Elementary course on atomic polarization and Hanle effect

1
Elementary course on atomic polarization and
Hanle effect

• Rev. 1.2 13 April 2009
• Saku Tsuneta
• (NAOJ)

2
Table of contents

• Quantum states
• Atomic polarization and quantization axis
• van Vleck angle
• He10830
• Role of magnetic field
• Hanle effect
• Formal treatment (in preparation)

3
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

4
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 ?0
• 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

5
1. Quantum states
6
A note on photon polarization

• Right-circularly polarized photon Rgt (spin 1)
• Left-circularly polarized photon Lgt (spin -1)
• defined in terms of rotation axis along direction
  of motion
• Linearly polarized photon is a coherent
  superposition of Rgt and Lgt
• xgt 1/v2 (Rgt Lgt),
• ygt -i/v2 (Rgt - Lgt),
• Or, equivalently
• Rgt 1/v2 (xgt iygt)
• Lgt 1/v2 (xgt - iygt)
• It is not zero angular momentum. It does not have
  a definite angular momentum.

7
Base states in energy and in angular momentum

• If B 0, base states representing magnetic
  sublevels m for J1, 1gt,0gt, and -1gt are
  degenerated states in energy H
• H1gtE1gt
• H0gtE0gt
• H-1gtE-1gt
• But, these base states ,1gt,0gt, and -1gt are
  also eigenstates for angular momentum, and are
  not degenerated in angular momentum JzM
• Jz1gt11gt
• Jz0gt00gt
• Jz-1gt-1-1gt
• Throughout this handout, ?1

8
If magnetic field B ?Zero
Quantization axis direction of B
No polarization with B?0
J1, m1gt
J1, m0gt
s
Observer
J1, m-1gt
s-
p
s-
s
Observer
No polarization only when different mgt state has
the same population! Polarization remains if with
population imbalance
J0, m0gt
9
Emission from 3-level atomwith magnetic field
Zeeman splitting Quantization axis direction of
B
J1, m1gt
J1, m0gt
s
Observer
J1, m-1gt
B
s-
p
B
s-
s
Observer
J0, m0gt
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2. Atomic polarization and quantization axis
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Atomic polarization is merely conservation of
angular momentum Example 1 1-0 system
J0, m0gt1,0gt
Take quantization axis to be direction of
Incident photons
1/2
1/2
J1, m-1gt-1gt
J1, m1gt1gt
J1, m0gt0gt
1gt, 0gt,-1gt B0 degenerated state
Unpolarized light from a star
Rgt
Lgt
A right-circularized photon carrying angular
momentum -1 Lgt causes transition to m1 state
of atom (1gt to 0gt). A left-circularized photon
carrying angular momentum 1 Rgt causes transition
to m-1 state of atom (-1gt to 0gt).
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Atomic polarization is merely conservation of
angular momentumExample 2 0-1 system
J1, m-1gt-1gt
J1, m1gt1gt
J1, m0gt0gt
1gt, 0gt,-1gt B0 degenerated state
1/2
1/2
Take quantization axis to be Direction of
Incident photons
J0, m0gt0,0gt
Unpolarized light from a star
Lgt
Rgt
A right-circularized photon carrying angular
momentum 1 Rgt causes transition to m1 state
of atom (0gt to 1gt). A left-circularized photon
carrying angular momentum -1 Lgt causes
transition to m-1 state of atom (0gt to -1gt).
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If unpolarized light comes in from horizontal
direction,
Take quantization axis to be direction of
Incident photons
J1, m-1gt-1gt
Lgt
Exactly the same atomic polarization take place
but in the different set of quantum base states
-1gt, 0gt,1gt Note that 1gt and 1gt are
different quantum states. For instance 1gt is
represented by linear superposition of 1gt,
0gt and -1gt.
1/2
J1, m0gt0gt
Un-polarized light from a side
J0, m0gt0,0gt
1/2
Rgt
J1, m1gt1gt
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What is the relation betweenJmgt and Jmgt base
states?
Rotation matrix for spin 1 (See Feynman section
17.5)
Normal to stellar surface
1gt,0gt,-1gt
1gt,0gt,1gt
?
If ? is 90 degree, 1gt (1cos?)/21gt
(1-cos?)/2-1gt 1/2 (1gt-1gt )
lt11gt1/4 0gt sin?/v2 1gt - sin?/v2
-1gt 1/v2 (1gt - -1gt) ) lt00gt1/2 -1gt
(1-cos?)/21gt (1cos?)/2-1gt 1/2
(1gt-1gt ) ) lt-1-1gt1/4 If ? is 0 degree,
lt11gt1/2 lt00gt0 lt-1-1gt1/2 Thus,
illumination from side provides different atomic
polarization!
15
Rotation matrix for spin 1some analogy..
Base states of the old frame
z
z
Ry(?)
?
Base states of the new frame
y
y
y
x
Rotation matrix for 2D space
?
x
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This means that
J1, m-1gt-1gt
J1, m-1gt-1gt
J1, m1gt1gt
J1, m0gt0gt
Lgt
1/4
1/4
1/2
1/2
Lgt
Un-polarized light from a side

