Section 2: Wavefunctions & The Uncertainty Principle

Question 1

A typical experiment that shows the wave-particle duality is

Answer 1

the double-slit experiment

Question 2

experiment with bullets - set up

Answer 2

a machine gun shooting bullets, a wall with two slits (each just big enough to let one bullet go through), and a backstop with a detector, where the bullets stop and accumulate.

Question 3

experiment with bullets - only slit #1 open

Answer 3

observe a distribution of bullets

gives probability P1(x)=N1(x)/N of finding a bullet from slit 1 at position x

Question 4

experiment with bullets - only slit #2 open

Answer 4

same kind of result as for phase 1, but now with the bell curve P2=N2(x)/N centered around the coordinate aligned with the second slit

Question 5

experiment with bullets = both slots open

Answer 5

total probability is the sum of the individual probabilities

ie no interference

Question 6

experiment 2 - replace the bullets with waves
if we cover one of the two slits

Answer 6

intensity bell curves

Question 7

waves instead of bullets - if both slits are open

Answer 7

the waves interfere and form a pattern with maxima and minima of intensity (corresponding to positions of constructive and destructive interference)

Question 8

with waves instead of bullets - we cannot consider

Answer 8

a sum of intensities, need to consider the waves amplitudes

I12 = |h1+h2|^2

interference observed

Question 9

experiment 3 - electrons instead of bullets or waves

do electrons split across the two slits

Answer 9

no
no half electrons
each is a whole localised entity with its own mass/charge

Question 10

experiment 3 - does each electron go through either slit 1 or 2

Answer 10

no
p1 does not = p1+p2
there is interference

Question 11

experiment 3 -how do electrons behave at the respective stages

Answer 11

electrons behave like waves while in transit, before they get detected, but behave like particles upon detection.

Question 12

interference pattern in electron experiment

Answer 12

analogous to waves

Question 13

We need to introduce a concept analogous to the classical wave amplitudes and to the classical waves, but for probabilities. So we come up with an analogous concept to describe the wave behaviour of matter:

Answer 13

the wavefunction

plays the role of probability amplitude

Question 14

Probability =

Answer 14

Φ*Φ = |Φ|^2

Question 15

If an event - like the detection of a particle on the screen - can happen following different possible paths, we associate a

Answer 15

wavefunction to each path

Question 16

since wavefunctions are complex numbers, they can be written as

Answer 16

Φ1 = |Φ1| e^i cos θ1

Question 17

The probabilities of detecting a particle at the screen, coming from slit 1 and slit 2 respectively (when one of the slits is closed), are

Answer 17

P1 = |Φ1|^2
P2 = |Φ2|^2

Question 18

When both slits are open, the total probability of observing a particle at the screen is not

Answer 18

the sum of the two probs

Question 19

the total probability is obtained from the

Answer 19

modulus squared of the total probability amplitude
Φ1+Φ2

Question 20

P12=

Answer 20

|Φ1+Φ2|^2
=(Φ1+Φ2)*(Φ1+Φ2)
then put in exp form

Question 21

we can define θ=

Answer 21

θ2-θ1

Question 22

final form of P12

Answer 22

|Φ1|^2 + |Φ2|^2 + 2 |Φ1||Φ2| cosθ

Question 23

we say the electron is in a

Answer 23

superposition state
Φ=Φ1+Φ2

Question 24

the term 2 |Φ1||Φ2| cosθ is an

Answer 24

interference term that makes the difference between this experiment and the classical case with no interference

Question 25

θ is the

Answer 25

phase difference between the two complex wavefunctions Φ1 and Φ2

Question 26

Electrons (and other matter) can behave like either particles or waves, depending on the kind of experiment performed on them, but never

Answer 26

behave like both waves and particles simultanously.

Question 27

Complementarity principle

Answer 27

pairs of complementary quantities, like wave and particle properties, can’t be measured simultaneously. The principle of complementarity is not limited to the wave-particle duality, but can be applied to other complementary quantities such as position and momentum, therefore extending to the uncertainty principle.

Question 28

Experiment 4: An experiment with electrons… but we monitor the electrons through the slits set up

Answer 28

we add a strong light source between the two slits so when the electron passes through a slit, it scatters some light and we can know which one it went through by seeing which aperture the “flash” comes from.

Question 29

experiment 4 - what do we observe

Answer 29

Every time we hear a “click” (we detect an electron at the screen), we also see a flash, either coming from slit 1 or slit 2. However, we never observe two flashes coming from both slits!

