Elementary course on atomic polarization and Hanle effect Rev. 1.2: 13 April 2009 Saku Tsuneta...

Elementary course on atomic polarization and Hanle effect

Rev. 1.2: 13 April 2009Saku Tsuneta

(NAOJ)

1

Table of contents

1. Quantum states 2. Atomic polarization and quantization axis3. van Vleck angle 4. He108305. Role of magnetic field6. Hanle effect7. Formal treatment (in preparation)

2

Introduction• Hanle effect or atomic polarization should be understood

with quantum mechanics. I do not find any merit to rely on the classical picture.

• It is a beautiful application of very fundamental concept of quantum mechanics such as quantum state and angular momentum, scattering, and conceptually should not be a difficult topic.

• This is an attempt to decode series of excellent papers by Javier Trujillo Bueno and people.

• The outstanding textbooks for basic quantum mechanics are• R. P. Feynman, Lectures on physics: Quantum Mechanics• J. J. Sakurai, Modern Quantum Mechanics

3

Required reading• R. P. Feynman, Lectures on physics: Quantum Mechanics is the

best introductory text book to understand the basic of atomic polarization etc in my opinion. (Please be advised not to use standard text books that you used in the undergraduate quantum mechanics course.)

• In particular read the following chapters:– Section 11-4 The Polarization State of the Photon– Section 17-5 The disintegration of the Λ₀

– Section 17-6 Summary of rotation matrices– Chapter 18 Angular Momentum – Chapter 5 Spin One– Chapter 6 Spin One Half– (Chapter 4 Identical Particles) – Chapter 1 Quantum Behavior– Chapter 3 Probability Amplitudes

4

1. Quantum states

5

A note on photon polarization

• Right-circularly polarized photon |R> (spin 1)• Left-circularly polarized photon |L> (spin -1)

– defined in terms of rotation axis along direction of motion

• Linearly polarized photon is a coherent superposition of |R> and |L>– |x> = 1/√2 (|R> + |L>), – |y> =-i/√2 (|R> - |L>), – Or, equivalently

• |R> =1/√2 (|x> + i|y>)• |L> =1/√2 (|x> - i|y>)

• It is not zero angular momentum. It does not have a definite angular momentum.

6

Base states in energy and in angular momentum

• If B = 0, base states representing magnetic sublevels m for J=1, |1>,|0>, and |-1> are degenerated states in energy H:– H|1>=E|1>– H|0>=E|0>– H|-1>=E|-1>

• But, these base states ,|1>,|0>, and |-1> are also eigenstates for angular momentum, and are not degenerated in angular momentum Jz=M:– Jz|1>=1|1>– Jz|0>=0|0>– Jz|-1>=-1|-1>

• Throughout this handout, ћ=17

No polarization only when different |m> statehas the same population!Polarization remains if with population imbalance

If magnetic field B →Zero

|J=1, m=1>|J=1, m=0>|J=1, m=-1>

|J=0, m=0>

Quantization axis = direction of B

Observerσ+

σ-

σ+σ-

π

Observer

No polarization with B→0

8

Emission from 3-level atomwith magnetic field

|J=1, m=1>

|J=1, m=0>

|J=1, m=-1>

|J=0, m=0>

Zeeman splittingQuantization axis = direction of B

Observerσ+

σ-

σ+σ-

π

Observer

B

B

9

2. Atomic polarization and quantization axis

10

Atomic polarization is merely conservation of angular momentum

Example #1; 1-0 system

|J=1, m=1>=|1> |J=1, m=0>=|0> |J=1, m=-1>=|-1>

|J=0, m=0>=|1,0>

|1>, |0>,|-1>B=0: degenerated state

|L> |R>

Take quantization axis to be direction of Incident photons

Unpolarized light from a star

A right-circularized photon carrying angular momentum -1 |L> causes transition to m=1 state of atom (|1> to |0>).A left-circularized photon carrying angular momentum 1 |R>causes transition to m=-1 state of atom (|-1> to |0>).

1/2 1/2

11

Atomic polarization is merely conservation of angular momentum

Example #2; 0-1 system |J=1, m=1>=|1> |J=1, m=0>=|0> |J=1, m=-1>=|-1>

|J=0, m=0>=|0,0>

|1>, |0>,|-1>B=0: degenerated state

|R> |L>

Take quantization axis to be Direction of Incident photons

Unpolarized light from a star

A right-circularized photon carrying angular momentum +1 |R> causes transition to m=1 state of atom (|0> to |1>).A left-circularized photon carrying angular momentum -1 |L>causes transition to m=-1 state of atom (|0> to |-1>).

