4.3. Evidence of Oscillations

A lot of experiments have been done to research on neutrino oscillations. In summary there are three types,

  1. Solar neutrinos,

  2. Reactor and accelerator neutrinos,

  3. Atmospheric neutrinos.

4.3.1. Results of Experiments

  1. Difference between masses from data

    |Δm212||Δm31(32)2|0.03.

    We also have

    |Δm212||Δm31(32)2|.

    By some convention, people would use numbers so that Δm212>0 or m1<m2.

4.3.1.1. Determine |Δm2| and θ

The neutrino experimantal data shows the mixing angles are 1

  1. θ23=39±2;

  2. θ13=8.9±0.5;

  3. θ12=34±1.

Experimental result of the δmij2 s are 1

  1. δm212=7.50.2+0.3×105eV2;

  2. |δm322|=2.40.1+0.1×103eV2.

Definition of Mass-squared Difference

δmij2=mi2mj2. Obviously, δm312=δm232δm212.

As |δm212||δm322|, we should have δm312δm322.

1(1,2)

Neutrino tomography by Margaret A. Millhouse & David C. Latimer, American Journal of Physics 81, 646 (2013); doi: 10.1119/1.4817314 .

4.3.1.2. Atmospheric Results

4.3.1.3. Accelerator Results

4.3.1.4. Reactor Results

4.3.2. Hamiltonian, Equation of Motion and Solution

Neutrinos evolve in mass eigenstates. So we need to describe flavour states |να using mass eigenstates |νj.

|να=jUαj|νj;p~j,

where Uαj is the element of neutrino mixing matrix.

PMNS Mixing Matrix

Pontecorvo-Maki-Nakagawa-Sakata (PMNS) mixing matrix is the product of three rotation matrices, in addition to an extra phase,

U=U23×U13,δ×U12=(1000cosθ23sinθ230sinθ23cosθ23)(cosθ130eiδsinθ13010eiδsinθ130cosθ13)(cosθ12sinθ120sinθ12cosθ120001)

The δ is the CP violation phase.

The origin of the phase is from the fact that we need 4 degrees of freedom for this mixing matrix while a convinient way is to write down the SO(3) rotation matrix then put this extra phase here.

More About Phase of Nutrinos

The mixing of mass eigenstates is

(νeνμντ)=(eiα1000eiα2000eiα3)Some Unitary Matrix(1000eiβ2000eiβ3)

Since the phase of neutrinos can be redefined, we have 3 phases for each flavour and a global phase being arbitary. The first matrix on the RHS can be eliminated. The third matrix on the RHS is not important for neutrino oscillations so it can be neglected. (Proof required)

In ultra relativistic case, we can simply find out the time evolution, which is equivalent to distance evolution,

|ψ(t)=jUαjGj(t,t0)|νj;p~j.

The survival probability means how much neutrinos of a flavour left after some time or distance, which is calculated by

P(νlνl)=|νlψ(t)|2.

We can see clearly that the survival probability depends on some parameters.

4.3.2.1. Two-flavour Vacuum Oscillations

To write down this clearly, we need to write down the mixing matrix and propagator. For simplicity, we calculate the example of two flavour (a, b) oscillation.

It’s easier to write down the propagation in mass eiginstates so the first thing to work out is the mixing matrix.

Suppose we have only a flavour neutrino initially,

|ψ(0)=|νa

4.3.2.1.1. Mixing Matrix

The mixing matrix is an rotation of eigenbasis.

Two Flavour Neutrino Mixing

Fig. 4.1 Two flavour neutrino mixing diagram with θ being the mixing angle

The flavour states can be expressed in terms of mass eigenstates,

(νaνb)=(cosθsinθsinθcosθ)(ν1ν2)

where the matrix

U=(cosθsinθsinθcosθ)

is the mixing matrix which is a rotation of basis geometrically. In other words, this matrix is the representation of the rotation eiθ^.

4.3.2.1.2. Survival Probability

Neutrinos are usually produced in electron flavor, which we choose as the initial condition for this example,

|ψ(t=0)=cosθ|ν1+sinθ|ν2.

With the mixing matrix, the propagation of an initial state of only flavour a is

|ψ(t)=cosθ|ν1eiE1t+sinθ|ν2eiE2t.

