Matter waves:

Probability density = P(x,t) = |Ψ|2 = Ψ*Ψ

P(x,t=0)

Ψ(x,t=0)

xL-L

L-L xProbability of electron being in interval dx = P(x)·dx

Wave function = Ψ(x,t)

dx

More general: Probability of finding electron between x1 and x2 at time t: P(x,t)dx = |Ψ(x,t)|

2dxx1

x2

x1

x2

à This requires ‘normalization’ of ψ(x)

Normalization Probability density = P(x,t) = |Ψ|2 = Ψ*Ψ

P(x,t=0)

L-L x

à “Normalized wave function”

The probability of finding the particle anywhere in space (i.e. between –∞ and +∞)must be 100%!

P(x,t)dx = |Ψ(x,t)|2dx ≡ 1-∞

∞

-∞

∞ , ∀t

(except if the particle decays!)

Example for a probability density

Probability density of grades:What is the probability of getting a score between 38-40?

0 20 50

0.06 What is Probability of getting between 42.5 - 45?

Probability = P(x)*dx = 0.07*2.5 = 0.16

P(x)42.5

45

(~16/100 students)

0

2

4

6

8

10

12

14

16

0 2.5 5 7.5 10 12.5 15 17.

.5 20 22.5 25 27.

5 30 32.5 35 37.

5 40 42.5 45 47.

5 50Score

Freque

ncy

Observed grade distribution

All sort of funky… ØParticles are described by wave functions (Ψ)

(generally Ψ can have both real and imaginary components)Ø|Ψ|2 tells us probability density ØElectron waves ~ waves of probability.

Electron double slit experiment. Display=Magnitude2 of wave function

Large Magnitude |Ψ|2 means: probability of detecting electron here is high

Small Magnitude |Ψ| 2 means:probability of detecting electron here is low

• The “blob” in Quantum Wave Interference is a 2D wave packet.

• In this simulation, intensity (i.e. |Ψ|2 or |E|2) is represented by brightness.

Electron Photon

|Ψ|2 |Ε|2

Review: Ψ, a wave of probability

What is the value of normalization constant ‘c’, such that P(x) integrated over all space =1?

A. LB. sqrt(1/L)C. 1/LD. sqrt(2/L) E. 2/L

Example:Ψ(x,t=0) =

0

c

0 L

Ψ(x,t=0)

x

Normalization of ψ:Integral of |ψ|2 over all space must be = 1(100% likely will find the particle somewhere in space)!

0, for x=0{

Ψ : a wave of probabilityΨ(x,t=0) = 0 for x

-L

α

0 L

Ψ(x,t=0)

a bdx

Say an electron were described by the following wave function:

How do the probabilities of finding the electron near of a and b compare (within an interval dx)?:

A. a > 2bB. a = 2bC. a < 2b

x

Ψ(x,t=0)= αx/L from -L to LΨ(x,t=0)=0 elsewhere

P(x,t=0) = |ψ(x,t=0)|2

0-L

α

0 L

Ψ(x,t=0)=αx/L from -L to L

a bdx

How do the probabilities of finding the electron near (within dx) of a and bA. a>2bB. a=2bC. a 2 * (P(x) at b)Probability density is what we detect!!

Say an electron were described by the following wave function:

Vibrations on a string: Electromagnetic waves:

Wave Equations

2

2

22

2 1ty

vxy

∂∂=

∂∂

v = speed of wave

2

2

22

2 1tE

cxE

∂∂=

∂∂

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)

What are matter waves?EM Waves (light/photons):Amplitude E = electric fieldE2 tells us probability of

finding photon.Maxwell’s Equations:

Solutions are realsin/cosine waves:

Matter Waves (electrons, etc.):Amplitude ψ = “wave function”|ψ|2 tells us probability of

finding particle.Schrödinger Equation:

Solutions are complexsine/cosine waves:

2

2

22

2 1tE

cxE

∂∂=

∂∂

ti

xm ∂∂=

∂∂− ψψ !! 222

2

)cos(),()sin(),(tkxAtxEtkxAtxE

ωω−=−=

))sin()(cos(),( )(

tkxitkxAAetx tkxi

ωωψ ω

−+−== −

Review of sinusoidal waves:

Wave in time: cos(2πt/T) = cos(ωt) = cos(2πf·t)

Wave in space: cos(2πx/λ) = cos(kx)

k is spatial analogue of angular frequency ω.(We use k because it’s easier to write sin(kx) than sin(2πx/λ).)

λ = Wavelength = length of one cycle

k = 2π/λ = wave number = number of radians per meter

x

T = Period = time of one cycle

ω = 2π/T = angular frequency = number of radians per second

t

Plane Waves• 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))

• etc…Different k’s correspond to different energies, since E = ½mv2 = p2/2m = h2/2mλ2 = !2k2/2m

Superposition principle• 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:

• We can make a “wave packet” by combining plane waves of different energies:

)sin()sin(2)sin()sin(),( tkxAtkxAtkxAtxE ωωω −=+−−= ???