1/2 1/2

12

If unpolarized light comes in from horizontal direction,

Take quantization axis to be direction of Incident photons

|J=1

, m=1

>=|1

’>|J

=1, m

=0>=

|0’>

|J=1

, m=-

1>=|

-1’>

|J=0

, m=0

>=|0

’,0’>

|R’>

|L’>

Un-

pola

rized

ligh

t fro

m a

sid

e

Exactly the same atomic polarization take place but in the different set of quantumbase states |-1’>, |0’>,|1’>

Note that |1> and |1’> are different quantum states. For instance |1> is represented by linear superposition of |1’>, |0’> and |-1’>.

1/2

1/2

13

What is the relation between|Jm> and |Jm’> base states?

|1>,|0>,|-1>

|1’>,|0’>,|1’>

|1’> |0’> |-1’>

<1| (1+cosθ)/2 -sinθ/√2 (1-cosθ)/2

<0| sinθ/√2 cosθ -sinθ/√2

<-1| (1-cosθ)/2 sinθ/√2 (1+cosθ)/2

θ If θ is 90 degree, |1> = (1+cosθ)/2|1’> + (1-cosθ)/2|-1’> = 1/2 (|1’>+|-1’> ) <1|1>=1/4 |0> = sinθ/√2 |1’> - sinθ/√2 |-1’> = 1/√2 (|1’> - |-1’>) ) <0|0>=1/2 |-1> = (1-cosθ)/2|1’> + (1+cosθ)/2|-1’> = 1/2 (|1’>+|-1’> ) ) <-1|-1>=1/4If θ is 0 degree, <1|1>=1/2 <0|0>=0 <-1|-1>=1/2Thus, illumination from side provides different atomic polarization!

Rotation matrix for spin 1 (See Feynman section 17.5)

Normal to stellar surface

14

Rotation matrix for spin 1some analogy..

15

|1> |0> |-1>

<1’| (1+cosθ)/2 sinθ/√2 (1-cosθ)/2

<0’| -sinθ/√2 cosθ sinθ/√2

<-1’| (1-cosθ)/2 -sinθ/√2 (1+cosθ)/2

x

x’

y

y’

θx y

X’ cosθ sinθ

y’ -sinθ cosθ

Rotation matrix for 2D space

z

y

θ

z’

Base states of the old frame

Base states of the new frame

Ry(θ)

|1’> |0’> |-1’>

<1| (1+cosθ)/2 -sinθ/√2 (1-cosθ)/2

<0| sinθ/√2 cosθ -sinθ/√2

<-1| (1-cosθ)/2 sinθ/√2 (1+cosθ)/2

This means that|J

=1, m

=1>=

|1’>

|J=1

, m=0

>=|0

’>|J

=1, m

=-1>

=|-1

’>

|J=0

, m=0

>=|0

’,0’>

|R’>

|L’>

Un-

pola

rized

ligh

t fro

m a

sid

e

1/2

1/2

=

|J=1, m=1>=|1> |J=1, m=0>=|0> |J=1, m=-1>=|-1>

|J=0, m=0>=|0,0>

1/4 1/4

Un-

pola

rized

ligh

t fro

m a

sid

e|L

’>|R

’>

1/2

quantization axis

quantization axis 16

Generation of coherence due to rotation

• In the previous page, |1’> and |-1’> are not coherent due to non-coherent illumination, namely <-1’|1’>=0

• But, in the new base states|1> and |-1> have coherency. If θ is 90 degree, – |1> = 1/2 (|1’>+|-1’> ) <1|1>=1/4– |0> = 1/√2 (|1’> - |-1’>) ) <0|0>=1/2– |-1> = 1/2 (|1’>+|-1’> ) ) <-1|-1>=1/4

• for instance <-1|1>=1/2 not zero!• Thus, rotation can introduce coherency in quantum

states!17

3. van Vleck angle

19

van Vleck angle concept is easy tounderstand with what we learned so far

• van Vleck angle is merely quantization-axis dependent change in population of each state!