Question 30

experiment 4 - if we keep shooting electrons

Answer 30

we see something unexpected: the interference now disappears! We obtain the same result as in experiment 1 with bullets

Question 31

experiment 4 - if the light is turned off

Answer 31

the interference pattern of experiment 3 is restored

Question 32

if electrons are not seen, we have

Answer 32

interference

Question 33

If we try the experiment 4 with increasingly lower frequencies, nothing seems to change, until we eventually try with

Answer 33

light of wavelength comparable to the distance between the slits cannot distinguish these as two separate spots

Question 34

the form of the uncertainty principle reached through these experiments

Answer 34

it is impossible to design an experiment to determine through which slit the electron passes without destroying the interference pattern by disturbing the electron.

Question 35

wavefunction, Ψ(r,t)

Answer 35

a complex number associated to a particle (in the matter wave description of QM) and it is used to calculate the probability of finding the particle in a small region of space at a given time

Question 36

wavefunctions are not measurable quantities, they are represented by

Answer 36

complex numbers so do not directly define probabilities but probability amplitudes used to find probabilities

Question 37

To have physically “meaningful” measurable quantities, we need to obtain real values, so we use

Answer 37

the modulus square of wavefunction to obtain the probabilities

Question 38

For a one-dimensional wavefunction the probability density of finding a particle in an infinitesimal region between x and x+dx is

Answer 38

p(x,t)dx = |Ψ(x,t)|^2dx =Ψ(x,t)* Ψ(x,t)dx

Question 39

The probability of finding the particle in a finite region between a and b is

Answer 39

integral between a and b of |Ψ(x,t)|^2dx

Question 40

since the particle must be somewhere along the x-axis where the wavefunction is defined, the total probability of finding the particle in that region is

Answer 40

integral between -inf and +inf of |Ψ(x,t)|^2dx = 1

Question 41

normalisation condition

Answer 41

integral between -inf and +inf of |Ψ(x,t)|^2dx = 1 the particle exists somewhere

Question 42

If you consider the wavefunction at a fixed time T, the probability of finding the particle in a certain interval along x is given by

Answer 42

the area under the curve described by |Ψ(x,T)|^2

Question 43

Because of the relation between the wavefunction and the probability, it is important to stress that a wavefunction function of x and t must be:

Answer 43

single valued continuous smooth normalised

Question 44

single valued

Answer 44

for each value of x and t there is only one corresponding value of Ψ

Question 45

continuous

Answer 45

you could draw it without lifting pen from paper

Question 46

smooth

Answer 46

no sudden changes in the derivative - continuous derivative unless there is a singularity in the potential

Question 47

In one dimension and in the complex plane, a plane wave has the form

Answer 47

Ψ(x,t)=Ae^i(kx-wt)

Question 48

from planck can write E=

Answer 48

hv =h bar w

Question 49

Dispersion relation for non-dispersive systems

Answer 49

For non-dispersive systems, the dispersion relation is linear w(k)=vk where v is speed of light

Question 50

Dispersion relation for light

Answer 50

c= λv = w/k = E/p w(k)=ck

Question 51

Dispersion relation for a non-relativistic particle

Answer 51

w(k)=h abr k^2/2m derived from p=h bar k and E=p^2/2m

Question 52

Dispersion relation for a relativistic particle

Answer 52

w(k)=sqrt (k^2c^2+m^2c^4/h bar^2) from E=sqrt (p^2c^2 +m^2c^4)

Question 53

Dispersion relation for a relativistic particle for p <

Answer 53

reduces to E = 1/2mv^2 w(k)= h bar k^2/2m

Question 54

If the plane wave describes the wavefunction for a free particle, it means that since this must be

Answer 54

continuous and extending to infinity with the same amplitude, we could find the particle everywhere in space with the same probability BIG UNCERTAINTY

Question 55

Issues with free particles as plane waves

Answer 55

the plane wave has the same intensity everywhere and is fully delocalsied integral cannot converge so wavefunction can't be normalised

Question 56

How do we obtain something that has zero amplitude somewhere and is contained in a wavepacket of length delta x

Answer 56

add more waves

Question 57

superposition - the resultant wave is now made of

Answer 57

beats resulting in wave groups hence more localised and reduces uncertainty in position

Question 58

by adding two waves of different wavelengths, because of the de Broglie wavelength, we have two possible values of momentum that we could observe, which means that

Answer 58

we increased the uncertainty of momentum, and this leads to a reduced uncertainty of the position.