J1, m0gt0gt
Un-polarized light from a side
J0, m0gt0,0gt
Rgt
J0, m0gt0,0gt
1/2
Rgt
J1, m1gt1gt
quantization axis
quantization axis
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Generation of coherence due to rotation

• In the previous page, 1gt and -1gt are not
  coherent due to non-coherent illumination, namely
  lt-11gt0
• But, in the new base states1gt and -1gt have
  coherency. If ? is 90 degree,
• 1gt 1/2 (1gt-1gt ) lt11gt1/4
• 0gt 1/v2 (1gt - -1gt) ) lt00gt1/2
• -1gt 1/2 (1gt-1gt ) ) lt-1-1gt1/4
• for instance lt-11gt1/2 not zero!
• Thus, rotation can introduce coherency in quantum
  states!

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Uniform radiation case
1gt (1cos?)/21gt (1-cos?)/2-1gt 0gt
sin?/v2 1gt - sin?/v2 -1gt -1gt
(1-cos?)/21gt (1cos?)/2-1gt sum over
0lt?lt p (dO2psin?/4 p) lt11gt ? (1cos?)²/8
(1-cos?)²/8 dO 1/3 lt00gt ? sin²?/4
sin²?/4 dO 1/3 lt-1-1gt ? (1-cos?)²/8
(1cos?)²/8 dO1/3 Thus, uniform irradiation
results in no atomic polarization!
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3. van Vleck angle
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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.

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Two quantization axes
Quantization axis normal to photosphere 1gt,0gt,-
1gt
Quantization axis along B 1gt,0gt,-1gt
Line of sight in general not in this plane (not
used in This section)
symmetry axis of pumping radiation field
Why horizontal field? We need B inclined to the
symmetry axis of the pumping radiation field to
redistribute the anisotropic population.
?
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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 lt11gtlt-1-1gt-2lt00gt
• (1cos?) 2/4 (1-cos?) 2/4 - sin2?/2
• (3cos2? -1)/4
• (use rotation matrix for spin1 in
  earlier page)
• Thus, LP changes sign at 3cos2?vv -10, i.e. ?vv
  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.

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The van Vleck Effect results in
Linear polarization direction

• van Vleck Angle
• ?vv 54.7 deg
• ? lt ?vv , then LP // B
• ? gt ?vv , then LP ? B

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

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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!
J1, m-1gt-1gt
J1, m1gt1gt
J1, m0gt0gt
Incoherent states(ie lt1-1gt0)
1/3
1/2
1/2
1/3
1/3
Prominence No Stokes Stokes LP even with zero B
J0, m0gt0,0gt
Lgt
Rgt
Unpolarised light from a star
LP Linear Polarization
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To understand Linear polarization from
prominence with zero horizontal B,

