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10 Summary of Quantum MechanicsVariational Method for the Ground State Consider an arbitrary state|ψ normalized to 1.. Consider an atomic transition treated as a two-level system, an exc

10 Summary of Quantum Mechanics

Variational Method for the Ground State

Consider an arbitrary state|ψ
normalized to 1 The expectation value of the energy in this state is greater than or equal to the ground state energy E0:

ψ| ˆ H |ψ
≥ E0 In order to find an upper bound to E0, one uses a set of trial wave functions which depend on a set of parameters, and one looks for the minimum ofE
for these functions This minimum always lies above E0

7 Identical Particles

All particles in nature belong to one of the following classes:

• Bosons, which have integer spin The state vector of N identical bosons

is totally symmetric with respect to the exchange of any two of these particles

• Fermions, which have half-integer spin The state vector of N identical

fermions is totally antisymmetric with respect to the exchange of any two

of these particles

Consider a basis {|n i
, i = 1, 2, } of the one particle Hilbert space Consider a system of N identical particles, which we number arbitrarily from

1 to N

(a) If the particles are bosons, the state vector of the system with N1

particles in the state|n1
, N2 particles in the state|n2
, etc., is:

|Ψ
= √1

N !

1

√

N1!N2!· · ·



P

|1 : n P (1) ; 2 : n P (2) ; ; N : n P (N)
,

where the summation is made on the N ! permutations of a set of N elements.

(b) If the particles are fermions, the state corresponding to one

parti-cle in the state |n1
, another in the state |n2
, etc., is given by the Slater

determinant:

|Ψ
= √1

N !











|1 : n1
|1 : n2
|1 : n N

|2 : n1
|2 : n2
|2 : n N

|N : n1
|N : n2
|N : n N











.

Since the state vector is antisymmetric, two fermions cannot be in the same quantum state (Pauli’s exclusion principle) The above states form a basis of

the N −fermion Hilbert space.

8 Time-Evolution of Systems

Rabi Oscillation

Consider a two-level system|±
, of Hamiltonian ˆ H0= ¯hω0|+
+| We couple

these two states with a Hamiltonian ˆH1:

ˆ

H1= ¯hω1 2

e−iωt |+
−| + e iωt |−
+|

We assume that the state of the system is|−
at time t = 0 The probability

to find the system in the state|+
at time t is:

P (t) = ω

2

Ω2 sin2(ΩT /2) with Ω2= (ω − ω0)2+ ω12 .

Time-Dependent Perturbation Theory

We consider a system whose Hamiltonian is ˆH(t) = ˆ H0+ ˆH1(t) We assume

the eigenstates |n
of ˆ H0 and the corresponding energies E n are known At

time t = 0, we assume that the system is in the eigenstate |i
of ˆ H0 To first order in ˆH1, the probability amplitude to find the system in another eigenstate

|f
at time t is:

a(t) = 1

i¯h

 t

0

ei(E f −E i )t/¯ h f| ˆ H1(t )|i
dt  .

In the case of a time-independent perturbation H1, the probability is:

P (t) = |a(t)|2= 1

¯

h2



f| ˆ H1|i
2 sin2(ωt/2)

(ω/2)2 ,

where we have set ¯hω = E f − E i

Fermi’s Golden Rule and Exponential Decay

Consider a system with an unperturbed Hamiltonian ˆH0 Initially, the system

is in an eigenstate|i
of energy E i We assume that this system is coupled to

a continuum{|f
} of eigenstates of ˆ H0by the time-independent perturbation ˆ

V For simplicity, we assume that the matrix elements f| ˆ V |i
only depend

on the energies E f of the states |f
.

To lowest order in ˆV , this coupling results in a finite lifetime τ of the state

|i
: the probability to find the system in the state |i
at time t > 0 is e −t/τ

with:

1

τ =

2π

¯

h |f| ˆ V |i
|2ρ(E i )

The matrix element f| ˆ V |i
is evaluated for a state |f
of energy E f = E i

The function ρ(E) is the density of final states For non relativistic particles

12 Summary of Quantum Mechanics

(E = p2/2m) or ultra-relativistic particles (E = cp, for instance photons), its

values are respectively:

ρ non rel. (E) = mL

3√ 2mE 2π2¯h3 ρ ultra rel. (E) =

L3E2

2π2¯h3c3 .

