particle in a box
the uncertainty principle
"As far as the laws of mathematics refer to reality, they are not certain,
and as far as they are certain, they do not refer to reality.“—A. Einstein
The equations you need to work the problems are not “new”
ones, but I’ll give you them to you as a reminder.
nλ = 2d sinθ
h
λ=
mv
1
KE = mv 2
2
3.6 Particle in a Box
Now you believe that particles “have” waves, right?
Could we digress a minute. Just what does this mean?
Is the particle the only reality, and the wave just something
physicists invented?
—ruled out by experiment—matter waves shown to interfere
Is it as Schrödinger first believed—the wave is real, not the
particle?
—abandoned for many reasons, including the fact that unlike
waves, particles do not “spread out”
Is the particle real, and the wave something that “guides” it (a
“pilot wave”)?
—no successful theory was found based on this idea
Is there some single reality which sometimes manifests itself
as particle and sometimes as wave but is really both at once?
—this has come to be the commonly accepted view—but it is
possible the whole story has not yet been told
Going much deeper into this discussion leads into philosophy
and away from physics, so back to the physics…
Oh…why study a particle in a box…
Anyway, supposing that you believe that particles have a wave
nature—and experiment certainly backs this up—let’s pursue
some of the consequences.
Remember your Physics 201
vibrating string experiment. You
couldn't set up arbitrary
vibrations of the string. Instead,
you saw patterns like this.
You get standing waves because the wave pulses traveling
down the string reflect (and change phase by 180°) when they
reach the ends.
The standing wave consists of a series of pulses moving up
and down the string. These pulses superpose, and when they
are in phase, you see maxima. When they are out of phase,
you see nodes.
You see standing waves only when
the pulse speeds are “just right,”
and the standing wave “fits” on the
length of the string. Here’s a static
picture:
(http://www.chem.uci.edu/education/unde
rgrad_pgm/applets/dwell/dwell.htm)
dead link January 2005—but I only used it for the picture anyway
What does this have to do with a particle in a box?
Place a particle, represented by a wave, in a box of length L.
The particle wave moves “with” the particle, reflects when it
hits the wall of the box, and then again when it hits the other
end of the box.
If the box is small enough (compared to the particle
wavelength), the particle wave "folds up" over and over again
every time it reflects off a wall. Here’s the “best” visualization
I could find:*
http://www.zbp.univie.ac.at/schrodinger/ewellenmechanik/simulation.htm
The segments of the particle wave bouncing back and forth
interfere. If they interfere constructively, our particle can fit in
the box. Otherwise -- no particle fits in the box.
*Caution! Intense material! content! Shows death of wave!
The folding of the particle wave packet
to add in phase works when the length
of the box is an integral number of half
wavelengths of the particle wave
(L=n/2), hence equation 3.17 in
Beiser for the de Broglie wavelengths of
a trapped particle:
2L
λn =
, n = 1,2,3...
n
Because* KE = mv2/2 and  = h / mv, the restriction on  also
places a restriction on the allowed particle energies:
n2h2
En =
, n = 1,2,3...
2
8mL
*So this is a nonrelativistic calculation.
2L
λn =
, n = 1,2,3...
n
n2h2
En =
, n = 1,2,3...
2
8mL
The permitted energies are called "energy levels" and the
number n is called a "quantum number."
Think of the box as a potential well,
inside which a particle is placed.
A free particle, outside the box, can have
any energy, and any wavelength.
When you put the particle in the box, only certain wavelengths
and energies are allowed (and note that zero is not one of the
allowed energies). You most likely will have to add or remove
energy to put a free particle into a box.
Comment: “quantum” implies something is quantized (energy,
momentum). Quantization of properties of matter is a
consequence of the wave nature of matter. Thus, the words
“wave” and “quantum” are closely associated in my
vocabulary.
All of us in this room have discrete energy levels and quantum
states.
However, as the example on page 107 shows, there are
enough energy levels to form, for all practical purposes, the
continuum that Newtonian mechanics supposes.
10 gram marble in a box 10 cm wide:
n2h2
En =
8mL2
n  6.63×10
2
En =
8 10×10
-3

