Modern Physics: Home work 5
Due date: 7 March. 2014
Problem Set 5: Measurement Probabilities,
Wave Packets and Uncertainty
Solution
1. Normalizing a Wave Packet and Finding Probabilities
Consider a particle whose wave function at a particular time is given by,
x
f (x) = Ae− 3 ,
x>0
x
= Ae 3 ,
x<0
= 0
otherwise
(1)
(a) We have seen in class that we normally want to normalize the wavefunction so
that the total probability of finding the particle anywhere is 1. Find the constant
A so that this wave function is normalized.
(b) Find the probability of finding this particle between x = 2 and x = 3. Also find
the probability of finding the particle between x = 3 and x = 4.
Answer 1 Consider the following piecewise defined function.
x
f (x) = Ae− 3 ,
x>0
x
= Ae 3 ,
x<0
= 0
otherwise
(a) We compute the constant A, so that f (x) is normalized. The normalization
condition is,
Z
∞
|f (x)|2 dx = 1,
−∞
From the definition of f (x) above, we can find,
2x
|f (x)|2 = |A|2 e− 3 ,
2x
Spring semester 2014
x>0
= |A|2 e 3 ,
x<0
= 0
otherwise
1
Modern Physics: Home work 5
Due date: 7 March. 2014
Use of this expression in the normalization condition gives,
Z ∞
Z 0
2x
2 2x
3
|A|2 e− 3 dx = 1
|A| e dx +
−∞
0
|A|2 × 3 = 1
1
|A| = √ .
3
Thus any complex number with the magnitude found above would do. Clearly
√
the simplest choice is A = 1/ 3.
(b) We now compute the probability of finding the particle between x = 2 and x = 3.
Z 3
|f (x)|2 dx
P2≤x≤3 =
Z2 3
1 − 2x
e 3 dx
=
3
2
∼
= 0.064.
lastly the probability of finding the particle between x = 3 and x = 4 is given by,
Z 4
P3≤x≤4 =
|f (x)|2 dx
Z3 4
1 − 2x
e 3 dx
=
3 3
∼
= 0.033.
2. Probabilities for Momentum Measurements
We have a beam of particles each of which is described by the wave function at t = 0,
r
r
1 ik1 x
2 ik2 x
Ψ(x) =
e
+
e .
3
3
Suppose we have some detectors which can measure their momenta. What possible
values of momentum our detectors will find? If we get a 1000 hits in our detectors,
what number of particles will be found with each possible value of momentum.
Answer 2
The wavefunction for each particle at the initial time is given by:
r
r
1 ik1 x
2 ik2 x
Ψ(x) =
e
+
e .
3
3
Spring semester 2014
2
Modern Physics: Home work 5
Due date: 7 March. 2014
This clearly is a superposition of two states of definite momenta, corresponding to
wavenumbers k1 and k2 . Upon a momentum measurement, we would observe either
of these momenta. The relation between the wavenumbers and momenta is given by,
Now suppose we perform the momentum measurement on 1000 particles. The probability associated with measuring the first value would be mod squared of the corresponding coefficients in the wavefunction,
1 2 1
P (p = p1 ) = √ = ,
3
3
and in a thousand particles, the number found in this momentum state would be,
N1 = 1000 ×
1
3
N1 = 333.
By a similar argument, we would have for the second momentum state:
r 2
2
=2
P (p = p2 ) = 3
3
2
and
N1 = 1000 ×
3
N1 = 667.
3. A Moving Wave Packet?
Consider a particle whose wave function is given by,
Ψ(x) = Ae−αx
2 +ik
0x
.
If we measure the momentum of this particle, what is the most probable value that
we will get? How much less likely are we to get a value around the momentum ~k0 /5
compared to the most probable value of the momentum?
Answer 3
The position space wavefunction of a particle is given by,
Ψ(x) = Ae−αx
Spring semester 2014
2 +ik
0x
.
