Classical and Quantum Waves Flashcards
(138 cards)
1
Q
longitudinal
A
if the oscillation of the medium’s particles is in the same direction as the propagation of the wave.
2
Q
transverse
A
if the oscillation of the medium’s particles is perpendicular to that of the wave propagation
3
Q
periodic waves
A
those which can be closely modeled by a simple harmonic oscillator
4
Q
properties to characterise waves by
A
wave speed
wavelength
period
frequency
angular frequency
wave number
amplitude
5
Q
for wave travelling along a straight line alligned along x-axis
if take photograph of wave at time t=0, transverse motion of any point is described by
A
y(x)=Asin(2pix/lambda)
6
Q
spatial frequency
A
term 2pi/lambda of the wave, commonly known as the wave number
SI unit radm^-1
7
Q
generalising transverse motion of wave equation to cover all potential times
A
y=Asin[2pi/lambda(x-vt)]
8
Q
most common form of wave equation
A
y=Asin(kx-wt)
9
Q
argument of the sin function in wave equation
A
phase of the wave
10
Q
wave speed
A
v=flambda = w/k = dx/dt
11
Q
definition of simple harmonic oscillator
A
transverse acceleration is proportional to displacement
12
Q
one-dimensional wave equation
A
d2y/dx2 = 1/v^2 d2y/dt2
13
Q
speed of a wave is determined by
A
the tension in the string and the mass per unit length (aka linear mass density)
14
Q
increasing the tension
A
increases the restoring forces that tend to straighten the string when it is disturbed, increasing the wave speed
15
Q
if you increase the mass per unit length
A
the motion becomes more sluggish and so speed drops
16
Q
derivation of specific wave disturbance - proving vy is constant.
A
mass string negligible
at t=0 constant Fy at LHS
wave constant speed, point p moves with p
all point left of p move with vy
impulse =Fyt
no initial momentum so Fyt=mvy
since p moving, total moving mass, m prop to t
change in momentum must be associated with increasing mass so vy is constant.
17
Q
derivation of a specific wave disturbance - proving v=sqrt(F/mu)
A
at t lhs moved up vyt, P moved horizontal distance vt
net tension lhs is sqrt(F^2+Fy^2) >F
2 similar triangles: Fy/F=vyt/vt
transverse impulse: Fyt=vy/v Ft
moving mass m=muvt
transverse momentum = mvy = muvtvy
equate with transverse impulse and rearrange
18
Q
derivation: generalised approach
A
mass of segment = mu delta x
horizontal forces equal and opposite
slope at end = F1y/F at other end = F2y/F
F1y/F = -dy/dx, F2y/F=dy/dx
Fy=F1y+F2y
apply newton 2
let delta x go to 0
19
Q
string to the left of point a exerts
A
a force on the string to the right of it
and vice versa
20
Q
Fy(x,t) must be negative when
A
the slope is positive
21
Q
when point a moves in the y directions, Fy(x,t)…
A
does work on this point and therefore transfers energy into the part of the string to the right of a
22
Q
power=
A
Fv
=Fy(x,t)vy(x,t)
23
Q
power in the string is the
A
instantaneous rate at which energy is transferred along the strong at position x and time t
24
Q
how to get to Pmax
A
from standard transverse wave expression
sub into eqn
max is whatever is infront of sin/cos terms
25
the average of a cos^2 term is
half of the max
26
reflection at a fixed end would result in
y=-Asin(kx+wt)
direction and displacement of the waves are reversed
27
if reflection is at end free to move in transverse direction, what wave function do we get
y=Asin(kx+wt)
28
if waves travelling in two directions along a string, when they meet the net result is
the mathematical sum of the interacting waves
29
when sinusoidal waves are reflected and interact
they will create standing waves
30
two waves of equal amplitude, frequency, wavelength and speed but travelling in opposite direction along string aligned along the x-axis
net effect is a wave which does not travel in the x-direction
disturbance remains in place with amplitude varying with time
31
anti-nodes
points where standing waves meet perfectly in phase
where you get maximum transverse displacement
32
nodes
points where standing waves meet perfectly out of phase
points of zero transverse motion
33
to derive an expression for the standing wave we
add the expressions for the individual travelling waves
34
standing waves do not transfer energy although
energy does oscillate back and forth between adjacent nodes and anti-nodes
35
locations of nodes
ysw=0 for all t
when sin(kx)=0 so kx=npi
x=npi/k = nlambda/2
36
plucking a string produces a wave which
reflects and re-reflects off the fixed ends, setting up a standing wave
37
standing wave then produces a sound wave in air with frequency determined by
the properties of the string
38
mathematical description for wave fixed at both ends
y=0 for x=0 and y=0 for x=L
39
adjacent nodes are separated by a distance of
lambda/2
40
length of string (fixed at both ends) must always be
integer multiple of half wavelengths
L=nlambda/2
f=v/lambdan = nv/2L
41
