Plane Waves
Review of sinusoidal waves:
Wave in time: cos(2πt/T) = cos(ωt) = cos(2πf·t)
T = Period = time of one cycle
t
ω = 2π/T = angular frequency = number of radians per second
Wave in space: cos(2πx/λ) = cos(kx)
λ = Wavelength = length of one cycle
x
• Most general kinds of waves are plane waves
(sines, cosines, complex exponentials) – extend
forever in space
• ψ1(x,t) = exp(i(k1x-ω1t))
• ψ2(x,t) = exp(i(k2x-ω2t))
• ψ3(x,t) = exp(i(k3x-ω3t))
• ψ4(x,t) = exp(i(k4x-ω4t))
k = 2π/λ = wave number = number of radians per meter
k is spatial analogue of angular frequency ω.
(We use k because it’s easier to write sin(kx) than sin(2πx/λ).)
• etc…
Different k’s correspond to different energies, since
E = ½mv2 = p2/2m = h2/2mλ2 = 2k2/2m
Quiz
Three deBroglie waves are shown for particles of equal mass.
I
A, 2f
II
2A, 2f
x
III
A, f
x
x
The highest speed and lowest speed are:
a. II highest, I & III same and lowest
b. I and II same and highest, III is lowest
c. all three have same speed
d. cannot tell from figures above
ans b. shorter wavelength means larger momentum =
larger speed. III is largest wavelength, I and II are same.
amplitude of wave is not related to speed.
A: Amplitude. f: frequency
E=hc/λ…
A. …is true for both photons and electrons.
B. …is true for photons but not electrons.
C. …is true for electrons but not photons.
D. …is not true for photons or electrons.
c = speed of light!
E = hf is always true but f = c/λ only applies
to light, so E = hf ≠ hc/λ for electrons.
Superposition principle
Superposition
• If ψ1(x,t) and ψ2(x,t) are both solutions to
wave equation, so is ψ1(x,t) + ψ2 (x,t). →
Superposition principle
• E.g. homework (HW8, Q7b) – superposition
of waves one traveling to the left and to the
right create a standing wave:
???
ψ (x,t) = ΣnAnexp(i(knx-ωnt))
• We can make a “wave packet” by combining
plane waves of different energies:

6
Plane Waves vs. Wave
Packets
Plane Wave: ψ(x,t) = Aexp(i(kx-ωt))
Wave Packet: ψ(x,t) = ΣnAnexp(i(knx-ωnt))
Which one looks more like a particle?
• In real life, matter waves are more like wave packets.
Mathematically, much easier to talk about plane waves, and
we can always just add up solutions to get wave packet.
• Method of adding up sine waves to get another function (like
wave packet) is called “Fourier Analysis.” You explored it with
simulation in the homework.
8
Plane Waves vs. Wave Packets
Plane Wave: Ψ(x,t) = Aei(kx-ωt) :
Plane Wave: Ψ(x,t) = Aei(kx-ωt)
– Wavelength, momentum, energy: well-defined.
– Position: not defined. Amplitude is equal everywhere,
so particle could be anywhere!
Wave Packet: Ψ(x,t) = ΣnAnei(knx-ωnt) :
For which type of wave are position x and
momentum p most well-defined?
A. p most well-defined for
plane wave, x most
well-defined for wave
packet.
B. x most well-defined for
plane wave, p most
well-defined for wave
packet.
Plane Waves vs. Wave Packets
C. p most well-defined for
plane wave, x equally
well-defined for both.
D. x most well-defined for
wave packet, p most
well-defined for both.
E. p and x equally welldefined for both.
Superposition
Wave Packet: Ψ(x,t) = ΣnAnei(knx-ωnt)
– λ, p, E not well-defined: made up of a bunch of different
waves, each with a different λ,p,E
– x much better defined: amplitude only non-zero in small
region of space, so particle can only be found there.
Heisenberg Uncertainty Principle
• In math: Δx·Δp ≥ /2 (or better: Δx·Δpx ≥ /2)
• In words: Position and momentum cannot both
be determined precisely. The more precisely
one is determined, the less precisely the other is
determined.
• Should really be called “Heisenberg
Indeterminacy Principle.”
• This is weird if you think about particles. But it’s
very clear if you think about waves.
11
Heisenberg Uncertainty Principle
A slightly different scenario:
Plane-wave propagating in x-direction.
Δy: very large  Δpy: very small
y
Tight restriction in y:
Small Δy  large Δpy
 wave spreads out
strongly in y direction!
Δx
small Δp – only one wavelength
x
Δx
medium Δp – wave packet made of several waves
ΔyΔpy ≥ /2
Δx
large Δp – wave packet made of lots of waves
Review ideas from matter waves:
Electron and other matter particles have wave properties.
See electron interference
If not looking, then electrons are waves … like wave of fluffy
cloud.
As soon as we look for an electron, they are like hard balls.
Each electron goes through both slits … even though it has mass.
(SEEMS TOTALLY WEIRD! Because different than our
experience. Size scale of things we perceive)
If all you know is fish, how do you describe a moose?
Electrons & other particles described by wave functions (Ψ)
Not deterministic but probabilistic
2
Physical meaning is in |Ψ| = Ψ*Ψ
|Ψ|2 tells us about the probability of finding electron in
various places. |Ψ|2 is always real, |Ψ|2 is what we measure
Weak restriction in y:
somewhat large Δy
 somewhat small Δpy
 wave spreads out
weakly in y direction!
