Elementary course on atomic
polarization and Hanle effect
Rev. 1.2: 13 April 2009
Saku Tsuneta
(NAOJ)
1
Table of contents
1.
2.
3.
4.
5.
6.
7.
Quantum states
Atomic polarization and quantization axis
van Vleck angle
He10830
Role of magnetic field
Hanle effect
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=1
7
If magnetic field B →Zero
Quantization axis = direction of B
No polarization with B→0
|J=1, m=1>
|J=1, m=0>
|J=1, m=-1>
σ+
Observer
σ-
Observer
|J=0, m=0>
π
σ- σ+
No polarization only when different |m> state
has the same population!
Polarization remains if with population imbalance
8
Emission from 3-level atom
with magnetic field
|J=1, m=1>
Zeeman splitting
Quantization axis = direction of B
|J=1, m=0>
σ+
Observer
|J=1, m=-1>
B
σπ
B
Observer
σ-
σ+
|J=0, m=0>
9
2. Atomic polarization and
quantization axis
10
Atomic polarization is merely
conservation of angular momentum
Example #1; 1-0 system
|J=0, m=0>=|1,0>
1/2
|J=1, m=1>=|1>
1/2
|J=1, m=0>=|0>
Take quantization axis to be
direction of Incident photons
|J=1, m=-1>=|-1> |1>, |0>,|-1>
B=0: degenerated state
|L>
|R>
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>
11
causes transition to m=-1 state of atom (|-1> to |0>).
Atomic polarization is merely
conservation of angular momentum
Example #2; 0-1 system
|J=1, m=1>=|1>
|J=1, m=0>=|0>
1/2
|J=1, m=-1>=|-1>
|1>, |0>,|-1>
B=0: degenerated state
1/2
Take quantization axis to be
Direction of Incident photons
|J=0, m=0>=|0,0>
|R>
|L>
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>
12
causes transition to m=-1 state of atom (|0> to |-1>).
1/2
|R’>
1/2
Un-polarized light from a side
|L’>
Take quantization axis to be
direction of Incident photons
|J=0, m=0>=|0’,0’>
|J=1, m=-1>=|-1’>
|J=1, m=1>=|1’> |J=1, m=0>=|0’>
If unpolarized light comes in
from horizontal direction,
Exactly the same atomic polarization
take place but in the different set of quantum
base 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’>.
13
What is the relation between
|Jm> and |Jm’> base states?
Rotation matrix for spin 1
(See Feynman section 17.5)
Normal to stellar surface
|1>,|0>,|-1>
|1’>,|0’>,|1’>
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
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>=1/4
If θ is 0 degree,
<1|1>=1/2
<0|0>=0
<-1|-1>=1/2
Thus, illumination from side provides different atomic polarization!
14
Rotation matrix for spin 1
some analogy..
Base states of the old frame
z
Ry(θ)
z’
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
Base states of the new frame
y
y
y’
|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’
Rotation matrix for 2D space
θ
x
x
y
X’
cosθ
sinθ
y’
-sinθ
cosθ
15
|J=1, m=1>=|1’> |J=1, m=0>=|0’>
1/2
1/2
|L’>
|R’>
Un-polarized light from a side
|J=0, m=0>=|0’,0’>
|J=1, m=-1>=|-1’>
|J=1, m=1>=|1> |J=1, m=0>=|0> |J=1, m=-1>=|-1>
1/4
=
1/2
1/4
|J=0, m=0>=|0,0>
quantization axis
|R’>
|L’>
Un-polarized light from a side
This means that
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
Uniform radiation case
|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’>
sum over 0<θ< π (dΩ=2πsinθ/4 π)
<1|1> = ∫ (1+cosθ)²/8 + (1-cosθ)²/8 dΩ = 1/3
<0|0> = ∫ sin²θ/4 + sin²θ/4 dΩ =1/3
<-1|-1> = ∫ (1-cosθ)²/8 + (1+cosθ)²/8 dΩ=1/3
Thus, uniform irradiation results in no atomic
polarization!
18
3. van Vleck angle
19
van Vleck angle concept is easy to
understand 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 sight
in general not
in this plane
(not used in
This section)
symmetry axis of
pumping radiation field
Quantization axis
normal to photosphere
|1>,|0>,|-1>
Quantization axis
along B
|1’>,|0’>,|-1’>
θ
Why horizontal field?
We need B inclined to the
symmetry axis of the pumping
radiation field to redistribute the
anisotropic population.
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:
π
• LP ≈ <1’|1’>+<-1’|-1’>-2<0’|0’>
B
2
2
2
= (1+cosθ) /4+ (1-cosθ) /4 - sin θ/2
σObserver
= (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+
The van Vleck Effect results in
 > vv
Linear polarization direction
 < vv
• van Vleck Angle
vv = 54.7 deg
–  < vv , then LP // B
–  > vv , then LP  B
Figure taken from H. Lin presentation for SOLAR-C meeting
4. He10830
24
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
Dark filament
No Stokes-V
No Stokes-LP Can not exist without B due to symmetry
(LP exists with horizontal B) Hanle effect!
|J=1, m=1>=|1>
|J=1, m=0>=|0>
|J=1, m=-1>=|-1>
Incoherent states(ie <1|-1>=0)
1/3
1/3
1/2
1/2
1/3
Prominence
No Stokes
Stokes LP even with zero B
|J=0, m=0>=|0,0>
|R>
|L>
Unpolarised light from a star
LP = Linear Polarization
26
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
What is the polarization state of a
photon emitted at any angle θ?
