h
6.63  1034
x 

 5.28  1034 m
(4 )(m)v (4 )(0.100)(1)
3. A measurement establishes the position of a proton with an accuracy of 1.50  1011m .
Find the minimum uncertainty in the proton’s position 1.00 s later.
[Ans: 2110m]
Using the position-momentum uncertainty principle,
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2009
h
4
h
6.63  1034
v 

(4 )(m )x (4 )(1.67  1027 )(1.50  1011 )
x p 
= 2110 m s-1
Hence, the uncertainty in the position after 1.00 s is (2110) x (1.00) = 2110 m.
4. What is the uncertainty in the location of a photon of wavelength 300 nm if this
wavelength is known to an accuracy of one part in a million?
[Ans: 23.9 mm]
Using de Broglie relation,
h
p

dp
h
h
  2  p   2 
d


h
h
 2
x 


h
(4 )p
(4 )(  )
(4 )(  2  )


9 2
(3.00  10 )
 0.0239m
(4 )(3.00  10 13 )
Hence, the uncertainty in the location is 23.9 mm
Heisenberg energy-time uncertainty principle
5. An electron remains in an excited state of an atom for typically 10-8s. What is the
minimum uncertainty in the energy of the state?
[Ans: 5.28 1027 J]
Minimum uncertainty in the energy of the state is
h
6.63  1034
E 

 5.38  1027 J
4t (4 )(1 108 )
6. The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it
typically decays very quickly. Its average rest mass energy is 91.2 GeV, but its short
lifetime shows up as an intrinsic width of 2.5GeV (rest mass energy uncertainty). What is
the approximate lifetime of this particle?
[Ans: 1.32 1025 s]
The approximate lifetime of the particle is
h
6.63  1034
t 

 1.32  1025 s
4E (4 )(2.5  1.60  1010 )
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2009
The wave function
7. An electron in an atom may be considered to be a potential well, as illustrated by the
sketch graph below.
(a) Explain how, by considering the wave function of the electron, rather than by
considering it as a particle, there is a possibility of the electron escaping from the
potential well by a process called tunnelling.
Classically, an electron could never exist outside the potential barrier imposed by the
atom because it does not have sufficient energy. If the electron is treated as a wave and
applying Schrodinger equation, its wave function
-is sinusoidal with large amplitude between the barrier
-decays exponentially within the barrier
-is sinusoidal with a much smaller amplitude outside the atom
Recalling that the square of the amplitude of the wave function gives the probability of
finding the electron at a point, that means that the electron has a non-zero probability of
existing outside the potential well.
(b) Sketch a possible wave function for the electron in the figure.
8. Consider the following triangular wave function. Note that ψ(x) = 0 for all x > 1m.

-1
1
x/m
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2009
(a) For the triangular wave function to be a valid, physical wave function in
quantum mechanics, deduce its maximum value at x = 0m.
[Ans: 1.22]
For the wave function to be valid, it has to be normalizable. In this case, since the
graph is symmetric about x = 0, we have
1
2 |  |2dx  1
0
The graph is described by   a  ax where a is the max. value. Thus,
1
1


2 (a  ax ) dx  2 a  a x  2a xdx  2 a 2 x  a 2 x 3  a 2 x 2 
3

0
0
0
2
 a2  1
3
1
1
2
2
2
2
2
Hence, a2 = 1.5
a = 1.22
(b) Assuming that the wave function describes a particle moving in some potential,
what is the average value of the particle’s position?
[Ans: 0m]
Since the graph is symmetric about x = 0, the probability density function is also
symmetric about x = 0. Hence the average value of the particle’s position is at x = 0
since it is equally likely to find the particle on either side of the vertical axis.
(c) Calculate the probability of locating the particle between x = 0.75m and x =
1.00m.
[Ans: 7.81 103 ]
Probability =
1

0.75
 7.81 103
1
1


a 2  a 2 x 2  2a 2 xdx  a 2 x  a 2 x 3  a 2 x 2 

3

 0.75
0.75
1
(a  ax )2 dx 
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2009
9. A one-dimensional harmonic oscillator wave function is 
A exp(-bx2), where A and b
are constants and A >1. Sketch, on the same graph, (a) the graph of 
x. (b)
probability density distribution. (You may use the graphing calculator to help you)


