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(1)

Quantum Mechanics 1 Lecture 1: Introduction

Massimiliano Grazzini

University of Zurich

(2)

Outline

Blackbody radiation

Compton effect

Atom stability and the Bohr model Bohr correspondence principle

Photoelectric effect

De Broglie hypothesis

(3)

Introduction

Between the end of the XIX century and the beginning of the XX century an impressive amount of experimental results demonstrated that classical physics was not adequate to describe microscopic phenomena

The first attempts to solve these problems initially led to some confusion, because it appeared that electromagnetic radiation and electrons behaved sometimes as waves and some times as particles

This apparent inconsistency was resolved in the years 1926-27 with the formulation of the theory called Quantum Mechanics

This theory states that there are experiments in which the outcome is essentially unpredictable and that sometimes we have to content ourselves to the

probability that a certain measurement gives a given result

More importantly, the laws of combining probabilities are not those of classical

probability theory

(4)

Black-body radiation

The radiation emitted by a body, as an effect of its temperature, is called thermal radiation

In thermal equilibrium conditions, the emitted and absorbed radiation are equal The radiation spectrum strongly depends on the temperature of the emitting

body, and, to a lesser extent, on the body itself

A body absorbing all the incident radiation is called black body

An example of black body is a cavity enclosed by metal walls in which there is a small hole

If the cavity is in thermal equilibrium at temperature T

the emitted spectrum R

T

(ν) is the black-body spectrum

The radiation entering the hole will be reflected and

absorbed by the walls and only a negligible fraction of

it will be reemitted

(5)

We want to compute the electromagnetic energy density in the interval

in a metallic cavity in equilibrium at temperature T ν, ν + d ν

Requiring that the electric field vanishes on the walls We obtain the condition ( c = λν , k = 2π / λ and ω = 2πν )

x y

z

k

y

L = n

y

π k

z

L = n

z

π n

y

, n

z

Positive integers

Analogous considerations for the other components will lead to

k = π

L (n

x

, n

y

, n

z

)

nx,ny,nz

positive integers

We now want to compute the number of modes in the interval in the

octant k

i

> 0 k, k + dk

k = k

x2

+ k

y2

+ k

z2

E

x

∼ cos(k

x

x)sin(k

y

y)sin(k

z

z)sin(ωt)

For example for the component of the electric field we have x

In such a cavity electromagnetic waves can exist as stationary waves

L

(6)

Positive octant

We have to multiply by 2 because of the two transverse polarisations

The energy density is computed multiplying the above equation by the average energy of each state at the temperature and dividing by the volume T L

3

The average energy can be computed by using the Boltzmann distribution

(1)

Multiplying Eq.(1) by k

B

T and dividing by we obtain for the energy density L

3

Rayleigh-Jeans law

(2)

This number is dN = 1

8 volume of the shell in k − space

unit volume in k − space = 4 πk

2

dk 8(π / L)

3

dN = L

3

k

2

dk

π

2

= 8 πL

3

ν

2

c

3

P(E ) = exp{− E /(k

B

T )}

k

B

T ⟨E⟩ = ∫

0

E P(E )dE = k

B

T

ρ

T

(ν)dν = k

B

T 8πν

2

c

3

By using k = 2π / λ = 2πν / c we find

k

k + dk

This is the equipartition theorem of energy (each mode contributes k

B

T )

(7)

…the integral of the energy density over the frequency is divergent !

