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Physics 2 > Classical and Quantum Waves > Flashcards

Classical and Quantum Waves Flashcards

1
Q

longitudinal

A

if the oscillation of the medium’s particles is in the same direction as the propagation of the wave.

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2
Q

transverse

A

if the oscillation of the medium’s particles is perpendicular to that of the wave propagation

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3
Q

periodic waves

A

those which can be closely modeled by a simple harmonic oscillator

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4
Q

properties to characterise waves by

A

wave speed
wavelength
period
frequency
angular frequency
wave number
amplitude

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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)

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6
Q

spatial frequency

A

term 2pi/lambda of the wave, commonly known as the wave number

SI unit radm^-1

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7
Q

generalising transverse motion of wave equation to cover all potential times

A

y=Asin[2pi/lambda(x-vt)]

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8
Q

most common form of wave equation

A

y=Asin(kx-wt)

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9
Q

argument of the sin function in wave equation

A

phase of the wave

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10
Q

wave speed

A

v=flambda = w/k = dx/dt

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11
Q

definition of simple harmonic oscillator

A

transverse acceleration is proportional to displacement

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12
Q

one-dimensional wave equation

A

d2y/dx2 = 1/v^2 d2y/dt2

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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)

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14
Q

increasing the tension

A

increases the restoring forces that tend to straighten the string when it is disturbed, increasing the wave speed

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15
Q

if you increase the mass per unit length

A

the motion becomes more sluggish and so speed drops

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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.

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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

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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

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19
Q

string to the left of point a exerts

A

a force on the string to the right of it

and vice versa

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20
Q

Fy(x,t) must be negative when

A

the slope is positive

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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

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22
Q

power=

A

Fv

=Fy(x,t)vy(x,t)

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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

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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

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25
Q

the average of a cos^2 term is

A

half of the max

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26
Q

reflection at a fixed end would result in

A

y=-Asin(kx+wt)

direction and displacement of the waves are reversed

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27
Q

if reflection is at end free to move in transverse direction, what wave function do we get

A

y=Asin(kx+wt)

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28
Q

if waves travelling in two directions along a string, when they meet the net result is

A

the mathematical sum of the interacting waves

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29
Q

when sinusoidal waves are reflected and interact

A

they will create standing waves

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30
Q

two waves of equal amplitude, frequency, wavelength and speed but travelling in opposite direction along string aligned along the x-axis

A

net effect is a wave which does not travel in the x-direction

disturbance remains in place with amplitude varying with time

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31
Q

anti-nodes

A

points where standing waves meet perfectly in phase

where you get maximum transverse displacement

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32
Q

nodes

A

points where standing waves meet perfectly out of phase

points of zero transverse motion

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33
Q

to derive an expression for the standing wave we

A

add the expressions for the individual travelling waves

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34
Q

standing waves do not transfer energy although

A

energy does oscillate back and forth between adjacent nodes and anti-nodes

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35
Q

locations of nodes

A

ysw=0 for all t
when sin(kx)=0 so kx=npi

x=npi/k = nlambda/2

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36
Q

plucking a string produces a wave which

A

reflects and re-reflects off the fixed ends, setting up a standing wave

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37
Q

standing wave then produces a sound wave in air with frequency determined by

A

the properties of the string

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38
Q

mathematical description for wave fixed at both ends

A

y=0 for x=0 and y=0 for x=L

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39
Q

adjacent nodes are separated by a distance of

A

lambda/2

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40
Q

length of string (fixed at both ends) must always be

A

integer multiple of half wavelengths

L=nlambda/2

f=v/lambdan = nv/2L

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41
Q

lowest possible frequency when n=1 corresponds to

A

the largest possible wavelength

f1=v/2L so lambda=2L

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42
Q

f1 is known as the

A

fundamental frequency

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43
Q

the frequencies fn are known as

A

harmonics

series of such frequencies is known as a harmonic series

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44
Q

difference in harmonics or overtones

A

f2 is second harmonic, first overtone

f3 is third harmonic, second overtone

etc

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45
Q

fundamental frequency is sometimes referred to as

A

the first harmonic

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46
Q

a normal mode of an oscillating system is a

A

motion in which all particles of the system move sinusoidally with the same frequency

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47
Q

for a system of string length L, fixed at both ends, each of the frequencies given by fn=nf1 corresponds to

A

a possible normal-mode pattern

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48
Q

if we displace a string so that its shape was the same as one of the normal mode patterns and then release it

A

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

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49
Q

actual generated sound is

A

a superposition of travelling sinusoidal waves which is heard as a rich, complex tone with fundamental frequency f1

