Gaitskell

PH0008Quantum Mechanics and Special Relativity

Lecture ?? (Quantum Mechanics)020516

TEST FILE

Prof Rick Gaitskell

Department of PhysicsBrown University

Main source at Brown Course Publisher

background material may also be available at http://gaitskell.brown.edu

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

Recommended ReadingRecommended Reading

PH0008 Gaitskell Class Spring2002 Rick GaitskellBackground reading only - not examined

Reading - Complete Summary

• Please note that Ch 13 is NOT on the list now• I have also indicated areas of background interest only

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

Wave Function - Starting Point…

• Proposition: A propagating particle has an associated wave function o This appears as a “reasonable” guess, given our previous studies of waveso Experimental evidence indicated matter has wave like properties

o Why is the complex amplitude necessary?• In order to extract the kinetic energy (p2/2m) and total energy (E) in the

non-relativistic Schrödinger equation from the wave function we require a second order derivative w.r.t. space, and a first order derivative w.r.t. time

• A expression formed from a linear combination of sin() & cos() does not have the desired behaviour

—We cannot form an eigen-equation for the Total Energy, which has to be first order derivative w.r.t. time in order that E (or ) drops out

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Ψ(x, t) = e−i(ωt−kx ) = e−

i

h(Et− px )

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Wave ↔ Particle characteristics

E = hω (Einstein - Planck relation)

and also

p = hk (de Broglie's generalisation)

λ =h

por with k =

2π

λ

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if Ψ(x, t) = Acos(( px − Et) h) there is no simple operator such that EopΨ = EΨ

whereas if Ψ(x, t) == e−

i

h(Et− px )

, if Eop = ih∂

∂t then EopΨ = EΨ

Empirically determined

He took relationship from photons, and generalised to massiveparticles

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

FAQ - Schrödinger Equation

• Why does the Sch. Eq. have the form it does?o As horrible as it sounds - because it works so well (for non-relativistic particles) when used to predict their behaviour in experiments

o If we assume that a free particle has the formthen the differential operators naturally provide expressions for the Kinetic, Potential and Total Energy

o The Sch. Eq. also has the desirable property of being linear, meaning that if Ψ1 and Ψ2 are separately solutions of the Sch. Eq. then aΨ1 + bΨ2 is also a solution

o If we consider the wave function Ψ to be a probability “amplitude”. |Ψ|2 is then interpreted directly as the probability of the particle being at (x,t). “Copenhagen Interpretation”

• This interpretation seems very natural and (again) works well in our formalism of quantum mechanics - therefore we use it !

• Remember we never know certain outcome, just the probability distribution of outcomes

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Ψ(x, t) = e−i(ωt−kx ) = e−

i

h(Et− px )

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

Heisenberg Uncertainty Principle

• Heisenberg proposed the Uncertainty Principle o “It is impossible to design an apparatus to determine which hole the electron passes through, that will not at the same time disturb the electrons enough to destroy the interference pattern”.

• The Uncertainty Principle is a necessary for Quantum Mechanics to stay intact

o Contradictions arise if we are able to measure both the position and the momentum of a particle with arbitrary accuracy

• e.g. See Double Slits discussions

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Δx Δp ≥ h & Δt ΔE ≥ h

or Δx Δk ≥ 1 & Δt Δω ≥ 1

Note : Dimensionally these expressions are correct

e.g. k has units of inverse length, ω inverse time

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

A few constants you should be comfortable using…

• You will be given constants, but make sure you know how to use them…

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h=1.06 ×10−34 Js = 0.658 eV fs

c = 3.00 ×108 ms-1 = 300 nm fs-1

hc =197 eV nm

For Photons ω = ck( )

E = hω = hc2π

λ=

1240

λ eV nm

For massive particles

KE =p2

2m=

(hk)2

2m=

(hc)2

2mc 2

2π

λ

⎛

⎝ ⎜

⎞

⎠ ⎟2

⇒ λ =2π hc

(2 mc 2 KE)12

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EXAMPLES

For Photons 2πν = ω = ck( )