This is called Ultraviolet catastrophe Experimental observations indeed suggest that

at large frequencies the quadratic behavior is cut off and we have

This result is in agreement with the observations for low frequencies but…

In 1900 Planck proposed that the energy of an electromagnetic wave of frequency can only take discrete values, integer multiples of , that is where

ν h ν E

n

= nhν

Rayleigh-Jeans

Wien ρ

T

(ν)

ν

0

d νρ

T

(ν)dν = ∞

ρ

T

(ν) ∼ ν

3

e

−gν/(kBT)

Wien law

is the Planck constant

h = 6.63 ⋅ 10

−34

J ⋅ sec

This constant has the dimension of energy time, that is of an action ×

(8)

The sum of the series can be easily computed and we obtain

Multiplying by Eq.(1) and dividing by the volume we now find for the energy density

Which is in good agreement with observations

At small we have ν h ν so we recover the Rayleigh-Jeans result

e

hν/(kBT)

− 1 ∼ k

B

T

Setting α = h ν we get

k

B

T ⟨E⟩ = k

B

T

0

nαe

−nα

0

e

−nα

ρ

T

(ν) = 8 πν

2

c

3

h ν

e

hν/(kBT)

− 1

⟨E⟩ = k

B

T

0

nαe

−nα

0

e

−nα

=

e

hν/(kBT)

− 1

Planck law

Then the energy density should not be computed as an integral as in but as a discrete sum

⟨E ⟩ = ∑

0

E

n

P(E

n

)

0

P(E

n

) (2)

(3)

(9)

Photoelectric effect

The photoelectric effect was first observed by H. Hertz in 1887 and then studied by many others in the the following years.

Matter emits electrons when absorbing electromagnetic radiation in the visible or UV spectrum.

Experimentally we find that:

The effect occurs only when the frequency of the incoming radiation is higher than a threshold frequency , which depends on the material. ν

0

ν

The number of electrons emitted per unit of time is proportional to the intensity of the incoming radiation.

The maximal kinetic energy T

kin

of the emitted electrons satisfies the relation T

max

= h(νν

0

) (4)

Planck constant

(10)

Indeed the energy flux is characterised by the Poynting vector

The continuum absorption of radiation would suggest that: S = E × B

There should be no upper limit on the kinetic energy of electrons There should be no lower limit on the frequency

Einstein’s proposal was to assume that electromagnetic energy is exchanged

between radiation and electrons in “packets” called photons, all carrying the same energy related to the radiation frequency with E

ν

= h ν

The constant is the same appearing in Planck’s formula for black-body radiation h Denoting by the energy necessary to pull out an electron from the metal, one deduces that the maximal kinetic energy of the emitted electron is W

Which agrees with Eq. for (4) ν

0

= W / h

T

max

= h νW = h(νW / h)

Classically this result cannot be explained

(11)

Compton effect

The particle interpretation of electromagnetic radiation emerging from Einstein’s explanation of the photoelectric effect provides us with a simple explanation of the effect observed by A.H.Compton (1923) by studying the scattering of

electromagnetic radiation off electrons at high frequencies

λ

λ′

p

e

Compton found that the wavelength of the scattered radiation was shifted by an amount depending on the scattering angle Δλ

Classically one expects a different result. Indeed a continuum momentum

transfer from the radiation to the electron should accelerate all electrons in the same way

Absorption and reemission of radiation from the electron should happen with the same wavelength

θ

(12)

Interpreting the electromagnetic radiation as photons immediately explains the effect: by using energy and momentum conservation one finds

Therefore, radiation which is diffused forward (with ) has the same

wavelength as the incident radiation but the wavelength of radiation which is diffused at is shifted by an amount which is proportional to the so-called Compton length of the electron

θ = 0 θ ≠ 0

Compton scattering shows that one can attribute to photons not only energy but also momentum and that they are both conserved in scattering processes.

λ′λ = h

mc (1 − cos θ)

λ

c

= h

mc ∼ 2.43 ⋅ 10

−10

cm

(13)

Atom stability

The idea of atoms was introduced to explain basic phenomena in

chemistry. Other phenomena (electric conduction, light emission from matter, electrolysis, etc) show that atoms, although neutral, contain

positive and negative electric charges. Negative charges inside atoms turn out to be all equal among themselves: they are called electrons. Their mass and charge were measured (Thomson, Millikan) to be

Atomic masses are much larger, for example, the mass of the hydrogen atom is about m

H

1836 m

e

Rutherford’s hypothesis is that an atom is made of a nucleus which contains positive charges and whose size is much smaller than the size of the atom:

negatively charged electrons orbit around the nucleus much like the planets of the solar system.