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50
Q

harmonic analysis

A

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

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51
Q

fourier series

A

sum of sinusoidal functions that represents a complex wave

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52
Q

in case of two waves of equal amplitude interacting, can rewrite standing wave function as

A

replacing 2A with An which is amplitusde of the nth standing wave

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53
Q

fourier’s therorem states

A

that any periodic function with period 2pi/w can be described by the addition of normal modes with varying amplitudes

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54
Q

fourier series at t=0

A

cos term disappears

this is spatial profile of string

can be analysed by an infinite series of sine functions

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55
Q

fourier series at x=0

A

sin term disapperas

infinite series of cosine functions

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56
Q

resultant wave of two waves moving along x-axis with same amplitude but different frequencies and wave numbers (hence different speeds)

A

y1+y2 = 2Asin[ ]cos[ ] using sinA+sinB=2sin (A+B/2)cos(A-B/2)

still get standing wave but with complex shape

57
Q

is waves differ only slightly in w and k

A

w=w1+w2
w1-w2=dw

same for k

58
Q

combined wave for slightly different w and k is the product of

A

the same form as before and the modifying envelope

59
Q

in combination of waves have two velocities to consider. These are

A

phase velocity and group velocity

60
Q

phase velocity

A

inner speed

bigger w so lots more up and downs

61
Q

group velocity

A

outer bit - ie velocity of the envelope
much longer period

62
Q

velocity of the envelope

A

dw/dk = vgroup

63
Q

derivation of the relationship between group and phase velocity

A

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
Q

non-dispersive medium

A

one where dvphase/dlambda=0

ie vgroup=vphase

65
Q

a dispersive medium is one where

A

dvphase/dlambda does not =0

66
Q

vgroup > vphase if

A

dvphase/dlambda <0

67
Q

vgroup<vphase if

A

dvphase/dlambda>0

68
Q

de broglie wavelength

A

h/p

69
Q

relativistic de broglie wavelength

A

h/gamma mv

70
Q

for non-relativistic speeds work done=

A

eVba = K = 1/2mv^2 = p^2/2m

so p=sqrt(2m eVba)

71
Q

photomultiplier

A

device used to detect individual photons

72
Q

distribution of photons as detector is moved up and down

A

peak in middle, then little bumps

73
Q

assumption made in diffraction experiment, passing through a slit

A

wavelength &laquo_space;width of slit

so vast majority go to central maximum

74
Q

if lambda«a then theta1

A

can be considered small so sintheta1 = theta1

theta1=lambda/a

75
Q

py=

A

theta1 px = lambda/a px

because py/px=tan theta1

76
Q

85% of photons arrive at the detector at angles

A

between lambda/a and -lambda/a

77
Q

heisenberg uncertainty principle

A

if reduce uncertainty in momentum, increase the uncertainty in position

> hbar/2

78
Q

Ey

A

rewriting E but subbing in
px=hbar k

E=hbar w

79
Q

photon is more likely to be found in the regions where

A

amplitude is greatest

therefore created a localised photon

80
Q

deriving vphase=E/P

A

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
Q

deriving Vgroup = dE/dP

A

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
Q

non-relativistic motion relationship between Vphase and Vgroup

A

Vgroup = 2 Vphase

83
Q

relativistic relationship between Vgroup and Vphase

A

VphaseVgroup=c^2

84
Q

if we had a function, we could test if it was valid by

A

substituting in into the wave equation and seeing what that yielded

85
Q

waves on a string - k propotional to

A

w

this is not true for particles

86
Q

free particle (no external force) moving along x-axis. If no for acting on particle then it’s potential energy

A

must be constant since

F=-dU/dx

87
Q

when U=0 the total energy of our free particle

A

purely kinetic

E=1/2mv^2=p^2/2m

88
Q

can rewrite
E=1/2mv^2=p^2/2m as…

A

hbarw=hbar^2k^2/2m

convenient to not cancel the h bars

89
Q

what does hbarw=hbar^2k^2/2m show

A

for particles w prop to k^”

different to k that was found for waves on a string

90
Q

best way to think about complex wave function to describe particle

A

the distribution of a particle in space

91
Q

the physics of interference and diffraction patterns shows that the intensity of the radiation at any point in a pattern is proportional to

A

the square of the electric field magnitude

92
Q

when looked at photon interpretation of interference/diffraction, intensity prop to