lowest possible frequency when n=1 corresponds to
the largest possible wavelength
f1=v/2L so lambda=2L
42
f1 is known as the
fundamental frequency
43
the frequencies fn are known as
harmonics
series of such frequencies is known as a harmonic series
44
difference in harmonics or overtones
f2 is second harmonic, first overtone
f3 is third harmonic, second overtone
etc
45
fundamental frequency is sometimes referred to as
the first harmonic
46
a normal mode of an oscillating system is a
motion in which all particles of the system move sinusoidally with the same frequency
47
for a system of string length L, fixed at both ends, each of the frequencies given by fn=nf1 corresponds to
a possible normal-mode pattern
48
if we displace a string so that its shape was the same as one of the normal mode patterns and then release it
the string would vibrate with the frequency of that mode
would displace surrounding air with same frequency producing travelling sinusoidal wave
ears would hear a pure tone
49
actual generated sound is
a superposition of travelling sinusoidal waves which is heard as a rich, complex tone with fundamental frequency f1
50
harmonic analysis
can represent every possible motion of the string as some superposition of normal modes motions
finding this representation for a given vibration pattern is the harmonic analysis
51
fourier series
sum of sinusoidal functions that represents a complex wave
52
in case of two waves of equal amplitude interacting, can rewrite standing wave function as
replacing 2A with An which is amplitusde of the nth standing wave
53
fourier's therorem states
that any periodic function with period 2pi/w can be described by the addition of normal modes with varying amplitudes
54
fourier series at t=0
cos term disappears
this is spatial profile of string
can be analysed by an infinite series of sine functions
55
fourier series at x=0
sin term disapperas
infinite series of cosine functions
56
resultant wave of two waves moving along x-axis with same amplitude but different frequencies and wave numbers (hence different speeds)
y1+y2 = 2Asin[ ]cos[ ] using sinA+sinB=2sin (A+B/2)cos(A-B/2)
still get standing wave but with complex shape
57
is waves differ only slightly in w and k
w=w1+w2
w1-w2=dw
same for k
58
combined wave for slightly different w and k is the product of
the same form as before and the modifying envelope
59
in combination of waves have two velocities to consider. These are
phase velocity and group velocity
60
phase velocity
inner speed
bigger w so lots more up and downs
61
group velocity
outer bit - ie velocity of the envelope
much longer period
62
velocity of the envelope
dw/dk = vgroup
63
derivation of the relationship between group and phase velocity
combine equations for group and phase velocity
use product rule
sub in for k=2pi/lambda
find dlambda/dk
d/dk=d/dlambda . dlambda/dk
64
non-dispersive medium
one where dvphase/dlambda=0
ie vgroup=vphase
65
a dispersive medium is one where
dvphase/dlambda does not =0
66
vgroup > vphase if
dvphase/dlambda <0
67
vgroup
dvphase/dlambda>0
68
de broglie wavelength
h/p
69
relativistic de broglie wavelength
h/gamma mv
70
for non-relativistic speeds work done=
eVba = K = 1/2mv^2 = p^2/2m
so p=sqrt(2m eVba)
71
photomultiplier
device used to detect individual photons
72
distribution of photons as detector is moved up and down
peak in middle, then little bumps
73
assumption made in diffraction experiment, passing through a slit
wavelength << width of slit
so vast majority go to central maximum
74
75
py=
theta1 px = lambda/a px
because py/px=tan theta1
76
85% of photons arrive at the detector at angles
between lambda/a and -lambda/a
77
heisenberg uncertainty principle
if reduce uncertainty in momentum, increase the uncertainty in position
>hbar/2
78
Ey
rewriting E but subbing in
px=hbar k
E=hbar w
79
photon is more likely to be found in the regions where
amplitude is greatest
therefore created a localised photon
80
deriving vphase=E/P
have vphase=w/k
E=h bar w so e=E/h bar and dw=dE/hbar
px=hbar k so k=px/h bar so dk=dp/h bar
vphase=w/k = E/k / p/hbar = E/p
81
deriving Vgroup = dE/dP
E=h bar w so e=E/h bar and dw=dE/hbar
px=hbar k so k=px/h bar so dk=dp/h bar
have Vgroup=dw/dk = dE/hbar / dp/hbar = dE/dP
82
non-relativistic motion relationship between Vphase and Vgroup
Vgroup = 2 Vphase
83
relativistic relationship between Vgroup and Vphase
VphaseVgroup=c^2
84
if we had a function, we could test if it was valid by
substituting in into the wave equation and seeing what that yielded
85
waves on a string - k propotional to
w
**this is not true for particles**
86
free particle (no external force) moving along x-axis. If no for acting on particle then it's potential energy
must be constant since
F=-dU/dx
87
when U=0 the total energy of our free particle
purely kinetic
E=1/2mv^2=p^2/2m
88
can rewrite
E=1/2mv^2=p^2/2m as...