The atomic model that emerged at the beginning of the XX century was the one proposed by Rutherford

e = − 1.6 ⋅ 10

−19

C m

e

= 0.9 ⋅ 10

−27

g

(14)

The atom typical dimension is of the order of 10

−8

cm

In Rutherford model the atom dimension is related to the one of the electron orbits, pretty much in the same way as it happens for the planets in the solar system

The equations of classical dynamics tell us that

Here is the reduced mass of the electron-proton system (since the proton is much heavier than the electron we have m m m

e

)

From the above equation we can write

And we obtain for the energy of the electron F = e

2

4πϵ

0

r

2

= ma = mv

2

r

r = e

2

4πϵ

0

mv

2

E = 1

2 mv

2

e

2

4πϵ

0

r = − 1

2 mv

2

E = − e

2

4πϵ

0

1

2r

(15)

We thus see that the radius of the atom is only determined when the energy is given, as it happens for the Kepler problem

But this is in contrast with the evidence that atoms of the same kind all have the same dimension, and in turn, it would imply that, for example, the

electrons of all hydrogen atoms would always have the same energy What should be the “preferred” energy ?

The other problem posed by the Rutherford model is that, according to classical electromagnetism an accelerated charge should emit radiation

The energy loss is provided by the Larmor formula (recall Electrodynamics !)

dE

dt = − 2

3 e

2

a

2

4πϵ

0

c

3

acceleration

By using a = v

2

/ r we obtain

dE

dt = − 2

3 e

2

4πϵ

0

c

3

v

4

r

2

= 2 3 e

2

4πϵ

0

c

3

(

e

2

4πϵ

0

mr )

2

1

r

2

= − 2 3 (

e

2

4πϵ

0

)

3

1

m

2

c

3

r

4

(16)

For r

0

= 10

−8

cm we obtain τ 10

−10

s !

dE

dt = − 2 3 (

e

2

4πϵ

0

)

3

1

m

2

c

3

r

4

= dE dr dr

dt

But since E = e

2

we have and then

4πϵ

0

1 2r

dE

dr = e

2

4πϵ

0

1 2r

2

dr

dt = − 2 3 (

e

2

4πϵ

0

)

3

1

m

2

c

3

r

4

2r

2

(

e

2

4πϵ

0

)

−1

= − 4 3 (

e

2

4πϵ

0

)

2

1

m

2

c

3

r

2

Separating the variables we have

3r

2

dr ( 4πϵ

0

e

2

)

2

c

3

m

2

= − 4dt ( 4πϵ

0

e

2

)

2

c

3

m

2

(r

3

r

03

) = − 4t

radius at

t = 0

The collapse time is thus

τ = 1 4 m

2

c

3

r

03

( 4πϵ

0

e

2

)

2

(17)

Bohr model

Already in late XIX century experiments had shown that atoms emit electromagnetic radiation at certain discrete frequencies

Einstein’s idea that the exchange of electromagnetic energy occurs in quanta was exploited by Bohr in 1913 to explain atomic spectra

Spectroscopists had managed to express the emission frequencies of some substances as differences of spectroscopic terms

To each there corresponds a spectral series n

Assuming Einstein relation between energy and frequency, the energy of the emitted or absorbed photon could be written as

By using energy conservation the “allowed” energy of the electrons is

1

λ = T(n)T(m)

E = = hc ( T(n)T(m) )

E

n

= − hc T(n)

(18)

Let us now focus on the hydrogen atom: an empirical formula for the spectroscopic term was found to be

Where R

H

= 109677.6 cm

−1

is the Rydberg constant Bohr hypothesis leads therefore to

If we now use the classical formula for the radius of the Rutherford atom, this implies that the radii of the allowed orbits are quantised

Where we have introduced the fine structure constant

T

H

(n) = R

H

/ n

2

n = 1,2....