3
Modern Physics: Home work 5
Due date: 7 March. 2014
The Fourier transform of this function gives us the momentum space (or the k space)
wavefunction.
Ψ(k) =
=
=
=
=
Z ∞
1
√
Ψ(x)e−ikx dx
2π −∞
Z ∞
1
2
√
Ae(−αx +ik0 x) e−ikx dx
2π −∞
Z ∞
A
2
√
e(−αx +i(k0 −k)x) dx
2π −∞
Z ∞
i(k0 −k) 2 (k−k0 )2
A
√
e(−α[x+ 2α ] − 4α ) dx
2π −∞
(k−k0 )2 Z
∞
i(k0 −k) 2
Ae− 4α
√
e(−α[x+ 2α ] ) dx,
2π
−∞
Thus the k-space wavefunction is,
A − (k−k0 )2
e 4α .
Ψ(k) = √
2α
The most probable k-value is where the wavefunction peaks. Clearly, this happens at
(k = k0 ), and the corresponding momentum value is p0 = ~k0 .
Moreover, a momentum value of ~k0 /5 corresponds to a k value of k0 /5. The probability associated with the k value is,
P (k0 /5) = |Ψ(k0 /5)|2 dk,
and the probability associated with the peak value is,
P (k0 ) = |Ψ(k0 )|2 dk.
The ratio can be carried out using the expression for Ψ(k),
2
−8k0
P (k0 /5)
= e 25α .
P (k0 )
The factor on the right hand side above, tells us how less likely we are.
4. Constructing Wave Packets When Amplitude Distribution Is Given
Waves with different amplitudes for different wave numbers superpose to give a resultant wave form. The amplitudes for different wave numbers (k) are given by,
A(k) = k0 − k
k0 > k > 0
= k0 + k
− k0 < k < 0
= 0
Spring semester 2014
otherwise
(2)
4
Modern Physics: Home work 5
Due date: 7 March. 2014
(a) What is the shape of this Amplitude function?
(b) What would be the resultant wave-packet?
(c) Define the spread in the wave numbers to be simply equal to k0 as the amplitudes
outside of this region are zero. Define a suitable width of the resultant wave
packet. Any definition which you think makes sense and is easy to read off from
the wave packet. Show that
∆k∆x ∼ 1,
within an order of magnitude.
Answer 4 Consider the amplitude distribution for wavenumbers:





∀ 0 < k < k0 


 k0 − k

A(k) =
k0 + k




0
∀ − k0 < k < 0
otherwise




(a) It is straight forward to sketch this function.
(b) The wavepacket in position space is just the Fourier transform of A(k).
Z ∞
1
A(k)e−ikx dk
Ψ(x) = √
2π −∞
Z ∞
Z ∞
1
= √
A(k) cos kx dk + i
A(k) sin kx dk .
2π −∞
−∞
Nota that A(k) is even and sin kx is odd, and thus the second integral above,
vanishes. Therefore,
Z ∞
1
Ψ(x) = √
A(k) cos kx dk
2π −∞
Z 0
Z k0
1
= √
(k + k0 ) cos kx dk +
(−k + k0 ) cos kx dk
2π −k0
0
Z 0
Z 0
Z k0
1
= √
k cos kx dk +
k cos kx dk + k0
cos kx dk
2π −k0
k0
−k0
2
= 2 (1 − cos k0 x)
x
4
2 k0 x
= 2 sin
.
x
2
Spring semester 2014
5
Modern Physics: Home work 5
Due date: 7 March. 2014
(c) We can recast the wavefunction into a form most suitable for sketching.
4
2 k0 x
Ψ(x) = 2 sin
x
2
2
sin(k0 x/2)
= k0
.
(k0 x/2)
From the class room discussion, the graph of the expression inside the square
brackets is easily seen to be,
The square then looks like the following.
We define the width to be π/k0 from this graph. Thus,
2π
and ∆k = k0
k0
⇒ ∆k ∆x = 2π ∼ 1.
