Physics 214
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Transcript of Physics 214
![Page 1: Physics 214](https://reader034.fdocuments.net/reader034/viewer/2022042900/56812ea7550346895d944353/html5/thumbnails/1.jpg)
Physics 214
5: Quantum Physics
•Photons and Electromagnetic Waves•The Particle Properties of Waves•The Heisenberg Uncertainty Principle•The Wave Properties of Particles•A Particle in a Box•The Schroedinger Equation•The Simple Harmonic Oscillator
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•Long wavelength electromagnetic radiation acts as “waves”
•consider a radio wave with = 2.5 Hz•E = ~10-8 eV too small be detected as a single photon......
a detectable single would require ~1010 such photons, which on the average act as a continuous wave
•Short wavelength electromagnetic radiation acts as “particles”
•consider a X ray wave with = 1018 Hz•E = ~103 eV, which can easily be detected as a single photon
Photons and Electromagnetic Waves
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How should we think of
light?
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Remember Beats
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•particle•localized•finite size
•wave•delocalized
Photon
A solitary pulse can be produced by mixing together waves of infinitely many different harmonic waves of
different frequencies, such pulses are
called wave-packets. These pulses
exhibit properties of both particles and
waves
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The wavepacket as whole moves with a velocity vG
---the group velocityThe waves in the wavepacket
move with a velocity vp----the phase velocity
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Beats
BEATSytotal x,t y1 x,t y2 x,t
ytotal x,t A1 sin k1x 1t 1 A2 sin k2 x 2t 2 let A1A2 A; 1 20
ytotal 2Acosk
2 k
1 2
x
2
1
2t
sin
k1k
2 2
x
1
2
2t
2Acos kbx
bt
modulated amplitude beat
sin ksx
st
interference wave
b
2
1
2
2; k
b k
2
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Dispersion
v
in a vacuum
in an another medium
vac
med
ckn k
k kc
kckn k
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Different Phase and Group Velocities
v
v
phase
group
kd
k dk
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Intensity
of Particle Streams
and
Probability
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The instantaneous energy density of a light wave u r,t 12
0 E r,t 2
If the frequency of the light is Eh, (E is energy) this must
correspond to u r,t E
photons per unit volume at that point and time
Thus the photon density at r,t is proportional to the square
of the amplitude of the electromagnetic wave amplitude at r,t
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x
If there is only one photon, the wave packet model implies that the photon density at r,t has to be interpreted as the probability that a photon is at
the position r at the time t. This interpretation also workseven if there are many photons
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A wave-packet has an average position, which corresponds to the average position of
the photon
x = x Prob(x) dxThe width of the wave-packet corresponds to the dispersion of
the positions of the photons
x x x 2= x x 2 Prob(x ) dx
but has no exact position that can be measured
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Prob
Noting that
x,t E(x,t) 2
and the normalization condition for probability distributionsProb
x ,t dx 1
we can define
Prob x,t = E(x, t) 2
E(x,t) 2 dx
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The Heisenberg Uncertainty Principle
A wave-packet has width x It is made up with a range of
wavelength waves
max - min =range of wavelengths of harmonic oscillators making up wave-packet proportional to the dispersion of the EXPECTED
observed value of a wavelength in the wave-packet UNCERTAINTY in the observed value of
i.e 2
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x xmax
xmin
range of spatial positions covered by wave-packet proportional to the dispersion of
the EXPECTED observed value of the position of the wave-packet UNCERTAINTY in the observed value of x
x x x2
The difference of the wavelengths, , of the waves contained in the wave-packet cannot be greater than the width of the
wave-packet, otherwise waves with wavelengths larger than that of the wave-packet would be in the wave-packet hence
x
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h p
xhp
x p hHeisenberg's Uncertainty PrincipleA more accurate analysis shows that
x p 2
= h4
m xp
p p
m
2 x
v
E
2 t E
2
E p 2
2 m E pp
m & v x
t
i. e. t E 2
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Complex Number Representation of WavesComplex number definitionii 2
z x iy
z1z
2 x
1 iy
1 x2 iy
2 x1x
2i y
1x
2 x
1y
2 y1y
2
x1x2 y1y2
x3
i y1x2 x1y2
iy3
complex conjugate z x iymagnitude squared z 2 z z x2 y2
-1
-1
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e i cos i sin
1
2 ie i e i sin
1
2e i e i cos
Thus
E k ,t Emax
sin kx t Emax
2 ie i kx t e i kx t
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The wave-packet is a linear combination (integral) of infinitely many waves, thus has a wave-function
E , x , t Emax
sin2
x ( ) t
in general is a complex valued functionand its magnitude squared at the point x at time t
x , t 2 x, t x , t Prob x , t
when K is chosen so that
x , t 2
dx 1
complex ;,,,max
min
cdtxEcKtx
tx,
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For massless particles using The Special Theory of Relativity
E pc p Ec
and Plancks Hypothesis
E h hc
Ec
h
p h
hp
& p hk2
k
de Broglie hypothesized that this was valid for particles with mass also
hp h
m vg & p k
The Wave Properties of Particles
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Eh
vp = Emvg
for a free particle E p 2
2m
=12
mvg2
m vg
vg
2
vp vg
2
comparison of group velocity to phase velocity for a free particle with mass
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de Broglies explanation of the Bohr modelusing matter waves
Electrons are waves, but they are restricted to the one dimensional Bohr orbits
They only way they can exist in such a restricted region of space is as standing wave patterns
In order to fit into orbits without destructive interference one has to have an integer number of standing
wave patterns in one orbit
2r n n hmv
mvr nh
2 n
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This condition is exactlyBohrs Angular Momentum Quantization!!!!