• Consider the transformation from quantization axis normal-to-photosphere to quantization axis along B (see figure in next slide).

• 90 degree ambiguity is explained in chapter 6.

20

Two quantization axes

Line of sightin general not in this plane(not used in This section) θ

Quantization axisnormal to photosphere|1>,|0>,|-1>

Quantization axisalong B|1’>,|0’>,|-1’>

Why horizontal field?We need B inclined to the symmetry axis of the pumping radiation field to redistribute the anisotropic population. sy

mm

etry

axi

s of

pu

mpi

ng r

adia

tion

field

21

Derive van Vleck magic angle • Degree of linear polarization LP is proportional to population

balance in base quantum states in saturation (Hanle) regime (to be explained in later sections), where there is no coherence among the states:• LP ≈ <1’|1’>+<-1’|-1’>-2<0’|0’> = (1+cosθ) 2/4+ (1-cosθ) 2/4 - sin2θ/2 = (3cos2θ -1)/4 (use ‘’rotation matrix for spin1’’ in earlier page)

• Thus, LP changes sign at 3cos2θvv -1=0, i.e. θvv =54.7 degree.• If the angle of the magnetic field with respect to normal to

photosphere is larger or smaller 54.7 degree, Stokes LP will change its sign.

22

σ+σ-

π

Observer

B

• van Vleck Anglevv = 54.7 deg– < vv , then LP // B

– > vv , then LP B

The van Vleck Effect results in > vv

< vv

Figure taken from H. Lin presentation for SOLAR-C meeting

Linear polarization direction

4. He10830

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He 10830

• Blue 10829.09A– J(low)=1 J(up)=0

• Red1 10830.25A– J(low)=1 J(up)=1

• Red2 10830.34A– J(low)=1 J(up)=2

25

He10830 red wing

|J=1, m=1>=|1> |J=1, m=0>=|0> |J=1, m=-1>=|-1>

|J=0, m=0>=|0,0>|R> |L>

Unpolarised light from a star

1/2 1/2

Dark filamentNo Stokes-V No Stokes-LP Can not exist without B due to symmetry(LP exists with horizontal B) Hanle effect!

ProminenceNo StokesStokes LP even with zero B

1/3

1/3

1/3

Incoherent states(ie <1|-1>=0)

LP = Linear Polarization26

To understand Linear polarization from prominence with zero horizontal B,

• |1> state is created by absorption of an |L> photon from below (photosphere).

• Consider the case of 90 degree scattering, we rotate the quantization axis normal to photosphere by 90 degree i.e. parallel to photosphere.

• With |1,1> to |0,0> transition, a photon with state ½|R> + ½ |L> is emitted (90 degree scattering).

• This is a linearly polarized photon with state |x> = 1/√2 (|R> + |L>) !

• Likewise, for |-1> state, -|x> = -1/√2 (|R> + |L>) 27

28

What is the polarization state of a photon emitted at any angle θ?

Spin -1 atom |atom, -1>

Spin 0 atom |atom, 0>

|L>

|L>

Spin 1 atom |atom, -1>

Spin 0 atom |atom, 0>

Emission of circular polarized light

θ

Z

Z’

?

ProcedureStep 1: Change the quantization axis from Z to Z’ and represent the atomic state with respect to new Z’ axis.Step 2: Then apply usual angular momentumConservation around Z’ axis!

Easy-to-understand case!

X

Importance of Quantization axisWhy do we need new axis Z’?

• For a particle at rest, rotations can be made about any axis without changing the momentum state.

• Particle with zero rest mass (photons!) can not be at rest. Only rotations about the axis along the direction of motion do not change the momentum state.

• If the axis is taken along the direction of motion for photons, the only angular momentum carried by photons is spin. Thus, take the axis that way to make the story simple.

29

σ transition |1,1> to |0,0> and |1,-1> to |0,0>

• |1’> = (1+cosθ)/2|1> +sinθ/√2|0>+(1-cosθ)/2|-1>• |-1’> = (1-cosθ)/2|1> -sinθ/√2|0>+(1+cosθ)/2|-1>

– <1|1’>= (1+cosθ)/2=1/2 probability to emit |R> photon– <1|-1’>= (1-cosθ)/2=1/2 probability to emit |L> photon

• Thus, for transition |J, m>=|1, 1> to |0,0> with θ=90 degree, a photon with state (|1’>+|-1’>)/2 = (|R’>+|L’>)/2 ≈|x>, x-linearly polarized light!