x
Note that the wave function is:
1. Continuous
2. Normalizable (tends towards 0 as x tends to positive and negative infinity)
3. Single-valued
Quantum tunnelling
10. A beam of electrons is incident on a barrier 6.00eV high and 0.200nm wide. Find the
energy they should have if 1.00 percent of them are to get through the barrier.
[Ans: 0.937eV]
Using the tunnelling equation T = exp (-2kL), k 
T  exp[ 2(
2 2m(U  E )
h
2 2m(U  E )
)L]
h
2 2(9.11 10 31 )(6.00  1.6  10 19  E )
)0.200  10 9 ]
6.63  10 34
 6.00  1.6  10 19  E
0.01  exp[ 2(
8.10  10 19
E  1.4999  10 19 J
E  0.937eV
11. An electron with energy E of 5.1 eV approaches a barrier of height U = 6.8 eV and
thickness L = 750 pm. (1 pm = 10-12 m). What is the transmission coefficient?
[Ans: 4.5 x 10-5]
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Using the tunnelling equation T = exp (-2kL), k 
2009
2 2m(U  E )
,
h
T  exp[ 2(
2 2m(U  E )
)L]
h
T  exp[ 2(
2 2(9.11 10 31 )(6.8  5.1)  1.6  10 19
)750  10 12 ]
34
6.63  10
T  4.5  10 5
12. Seng, 60 kg, learnt from his physics tutor about tunnelling and began to fantasize that
perhaps he could tunnel through the classroom walls to a nearby arcade during his
physics lesson. One day, in the middle of a physics test, Seng moves at a stealthy
speed of 1 m/s towards the 1m wide wall which requires a moving object of 1 million
Joules of kinetic energy to be knocked down. Calculate the probability of Seng being
allowed by the laws of quantum mechanics to tunnel through.
[Ans: 0]
The probability of Seng tunnelling through the classroom wall is
2 2m(U  E )
T  exp[ 2(
)L]
h
T  exp[ 2(
2 2(60)[1 106  (0.5)(60)(1)2 ]
)1]
6.63  1034
T 0
13. A scanning tunnelling microscope (STM) image that Lawrence Livermore National
Laboratory of Silicon (100) took for a deposition of one monolayer of molybdenum has a
resolution of 1 1010 m.
(a) Briefly describe the application of quantum tunnelling to the probing tip of a STM and
how this is used to obtain atomic-scale images of surfaces.
Classically, since electrons are attracted to the positive ions on the tip and the material,
they cannot move across the vacuum gap between the sample and tip due to a lack of
energy. Quantum mechanically, electrons can tunnel through the empty space barrier. A
voltage is applied between the surface and tip to make the electrons tunnel
preferentially. The electron wave function falls off exponentially with a decay length of
order 0.1 nm. This exponential behaviour causes the current of electrons to depend
strongly on the tip-to-surface distance. By either using the constant current or constant
height mode, the images of surfaces on the atomic scale may be obtained.
(b) A realistic scenario for the STM can be modelled as shown in a rough outline as
depicted below:
Electron
V = 8 eV
d = 0.15 nm
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2009
Explain what the quantities d and V might mean, if the diagram is depicting the STM.
d refers to the width of the barrier and V refers to the potential height.
(c) Calculate the probability for an electron of energy 5 eV to tunnel through.
[Ans: 0.0699]
(2 )


T  exp( 2kL)  exp  2(0.15  109 )
2(9.11 1031 )(8eV  5eV ) 
h


0.0699
(d) Calculate the distance by which d has to be increased for T to drop to 1/e of its
value.
[Ans: 5.64 1011 m]
T2 exp( 2d 2k )

 exp  2k (d2  d1 )  exp( 1)
T1 exp( 2d1k )
d2  d1 
1
 5.64  1011 m
2k
Beyond A-level: Strictly Independent Work
14. Quantum mechanics seems to give us statistical information about possible trajectories of
particles. Suppose that one measured the position of the particle, and found it to be at a certain
point. At that moment, what happens to the wave function? Note that prior to the measurement,
the wave function specifies the probability density function, yet how does it change (if it does)
when we make the measurement and determine the position of the particle?
15. The diagram below sketches the wave function corresponding to a particle subject to a
restoring force, and thus experiencing simple harmonic motion.

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(a)
Interpret the wave function above, paying attention to (i)why it would represent the
trajectory of a particle in simple harmonic motion (ii)the locations where the wave
function is zero
(b)
From the wave function, we can obtain the probability density function of the particle
quantum mechanically. Derive the probability density function that describes the
particle following simple harmonic motion classically. Assume that x  x0 cos t where
 is the angular frequency, and sketch the probability density function you derived on
the sketch above. Take clear note of any assumption you make.
(c)
Bohr’s correspondence’s principle states that quantum mechanics is in agreement with
classical physics when the ‘quantum number’ is high. The plot above is an example of
such a quantum mechanical solution. Compare your derived classical probability
density function with the plot, and relate to the correspondence principle.
Q14.
Quantum theory of measurement process is a widely debated topic. The conventional view is as
follows: the wave function evolves according to Schrodinger’s equation before the
measurement, but upon measurement, the wave function collapses to a spike at the measured
value. In other words, the measurement causes our system to jump into an eigenstate of the
dynamical variable being measured. This is sometimes known as the Copenhagen
interpretation. (Interested students can search along the lines of ‘many-world interpretation’ as
well which speaks of parallel universes but this argument is not easy to understand, and it only
receives active revival recently following some results in string theory.)
Q15.
(a)(i) Notice that if we square the real wave function to obtain the probability density function, it
is higher towards each end of the significant part of the wave function. It is lowest at the centre.
This agrees with our intuitive notion of the classical oscillator which moves fastest at equilibrium
and has zero velocity at the amplitudes, and hence greater probability of being located there.
(ii)Quantum mechanics introduces the novel feature of ‘nodes’ points where the probability
density function is zero. The particle can never be found at these exact points.
(b)
1. Consider one interval for 0  t  T / 2 . Invoke the assumption:
t t
P( x1  x  x2 )  P(t2  t  t1 )  1 2
(T / 2)
This is true if every infinitesimal but similar time-interval holds equal probability for the xmeasurement to occur  a uniform distribution for the time of measurement. Invert the simple
 x
1
harmonic motion equation to obtain t  cos 1   .

 x0 
2. We can now construct the probability density function f (x) as follows: from our argument in 1,
we have:
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x2
2 
x1

 f ( x)dx  T   cos
 f ( x) 
1
2009

 x2 
1  x1 
   cos   
 x0 
 x0  
 1
2 d 
1  x 
  cos    
T  dx 
 x0   
1
x  x2
2
0
(c) The classical probability density function is resembles the quantum counterpart – for
example, the classical pdf has vertical asymptotes at the amplitudes while quantum
mechanically, the peaks of the wave function are very near to the amplitudes. Both also
decrease towards the equilibrium position. This resemblance increases as we move towards
solutions of higher quantum numbers, in accordance with the correspondence principle.