Plane Waves vs. Wave Packets

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.

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 well-defined for both.

Plane Wave: Ψ(x,t) = Aei(kx-ωt) :

Wave Packet: Ψ(x,t) = ΣnAnei(knx-ωnt) :

For which type of wave are position x and

momentum p most well-defined?

Q1:

Plane Waves vs. Wave PacketsPlane 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)

– 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.

3

SuperpositionThree deBroglie waves are shown for particles of equal mass.

I II III

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

A, 2f 2A, 2f A, f

xx

Review question

A: Amplitude. f: frequency

Q2:

E=hc/λλλλ�

A. 6is true for both photons and electrons.

B. 6is true for photons but not electrons.

C. 6is true for electrons but not photons.

D. 6is not true for photons or electrons.

Q3:

Today:

• Heisenberg’s

uncertainty relation

• Intro to Schrödinger's

equation

Heisenberg Uncertainty Principle

• In math: ∆x�∆p ≥ h/2 (or better: ∆x�∆px ≥ h/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.

Heisenberg Uncertainty Principle

∆xsmall ∆p – only one wavelength

∆xlarge ∆p – wave packet made of lots of waves

∆xmedium ∆p – wave packet made of several waves

A slightly different scenario:

x

y

Plane-wave propagating in x-direction.

∆y: very large � ∆py: very small

Tight restriction in y:

Small ∆y � large ∆py

� wave spreads out

strongly in y direction!

Weak restriction in y:

somewhat large ∆y

�somewhat small ∆py

� wave spreads out

weakly in y direction!

∆y∆py ≥ h/2

Review ideas from matter waves:Electron and other matter particles have wave properties.

See electron interference

If not looking, then electrons are waves 6 like wave of fluffy cloud.

As soon as we look for an electron, they are like hard balls.

Each electron goes through both slits 6 even though it has mass.

(SEEMS TOTALLY WEIRD! Because different than our experience. Size scale of things we perceive)

Electrons & other particles described by wave functions (Ψ)Not deterministic but probabilistic

Physical meaning is in |Ψ|2 = 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

If all you know is fish, how do you describe a moose?

Reading Quiz

For the “Particle in a Rigid Box” problem:

Which of the following is correct if the particle is in

the ground state inside the box with V=0 inside?

A) E > V

B) E = V

C) E < V

D) E can be more than one of the above!!

E: “total energy of the particle”

V = 0: “Potential energy of the particle inside the box”

Q4:

Up next:

The Schrödinger Equation

−h

2

2m

∂2Ψ(x,t)

∂x 2+ V (x,t)Ψ(x,t) = ih

∂Ψ(x,t)

∂t

The Schrodinger Equation

Once at the end of a colloquium Felix Bloch heard Debye saying

something like: “Schrödinger, you are not working right now on

very important problems6why don’t you tell us some time about

that thesis of deBroglie, which seems to have attracted some

attention?” So, in one of the next colloquia, Schrödinger gave a

beautifully clear account of how deBroglie associated a wave

with a particle, and how he could obtain the quantization

rules6by demanding that an integer number of waves should be

fitted along a stationary orbit. When he had finished, Debye

casually remarked that he thought this way of talking was rather

childish6To deal properly with waves, one had to have a wave

equation.

�Work towards finding an equation that describes/predicts

the probability wave for matter in any situation.

Solving this (differential) equation will give incredible

insight to the inner workings of nature and technology

Look at general aspects of wave equations 6

apply to classical and quantum wave equations

“Stop talking childish!” (as Debye put it)(and get a Nobel Prize in Physics6: Schrödinger, 1933)

Vibrations on a string: Electromagnetic waves:

Review: classical wave equations

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂

v = speed of wave

2

2

22

2 1

t

E

cx

E

∂

∂=

∂

∂

c = speed of light

x

yE

x

Magnitude is non-spatial:

= Strength of Electric field

Magnitude is spatial:

= Vertical displacement of String

Solutions: E(x,t) Solutions: y(x,t)

2

2

22

2 1

t

E

cx

E

∂

∂=

∂

∂

What does mean?2

2

x

E

∂

∂

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 clue6.

‘w.r.t’ = “with respect to”

Q5:

In DiffEq class, learn lots of algorithms for solving DiffEq’s.

In Physics, only ~8 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)=f12 derivatives: need 2 soln � f(x,t) = f1 + f2

3) Apply all boundary conditions

(find any constraints on constants)

How to solve?