• For |J, m>=|1, -1> in z-coordinate– <-1|1’>= (1-cosθ)/2=1/2 probability to emit |R> photon– <-1|-1’>= (1+cosθ)/2=1/2 probability to emit |L> photon

• Thus, for transition |J, m>=|1, -1> to |0,0> with θ=90 degree, a photon with state (|1’>+|-1’>)/2 = (|R’>+|L’>)/2 ≈|x> , x-linearly polarized light!

30

π transition |1,0> to |0,0>• |1’> = (1+cosθ)/2|1> +sinθ/√2|0>+(1-cosθ)/2|-1>• |-1’> = (1-cosθ)/2|1> -sinθ/√2|0>+(1+cosθ)/2|-1>

– <0|1’> = sinθ/√2 = 1/ √ 2 probability to emit |R> photon– <0|-1’> = -sinθ/√2 = -1/ √ 2 probability to emit |L> photon

• Thus, for transition |J, m>=|1, 0> to |0,0> with θ=90 degree, a photon with state

(|1’>-|-1’>) = |R’>-|L’> ≈√ 2i|y>, y-linearly polarized light!

31

He10830 blue wingsDark filamentNo Stokes-V No Stokes-LP Can not exist without B due to symmetry(LP exists with horizontal B) Hanle effect!

ProminenceNo Stokes-VNo Stokes-LP(even with B)

|J=1, m=1>=|1> |J=1, m=0>=|0> |J=1, m=-1>=|-1>

|J=0, m=0>=|1,0>

|L> |R>

1/2 1/2

1/3 1/3 1/3

32

He 10830 with horizontal B

prominence filament• Blue 10829.09A no LP LP<0

– J(low)=1 J(up)=0

• Red1 10830.25A LP>0 LP>0– J(low)=1 J(up)=1

• Red2 10830.34A LP>0 LP>0– J(low)=1 J(up)=2

• LP=Stokes Q (plus for B-direction)

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5. Role of magnetic field

34

Atomic coherence|J

=1, m

=1>=

|1’>

|J=1

, m=0

>=|0

’>|J

=1, m

=-1>

=|-1

’>

|J=0

, m=0

>=|0

’,0’>

|R’>

|L’>

Un-

pola

rized

ligh

t fro

m a

sid

e

1/2

1/2

=

|J=1, m=1>=|1> |J=1, m=0>=|0> |J=1, m=-1>=|-1>

|J=0, m=0>=|0,0>

1/4 1/4

Un-

pola

rized

ligh

t fro

m a

sid

e|L

’>|R

’>

1/2

quantization axis

quantization axis

|1’> and |-1’> are not coherent, namely <-1’|1’>=0

|1> , |0>, and |-1’> are coherent, simply due to |1> = (1+cosθ)/2|1’> + (1-cosθ)/2|-1’> |0> = -sinθ/√2 |1’> + sinθ/√2 |-1’> |-1> = (1-cosθ)/2|1’> + (1+cosθ)/2|-1’> and thus eg. <0|1>≠0

35

Creation and destruction of atomic coherence

Line of sightin general not in this plane(not used in This section)

the most Important angle

Quantization axisnormal to photosphere|1>,|0>,|-1>

Quantization axisalong B|1’>,|0’>,|-1’>

Why horizontal field?We need B inclined to the symmetry axis of the pumping radiation field to redistribute the anisotropic population.

sym

met

ry a

xis

of

pum

ping

rad

iatio

n fie

ld

No coherence

Strong coherence

Relaxation of coherence(Hanle effect)

If B is strong (2πνLgJ>>Alu)

36

Multiple roles of magnetic field!