Question 59

adding two waves with the same amplitude and different wavenumbers, we obtain

Answer 59

a series of wave “envelopes” travelling at a group velocity that is different from the velocity of the individual waves

Question 60

The wave within the envelope moves at the

Answer 60

phase velocity give by term cos[kx-wt]

Question 61

The other cosine term gives the velocity of

Answer 61

the whole envelope aka group velocity

Question 62

phase velocity =

Answer 62

w/k

Question 63

group velocity =

Answer 63

delta w / delta k

Question 64

how do we create a single localised wavepacket and reduce the surrounding noise?

Answer 64

Add more waves of different frequencies (more values of momentum/wavenumber) choose conveniently the distribution of the amplitudes of each wave component

Question 65

adding more waves increases the wavepackets, while changing their weights in the sum (by changing their amplitudes) reduces the

Answer 65

noise between wavepackets

Question 66

We can keep increasing the number of waves to eventually

Answer 66

isolate a wavepacket

Question 67

particle represented by a localised wavepacket Φ(k) are the

Answer 67

amplitudes in the k-space and provide the so called spectral content of the wavepackets

Question 68

particle represented by a localised wavepacket Ψ(x) is the

Answer 68

fourier transform of Φ(k)

Question 69

the broader the function Φ(k)...

Answer 69

the narrower is the wavepacket Ψ(x,0) satisfying delta k delta x =1 if Φ(k) and Ψ(x) are Gaussians of width delta k and delta x

Question 70

position and wavenumber (or momentum) are

Answer 70

conjugate variables so increasing one will decrease the other

Question 71

For the wavepacket to travel without dispersion (i.e. no distorsion)...

Answer 71

the phase and group velocities must be equal vp=vg

Question 72

For a non-dispersive medium the dissipation relation gives a

Answer 72

constant phase velocity equal to the group velocity vp=v=vg and the wavekpacket does not disperse

Question 73

for a dispersive medium with w(k) prop to l^2

Answer 73

vp=k bar k/2m the components of waves with smaller wavenumbers would move more slowly

Question 74

we can use the Parseval identity to

Answer 74

relate the probabilities in position space and k space

Question 75

Starting from the total probability in position space, we can demonstrate

Answer 75

that the normalization in the wavenumber (or momentum) -space is preserved even if we have time evolution:

Question 76

Note that the probability of finding the particle at a certain value of k is not

Answer 76

changing with time the evolution of each k is just a phase that cancels out using Dirac delta properties

Question 77

the best function of Φ(k) that can be used to obtain a good wavepacket that minimizes uncertainty in both position and momentum space is

Answer 77

a gaussian

Question 78

why a gaussian

Answer 78

the FT of a guassian is still a gaussian

Question 79

The product of the widhts of the two Gaussian in k-space and x-space is

Answer 79

a constant independent of a showing that the relation is reciprocal

Question 80

Heisenberg uncertainty relation

Answer 80

The same kind of relation can be applied to any other conjugate variables delta x delta p > or = h abr/2

Question 81

gaussian gives minimum uncertainty in position so yields

Answer 81

something that looks like heisenberg uncertainty principle but has an equality sign (instead of > or=)

Question 82

Heisenberg's uncertainty relation is central in QM and explains things such as

Answer 82

the breakdown of the double slit on observation natural spectral linewidths

Question 83

heseinberg tells us that if position is constrained

Answer 83

momentum must become uncertain and vice versa

Question 84

how is the uncertainty principle responsible for avoiding the in-spiral and collapse of the Bohr atom

Answer 84

as an electron falls in and becomes localised near the nucleus at r=0, its root-mean-square momentum increases, and hence so does its kinetic energy: the ground state of minimum energy is not located at r=0, but instead at a finite distance and electrostatic potential.

Question 85

how the uncertainty principle explains the double-slit experiment issues

Answer 85

the observation of the position (which reduces the uncertainty delta x) will increase the uncertainty in the momentum sufficient to destroy the interference pattern

Question 86

why the ground state of the quantum harmonic oscillator has non-zero energy

Answer 86

due to Heisenberg's uncertainty principle since classically, E=T+V if the particle was exactly at the bottom of the potential with zero uncertainty, paradoxically the kinetic term related to the momentum would have to go to infinity

Question 87

A single plane wave is not physically realistic to describe a free particle as this would be

Answer 87

infinitely delocalised and not normalisable; we need to add an infinite number of weighted plane waves to form a wavepacket localised in space, obtaining the FT of a wavefunction in momentum space.