∆x =
Hence proved.
5. A (not so) long Wave
Consider a wave-packet which is given by,
(x) = eik0 x
= 0
π
π
<x<N
k0
k0
otherwise.
for − N
(3)
(a) Find the amplitude distribution for different wave numbers that are present.
(b) Make a sketch of this wave packet when N = 5.
(c) What would be suitable width for N = 5 wave packet?
(d) What would be suitable width for the wave numbers that are present (width of
the amplitude distributions)?
(e) Find the uncertainty relation for the two widths?
(f) Now make this wave very long. Say make N = 1000. How does the wave look
like? What happens to the width of the Amplitude distribution? Find it. Find
the product of the two widths again. How do you compare this wave with a pure
Spring semester 2014
6
Modern Physics: Home work 5
Due date: 7 March. 2014
harmonic wave of wave number k0 in the light of above fact about the width of
the wavenumbers?
Answer 5
We are given a wave packet defined by,

 eik0 x
∀ −
Ψ(x) =
 0
Nπ
k0

< x < + Nk0π 
otherwise 
(a) The amplitude distribution for the various wavenumbers is just the Fourier transform of the wavefunction above.
A(k) =
=
=
=
=
Z ∞
1
√
Ψ(x)e−ikx dx
2π −∞
Z +N π/k0
1
√
eik0 x · e−ikx dx
2π −N π.k0
Z +N π/k0
1
√
ei(k0 −k)x dx
2π −N π.k0
r Z +N π/k
0
2
cos[(k0 − k)x] dx
π 0
r
2 N π sin[(k − k0 ) · N π/k0 ]
,
π k0
[(k − k0 )N π/k0 ]
where the function has been written out in a form that is most convenient for
sketching.
(b) We are given that N = 5,
r
⇒ A(k) =
2 5π sin[(k − k0 ) · 5π/k0 ]
.
π k0
[(k − k0 )5π/k0 ]
From the class room discussion, it is very straight forward to sketch the function.
(c) Clearly a suitable width for this function is,
∆k =
6k0
.
5
Moreover the position space wave packet has oscillations which do not die out
unless x >
5π
k0
or x < − 5π
. Thus a suitable width for the wave packet would be,
k0
∆x =
Spring semester 2014
5π
.
k0
7
Modern Physics: Home work 5
Due date: 7 March. 2014
(d) Answer for this part is same as that of part (c).
(e) The product of two uncertainties found above would be,
∆k ∆x =
6k0 5π
·
= 6π.
5 k0
(f) In the case when N = 1000, the position space wave packet looks exactly the
same as before. Its real and imaginary parts are the usual sinusoids. The only
difference is that now, a sketch of each of these parts includes a large number
of cycles. So the shape of the wavefunction remains exactly the same, while its
length becomes very large.
The width of the amplitude distribution is given by equating argument of the
sine function to π.
(k − k0 )
Nπ
= π
k0
1
⇒ ∆k = k0 1 +
.
N
Clearly as N grows without bound, the width approaches k0 , and for the case
when N = 1000, the width is,
∆k = 1.001k0 .
The width of the position space wavefunction,
∆x =
Nπ
1000π
=
,
k0
k0
and so the product of the uncertainties for N = 1000 is,
1001
1000π
∆k∆x =
k0
= 1001π.
1000
k0
We see that for large N , the width of the wavenumbers approaches k0 . This is
exhibited with the choice N = 1000, and this is exactly true for a pure harmonic
wave with wavenumber k0 . ∆k = k0 , since N → ∞. Thus our wave with
N = 1000, accurately models a pure harmonic wave as far as the width of the
amplitude distribution is concerned.
6. A Radio Detector
Spring semester 2014
8
Modern Physics: Home work 5
Due date: 7 March. 2014
A radio detector observes a pulse sent from somewhere with the following form for the
amplitudes of various frequencies,
2 ω − ω0
.