A

number of photons striking that point

ie probability of any individual photon striking the point

93
Q

square of electric field magnitude at each point is prop to

A

probability of finding a photon around that point

94
Q

the square of the wavefunction of a particle at each points tells us

A

the probability of finding the particle around that point

95
Q

since we are dealing with complex numbers need to strictly consider |Ψ|^2=

A

Ψ* Ψ

ie multiplied by complex conjugate

96
Q

assuming moving only in x-direction then |Ψ(x,t)|^2 dx is

A

the probability that the particle will be found at time t in a position somewherebetween x and x+dx

97
Q

If |Ψ|^2 is large then

A

more probable that a particle will be found there

98
Q

Ψ needs to be

A

normalised

ie integral from - to + infinity of |Ψ(x,t)|^2dx=1

ie particle must exist somewhere

99
Q

|Ψ(x,t)|^2 is

A

the probability distribution function or probability density

100
Q

|Ψ(x,t)|^2 dx is

A

the probability

101
Q

the probability distribution function does not depend on

A

position

ie equally likely to find particle anywhere along x-axis

102
Q

if want to create a truly localiased function need to

A

continue to superimpose more and more functions with wave numbers and amplitudes chosen such that they reinforce alternate maxima whilst cancelling others

103
Q

when superimpsoe more and more functions we create something that looks like

A

a particle and a wave

particle in a sense that it is localised in space

wave as it still has periodic properties

104
Q

wave packet

A

a localised pulse

105
Q

wave packet mathematically

A

Ψ(x,t)=ntegral of A(k)e^i(kx-wt)dk

106
Q

wave packet represents the

A

superposition of a very large number of waves each with different k and w and with A that is dependent on k

107
Q

expanding schrodinger to account for U

A

simply add U(x)Ψ(x,t) term in

108
Q

schrodinger equation

A

kinetic energy+potential energy=total energy

just energy conservation

109
Q

stationary state

A

a state of definite energy

probabiltiy distribution does not move

110
Q

time independent schrodinger equation

A

plugging Ψ(x)e^-iEt/hbar into schrodinger eqn

111
Q

potential in a box

A

potential of a rigid wall is infinite

between the walls the potential is zero.

112
Q

wave functions for particle in a box - boundary conditions

A

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
Q

inside the box, energy is

A

purely kinetic since u(x) is defined as zero in this region

114
Q

each energy level has its own

A

corresponding wave function Ψn

115
Q

a wave function can be normalised if it contains

A

a constant that allows the total probability to equal 1

116
Q

can find normalisation constant C by

A

integral between 0 and L of probabiltiy distribution function

use trig identity sin^2theta=1/2[1-costheta]

117
Q

potential well

A

situation where U(x)=0 inside region but finite outside region

118
Q

simplest model of potential well

A

outside region takes one specific value U0, known as a square well potential

119
Q

in mechanics, particle is trapped in a well if

A

total mechanical energy is less than U0

120
Q

in quantum mechanics a trapped state in known as

A

bound state

121
Q

for a finite situation there exists the possibility that

A

E>U0

122
Q

inside square well (U=0) schrodinger equation is

A

same as particle in box

123
Q

outside square well schrodinger equation is

A

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
Q

Ψ cannot be allowed to approach infinity as x tends to +/- infinity because

A

if it did, couldn’t normalise

so D=0 for x<0 and C=0 for x>L

125
Q

inside well the wave functions are

A

sinusoidal

126
Q

outside well the wave functions are

A

exponential

127
Q

matching sinusoids and exponentials is only possible for

A

certain specific values of energy

128
Q

wavelength of sinusoidal part of each wave function is longer than

A

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
Q

finite well - only finite number of possible bound states and therefore

A

energy levels

number of levels depends on magnitude of U0 compared to ground level energy

130
Q

potential finite so entirely possible for

A

total energy to exceed potential and particle becomes free

131
Q

potential well

A

opposite of a potential well

potential energy function that has a maximum rather than a minimum

132
Q

square potential barrier

A

potential energy zero everywhere except between 0 and L where it has U0

133
Q

square potential well boundary conditions

A

two solutions need to join smoothly at boundaries so Ψ(x) and dΨ(x)/dx will have to be continuous

134
Q

square potential barrier boundary conditions results in

A

function not zero inside barrier where classical machanics would forbid particle

and possibility of particle being on other side

135
Q

how great the probability of particle existing on other side of barrier depends on

A

the width of the barrier and how the particle’s energy compares to height of barrier

136
Q

tunnelling probability is proportional to the

A

square of the ratio of the amplitudes of the sinusoidal wave functions on either sides of the barrier

137
Q

tunnelling amplitudes are determined by

A

matching wave functions and their derivatives at the boundary points.

138
Q
A

Physics 2 (9 decks)

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