Violet λ = 400 nm E = 3.10 eV ν = 7.49 ×1014 Hz

Red λ = 700 nm E =1.77 eV ν = 4.28 ×1014 Hz

For massive particles (e.g. Electron)

KE =10 keV ⇒ λ =2π hc

(2 mec2 KE)

12

=2π 197 eV nm

(2 511 keV 10 keV)12

=1.24 keV nm

101 keV= 0.012 nm

KE =10 eV ⇒ λ =2π 197 eV nm

(2 511 keV 0.01 keV)12

=1.24 keV nm

3.20 keV= 0.387 nm

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Note h is in units of angular momentum

1 eV =1.6 ×10−19J

1 fs =10−15s ("femtosecond")

1 nm =10−9m ("nanometer")

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Note use of mc 2 so that mass can be

entered directly as an energy equivalent

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Consistent use of

h and 2π h reduces

accidental confusion of

h and h.

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

Solving Sch. Eq. in a Infinite Square Potential (2)

• Solutions:-

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Ψ(x, t) = ′ A e−

i

hEn t

sinnπ

Lx

⎛

⎝ ⎜

⎞

⎠ ⎟ with En =

n2π 2h2

2mL2

Ψ(x, t)2

= ′ A ( )2sin2 nπ

Lx

⎛

⎝ ⎜

⎞

⎠ ⎟

x=0 x=L

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∞

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∞

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Re Ψ(x, t = 0)( )

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Ψ(x, t)2

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

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Region I

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Region II

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x = 0

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E

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V0 < E

Reflection at Step Up or Down - Review

• Wave Incident on step up

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ψI (x) = e ik1x + Ae−ik1x

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ψII (x) = Be ik2x

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⇒ A =k1 − k2

k1 + k2

, B =2k1

k1 + k2

The Reflection Coefficient is given by

R = A2

=k1 − k2( )

2

k1 + k2( )2

The Transmission Coefficient must be T =1− R

=k1 + k2( )

2− k1 − k2( )

2

k1 + k2( )2 =

4k1k2

k1 + k2( )2 ≠ B

2

You need to know why this naive guess is wrong (see L13 - currents)

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

Superposition Demonstration - Review

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Ψn(x,t) = A sinnπ x

L

⎛

⎝ ⎜

⎞

⎠ ⎟e−iωn t

Consider some arbitrary combination

Ψ(x,t) = 2Ψ1(x,t)+ Ψ2(x,t)

= A 2sinπ x

L

⎛

⎝ ⎜

⎞

⎠ ⎟e−iω1 t + sin

2π x

L

⎛

⎝ ⎜

⎞

⎠ ⎟e−iω2 t

⎡

⎣ ⎢

⎤

⎦ ⎥

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

• Planck 1900o Suggest that “if” it is assumed that energy of normal mode is quantised such that E=h (h is

an arbitrary constant, Planck’s arbitrary constant, experimentaly determined so that theory fits data) then higher frequency (shorter wavelength) modes will be suppressed/eliminated.

o Planck suggests ad hoc that the radiation emitted from the walls must happen in discrete bundles (called quanta) such that E=h . Mathematically this additional effect generates an expression for spectrum that fits data well.

• The Planck constant is determined empirically from then existing data• The short wavelength modes are eliminated

o In a classical theory, the wave amplitude is related to the energy, but there is no necessary link between the frequency and energy

• Classically one can have low freq. waves of high energy and vise versa without constraint• Planck is unable to explain how such an effect could come about in classical physics

• Einstein 1905o Based on Photoelectric effect, Einstein proposed quantisation of light (photons)

• Photons are both emitted and absorbed in quanta

Resolving Crisis: The beginning…

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

Watching the Electrons (6)• Let’s repeat the previous 2 slit experiment, but we will include a strong light source so that we can see which slit the electrons go through…

• Electrons are charged and so scatter light• Every time we detect a “click” on the far right wall

o We will also see a flash of light from near the slitso If we tabulate the results we see P1 and P2 distns as for the case of single slit

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′ P 1

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′ P 2

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′ P 12

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′ P 12 = ′ P 1 + ′ P 2

ElectronGun

• What about the combined probability distn?