To find out the amplitude of flavour a, we need to project the state |ψ(t) onto a flavour eigenstate, say, |νa,

νaψ(t)=νa|(cosθ|ν1eiE1t+sinθ|ν2eiE2t)=(cosθ|ν1+sinθ|ν2)(cosθ|ν1eiE1t+sinθ|ν2eiE2t)=cos2θeiE1t+sin2θeiE2t

The survival probability is the amplitude squared,

Paa=|νaψ(t)|2=|cos2θeiE1t+sin2θeiE2t|2=(cos2θeiE1t+sin2θeiE2t)(cos2θeiE1t+sin2θeiE2t)=cos4θ+sin4θ+cos2θsin2θei(E1E2)t+sin2θcos2θei(E1E2)t=cos4θ+sin4θ+cos2θsin2θeiΔEt+sin2θcos2θeiΔEt=cos4θ+sin4θ+2cos2θsin2θcos(ΔEt)=(cos2θ+sin2θ)22cos2θsin2θ+2cos2θsin2θcos(ΔEt)=12cos2θsin2θ(1cos(ΔEt))=1sin2(2θ)sin2(ΔEt2)

with the definition ΔE=E1E2p1+12m12p1p212m22p2. We usually calculate the case p1=p2=p , which takes us to

ΔEm12m222p=δm22p.

with δm2=m12m22. Most of the time we would like to know the oscillation with respect to distance. Using the approximation t=L and ΔEm12m222p, we have

Paa=1sin2(2θ)sin2(ΔEL2)=1sin2(2θ)sin2(δm2L4p).

This is the survival probability of flavour a neutrino with an initial state of flavour a.

There are several things to be noticed,

  1. θ=0 leads to oscillation free neutrinos.

  2. ΔE=0 or δ2m=0 (in the case of same momentum) also gives us no oscillation.

  3. At L=0 the survival probability is 1, which means no oscillation is done.

4.3.2.1.3. Hamiltonian in Flavor Basis

It’s easy to write down the Hamiltonian for the mass state stationary Schrodinger equation. As we have proven, to first order approximation,

E=p+12m2p
Hj=(p+12m12p00p+12m22p)=pI+12p(m1200m22)

However, the Hamiltonian we prefer is the one for flavour eigenstates. To achieve this, we only need to rotate this previous Hamiltonian using the mixing matrix U.

Hα=UH^jUT=(cosθsinθsinθcosθ)(pI+12p(m1200m22))(cosθsinθsinθcosθ)=pI+12p(cos2θm12+sin2θm22sinθcosθm12+sinθcosθm22sinθcosθm12+sinθcosθm22sin2θm12+cos2θm22)=pI+12p(m12δm2sin2θ12sin2θδm212sin2θδm2m22+δm2sin2θ)=pI+12p(12(m12+m22)I12(δm2cos2θδm2sin2θδm2sin2θδm2cos2θ))=(p+m12+m224p)I14p(δm2cos2θδm2sin2θδm2sin2θδm2cos2θ)

Again we see clearly, no oscillation will apear as long as mixing angle θ=0 or δm2=0.

Note

The reason we can do this is that this mixing matrix is time and space independent. To see this, we first write down the Schrodinger equation for mass eigenstates,

idt|Φj=H^j|Φj.

Applying the mixing matrix,

idtU1|Φα=H^jU1|Φα.

Notice that the mixing matrix, which is a rotation, is orthonormal, UUT=I. Then we have inverse of this matrix is the same as the transpose.

idtUT|Φα=H^jUT|Φα.

Multiply on both sides U and remember the fact that the mixing matrix is orthonormal, we have

idt|Φα=UH^jUT|Φα.

Now we can define the Hamiltonian for flavour states,

Hα=UHjUT.

Since Pauli matrices plus identity forms a complete basis for all 2 by 2 matrices, it our Hamiltonian can be written as

H=δm24E(cos2θsin2θsin2θcos2θ)=δm24E(cos2θσz+sin2θσx).

To be explicit, δm2=δm122=m12m22. The trandition is always

δmi,j2=mi2mj2.

In two flavor oscillations, most literature use Δm2δm212 so that we don’t have the negative sign in front of the Hamiltonian here. So be it,

H=Δm24E(cos2θsin2θsin2θcos2θ)=Δm24E(cos2θσz+sin2θσx).

Note

Pauli matrices are

σx=(0110)σy=(0ii0)σx=(1001).

In a more compact way,

σj=(δj3δj1iδj2δj1+iδj2δj3).

4.3.3. Matter Effect - An Introduction

4.3.3.1. Hamiltonian

We have already derived the Hamiltonian for vacuum oscillatioin,

Hv=δm22E12(cos2θvsin2θvsin2θvcos2θv),

where we would like to define a new matrix,

B=12(cos2θvsin2θvsin2θvcos2θv),

so that the vacuum Hamiltonian can be written as

Hv=δm22EB

The effect of matter, as we have already discussed before, adds an extra term

Hm=2GFneL.

Here we have

L=(1000).

Note

Previously in the MSW effect section, we have L=12σ3. The reason, as explained there, is that we can always write down a 2 by 2 matrix using Pauli matrices and indentity matrix and identity matrix only shifts the overall eigenvalue not the eigenvector so we can just drop the identity term.