• 1gt state is created by absorption of an Lgt
  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 1,1gt to 0,0gt transition, a photon with
  state ½Rgt ½ Lgt is emitted (90 degree
  scattering).
• This is a linearly polarized photon with state
  xgt 1/v2 (Rgt Lgt) !
• Likewise, for -1gt state, -xgt -1/v2 (Rgt
  Lgt)

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What is the polarization state of a photon
emitted at any angle ??
Easy-to-understand case!
Z
Z
?
?
Lgt
X
Spin -1 atom atom, -1gt
Spin 0 atom atom, 0gt
Spin 1 atom atom, -1gt
Spin 0 atom atom, 0gt
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!
Lgt
Emission of circular polarized light
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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.

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s transition 1,1gt to 0,0gt and 1,-1gt to 0,0gt

• 1gt (1cos?)/21gt sin?/v20gt(1-cos?)/2-1gt
• -1gt (1-cos?)/21gt -sin?/v20gt(1cos?)/2-1gt
  
• lt11gt (1cos?)/21/2 probability to emit Rgt
  photon
• lt1-1gt (1-cos?)/21/2 probability to emit Lgt
  photon
• Thus, for transition J, mgt1, 1gt to 0,0gt with
  ?90 degree, a photon with state (1gt-1gt)/2
  (RgtLgt)/2 xgt, x-linearly polarized light!
• For J, mgt1, -1gt in z-coordinate
• lt-11gt (1-cos?)/21/2 probability to emit Rgt
  photon
• lt-1-1gt (1cos?)/21/2 probability to emit Lgt
  photon
• Thus, for transition J, mgt1, -1gt to 0,0gt with
  ?90 degree, a photon with state (1gt-1gt)/2
  (RgtLgt)/2 xgt , x-linearly polarized light!

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p transition 1,0gt to 0,0gt

• 1gt (1cos?)/21gt sin?/v20gt(1-cos?)/2-1gt
• -1gt (1-cos?)/21gt -sin?/v20gt(1cos?)/2-1gt
  
• lt01gt sin?/v2 1/ v 2 probability to emit Rgt
  photon
• lt0-1gt -sin?/v2 -1/ v 2 probability to emit
  Lgt photon
• Thus, for transition J, mgt1, 0gt to 0,0gt with
  ?90 degree, a photon with state
• (1gt--1gt) Rgt-Lgt v 2iygt, y-linearly
  polarized light!

32
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!
J0, m0gt1,0gt
1/2
1/2
1/3
1/3
1/3
J1, m-1gt-1gt
J1, m1gt1gt
J1, m0gt0gt
Prominence No Stokes-V No Stokes-LP (even with B)
Rgt
Lgt
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He 10830 with horizontal B

• prominence filament
• Blue 10829.09A no LP LPlt0
• J(low)1 J(up)0
• Red1 10830.25A LPgt0 LPgt0
• J(low)1 J(up)1
• Red2 10830.34A LPgt0 LPgt0
• J(low)1 J(up)2
• LPStokes Q (plus for B-direction)

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5. Role of magnetic field
35
Atomic coherence
1gt , 0gt, and -1gt are coherent, simply due to
1gt (1cos?)/21gt (1-cos?)/2-1gt 0gt
-sin?/v2 1gt sin?/v2 -1gt -1gt
(1-cos?)/21gt (1cos?)/2-1gt and thus eg.
lt01gt?0
1gt and -1gt are not coherent, namely
lt-11gt0
J1, m-1gt-1gt
J1, m-1gt-1gt
J1, m1gt1gt
J1, m0gt0gt
Lgt
1/2