Easy-to-understand case!
Z
Z’
θ
|L>
?
X
Spin -1 atom
|atom, -1>
Spin 0 atom
|atom, 0>
|L>
Emission of circular
polarized light
Spin 1 atom
|atom, -1>
Spin 0 atom
|atom, 0>
Procedure
Step 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 momentum
Conservation around Z’ axis!
28
Importance of Quantization axis
Why 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>, xlinearly 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> , xlinearly 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 wings
Dark filament
No Stokes-V
No Stokes-LP Can not exist without B due to symmetry
(LP exists with horizontal B) Hanle effect!
|J=0, m=0>=|1,0>
1/2
1/2
1/3
1/3
1/3
|J=1, m=-1>=|-1>
|J=1, m=1>=|1> |J=1, m=0>=|0>
Prominence
No Stokes-V
No Stokes-LP
(even with B)
|L>
|R>
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
LP>0
LP>0
– J(low)=1 J(up)=1
• Red2 10830.34A
– J(low)=1 J(up)=2
• LP=Stokes Q (plus for B-direction)
33
5. Role of magnetic field
34
1/2
|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
|R’>
|L’>
Un-polarized light from a side
1/2
|L’>
|R’>
Un-polarized light from a side
|1’> and |-1’> are not
coherent, namely
<-1’|1’>=0
|J=0, m=0>=|0’,0’>
|J=1, m=1>=|1’> |J=1, m=0>=|0’>
|J=1, m=-1>=|-1’>
Atomic coherence
|J=1, m=1>=|1> |J=1, m=0>=|0> |J=1, m=-1>=|-1>
=
1/4
1/2
1/4
|J=0, m=0>=|0,0>
quantization axis
quantization axis
35
Creation and destruction
of atomic coherence
Quantization axis
normal to photosphere
|1>,|0>,|-1>
Line of sight
in general not
in this plane
(not used in
This section)
Why horizontal field?
We need B inclined to the
symmetry axis of the pumping
radiation field to redistribute the
anisotropic population.
symmetry axis of
pumping radiation field
No coherence
Quantization axis
along B
|1’>,|0’>,|-1’>
Strong coherence
If B is strong
(2πνLgJ>>Alu)
Relaxation of coherence
(Hanle effect)
the most
Important angle
36
Multiple roles of magnetic field!
First quantization axis
always start with symmetry
axis of radiation field
Nothing to do with
magnetic field B
Have to move to second
quantization axis parallel to B
to calculate radiation field
First role of B is to
Change the axis
Strong B field removes the
coherence of the base states
Third quantization axis
Is the direction of emitted
photon
Hanle regime
Second role of B
37
Density matrix
and atomic coherence
• Atomic (quantum) coherence is non-diagonal elements <m|ρ|m’> of
atomic density matrix ρ = ∑ |m> Pm <m|, where Pm is the probability of
having |m>, not the amplitude of |m>!
• If we have complete quantum mechanical description on the whole
system namely atoms and radiation field |radiation field, atom>, we will
not need the density matrix.
• But, if the radiation field |radiation field> and atoms |atom> are
separately treated, information on the population in the state |atom, m>
is represented by the probability Pm.
• If atomic coherence is zero, then <n| ρ|n>=Pn, <n| ρ|m>=0 (n≠m).
• Atomic coherence is non zero if <m| and |m’> are not orthogonal. Then, ,
<n| ρ|m> ≠ 0 (n≠m).
38
Multipole components
of density matrix ρQK
• is the linear superposition of density matrix.
– Total population √(2J+1) ρ00 ↔Stokes I
– Population imbalance ρ0K
• ρ02 (Ju=1)=(N1-2N0+N-1)/√6 (Alignment coefficient)
– ρQ2 :Stokes Q and U
• ρ01(Ju=1)=(N1-N-1)/√2 (orientation coefficient)
– ρQ1: 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) ρQK (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
w
Hanle conditions
B
A
• 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 wB is much larger than the natural line
width of the spectral line (wB >> 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
In the case of He10830,
• 0-1 atom (red wing)
– wB >> A10
– Saturated B=10-100G: except for sunspots, in
saturated.
• 1-0 atom (blue wing)
– wB >> B10J00(w01)
– B10J00(w01)/ A10≈10-4
– Saturated B=1mG: completely saturated in any
case!
43
K
ρQ
(Ju)=1/(1+iQΓu)
K
ρQ
(Ju, B=0)
• 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
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 ρ02
(Ju=1)
• This also means that the linear polarization signal
does not have any sensitivity to the strength of
magnetic fields.
45
More concretely,
In He-10830 saturated
regime, Stokes profile
determines only the
cone angle θB and
azimuth ΦB
(Casini and Landi Degl’Innocenti)
46
To summarize….
y
• 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.
z
– No sensitivity to |B |
– the van Vleck effect
90 degree ambiguity in the
azimuth direction of B,
depending on 
B


B

B
x
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
Van-Vleck 90 degree ambiguity!
sin 2(ΦB+90)=-sin2 ΦB
σ 02 (Ju=1) ≡ρ02 (Ju=1) see section on multipole component
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.
49
To continue
50