One other term is the self-interaction of neutrinos, i.e., neutral-current neutrino-neutrino forward exchange scattering,

Hν=2GFd3p(1p^p^)(ρpρ¯p).

The overall Hamiltonian is

H=H0+Hm+Hν,

where the vacuum Hamiltonian is

H0=δm22EB=δm22EU(12σ3)U.

4.3.3.2. Equation of Motion

From the Hamiltonian, Von Neumann equation is

itρ=[H,ρ]

In Picture chapter we have seen the definition of a polarization matrix. The components of a polarization vector (for neutrinos) is given by

Pω,iTr(ρEσi)=1nν|δm2|2ω2×Tr(ρEσi).

For anitneutrinos, we have a negative ω which is defined as ω=δm22E (neutrinos) and ων¯=δm22E (anitneutrinos). The polarization is defined as

Pω,i=1nν|δm2|2ω2×Tr(ρ¯Eσi).

With all these definitions, Von Neumann equation multiply by σ=σ1e^1+σ2e^2+σ3e^3, we have

iρ˙iσie^i=[H,ρ]iσie^i.

Notice that Pauli matrices are Hermitian and Unitary, we can alway insert the identity I=σjσj.

Commutator and Cross Product

Commutator of two vectors,

Misplaced &

Trace of Pauli Matrices

All Pauli matrices have vanishing trace. And what makes our calculation more convinient is that the trace of matrices is invariant under cyclic permutation, that is

Tr(σiHσj)=Tr(Hσjσi)

Notice that to have a non-vanishing trace we need i=j. This property really saves our life.

As the definition, we have

H=Hσρ=ρσ

Using these we can rewrite the commutator

[H,ρ]=[Hσ,ρσ]=ik(HiσiρkσkρkσkHiσi)=ik(HiρkσiσkρkHiσkσi)=ikHiρk(σiσkσkσi)=ikHiρk[σi,σk]=ikHiρk2iϵiknσn=2iikϵiknσnHiρk

Multiply by σj and take the trace, we get,

Tr(σj[H,ρ])=2iTr(ikϵiknσjσnHiρk)=2iikTr(ϵikjIHiρk)=2iikϵjikHiρkTr(I)=4iϵjikHiρk.

The corresponding LHS after these work becomes

iTr(σjρ˙iσi)=itρjTr(I)=2iPj˙

The Von Neuman equation becomes

P˙=2H×P.

We know explicitly what polarization vector is

Pj=ConstantTr(ρσj)

for neutrinos while

P¯j=ConstantTr(ρ¯σj).

The vectorized Hamiltonian is

H=Hiσi.

Multiply by σj and take the trace,

Tr(Hσj)=HjTr(I),

that is,

Tr(Hσj)=2Hj.

Hamiltonian

The Hamiltonian for homogeneous isotropic environment is

H=H0+Hm+Hν=ωB+λL+GF0dE(ρEρ¯E).

Then the equation we need becomes

Pω˙=(ωB+λL+μD)×Pω.

where B=Tr(Bσ), L=Tr(Lσ), D=dωPω.

4.3.4. Q&A

Question

What are some of the conventions used in liturature?

Answer

  1. Δmij2=mi2mj2.

  2. Flavours of left hand neutrinos are mixing of mass eigen states, νlL=j=13UljνjL(x).

Question

Why can we use just quantum mechanics on relativistic neutrinos? In principle one should use quantum field theory or at least relativistic quantum mechanics?

Answer

To be answered.

Question

What does the mixing angle mean exactly both in vacuum and matter environment?

Answer

There are several ways to illustrate this.

  1. Rotation angle in flavour space. For simplicity I use a two component neutrino model.

|ν1=cosθ|νe+sinθ|νμ|ν2=sinθ|νe+cosθ|νμ

This is a rotation in a plane with a generator eiθ^. (Make a figure for this.) + (Write down the 3 components model.)

  1. Oscillation probability involves this angle too. It is a suppression of the oscillation probability.

  2. From the view of quantum states, this angle determines how the flavour states are composed with mass eigenstates, i.e., the fraction or probability of each mass eiginstates in a flavour state.

Question

What does wave packet in neutrino oscillation mean?

Answer

To Be Answered.

Question

How would a wave packet spread?

Answer

A Gaussian wave packet would spread or shrink. The key of this spreading or shrinking is the dispersion relation.

For non-relativistic Gaussian wave packet ψ(x,t)=eα(kk0)2 in momentum basis with dispersion relation ω=2k22m, the expansion of packet is

Δx=α2+(t2m)2.

Obviously, the RMS width spreads according to group velocity vg=0/m.

However, the situation could be different for a relativistic neutrino.

Question

What will scattering do to a wave packet.

Answer

Momentum transfer for a plan wave case in Born approximation is


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