Localization of waves
Standing waves
Quantized energies
Electron diffraction from Nickel crystals
confirmed de Broglies ideas
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The diffraction pattern produced by electrons passing through 2 slits can be viewed as the probability distribution of
the electrons hitting the screen
•If << distance between slits then •Slits act as single slits
•If << slit width then •Electrons act as particles
•If ~ slit width or > slit width electron acts as WAVE !!!!
photons display the same behavior
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A Particle in a Box
U U=0
U
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L 12 1 2 3
2 3 2 4
Ln2 n n
2 Ln
Stationary States for Electron in Box
n x , t A sin kn x cos t A sin nL
x
cos t
as kn 2 n
nL
Boundary Conditions 0, t 0 L , t
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Prob x , t x , t 2 A2 sin 2 n L
x
cos 2 t
pn h
n
h2L
n nh
2L, n 1 , ,
i .e. Momentum is Quantized
Kn p 2
2m
n 2 h 2
4 L2
2m h 2
8 mL2
n 2 E
n
Kinetic energy = Total energy as electric potential is zero inside box
Kinetic & Total energy are QuantizedEn n2 E 1
E1 h2
8 mL2
0Zero Point Energy
If electron drops from energy level b to energy level a the frequency of light emitted is
ba h
8 mL2
b2 a2 Hz
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The Schrödinger Equation
Schrödinger first guessed that the matter wave x, t would satisfy the
linear wave equation2t2 2
k22x2 v2
2x2
just as string waves do, however this did not give the correct non relativistic
energy for a free (traveling) electron
i.e. E p2
2 m= K
nor did it give the correct spectrum for the hydrogen spectrum (bound localized electron)
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The Equation for matter (in particular electrons) that does give the correct energy is
i x, t
t=- 2
2m2 x, t
x2 U( x) x, t
which is called Schrödinger's Time Dependent Wave Equation
U (x) is the P.E. of the particle
For free particles U(x)=0 this equation has solutions of the form
x,t e ikx it ei px Et
E; p= k
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substituting x , t e ikx i t ei px Et
into
i x, t
t= - 2
2m 2 x,t
x 2 gives
iiE
e
i px Et
= - 2
2m
ip
2
ei px Et
Eei px Et
p2
2 me
i px Et
E p 2
2m
the non relativistic K.E. for a free particle
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Notice that x , t e ikx i t = x t
plugging this product into
i x , t
t= - 2
2m
2 x, t x2
+ U(x) x , t gives
x d t
dt - 2
2 m
2 x x2
+ U(x) x
t
1
t d t
dt
1
x - 2
2m
2 x x2
+ U(x) x
This equation can only have a solution if both sides are constant
1
t d t
dt
E
& 1
x - 2
2m
2 x x 2
+ U(x) x
E
i
i
i
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1 t
d t dt
E
d t dt
E t
t Ce iEt ; C is a constant
1 x
-2
2m2 x x2
+ U(x ) x
E
-2
2m2
x2+ U(x )
x E x
Time Independent Schrödinger Equation
i
i
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Solutions of differential equations arenot completely characterized by the equation
alone. The functions must also satisfy some boundary conditions, such as having a specified
value at t 0. For the Schrödinger Equation the boundary conditions are
(1) x, t 2
-
dx =1
(2) x, t is a continuous function in x
(3) d x, t dx is a continuous function in x
(4) x, t 0 where U(x)=
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Particle in a Box -- Again
The potential energy of a particle in a box is
U (x) = for x 00 for x 0 , L for x L
This gives the Time Independent Schroedinger Equation
- h2
2m 2
x2 U ( x )
x E x
Which has solutions when En h2
8mL2
n2 ; n 1, 2 ,
For n 1 one can then solve the equation
- h2
2 m2
x 2 U ( x )
1 x h2
8mL 2
1 x
1 x A sinnx
L
; A is a constant that can be chosen to
normalize 1 x
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The Simple Harmonic Oscillator
The potential energy of a particle that moves in SHM is
U (x) = 12
kx 2 12
m 2x 2
where = km
& k is the force constant
This gives the Time Independent Schroedinger Equation
- 2
2 m2
x 2+ 1
2m2 x 2
x E x
Which has solutions when En n 12
; n 0 ,1, 2 ,
For n 0 one can then solve the equation
- 2
2 m 2
x 2+ 1
2 m 2 x 2
x
2 x
x Be m
x 2
; B is a constant that can be chosen to normalize x
2