ψ (x,t) = ΣnAnexp(i(knx-ωnt))

à

35

Superposition

Plane Waves vs. Wave PacketsPlane 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 will explore it with simulation in the homework.

37

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?

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

Superposition

Three deBroglie waves are shown for particles of equal mass.

I II IIIx

The highest speed and lowest speed are:a. II highest, I & III same and lowestb. I and II same and highest, III is lowestc. all three have same speedd. 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, 2f 2A, 2f A, f

xx

Review question

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.

E = hf is always true but f = c/λ only applies to light, so E = hf ≠ hc/λ for electrons.

c = speed of light!

Today:

• Heisenberg’s uncertainty relation

• Intro to Schrödinger's equation

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.

Heisenberg Uncertainty Principle

Δxsmall Δp – only one wavelength

Δxlarge Δp – wave packet made of lots of waves

Δxmedium Δp – wave packet made of several waves

A slightly different scenario:

x

y

Plane-wave propagating in x-direction.Δy: very large à Δpy: very small

Tight restriction in y:Small Δy à large Δpyà wave spreads out strongly in y direction!

Weak restriction in y:somewhat large Δy àsomewhat small Δpyà wave spreads out weakly in y direction!

ΔyΔpy ≥ !/2

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 > VB) E = VC) E < VD) 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”

Up next:The Schrödinger Equation

   

− h2

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 problems…why 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 rules…by 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 childish…To deal properly with waves, one had to have a wave equation.

à 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 …apply to classical and quantum wave equations

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

Vibrations on a string: Electromagnetic waves:

Review: classical wave equations

2

2

22

2 1ty

vxy

∂∂=

∂∂

v = speed of wave

2

2

22

2 1tE

cxE

∂∂=

∂∂

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 1tE

cxE

∂∂=

∂∂

What does mean?2

2

xE

∂∂

a)Take the second derivative of E w.r.t. x onlyb)Take the second derivative of E w.r.t. x,y,z onlyc)Take the second derivative of E w.r.t both x and td) Take the second derivative of E w.r.t x,y,z and te) I don’t have a clue….

‘w.r.t’ = “with respect to”

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 solution2) Make sure functional form satisfies Diff EQ

(find any constraints on constants) 1 derivative: need 1 soln à f(x,t)=f12 derivatives: need 2 soln à f(x,t) = f1 + f2

3) Apply all boundary conditions (find any constraints on constants)

How to solve?

2

2

22

2 1ty

vxy

∂∂=

∂∂Wave Equation Reminder of this class:

only 2 Diff Eq’s:

andψψ 222

kx

−=∂∂ ψαψ 22

2

=∂∂x

X

(k & α ~ constants)

(You did this in HW6)

2

2

22

2 1ty

vxy

∂∂=

∂∂

Answer is C. Only: y(x,t)=Acos(Bx)sin(Ct)

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

2

2

22

2 1ty

vxy

∂∂=

∂∂

)sin()cos(222

CtBxABxy −=

∂∂

)sin()cos(1 22

2

2

2 CtBxvAC

ty

v−=

∂∂

1) Guess functional form for solution

y(x, t) = Asin(Bx)

LHS:

RHS:

2

22

2

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

vCB

CtBxvACCtBxAB

=

−=−

OK! B and C are constants. Constrain them so satisfy this.

)sin(

)sin(),(

22

2

BxABxy

BxAtxy

−=∂∂

=

01 22

2 =∂∂ty

v

0)sin(0)sin(2

==−

BxBxAB

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 … àHow 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+2πkλ=2πà k=2π

λ

k=wave number (radians-m-1)

x

yt=0

What is the period of this wave? Ask yourself … àHow much does t need to increase to increase kx-ωt by 2πsin(kx-ω(t+Τ)) = sin(kx – ωt + 2π )ωΤ=2π à ω=2π/Τ ω= 2πf

ω= angular frequency

2

2

22

2 1ty

vxy

∂∂=

∂∂

Speed:

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

What functional form works? Two examples:

2

2

22

2 1ty

vxy

∂∂=

∂∂

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(22

tkxvC ωω −−

ν= speed of wave

k, ω, A, B, C, D are constants

λων fk==

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

Boundary conditions?y(x,t) = 0 at x=0

2

2

22

2 1ty

vxy

∂∂=

∂∂

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)

à 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=LIs there another boundary condition?

2

2

22

2 1ty

vxy

∂∂=

∂∂

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 … quantization of λ and ω

λ=2L n

à sin(kL)=0à kL = nπ (n=1,2,3, … )à k=nπ/L

y(x,t) = Asin(kx)cos(ωt)