37

First quantization axisalways start with symmetry

axis of radiation field

Have to move to second quantization axis parallel to B

to calculate radiation field

Nothing to do withmagnetic field B

First role of B is to Change the axis

Strong B field removes the coherence of the base states

Hanle regimeSecond role of B

Third quantization axis Is the direction of emitted

photon

Multipole components of density matrix ρQ

K

• is the linear superposition of density matrix.– Total population √(2J+1) ρ0

0 ↔Stokes I

– Population imbalance ρ0K

• ρ02 (Ju=1)=(N1-2N0+N-1)/√6 (Alignment coefficient)– ρQ

2 :Stokes Q and U

• ρ01(Ju=1)=(N1-N-1)/√2 (orientation coefficient)– ρQ

1: Stokes V

– ρQK (Q ≠0) =complex numbers given by linear combinations

of the coherences between Zeeman sublevels whose quantum numbers differ by Q

• ρ22 (J=1) = ρ(1,-1)

39

6. Hanle effect

40

Hanle effect

• Relaxation (disappearance) of atomic coherences for increasing magnetic field strength

• ρQK (Ju)=1/(1+iQΓu) ρQ

K (Ju, B=0)– Γu=(Zeeman separation for B)/(natural width) =(2πνLgJ=8.79x106BHgJ)/(1/tlife=Alu)

– Quantization axis for ρQK (Ju, B=0) is B.

– Population imbalance ρ0K not affected by B

– ρQK (Q ≠0) reduced and dephased

41

Hanle conditions

• The natural line width of the spectral line (in the rest frame of the atom) is proportional to A (Einstein’s A coefficient)

• If the Zeeman splitting B is much larger than the natural line width of the spectral line (B >> A)

• , then there is no coherency between the magnetic substates– For forbidden transition e.g., He I 1083.0 nm blue wing, Fe XIII 1074.7

nm, Si IX 3934.6 nm • A ≈ 101 to 102/sec• B0 ~ mG satisfies the strong field condition.

– For permitted lines e.g., He I 1083.0 nm red wing, O VI 103.2 nm • A ≈ 106 to 108/sec• B0 ~ 10 – 100 G, depending on the spectral line

42

B

A

In the case of He10830,

• 0-1 atom (red wing)– B >> A10

– Saturated B=10-100G: except for sunspots, in saturated.

• 1-0 atom (blue wing)– B >> B10J0

0(01– B10J0

0(01 A10≈10-4

– Saturated B=1mG: completely saturated in any case!

43

• This equation is the Hanle effect, which is the decrease of coherence among coherent states (magnetic sublevels) due to change in the quantization axis.

• This coherency disappearance with strong magnetic field is an unexplained issue in this handout. (I do not understand this yet!)

44

ρQK (Ju)=1/(1+iQΓu) ρQ

K (Ju, B=0)

Properties of the Hanle regime

• In the solar field strengths, we are most probably in the Hanle regime ie saturated regime, where coherence due to the changes in the quantization axis from the symmetric axis of radiation field.

• Emitted radiation only depends on the population imbalance as represented by the density matrix ρ0

2

(Ju=1)• This also means that the linear polarization signal

does not have any sensitivity to the strength of magnetic fields.

45

In He-10830 saturated regime, Stokes profiledetermines only thecone angle θB and azimuth ΦB

(Casini and Landi Degl’Innocenti)

More concretely,

46

To summarize….• Circular Polarization

– B – line-of-sight magnetic field strength…with an alignment effect correction

• Linear Polarization– –Azimuth direction of B

Direction of B projected in the plane of the sky containing sun center.

– No sensitivity to |B |– the van Vleck effect

90 degree ambiguity in the azimuth direction of B, depending on

x

y

zline of s

ight dire

ction

BB

B

Chart taken from H. Lin presentation for SOLAR-C meeting

In 10803 red wing Hanle (saturated) regime

(Casini and Landi Degl’Innocenti)

cos2(ΦB+90)=-cos2 ΦB

sin 2(ΦB+90)=-sin2 ΦB

σ 02 (Ju=1) ≡ρ0

2 (Ju=1) see section on multipole component

Van-Vleck 90 degree ambiguity!

48

Chapter 6 Final comment: In my opinion,

• Population imbalance ρ02 (Ju=1), in other

word, cone angle θB and azimuth ΦB are coupled. This is the van Vleck ambiguity in the Hanle regime. Because of this,

• We should not jump to the inversion routine, instead try to manually obtain the candidate solutions by looking at Stokes profiles, and then go to the inversion routine.

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To continue

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Elementary course on atomic polarization and Hanle effect Rev. 1.2: 13 April 2009 Saku Tsuneta...

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Transcript of Elementary course on atomic polarization and Hanle effect Rev. 1.2: 13 April 2009 Saku Tsuneta...