PH0008
Quantum Mechanics and Special Relativity
Lecture ?? (Quantum Mechanics)
020516
TEST FILE
Prof Rick Gaitskell
Department of Physics
Brown University
Main source at Brown Course Publisher
background material may also be available at http://gaitskell.brown.edu
Gaitskell
Recommended Reading
PH0008 Gaitskell Class Spring2002
Rick Gaitskell
Reading - Complete Summary
• Please note that Ch 13 is NOT on the list now
• I have also indicated areas of background interest only
Background reading only - not examined
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 waves
Empirically
o Experimental evidence indicated matter has wave like properties
determined
i(wtkx )
(x,t)  e
o Why
Wave  Particle characteristics
i
 (Et px)
e
is the complex amplitude necessary?
E  w (Einstein - Planck relation)
and also
p  k (de Broglie' s generalisation)
h
2 He took relationship
  or with k 
p
 from photons, and
• In order to extract the kinetic energy (p2/2m) and total energy (E) in the
generalised to massive
particles
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 w) drops out
if (x,t)  Acos(( px  Et) ) there is no simple operator such that
whereas if (x,t)  e
PH0008 Gaitskell Class Spring2002
i
 (Et px)
, if E op  i
E op   E

then E op   E
t
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
i
 (Et px)
we assume that a free particle has the form (x,t)  ei(wtkx )  e
then the differential operators naturally provide expressions for the Kinetic,
Potential and Total Energy
o If

Sch. Eq. also has the desirable property
of being linear, meaning that if 1 and
2 are separately solutions of the Sch. Eq. then a1 + b2 is also a solution
o The
we consider the wave function  to be a probability “amplitude”. ||2 is then
interpreted directly as the probability of the particle being at (x,t). “Copenhagen
Interpretation”
o If
• 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
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
x p 
&
t E 
or
x k  1 & t w  1
Note : Dimensionally these expressions are correct
e.g. k has units of inverse length,
PH0008 Gaitskell Class Spring2002
w inverse time
Rick Gaitskell
A few constants you should be comfortable using…
• You will be given constants, but make sure you know how to use them…
 1.06 10
34
Note is in units of angular momentum
Js  0.658 eV fs
1 eV  1.6 1019J
c  3.00 10 8 ms -1  300 nm fs -1
1 fs  1015s ("femtosecond" )
c  197 eV nm
1 nm  109m (" nanometer" )
EXAMPLES
For Photons w  ck 
E w c
2


1240


eV nm
Violet   400 nm E  3.10 eV   7.49 1014 Hz
Red
For massive particles
p 2 ( k) 2 ( c) 2 2 
KE 



2 

2m
2m
2mc  
2 c

1
(2 mc 2 KE) 2
2
PH0008
For Photons 2  w  ck 
Consistent use of
and 2  reduces
accidental confusion of
h and .
Note use of mc 2 so that mass can be

entered
directly
as an energy equivalent
Gaitskell
Class
Spring2002
  700 nm E  1.77 eV   4.28 1014 Hz
For massive particles (e.g. Electron)
2 c
KE  10 keV   
1
(2 m e c 2 KE) 2
2 197 eV nm

1
(2 511 keV 10 keV) 2

KE  10 eV   
1.24 keV nm
 0.012 nm
101 keV
2 197 eV nm
1
(2 511 keV 0.01 keV) 2
1.24 keV nm

 0.387 nm
3.20 keV
Rick Gaitskell
Solving Sch. Eq. in a Infinite Square Potential (2)
n 
n 2 2 2
(x,t)  Ae
sin  x  with E n 
 L 
2mL2
2
2
2 n 
(x,t)  A sin  x 
 L 
i
 En t
• Solutions:-

Re (x,t  0)

(x,t)
2




x=0
PH0008 Gaitskell Class Spring2002
x=L
Rick Gaitskell
Reflection at Step Up or Down - Review
• Wave Incident on step up
Region I
Region II
E

V0  E

 II (x)  Be ik x
 I (x)  e ik x  Aeik x

1
2
1
x 0

k1  k 2
2k1
 A
, B
k1  k 2 
k1  k 2
The Reflection
Coefficient is given by
k  k 
 1 2 2
k1  k2 
2
R A
2
The Transmission Coefficient must be
k1  k2   k1  k2 
2

PH0008 Gaitskell Class Spring2002
2
k1  k2 
2

T  1 R
You need to know
why this naive
guess is wrong
(see L13 currents)
4k1k2
2

B
2
k

k
 1 2
Rick Gaitskell
Superposition Demonstration - Review
n x  iw n t
n (x,t)  A sin 
e
 L 
Consider some arbitrary combination
(x,t)  21(x,t) 2 (x,t)

 x  iw1 t
2 x  iw 2 t 
 A 2sin  e
 sin 
e

 L 
 L 



PH0008 Gaitskell Class Spring2002
Rick Gaitskell
Resolving Crisis: The beginning…
• Planck 1900
o
o
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.
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 1905
o
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 strong
light source so that we can see which slit the electrons go through…
P12
P1
Electron
Gun

• Electrons are charged and so scatter light
• Every time we detect a “click” on the far rightwall

P2
P12  P1 P2
o We
will also see a flash of light from near the slits
o If we tabulate the results we see P1 and P2 distns as for the case of single slit
• What about the combined probability distn? 
PH0008 Gaitskell Class Spring2002
Rick Gaitskell