-34 2
10 
-1 2
En = 5.5×10-64 n2 Joules
Minimum energy and speed are indistinguishable from zero,
and a marble of reasonable speed has a quantum number on
the order of 1030. In other words, we can't perceive the
quantum features of the marble in the box.
Electron in a “box” 0.1 nm (10-10 m) wide (the size of an
atom):
n2h2
En =
8mL2
n  6.63×10
2
En =
8  9.11×10
-31

-34 2
10 
-10 2
En = 6.0×10-18 n2 Joules = 38 n2 eV
The minimum energy is 38 eV, a significant amount, and the
energy levels are far enough apart to be measurable.
Is this information useful, or is it just physics trash talk?
Quantum well lasers (183000 pages found in Google search,
September 2002):
http://nsr.mij.mrs.org/3/1/
http://www.iis.ee.ethz.ch/research/tcad/laser_sim.en.html
http://www.chem.wisc.edu/courses/801/Spring00/Ch1_3.html
Quantum well memory (209000 pages found in Google search,
September 2002):
http://www.physicstoday.org/pt/vol-54/iss-5/p46.html
http://www.sciencenews.org/sn_arc99/2_27_99/fob5.htm
http://theorie5.physik.unibas.ch/qcomp/qcomp.html
OK, so maybe we should pay attention to this wave/quantum
stuff. Is there anything else important “hidden” in the wave
nature of particles?
Quantum well lasers, 388000 pages Sept. 2003; 450000 pages Jan. 2005.
Quantum well memory, also 388000 pages Sept. 2003; 1100000 pages jan. 2005.
3.7 Uncertainty Principle I -- derivation based on the
wave properties of particles
Consider the particle represented
by this wave group.
Where is the particle?
What is its wavelength?
The position is well-defined.
But the wavelength is poorly defined, and
therefore there is a large uncertainty in the
particle’s momentum (remember--wavelength
and momentum are related).
Now consider the particle
represented by this wave
group.
Where is the particle?
What is its wavelength?
The wavelength seems to be rather well-defined, but the
position is poorly defined. There is a large uncertainty in the
particle’s position.
To quantify the uncertainties in the wave group's position and
momentum, we need to go into much more detail about
Fourier transforms and representation of wave groups by
summations of individual waves.
Beiser does this on pages 108-111. Please read this material.
You may be tested or quizzed on major concepts (but not
“trivial” details).
What I want you to know (backwards and
forwards), comes out of this derivation,
and is called Heisenberg’s* Uncertainty
Principle:
h
ΔxΔp x 
4
It is not possible to simultaneously measure, with
arbitrary precision, both the position and momentum
of a particle.
*1932 Nobel Prize for creation of quantum mechanics.
The quantity h/2 appears over and over again in modern
physics, so we give it a special symbol: ħ = h /2. The
uncertainty principle can then be written
ΔxΔp x 
2
.
There are fundamental limits on how precisely we can
simultaneously measure a particle's position and momentum.
Because of the wave nature of matter, there are fundamental
limits on how precisely we can know things.
These limits have nothing to do with our measurement
techniques; they are built into nature.
“Marvelous what ideas young people have these days. But I don’t believe a
word of it.”—A. Einstein, referring to the uncertainty principle
Example 3.6
A measurement establishes the position of a proton with an
accuracy of 1.00x10-11 m. Find the uncertainty in the
proton’s position 1 s later. Assume v << c.
At the time of measurement, the position uncertainty is x1,
and
Δx1 Δp x 
2
Δp x 
2Δx1
Δpx = Δ(mv x ) = m Δv x
Δv x =
Δp x

m
2m Δx1
it’s OK to do this for vc
A time t later, the position uncertainty x2 is
Δx 2 = t Δv x
Δx 2 
t