When the spin degree of freedom of the particle comes into play, this density of

state must be multiplied by the number of possible spin states 2s + 1, where

s is the spin of the particle The quantity L3 represents the normalization volume (and cancels identically with the normalization factors of the states

|i
and |f
) Consider an atomic transition treated as a two-level system,

an excited state |e
and a ground state |g
, separated by an energy ¯hω and coupled via an electric dipole interaction The lifetime τ of the excited state

due to this spontaneous emission is given by:

1

τ =

ω3

3π0¯hc3



e| ˆ D|g
2

,

where ˆD is the electric dipole operator.

9 Collision Processes

Born Approximation

We consider an elastic collision process of a non-relativistic particle of mass

m with a fixed potential V (r) To second order in V , the elastic scattering cross-section for an incident particle in the initial momentum state p and the final momentum state p  is given by:

dσ

dΩ =

m

2π¯ h2

2

| ˜ V (p − p )|2 , with V (q) =˜



eiq·r/¯h V (r) d3r

Example: the Yukawa potential We consider

V (r) = g ¯hc

r e

−r/a ,

which gives, writing p = ¯ hk:

dσ

dΩ =

2mgca2

¯

h

2

1

1 + 4a2k2sin2(θ/2) 2

(Born) ,

where θ is the scattering angle between p and p  The total cross-section is then:

σ(k) = 2 mgca

¯

h

2

4πa2

1 + 4k2a2 (Born)

In the case where the range a of the potential tends to infinity, we recover the

Coulomb cross section:

dσ

dΩ =

g¯

4E

2

1 sin4(θ/2) (exact) , where E = p2/(2m).

Scattering by a Bound State

We consider a particle a of mass m undergoing an elastic scattering on a system composed of n particles b1, , b n These n particles form a bound state whose wave function is ψ0(r1, , r n) In Born approximation, the cross section is

dσ

dΩ =

m 2π¯ h2

2

|V(p − p )|2 with V(q) =

j

˜

V j (q) F j (q)

The potential V j represents the interaction between particles a and b j The

form factor F j is defined by:

F j (q) =



eiq·r j /¯ h |ψ0(r1, , r j , , r n)|2d3r1 d3r j d3r n

In general, interference effects can be observed between the various q

con-tributing to the sum which definesV(q) In the case of a charge distribution,

˜

V is the Rutherford amplitude, and the form factor F is the Fourier transform

of the charge density

General Scattering Theory

In order to study the general problem of the scattering of a particle of mass

m by a potential V (r), it is useful to determine the positive energy E =

¯

h2k2/(2m) eigenstates of ˆ H = ˆ p2/(2m) + V (r) whose asymptotic form is

ψ k (r) ∼

|r|→∞e

ik·r + f (k, u, u )e

ikr

r .

This corresponds to the superposition of an incident plane wave eik·r and

a scattered wave Such a state is called a stationary scattering state The

scattering amplitude f depends on the energy, on the incident direction u = k/k, and on the final direction u  = r/r The differential cross section is given

by:

dσ

dΩ =|f(k, u, u )|2 .

The scattering amplitude is given by the implicit equation

f (k, u, u ) =− m

2π¯ h2



e−ik
·r

V (r  ) ψ k (r  ) d3r  with k  = ku 

We recover Born’s approximation by choosing ψ k (r ) eik·r.

14 Summary of Quantum Mechanics

Low Energy Scattering

When the wavelength of the incident particle λ ∼ k −1 is large compared to

the range of the potential, the amplitude f does not depend on u and u  (at

least if the potential decreases faster than r −3 at infinity) The scattering is

isotropic The limit a s=− lim k→0 f (k) is called the scattering length.

Elementary Particles, Nuclei and Atoms

Neutrino Oscillations

In β decay or, more generally, in Weak interactions, the electron is always associated with a neutral particle, the neutrino ν e There exists in nature

another particle, the µ lepton, or muon, whose physical properties seem com-pletely analogous to those of the electron, except for its mass m µ  200 m e The muon has the same Weak interactions as the electron, but it is associated

to a different neutrino, the ν µ

A neutrino beam produced in an accelerator can interact with a neutron

(n) in a nucleus and give rise to the reactions

ν e + n → p + e and ν µ + n → p + µ , (1.1)

whereas the reactions ν e + n → p + µ or ν µ + n → p + e are never observed.