A(ω) = exp −
∆
where ω0 and ∆ are constants. Approximately how long did this pulse last? (The time
of the beep in your circuit?)
Answer 6
We are given the amplitude distribution interms of the frequencies,
2 ω − ω0
A(ω) = exp −
.
∆
The time domain wavefunction for the wave can be found by taking the Fourier transform of the above expression.
Z ∞
1
f (t) = √
A(ω)eiωt dω
2π −∞
2
Z ∞
1
ω − ω0
= √
exp −
+ iωt dω.
∆
2π −∞
After completing the square and pulling out terms which are constant with respect to
the integration, last expression can be written as,
2 Z ∞
2 1
1
1
i∆2 t
i∆2 t
2
2
exp − 2 ω0 − ω0 +
·
dω
f (t) = √ exp − 2 ω0 − ω0 +
∆
2
∆
2
2π
−∞
2 √
1
1
i∆2 t
2
·∆ π
= √ exp − 2 ω0 − ω0 +
∆
2
2π
2 2
∆
−∆ t
= √ exp
exp (iωt).
4
2
The real part is a cosine function that fades off exponentially with time. It is zero for
the first time when ω0 t = π/2, ⇒ t = π/2ω0 , and we claim that the beep lasts nearly
this long.
7. Uncertainties in Position
A proton has a kinetic energy of 1.0 MeV . If its momentum is measured/known with
an uncertainty of 5%, what is the minimum uncertainty, in meters, in its position?
Answer 7
Spring semester 2014
9
Modern Physics: Home work 5
Due date: 7 March. 2014
We can use formula for the kinetic energy to solve for the momentum.
k=
p2
2m
⇒ p=
√
2mk.
(4)
Now from the problem statement,
∆p
= 0.05
p
⇒ ∆p = 0.05p
√
= 0.05 2mk
(5)
From the uncertainty principle,
∆x∆p ≥
⇒ (∆x)min. =
~
2
~
2∆p
~
√
2(0, 05 2mk)
∼
= 4.6 × 10−14 m.
=
8. Uncertainties in Masses of Particles
A π 0 meson is an unstable particle, with a rest mass of 135 MeV/c2 , which decays
into other particles with a life time of 8.7 × 10−17 seconds. Since we only have a finite
time available, in principle, to measure its mass, which is related to its energy, there
must be uncertainties in its mass and various π 0 ’s will show a range of masses when
we measure them. What is the spread in masses of this particle?
Answer 8
For the meson, the rest mass is m0 = 135 MeV/c2 , the mean life time is, ∆t =
8.7 × 10−17 s. We determine the uncertainty in the rest mass.
From mass energy equivalence,
E = m0 c2 ⇒ ∆E = ∆m0 c2
(6)
From the uncertainty relation,
∆E∆t ≥
Spring semester 2014
~
2
(7)
10
Modern Physics: Home work 5
Due date: 7 March. 2014
Using equations (6) and (7),
∆m0 ≥
~
2e2 ∆t
∼
= 6.7 × 10−36 kg
⇒ (∆m0 )min. ∼
= 6.7 × 10−36 kg.
9. Ground State or Minimum Energy due to Quantum Uncertainties
The wave nature of particles causes uncertainties in simultaneous knowledge of position
and momentum of a particle. (This uncertainty is sometimes thought of as fluctuations
in the variables of position and momentum, as we fail to prepare the states which have
a precise values for these variables. Making many such systems, each one shows some
variance around the central value we tried to create.) This uncertainty causes many
systems to have some lowest amount of energy above zero, at least if thought of as
a typical value of energy measured, due to fluctuations above the minimum possible
classical value. This is what we did for harmonic oscillators in class.