2m Δx1
1 1.054×10-34 
About 1/10 the size of
an atom, but much
larger than a nucleus.
2 1.67×10-27  1.00×10-11 
Δx 2  3.15×103 m , or 1.96 miles.
Is vc?
To put it bluntly, we have no clue where the proton is 1 second
later.
The proton didn’t spread out, because it is “somewhere,” but its wave
certainly did!
3.8 Uncertainty Principle II -- derivation based on the
particle properties of waves*
I claimed above that the limits implied by the uncertainty
principle are fundamental to nature, and are due to the wave
properties of matter. This follows cleanly and logically from
the mathematics of waves.
As humans, we are left with nagging doubts about the
uncertainty principle. How dare nature tell us there are things
we cannot know! Surely this is just technical glitch that
human cleverness can overcome.
Heisenberg (although a theorist first, last, and always) believed
he had to specify “definite experiments” for measuring an
object’s position in order to validate the uncertainty principle.
*Caution: reading this section may be hazardous to your grade!
Heisenberg proposed a thought experiment (which can be
realized in fact): let’s suppose we want to measure the
position of this electron very precisely. How do we do it?
Visible light?
The sphere doesn’t represent the
size of the electron; it represents
the size of the region in which we
wish to locate the electron.
A “real” red light wave would have a
much longer wavelength than this!
The wavelength of visible light is far too large to allow us to
detect the position of the electron. The wavelength needs to
be comparable to the position precision we seek.
You might say the electron is somewhere along the wave, but the
wavelength is so long that the imprecision in position is enormous.
The last sentence contains a clue: find some kind of radiation
that has a much shorter wavelength.
Gamma rays have short wavelengths. They should work.
But short wavelength gamma radiation carries
lots of energy and momentum.
http://www.aip.org/history/heisenberg/p08b.htm
Our gamma-ray microscope can tell us where the electron
was, but it can’t tell us where it came from or where it is
heading (its momentum).
So we can forget about position (but measure momentum), or
forget about momentum (but measure position).
I find this an interesting thought experiment, but it implies
that the uncertainty principle is really only a measurement
difficulty after all.
A surprising number of students tell me on tests (through their
answers to multiple choice questions) that the uncertainty
principle is just a result of experimental difficulties.
No!------------------------------------------------------------------------------------------------------------No!--------------------------------------------------------------------------------------------------------- No!
“The indeterminacy (of position and momentum) is inherent in
the nature of a moving body.”
Let me repeat: Heisenberg’s thought experiment is “bad”
because it implies the uncertainty is merely some technical
measurement difficulty.*
3.9 Applying the Uncertainty Principle
Planck's constant is so small that we never encounter the
uncertainty principle in Newtonian mechanics…
…but its consequences are manifested in materials we
constantly use in everyday life! You’ll hear about it repeatedly
in this course.
*To placate his critics and get his uncertainty principle paper published, Heisenberg
had to include this thought experiment in it. The thought experiment has appeared
in (probably) most texts, and confused thousands (millions?) of students. All to get
a paper published.
Frequency and time are related, and velocity and energy are
related, so we can derive an alternate expression of the
uncertainty principle:
h
ΔEΔt 
4π
ΔEΔt 
2
.
It is not possible to simultaneously measure, with
arbitrary precision, both a time for an event and the
energy associated with that event.
“Prediction is very difficult. Especially about the future.”—Neils Bohr and/or
Mark Twain (we are not certain who said this)
http://www.nearingzero.net
Example 3.7
A typical atomic nucleus is about 5x10-15 m in radius. Use the
uncertainty principle to place a lower limit on the energy an
electron must have if it is to be part of a nucleus.
The problem is asking you something about the energy of an
electron confined to a region 5x1015 m in size. Obviously, the
starting point is
ΔEΔt 
2
.
NOT!
You are given information about the electron’s x. In fact, you
are implicitly told to use x = 5x10-15 m. You must use
ΔxΔp x 
2
.
Proceeding trustingly with the math…
Δp x 
2 Δx
You could plug numbers in now. But let’s keep it symbolic and
think for a bit.
We’ve calculated a momentum uncertainty p. Does it make
sense to claim the electron has less momentum than p?
This discussion was designed to get you to agree with the
statement that if an electron has a momentum uncertainty
given by
Δp x 
2 Δx
,
then it doesn’t make sense to talk about a momentum for the
electron which is less than px.
If we agree that
p x,min = Δp x 
2 Δx
,
then the minimum electron momentum is
px =
2 Δx
.
Classically, KE = p2 / 2m, so*
2


