Gaitskell
PH0008
Quantum 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 functiono 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 requirea 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 desiredbehaviour
—We cannot form an eigen-equation for the Total Energy, which has to be first order derivativew.r.t. time in order that E (or w) drops out
†
Y(x, t) = e-i(wt-kx ) = e-
ih
(Et- px )
†
Wave ´ Particle characteristics E = hw (Einstein - Planck relation)
and alsop = hk (de Broglie's generalisation)
l =hp
or with k =2pl
†
if Y(x, t) = Acos((px - Et) h) there is no simple operator such that EopY = EY
whereas if Y(x, t) == e-
ih
(Et- px ), if Eop = ih ∂
∂t then EopY = EY
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 Y1 andY2 are separately solutions of the Sch. Eq. then aY1 + bY2 is also a solution
o If we consider the wave function Y to be a probability “amplitude”. |Y|2 is theninterpreted directly as the probability of the particle being at (x,t). “CopenhagenInterpretation”
• This interpretation seems very natural and (again) works well in our formalism of quantummechanics - therefore we use it !
• Remember we never know certain outcome, just the probability distribution of outcomes
†
Y(x, t) = e-i(wt-kx ) = e-
ih
(Et- px )
PH0008 Gaitskell Class Spring2002 Rick Gaitskell
Heisenberg Uncertainty Principle
• Heisenberg proposed the Uncertainty Principleo “It is impossible to design an apparatus to determine which hole the electronpasses through, that will not at the same time disturb the electrons enough todestroy the interference pattern”.
• The Uncertainty Principle is a necessary for Quantum Mechanics tostay intact
o Contradictions arise if we are able to measure both the position and the momentumof a particle with arbitrary accuracy
• e.g. See Double Slits discussions
†
Dx Dp ≥ h & Dt DE ≥ h
or Dx Dk ≥ 1 & Dt Dw ≥ 1Note : Dimensionally these expressions are correct e.g. k has units of inverse length, w 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…
†
h =1.06 ¥10-34 Js = 0.658 eV fsc = 3.00 ¥108 ms-1 = 300 nm fs-1
hc =197 eV nm
For Photons w = ck( )
E = hw = hc 2pl
=1240
l eV nm
For massive particles
KE =p2
2m=
(hk)2
2m=
(hc)2
2mc 22pl
Ê
Ë Á
ˆ
¯ ˜
2
fi l =2p hc
(2 mc 2 KE)12
†
EXAMPLESFor Photons 2pn = w = ck( )
Violet l = 400 nm E = 3.10 eV n = 7.49 ¥1014 HzRed l = 700 nm E =1.77 eV n = 4.28 ¥1014 Hz
For massive particles (e.g. Electron)
KE =10 keV fi l =2p hc
(2 mec2 KE)
12
=2p 197 eV nm
(2 511 keV 10 keV)12
=1.24 keV nm
101 keV= 0.012 nm
KE =10 eV fi l =2p 197 eV nm
(2 511 keV 0.01 keV)12
=1.24 keV nm
3.20 keV= 0.387 nm
†
Note h is in units of angular momentum1 eV =1.6 ¥10-19J1 fs =10-15s ("femtosecond")1 nm =10-9m ("nanometer")
†
Note use of mc 2 so that mass can be entered directly as an energy equivalent
†
Consistent use ofh and 2p h reducesaccidental confusion ofh and h.
PH0008 Gaitskell Class Spring2002 Rick Gaitskell
Solving Sch. Eq. in a Infinite Square Potential (2)
• Solutions:-
†
Y(x, t) = ¢ A e-
ih
En tsin np
Lx
Ê
Ë Á
ˆ
¯ ˜ with En =
n2p 2h2
2mL2
Y(x,t) 2= ¢ A ( )2 sin2 np
Lx
Ê
Ë Á
ˆ
¯ ˜
x=0 x=L
†
•
†
•
†
Re Y(x,t = 0)( )
†
Y(x,t) 2
PH0008 Gaitskell Class Spring2002 Rick Gaitskell
†
Region I
†
Region II
†
x = 0
†
E
†
V0 < E
Reflection at Step Up or Down - Review
• Wave Incident on step up
†
yI (x) = eik1x + Ae- ik1x
†
yII (x) = Beik2x
†
fi A =k1 - k2
k1 + k2
, B =2k1
k1 + k2
The Reflection Coefficient is given by
R = A 2=
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 knowwhy this naiveguess is wrong(see L13 -currents)
PH0008 Gaitskell Class Spring2002 Rick Gaitskell
Superposition Demonstration - Review
†
Yn(x,t) = A sin np xL
Ê
Ë Á
ˆ
¯ ˜ e-iwn t
Consider some arbitrary combinationY(x,t) = 2Y1(x,t)+ Y2(x,t)
= A 2sin p xL
Ê
Ë Á
ˆ
¯ ˜ e-iw1 t + sin 2p x
LÊ
Ë Á
ˆ
¯ ˜ e-iw2 tÈ
Î Í
˘
˚ ˙
PH0008 Gaitskell Class Spring2002 Rick Gaitskell
Resolving Crisis: The beginning…• Planck 1900
o Suggest that “if” it is assumed that energy of normal mode is quantised such that E=hn (h isan arbitrary constant, Planck’s arbitrary constant, experimentaly determined so that theoryfits 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 discretebundles (called quanta) such that E=hn . Mathematically this additional effect generates anexpression 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 necessarylink 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
PH0008 Gaitskell Class Spring2002 Rick Gaitskell
Watching the Electrons (6)
• Let’s repeat the previous 2 slit experiment, but we will include a stronglight 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
†
¢ P 1
†
¢ P 2†
¢ P 12
†
¢ P 12 = ¢ P 1 + ¢ P 2
ElectronGun
• What about the combined probability distn?