E

n

= − hcR

H

/ n

2

r

n

= e

2

8πϵ

0

| E | = e

2

n

2

8πϵ

0

hcR

H

αn

2

4πR

H

α = e

2

4πϵ

0

ℏc ∼ 1

137

(19)

For n=1 we obtain the smallest orbit for the hydrogen atom which corresponds to Bohr radius

Such value is indeed in agreement with the expected atomic size

The model can be developed further: since the Planck constant has the dimension of an angular momentum, one can compute the angular momentum of the

electron on a stationary orbit and require it to be an integer multiple of = h /(2π) We have seen that the classical energy is E

n

= mv

2

/2 v = 2 | E

n

| / m

The angular momentum is then

Setting L

n

= nℏ we obtain an expression for the Rydberg constant which is in good agreement with the experimental data

a

B

= r

1

∼ 0.53 ⋅ 10

−8

cm

L

n

= m

e

v

n

r

n

= m 2e

2

8πϵ

0

r

n

m

e

r

n

= e

2

mr

n

4πϵ

0

= n e

4

m

(4πϵ

0

)

2

2hcR

H

R

H

= e

4

m

(4πϵ

0

)

2

4πcℏ

3

(20)

The final expression of the energy levels is thus

The lowest energy level is the one with n = 1

which is the ionisation energy

The Bohr radius can now be written as

One can show that at large n the frequency of the photon emitted by

an electron of the Bohr hydrogen atom which goes from the level n+1 to the level n is equal to the classical frequency

Classical electromagnetism predicts that a charge rotating on a circular orbit with angular orbital frequency ω emits electromagnetic radiation

with the same frequency.

E

n

= − hcR

H

/ n

2

= − e

4

m

e

(4πϵ

0

)

2

2

1 2n

2

E

1

= − e

4

m

e

(4πϵ

0

)

2

2ℏ

2

∼ − 13.6 eV

a

B

= 4πϵ

0

2

e

2

m

(21)

In summary:

The electron is able to revolve in certain stable (stationary) orbits around the nucleus without radiating any energy , contrary to what classical

electromagnetism suggests.

The stationary orbits correspond to distances for which the angular

momentum of the revolving electron is an integral multiple of the reduced Planck's constant

The lowest value of is ; this gives a smallest possible orbital radius of known as the Bohr radius. Starting from the angular momentum quantum rule we obtain the energies of the allowed orbits of the hydrogen atom.

n n = 1 a

B

= r

1

∼ 0.53 ⋅ 10

−8

cm

L

n

= nℏ

Electrons can only gain or lose energy by jumping from one allowed orbit to another, absorbing or emitting electromagnetic radiation with a

frequency determined by the energy difference of the levels according to the Planck relation ν ΔE = E

2

E

1

=

ℏ = h /(2π)

(22)

Bohr correspondence principle

The ideas leading to the quantisation of angular momentum were extended by Bohr (and Sommerfeld) to more general systems as

Where and are conjugated variables and the integral is computed on the

“allowed” classical orbits p q

Consider a particle of mass moving in a box of size (infinite potential well) m L The Bohr-Sommerfeld condition gives

and

Exact result ! In general the result obtained with the BS

rule are exact only in the large n-limit

pdq = nh

2pL = nh

E

n

= p

n2

2m = 1 2m n

2

h

2

4L

2

= n

2

π

2

2

2mL

2

V = 0

x

V = ∞ V = ∞

0 L

(23)

De Broglie hypothesis

The Einstein relation which expresses the corpuscular nature of

electromagnetic waves involves the same Planck constant entering the Born- Sommerfeld quantisation condition which is applied to particles

E = h ν

This suggested to De Broglie that particles could in turn behave as waves For photons one can write and

which implies that

electrons behave like waves with wavelength

p = E / c λ = c

ν = hc

E = h p

λ = h / p

The Bohr-Sommerfeld quantisation condition can be reformulated through the de Broglie wavelength by writing for the length of the allowed classical orbits

L =

Lp = nh = nλp

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