2


2
px 

2
Δx
 =
KE =
=
.
2
2m
2m
8 m  Δx 
What’s this business about calculating KE. Didn’t the problem ask for
“energy?” Doesn’t that mean E? Well… in this context the problem was
really asking for kinetic energy. Ask me if you are unsure about which
energy an exam question is asking for.
So the minimum electron (kinetic) energy is
KE =
2
8 m  Δx 
2
.
1.055×10 

KE =
 8   9.11×10  5×10 
-34 2
-31
-15 2
.
KE = 6.11×10-11 joules .
Units work out correctly if you keep everything in SI. However, I
encourage you to show units in your calculations on exams, to prevent
mistakes and help me give you more points if you make a simple math
error.
So, KE=6.11x10-11 joules. Any comments?
Hey! Isn’t that kind of a big energy? I mean for an electron?
6.11x10-11 joules x 1 eV / (1.6x10-19 joules) = 3.82x108 eV.
3.82x108 eV = 3.82x102x106 eV = 382 MeV.
Now what’s that electron “rest mass” again… 0.511 MeV/c2.
KE = p2 / 2m
2

E = mc
2

2
+ p2 c 2 .
For extremely high speeds and energies, pc >> mc2 so
2

E = mc
≈0
2

2
+p2 c 2
c
E=
2 Δx
p
E=
Remind me again: how’d
the = sign slip in?
-34
8
1.055×10
3×10



2 5×10-15 
E = 3.17×10-12 joules = 19.8 MeV
Here we go again replacing KE by E, except for this large an energy, KE≈E.
Four things to notice:
The classical calculation far overestimated the kinetic
energy (makes sense—classical calculations can come up
with speeds greater than c).
The kinetic energy is about 20 MeV, which is much much
greater than 0.511 Mev, so E ≈ pc was a reasonable
approximation.
 The “p” in the uncertainty principle equation is the
relativistic momentum.
 If you want to confine an electron to a nucleus, its wave
nature requires that it have at least 20 MeV of energy.
This concludes (finally) example 3.7. Example 3.8 is similar
except that a nonrelativistic calculation is OK.
Example 3.9
An “excited” atom gives up its excess energy by emitting a
photon, as described in chapter 4. The average period that
elapses between the excitation of the atom and the time it
radiates is 10-8 s. Find the inherent uncertainty in the
frequency of the photon.
We have time, want f, but E and f are related, so
h
ΔEΔt 
4π
Beiser uses the ħ version; you’ll
see why I use this in a minute.
E = hf  ΔE = hΔf
h
h Δf Δt 
4π
1
Δf 
4 πΔt
By not calculating numerical values
right away, I simplified the math!
1
Δf 
4 π 10-8 
Δf  7.96×106 Hz
Application: it is usually desirable to have
laser lines be very “sharp,” i.e., the laser
emits only a single color of light. The width
of the laser line to the right depends on the
design of the laser, but not even the
cleverest design can produce a line narrower
than that given by the uncertainty principle.
intensity
If you measure the intensity vs. frequency of the light emitted
from this atom, the spectrum will have at least this intrinsic
linewidth.
frequency