Up next:
The Schrödinger Equation
KE
Mass of
particle
PE
Potential
space and time
coordinates
Etot
Complex i,
with i2 = -1
Review: classical wave equations
Electromagnetic waves:
Vibrations on a string:
y
E
x
x
v = speed of wave
c = speed of light
Solutions: y(x,t)
Solutions: E(x,t)
Magnitude is non-spatial:
= Strength of Electric field
Magnitude is spatial:
= Vertical displacement of String
Wave Equation
How to solve?
What does
mean?
a) Take the second derivative of E w.r.t. x only
b) Take the second derivative of E w.r.t. x,y,z only
c) Take the second derivative of E w.r.t both x and t
d) Take the second derivative of E w.r.t x,y,z and t
e) I don’t have a clue….
‘w.r.t’ = “with respect to”
Reminder of this class:
only 2 Diff Eq’s:
1) Guess functional form for solution
and
(k & α ~ constants)
In DiffEq class, learn lots of algorithms for solving DiffEq’s.
In Physics, only ~8
X differential equations you ever need to solve,
solutions are known, just guess them and plug them in.
How to solve a differential equation in physics:
1) Guess functional form for solution
2) Make sure functional form satisfies Diff EQ
(find any constraints on constants)
1 derivative: need 1 soln  f(x,t)=f1
2 derivatives: need 2 soln  f(x,t) = f1 + f2
3) Apply all boundary conditions
(find any constraints on constants)
(You did this in HW6)
Which of the following functional forms works as a possible
solution to this differential equation?
A.
B.
C.
D.
E.
y(x, t) = Ax2t2,
y(x, t) = Asin(Bx)
y(x,t) = Acos(Bx)sin(Ct)
Both, B&C work!
None or some other combo
Test your idea. Does it satisfy Diff EQ?
Answer is C. Only: y(x,t)=Acos(Bx)sin(Ct)
y(x,t)=Asin(kx)cos(ωt) + Bcos(kx)sin(ωt)
y(x,t)=Csin(kx-ωt) + Dsin(kx+ωt)
y
t=0
1) Guess functional form for solution
x
New guess: y(x,t) = Acos(Bx)sin(Ct)
y(x, t) = Asin(Bx)
LHS:
LHS:
RHS:
RHS:
Not OK! x is a
variable. There are
many values of x for
which this is not true!
OK! B and C are constants.
Constrain them so satisfy this.
What is the wavelength of this wave? Ask yourself …
How much does x need to increase to increase kx-ωt by 2π&
sin(k(x+λ)-ωt) = sin(kx + 2π – ωt)
k(x+λ)=kx+2π
k=wave number (radians-m-1)
kλ=2π ◊ k=2π&
&
&
λ&
What is the period of this wave? Ask yourself … How much
does t need to increase to increase kx-ωt by 2π
Speed:
sin(kx-ω(t+Τ)) = sin(kx – ωt + 2π )
ωΤ=2π ◊ ω=2π/Τ &
ω= angular frequency
ω = 2πf
(For your notes: Don’t go over this during class)
ν = speed of wave
What functional form works? Two examples:
y(x,t)=Asin(kx)cos(ωt) + Bcos(kx)sin(ωt)
y(x,t)=Csin(kx-ωt) + Dsin(kx+ωt)
k, ω, A, B, C, D are constants
Satisfies wave eqn if:
(For your notes: Don’t go over this during class)
0
L
Functional form of solution:
y(x,t) = Asin(kx)cos(ωt) + Bcos(kx)sin(ωt)
Boundary conditions?
y(x,t) = 0
at x=0
At x=0: y(x=0,t) = Bsin(ωt) = 0  only works if B=0
y(x,t) = Asin(kx)cos(ωt)
Is that it? Does this eqn. describe the oscillation
of a guitar string? What is k?
With Wave on Violin String:
y(x,t) = Asin(kx)cos(ωt)
Find: Only certain values of k (and thus λ, ω) allowed
 because of boundary conditions for solution
L
0
λ=2π&
&
k&
λ=2L&
& n &
Is there another boundary condition?
y(x,t) = 0 at x=L
n=1
At x=L:
y= Asin(kL)cos(ωt)= 0
n=2
 sin(kL)=0
 kL = nπ (n=1,2,3, … )
 k=nπ/L
n=3
Exactly same for Electrons in atoms:
Find: Quantization of electrons energies (wavelengths) …
 from boundary conditions for solutions
to Schrodinger’s Equation.
y(x,t) = Asin(nπx/L)cos(ωt)
Quantization of k … quantization of λ and ω&
With Wave on Violin String:
Find: Only certain values of k, λ, ω  I.e., the
frequencies of the string are quantized.
Same as for electromagnetic wave in microwave oven:
Exactly same for Electrons in atoms:
Find: Quantization of electrons energies (wavelengths) …
 from boundary conditions for solutions
to Schrodinger’s Equation. (next lecture)
Wow! Wait a minute! We just said that k, λ, and ω
are quantized but yet we can tune a violin by
changing the string tension. What gives?
A) The model we talked about is inconsistent with
the ability to “tune” a violin. But violins are
classical objects anyway, so no problem!
B) You can discuss the violin in terms of quantized
standing waves yet still tune a violin.
 k=nπ/L, and λ= 2L&
n &
Can tune!!
But:
 f=v/λ