hbarw=hbar^2k^2/2m
*convenient to not cancel the h bars*
89
what does hbarw=hbar^2k^2/2m show
for particles w prop to k^"
different to k that was found for waves on a string
90
best way to think about complex wave function to describe particle
the distribution of a particle in space
91
the physics of interference and diffraction patterns shows that the intensity of the radiation at any point in a pattern is proportional to
the square of the electric field magnitude
92
when looked at photon interpretation of interference/diffraction, intensity prop to
number of photons striking that point
ie probability of any individual photon striking the point
93
square of electric field magnitude at each point is prop to
probability of finding a photon around that point
94
the square of the wavefunction of a particle at each points tells us
the probability of finding the particle around that point
95
since we are dealing with complex numbers need to strictly consider |Ψ|^2=
Ψ* Ψ
ie multiplied by complex conjugate
96
assuming moving only in x-direction then |Ψ(x,t)|^2 dx is
the probability that the particle will be found at time t in a position somewherebetween x and x+dx
97
If |Ψ|^2 is large then
more probable that a particle will be found there
98
Ψ needs to be
normalised
ie integral from - to + infinity of |Ψ(x,t)|^2dx=1
ie particle must exist somewhere
99
|Ψ(x,t)|^2 is
the probability distribution function or probability density
100
|Ψ(x,t)|^2 dx is
the probability
101
the probability distribution function does not depend on
position
ie equally likely to find particle anywhere along x-axis
102
if want to create a truly localiased function need to
continue to superimpose more and more functions with wave numbers and amplitudes chosen such that they reinforce alternate maxima whilst cancelling others
103
when superimpsoe more and more functions we create something that looks like
a particle and a wave
particle in a sense that it is localised in space
wave as it still has periodic properties
104
wave packet
a localised pulse
105
wave packet mathematically
Ψ(x,t)=ntegral of A(k)e^i(kx-wt)dk
106
wave packet represents the
superposition of a very large number of waves each with different k and w and with A that is dependent on k
107
expanding schrodinger to account for U
simply add U(x)Ψ(x,t) term in
108
schrodinger equation
kinetic energy+potential energy=total energy
just energy conservation
109
stationary state
a state of definite energy
*probabiltiy distribution does not move*
110
time independent schrodinger equation
plugging Ψ(x)e^-iEt/hbar into schrodinger eqn
111
potential in a box
potential of a rigid wall is infinite
between the walls the potential is zero.
112
wave functions for particle in a box - boundary conditions
particle can only exist between 0 and L so outside range Ψ(x)=0
Ψ(x) must be continuous
(identical to boundary conditions for normal modes of a vibrating string)
also need dΨ/dx to be continuous except at points where U becomes infinite
113
inside the box, energy is
purely kinetic since u(x) is defined as zero in this region
114
each energy level has its own
corresponding wave function Ψn
115
a wave function can be normalised if it contains
a constant that allows the total probability to equal 1
116
can find normalisation constant C by
integral between 0 and L of probabiltiy distribution function
use trig identity sin^2theta=1/2[1-costheta]
117
potential well
situation where U(x)=0 inside region but finite outside region
118
simplest model of potential well
outside region takes one specific value U0, known as a square well potential
119
in mechanics, particle is trapped in a well if
total mechanical energy is less than U0
120
in quantum mechanics a trapped state in known as
bound state
121
for a finite situation there exists the possibility that
E>U0
122
inside square well (U=0) schrodinger equation is
same as particle in box
123
outside square well schrodinger equation is
solutions are exponential in form
C and D constants with different values depending on whether we are in regions where x<0 or x>L
124
Ψ cannot be allowed to approach infinity as x tends to +/- infinity because
if it did, couldn't normalise
so D=0 for x<0 and C=0 for x>L
125
inside well the wave functions are
sinusoidal
126
outside well the wave functions are
exponential
127
matching sinusoids and exponentials is only possible for
certain specific values of energy
128
wavelength of sinusoidal part of each wave function is longer than
it would be if we were dealing with an infinite well
increasing wavelength decreases momentum which reduces energy
so energy levels are lower for finite well
129
finite well - only finite number of possible bound states and therefore
energy levels
number of levels depends on magnitude of U0 compared to ground level energy
130
potential finite so entirely possible for
total energy to exceed potential and particle becomes free
131
potential well
opposite of a potential well
potential energy function that has a maximum rather than a minimum
132
square potential barrier
potential energy zero everywhere except between 0 and L where it has U0
133
square potential well boundary conditions
two solutions need to join smoothly at boundaries so Ψ(x) and dΨ(x)/dx will have to be continuous
134
square potential barrier boundary conditions results in
function not zero inside barrier where classical machanics would forbid particle
and possibility of particle being on other side
135
how great the probability of particle existing on other side of barrier depends on
the width of the barrier and how the particle's energy compares to height of barrier
136
tunnelling probability is proportional to the
square of the ratio of the amplitudes of the sinusoidal wave functions on either sides of the barrier
137
tunnelling amplitudes are determined by
matching wave functions and their derivatives at the boundary points.
138
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