The reactions (1.1) are used in practice in order to detect neutrinos

Similarly, a π − meson can decay via the modes

π − → µ + ¯ν µ (dominant mode) and π − → e + ¯ν e , (1.2)

whereas π − → µ + ¯ν e or π − → e + ¯ν µ are never observed This is how one can

produce neutrinos abundantly (it is easy to produce π mesons) In (1.2) we have introduced the antiparticles ¯ ν µ et ¯ν e There is a (quasi) strict symmetry between particles and their antiparticles, so that, in the same way as the

electron is associated with the neutrino ν e , the antielectron, or positron, e+

is associated with the antineutrino ¯ν e One observes the “charge-conjugate” reactions of (1.1) and (1.2)

¯e + p → n + e+, ¯µ + p → n + µ+ and π+→ µ++ ν µ (1.3)

In all what follows, what we will say about neutrinos holds symmetrically for antineutrinos

In 1975, a third lepton, the τ , was discovered It is much more massive,

m µ  3500 m e , it is associated with its own neutrino ν τ, and it obeys the same physical laws as the two lighter leptons, except for mass effects Since the

1990’s, the experimental measurements at the LEP colliding ring in CERN

have shown that these three neutrinos ν e , ν µ , ν τ (and their antiparticles) are

the only ones of their kinds (at least for masses less that 100 GeV/c2) For a long time, physicists believed that neutrinos were zero-mass particles,

as is the photon In any case, their masses (multiplied by c2) are considerably smaller than the energies involved in experiments where they are observed Therefore, many experimental limits on these masses are consistent with zero However, both theoretical and cosmological arguments suggested that this might not be the case The proof that neutrino masses are not all zero is a great discovery of the last ten years

In the present study, we show how the mass differences of neutrinos can

be measured by a quantum oscillation effect The idea is that the “flavor”

neutrinos ν e , ν µ and ν τ, which are produced or detected experimentally are

not eigenstates of the mass, but rather linear combinations of mass eigenstates

ν1, ν2, ν3, with masses m1, m2, m3

The neutrinos observed on earth have various origins They can be pro-duced in accelerators, in nuclear reactors, and also in the atmosphere by cos-mic rays, or in thermonuclear reaction inside stars, in particular the core of the sun, and in supernovae explosions

1.1 Mechanism of the Oscillations; Reactor Neutrinos

In this first part, we consider oscillations between two types of neutrinos, the

ν e and the ν µ This simple case will allow us to understand the underlying physics of the general case We will analyze the data obtained with nuclear reactors The average energy of the (anti-)neutrinos produced in reactors is

E = 4 MeV, with a dispersion of the same order.

In all what follows, we will assume that if m is the neutrino mass and p and E its momentum and energy, the mass is so small that the energy of a neutrino of mass m and momentum p is

E =

p2c2+ m2c4 pc + m2c4

and that the neutrino propagates to very good approximation at the velocity

of light c.

Let ˆH be the Hamiltonian of a free neutrino of momentum p, which we

assume to be well defined We note |ν1
and |ν2
the two eigenstates of ˆ H:

ˆ

H |ν j
= E j |ν j
, E j = pc + m

2

j c4

2pc , j = 1, 2

m1and m2are the respective masses of the states|ν1
and |ν2
, and we assume

m1= m2

1.1 Mechanism of the Oscillations; Reactor Neutrinos 19 The oscillations of freely propagating neutrinos come from the following quantum effect If the physical states of the neutrinos which are produced (reactions (1.2)) or detected (reactions (1.1)) are not|ν1
and |ν2
, but linear

combinations of these:

|ν e
= |ν1
cos θ + |ν2
sin θ , |ν µ
= −|ν1
sin θ + |ν2
cos θ (1.5)

where θ is a mixing angle to be determined, these linear combination of energy

eigenstates oscillate in time and this leads to measurable effects

1.1.1 At time t = 0, one produces a neutrino of momentum p in the state

|ν e
Calculate the state |ν(t)
at time t in terms of |ν1
and |ν2
.