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂

Wave Equation Reminder of this class:

only 2 Diff Eq’s:

andψψ 2

2

2

kx

−=∂

∂ψα

ψ 2

2

2

=∂

∂

x

X

(k & α ~ constants)

(You did this in HW6)

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂

Test your idea. Does it satisfy Diff EQ? (Just like HW6 prob 2)

1) Guess functional form for solution

Which of the following functional forms works as a possible solution to this differential equation?

A. y(x, t) = Ax2t2,

B. y(x, t) = Asin(Bx)

C. y(x,t) = Acos(Bx)sin(Ct)

D. Both, B&C work!

E. None or some other combo

Q6:

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂

)sin()cos(2

2

2

CtBxABx

y−=

∂

∂

)sin()cos(1

2

2

2

2

2CtBx

v

AC

t

y

v−=

∂

∂

1) Guess functional form for solution

y(x, t) = Asin(Bx)

LHS:

RHS:

2

22

2

22 )sin()cos()sin()cos(

v

CB

CtBxv

ACCtBxAB

=

−=−

OK! B and C are constants.

Constrain them so satisfy this.

)sin(

)sin(),(

2

2

2

BxABx

y

BxAtxy

−=∂

∂

=

01

2

2

2=

∂

∂

t

y

v

0)sin(

0)sin(2

=

=−

Bx

BxAB

LHS:

RHS:

Not OK! x is a

variable. There are

many values of x for

which this is not true!

New guess: 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)

kTv

ωλ==

What is the wavelength of this wave? Ask yourself 6

�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

kλ=2π � k=2πλλλλ

k=wave number (radians-m-1)

x

yt=0

What is the period of this wave? Ask yourself 6 �How much

does t need to increase to increase kx-ωt by 2π

sin(kx-ω(t+Τ)) = sin(kx – ωt + 2π )

ωΤ=2π � ω=2π/Τ

ω = 2πf

ω= angular frequency

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂

Speed:

(For your notes: Don’t go over this during class)

What functional form works? Two examples:

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂

y(x,t)=Asin(kx)cos(ωt) + Bcos(kx)sin(ωt)

y(x,t)=Csin(kx-ωt) + Dsin(kx+ωt)

2

22

vk

ω=

=−−

−=

)sin(

)sin(),(

2 tkxCk

tkxCtxy

ω

ω

Satisfies wave eqn if:

)sin(2

2

tkxv

Cω

ω−−

ν = speed of wave

k, ω, A, B, C, D are constants

λω

ν fk

==

(For your notes: Don’t go over this during class)

Boundary conditions?

y(x,t) = 0 at x=0

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂

0 L

y(x,t) = Asin(kx)cos(ωt) + Bcos(kx)sin(ωt)Functional form of solution:

At x=0: y(x=0,t) = Bsin(ωt) = 0y(x,t) = Asin(kx)cos(ωt)

� only works if B=0

Is that it? Does this eqn. describe the oscillation

of a guitar string? What is k?

y(x,t) = 0 at x=L

Is there another boundary condition?

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂

0 L

At x=L:

y= Asin(kL)cos(ωt)= 0

y(x,t) = Asin(nπx/L)cos(ωt)

λ=2π

k

n=1

n=2

n=3

Quantization of k 6 quantization of λ and ω

λ=2L

n

� sin(kL)=0� kL = nπ (n=1,2,3, 6 )� k=nπ/L

y(x,t) = Asin(kx)cos(ωt)

With Wave on Violin String:

Find: Only certain values of k (and thus λ, ω) allowed

� because of boundary conditions for solution

Exactly same for Electrons in atoms:

Find: Quantization of electrons energies (wavelengths) 6

� from boundary conditions for solutions

to Schrodinger’s Equation.

With Wave on Violin String:

Find: Only certain values of k, λ, ω� I.e., the

frequencies of the string are quantized.

Exactly same for Electrons in atoms:

Find: Quantization of electrons energies (wavelengths) 6

� from boundary conditions for solutions

to Schrodinger’s Equation.

Same as for electromagnetic wave in microwave oven (HW2):

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.

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?

kTv

ωλ==

2

2

22

2 1

t

y

vx

y

∂

∂=

∂

∂ 2L

n� k=nπ/L, and λ=

But:

� f = v / λ

Can tune!!

Q7:

Answers to clicker questions

Q1: A

Q2: B

Q3: B

Q4: A

Q5: A

Q6: C

Q7: B