Consider the following systems and obtain the minimum possible ground state energy
in each case,
(a) A particle confined in a one dimensional box of length L. The particle can only
be within this length and not outside it. What is the minimum energy
(b) An electron around a positively charge nucleus. The potential energy is given by,
e2
U (r) = −k .
r
Just consider this problem in radial direction as energy only depends on the radial
motion. (Motion in the direction towards or away from nucleus). Assume the
Uncertainty principle between momentum and r. And estimate the minimum
energy the electron can have. Notice that, classically the minimum is achieved
when electron is sitting on the nucleus and that causes the catastrophe. What
is the minimum energy now? What is the uncertainty in r corresponding to this
energy? Compare the minimum energy with Bohrs ground state energy and the
uncertainty in r with Bohrs radius of Hydrogen atom.
A photon is emitted as an atom makes a transition from n = 4 to n = 2 level.
What is the frequency, wavelength and energy of the emitted photon?
Spring semester 2014
11
Modern Physics: Home work 5
Due date: 7 March. 2014
Answer 9
(a) The position uncertainty is ∆x = L. We have from uncertainty principle,
∆x∆p ≥
~
2
⇒ ∆p ≥
~
~
=
2∆x
2L
(8)
Now when the particle inside the box, the potential energy is zero. Hence the
total energy is purely kinetic.
E =K +U =K +0=K =
p2
.
2m
Now corresponding to the momentum uncertainty, there is clearly a minimum
energy that can be associated to the particle.
~2
(∆p)2
=
.
Emin. ∼
=
2m
8mL2
(b) We consider the hydrogen atom in the radial direction only. The potential energy
is,
U (r) = −
ke2
,
r
and clearly the total energy of the electron would be,
E=
ke2
p2
−
2m
r
(9)
From the uncertainty relation,
∆p∆r ≥
∆p ≥
~
2
~
.
2∆r
(10)
To find the minimum energy, we need to minimize the expression in (9). The
only way to do this is to make the first term on the right hand side as small
as possible, and the second term as large as possible (in absolute values). This
clearly minimize E, and the obvious choice is to replace p with ∆p and r with
∆r. Thus
(∆p)2 ke2
E =
−
2m
∆r
~2
ke2
≥
−
.
8m(∆r)2 (∆r)
Spring semester 2014
12
Modern Physics: Home work 5
Due date: 7 March. 2014
We next compute the derivative of this function with respect to ∆r and equate
that to zero.
⇒
dE
~2
ke2
= −
+
=0
d(∆r)
4m(∆r)3 (∆r)2
~2
(∆r)0 =
.
4mke2
(11)
An evaluation of the second derivative at this point shows it to be a minimum.
4
3
d2 E
1
3~2 1
2
=
− 2ke
2
d(∆r)
4m ∆r
∆r
2
3
3
4
8
d E 4mk e
⇒
=
> 0.
d(∆r)2 (∆r)0
~6
Thus (∆r)0 is a minimum point, and he minimum energy is,
~2
ke2
−
8m(∆r)20 (∆r)0
2mk 2 e4
= −
.
~2
Emin. =
(12)
The corresponding radial uncertainty was found in (11),
(∆r)0 =
~2
.
4mke2
Using the notion of angular momentum and centripetal force, one can show that
the Bohr energy levels of the hydrogen atom are,
En = −
mk 2 e4
.
2n2 ~2
(13)
Setting n = 1 gives the ground state energy,
Eg = −
mk 2 e4
.
2~2
(14)
Using (12) and (14), we can compute the ratio,
Emin.
= 4
Eg
⇒ Emin. = 4Eg .
This yields the required comparison.
Moreover the orbital radii for the electron in the Bohr model can be shown to be,
rn =
Spring semester 2014
n2 ~2
.
mke2
13
Modern Physics: Home work 5
Due date: 7 March. 2014
Choosing n = 1 gives the Bohr radius.
a0 =
~2
.
mke2
(15)
Next using (11) and (15), we can perform the required comparison,
1
(∆r)0
=
a0
4
a0
⇒ (∆r)0 =
.
4
Note that in our approximation, the energy is four times larger and the position
uncertainty is four times smaller, in comparison with the Bohr model.
Spring semester 2014
14