1.1.2 What is the probability P efor this neutrino to be detected in the state

|ν e
at time t? The result will be expressed in terms of the mixing angle θ and

of the oscillation length L

L = 4π¯ hp

|∆m2| c2 , ∆m2= m2

1− m2

1.1.3 Calculate the oscillation length L for an energy E  pc = 4 MeV and

a mass difference ∆m2c4= 10−4 eV2

1.1.4 One measures the neutrino fluxes with a detector located at a distance

 from the production area Express the probability P e as a function of the

distance  = ct.

1.1.5 The mass of the muon satisfies m µ c2 = 106 MeV Conclude that in

such an experiment one cannot detect muon neutrinos ν µ with the reaction

(1.1) We recall that m p c2= 938.27 MeV and m n c2= 939.57 MeV.

1.1.6 The detectors measure neutrino fluxes with an accuracy of∼ 10% (a) Assuming ∆m2c4 = 10−4 eV2, determine the minimal distance min

where to put a detector in order to detect an oscillation effect For this

calculation, assume the mixing in (1.5) is maximum, i.e θ = π/4 (b) How does minchange if the mixing is not maximum?

1.1.7 Several experiments on neutrinos produced by nuclear energy plants

have been performed in Chooz and in Bugey in France The most recent data comes from the KamLAND collaboration, in Japan The results are given on Fig 1.1

(a) Explain the results of Fig 1.1, except that of KamLAND

(b) The KamLAND experiment, which was performed in 2002, consisted

in measuring the neutrinos coming from all the (numerous) reactors in Japan and neighboring countries, which amounts to taking an average

distance of  = 180 km Putting together that data and the results of

numerous experiments performed on solar neutrinos, the physicists of Kamland come to the following results:

Chooz KamLAND

Bugey Rovno Goesgen

1,0 1,2

0,8 0,6 0,4 0,2 0

distance (meters)

Ndetected

N expected

Fig 1.1 Ratio between the numbers of observed electron neutrinos and those

expected in the absence of oscillations as a function of the distance
to the reactor

|∆m2| c4= 7.1 ( ± 0.4) × 10 −5eV2, tan2θ = 0.45 ( ± 0.02) (1.7) Show that these values are consistent with the result P e = 0.61 ( ± 0.10)

of Fig 1.1

1.2 Oscillations of Three Species; Atmospheric Neutrinos

We now consider the general formalism with three neutrino species We denote

|ν α
, α = e, µ, τ the “flavor” neutrinos and |ν i
, i = 1, 2, 3 the mass

eigen-states These two bases are related to one another by the Maki-Nagawaka-Sakata (MNS) matrix ˆU ,

|ν α
=

3



i=1

U αi |ν i
, U =ˆ

⎛

⎜

⎝

U e1 U e2 U e3

U µ1 U µ2 U µ3

U τ 1 U τ 2 U τ 3

⎞

⎟

This matrix is unitary (

i U ∗

βi U αi = δ αβ) and it can be written as:

ˆ

U =

⎛

⎝10 c023 s023

0 −s23 c23

⎞

⎠

⎛

⎝ c13 0 s13e

−iδ

−s13eiδ 0 c13

⎞

⎠

⎛

⎝−s c1212 s c1212 00

⎞

⎠

where c ij = cos θ ij and s ij = sin θ ij The complete experimental solution of

the problem would consist in measuring the three mixing angles θ12, θ23, θ13,

the phase δ, and the three masses m1, m2, m3 We consider situations such that (1.4) is still valid

1.2.1 At time t = 0 a neutrino is produced with momentum p in the state

|ν(0)
= |ν α
Express, in terms of the matrix elements U αi, its state at a

s is the spin of the particle The quantity L3< /small> represents the normalization volume (and cancels identically with the normalization factors of the states

|i
... to the sum which definesV(q) In the case of a charge distribution,

˜

V is the Rutherford amplitude, and the form factor F is the Fourier transform

of the. .. data-page="9">

1.1 Mechanism of the Oscillations; Reactor Neutrinos 19 The oscillations of freely propagating neutrinos come from the following quantum effect If the physical states of the neutrinos which