1/4
1/4
1/2
Lgt
Un-polarized light from a side
J1, m0gt0gt
Un-polarized light from a side
J0, m0gt0,0gt
Rgt
J0, m0gt0,0gt
1/2
Rgt
J1, m1gt1gt
quantization axis
quantization axis
36
Creation and destruction of atomic coherence
Quantization axis normal to photosphere 1gt,0gt,-
1gt
No coherence
Quantization axis along B 1gt,0gt,-1gt
Line of sight in general not in this plane (not
used in This section)
symmetry axis of pumping radiation field
Strong coherence
If B is strong (2p?LgJgtgtAlu)
Why horizontal field? We need B inclined to the
symmetry axis of the pumping radiation field to
redistribute the anisotropic population.
Relaxation of coherence (Hanle effect)
the most Important angle
37
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
Hanle regime Second role of B
Third quantization axis Is the direction of
emitted photon
38
Density matrix and atomic coherence

• Atomic (quantum) coherence is non-diagonal
  elements ltm?mgt of atomic density matrix ? ?
  mgt Pm ltm, where Pm is the probability of
  having mgt, not the amplitude of mgt!
• If we have complete quantum mechanical
  description on the whole system namely atoms and
  radiation field radiation field, atomgt, we will
  not need the density matrix.
• But, if the radiation field radiation fieldgt and
  atoms atomgt are separately treated, information
  on the population in the state atom, mgt is
  represented by the probability Pm.
• If atomic coherence is zero, then ltn ?ngtPn,
  ltn ?mgt0 (n?m).
• Atomic coherence is non zero if ltm and mgt are
  not orthogonal. Then, , ltn ?mgt ? 0 (n?m).

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Multipole components of density matrix ?QK

• is the linear superposition of density matrix.
• Total population v(2J1) ?00 ?Stokes I
• Population imbalance ?0K
• ?02 (Ju1)(N1-2N0N-1)/v6 (Alignment
  coefficient)
• ?Q2 Stokes Q and U
• ?01(Ju1)(N1-N-1)/v2 (orientation coefficient)
• ?Q1 Stokes V
• ?QK (Q ?0) complex numbers given by linear
  combinations of the coherences between Zeeman
  sublevels whose quantum numbers differ by Q
• ?22 (J1) ?(1,-1)

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6. Hanle effect
41
Hanle effect

• Relaxation (disappearance) of atomic coherences
  for increasing magnetic field strength
• ?QK (Ju)1/(1iQGu) ?QK (Ju, B0)
• Gu(Zeeman separation for B)/(natural width)
• (2p?LgJ8.79x106BHgJ)/(1/tlifeAlu)
• Quantization axis for ?QK (Ju, B0) is B.
• Population imbalance ?0K not affected by B
• ?QK (Q ?0) reduced and dephased

42
Hanle conditions

• The natural line width of the spectral line (in
  the rest frame of the atom) is proportional to A
  (Einsteins A coefficient)
• If the Zeeman splitting wB is much larger than
  the natural line width of the spectral line (wB
  gtgt 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

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In the case of He10830,

• 0-1 atom (red wing)
• wB gtgt A10
• Saturated B10-100G except for sunspots, in
  saturated.
• 1-0 atom (blue wing)
• wB gtgt B10J00(w01)
• B10J00(w01)/ A1010-4
• Saturated B1mG completely saturated in any
  case!

44
?QK (Ju)1/(1iQGu) ?QK (Ju, B0)

• 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!)

45
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
  ?02 (Ju1)
• This also means that the linear polarization
  signal does not have any sensitivity to the
  strength of magnetic fields.

46
More concretely,
In He-10830 saturated regime, Stokes
profile determines only the cone angle ?B and
azimuth FB
(Casini and Landi DeglInnocenti)
47
To summarize.

• Circular Polarization
• B?? line-of-sight magnetic field strengthwith
  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 ?

Chart taken from H. Lin presentation for
SOLAR-C meeting
48
In 10803 red wing Hanle (saturated) regime
(Casini and Landi DeglInnocenti)
cos2(FB90)-cos2 FB sin 2(FB90)-sin2 FB s 02
(Ju1) ?02 (Ju1) see section on multipole
component
Van-Vleck 90 degree ambiguity!
49
Chapter 6 Final comment In my opinion,

• Population imbalance ?02 (Ju1), in other word,
  cone angle ?B and azimuth FB 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.

50
To continue