1. Elementary course on atomic polarization and Hanle effect Rev. 1.2: 13 April 2009 Saku Tsuneta (NAOJ) 1

2. Table of contents 1.Quantum states 2.Atomic polarization and quantization axis 3.van Vleck angle 4.He10830 5.Role of magnetic field 6.Hanle effect 7.Formal treatment (in preparation) 2

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 3

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 Λ 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 4

5. 1. Quantum states 5

6. 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

7. 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

8. No polarization only when different |m> state has 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 Observer σ+σ+ σ-σ- σ+σ+ σ-σ- π No polarization with B→0 8

9. Emission from 3-level atom with magnetic field |J=1, m=1> |J=1, m=0> |J=1, m=-1> |J=0, m=0> Zeeman splitting Quantization axis = direction of B Observer σ+σ+ σ-σ- σ+σ+ σ-σ- π B B 9

10. 2. Atomic polarization and quantization axis 10

11. 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 11

12. 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 12

13. 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-polarized light from a side 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’>. 1/2 13

14. 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/√2cosθ-sinθ/√2 <-1|(1-cosθ)/2sinθ/√2(1+cosθ)/2 θ If θ is 90 degree, |1> = (1+cosθ)/2|1’> + (1-cosθ)/2|-1’> = 1/2 (|1’>+|-1’> ) =1/4 |0> = sinθ/√2 |1’> - sinθ/√2 |-1’> = 1/√2 (|1’> - |-1’>) ) =1/2 |-1> = (1-cosθ)/2|1’> + (1+cosθ)/2|-1’> = 1/2 (|1’>+|-1’> ) ) =1/4 If θ is 0 degree, =1/2 =0 =1/2 Thus, illumination from side provides different atomic polarization! Rotation matrix for spin 1 (See Feynman section 17.5) Normal to stellar surface 14

15. Rotation matrix for spin 1 some analogy.. 15 |1>|0>|-1> <1’|(1+cosθ)/2sinθ/√2(1-cosθ)/2 <0’|-sinθ/√2cosθsinθ/√2 <-1’|(1-cosθ)/2-sinθ/√2(1+cosθ)/2 x x’ y y’ θ xy X’cosθsinθ y’-sinθcosθ 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/√2cosθ-sinθ/√2 <-1|(1-cosθ)/2sinθ/√2(1+cosθ)/2

16. 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-polarized light from a side 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 Un-polarized light from a side |L’> |R’> 1/2 quantization axis 16

17. Generation of coherence due to rotation In the previous page, |1’> and |-1’> are not coherent due to non-coherent illumination, namely =0 But, in the new base states|1> and |-1> have coherency. If θ is 90 degree, – |1> = 1/2 (|1’>+|-1’> ) =1/4 – |0> = 1/√2 (|1’> - |-1’>) ) =1/2 – |-1> = 1/2 (|1’>+|-1’> ) ) =1/4 for instance =1/2 not zero! Thus, rotation can introduce coherency in quantum states! 17

18. 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+cosθ)²/8 + (1-cosθ)²/8 dΩ = 1/3 = ∫ sin²θ/4 + sin²θ/4 dΩ =1/3 = ∫ (1-cosθ)²/8 + (1+cosθ)²/8 dΩ=1/3 Thus, uniform irradiation results in no atomic polarization! 18

19. 3. van Vleck angle 19

20. 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

21. Two quantization axes Line of sight in general not in this plane (not used in This section) θ Quantization axis normal to photosphere |1>,|0>,|-1> Quantization axis along 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. symmetry axis of pumping radiation field 21

22. 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 ≈ + -2 = (1+cosθ) 2 /4+ (1-cosθ) 2 /4 - sin 2 θ/2 = (3cos 2 θ -1)/4 (use ‘’rotation matrix for spin1’’ in earlier page) Thus, LP changes sign at 3cos 2 θvv -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+ σ-σ- π Observer B

23. van Vleck Angle  vv = 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

24. 4. He10830 24

25. 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

26. 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 Dark filament No Stokes-V No Stokes-LP Can not exist without B due to symmetry (LP exists with horizontal B) Hanle effect! Prominence No Stokes Stokes LP even with zero B 1/3 Incoherent states(ie =0) LP = Linear Polarization 26

27. 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. 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> Spin 1 atom |atom, -1> Spin 0 atom |atom, 0> Emission of circular polarized light θ Z Z’ ? 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! Easy-to-understand case! X

29. 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

30. 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+cosθ)/2=1/2 probability to emit |R> photon – = (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-cosθ)/2=1/2 probability to emit |R> photon – = (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

31. 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> – = sinθ/√2 = 1/ √ 2 probability to emit |R> photon – = -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

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! Prominence No Stokes-V No 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 1/3 32

33. 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) 33

34. 5. Role of magnetic field 34

35. 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-polarized light from a side 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 Un-polarized light from a side |L’> |R’> 1/2 quantization axis |1’> and |-1’> are not coherent, namely =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 35

36. Creation and destruction of atomic coherence Line of sight in general not in this plane (not used in This section) the most Important angle Quantization axis normal to photosphere |1>,|0>,|-1> Quantization axis along 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. symmetry axis of pumping radiation field No coherence Strong coherence Relaxation of coherence (Hanle effect) If B is strong (2πν L g J >>A lu ) 36

37. Multiple roles of magnetic field! 37 First quantization axis always start with symmetry axis of radiation field Have to move to second quantization axis parallel to B to calculate radiation field Nothing to do with magnetic field B 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 of atomic density matrix ρ = ∑ |m> Pm, 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 =Pn, =0 (n≠m). Atomic coherence is non zero if are not orthogonal. Then,, ≠ 0 (n≠m). 38

39. Multipole components of density matrix ρ Q K is the linear superposition of density matrix. – Total population √(2J+1) ρ 0 0 ↔Stokes I – Population imbalance ρ 0 K ρ 0 2 (Ju=1)=(N 1 -2N 0 +N -1 )/√6 (Alignment coefficient) – ρ Q 2 :Stokes Q and U ρ 0 1 (Ju=1)=(N 1 -N -1 )/√2 (orientation coefficient) – ρ Q 1 : Stokes V – ρ Q K (Q ≠0) =complex numbers given by linear combinations of the coherences between Zeeman sublevels whose quantum numbers differ by Q ρ 2 2 (J=1) = ρ(1,-1) 39

40. 6. Hanle effect 40

41. Hanle effect Relaxation (disappearance) of atomic coherences for increasing magnetic field strength ρ Q K (Ju)=1/(1+iQΓu) ρ Q K (Ju, B=0) – Γu=(Zeeman separation for B)/(natural width) =(2πν L g J =8.79x106B H g J )/(1/t life =A lu ) – Quantization axis for ρ Q K (Ju, B=0) is B. – Population imbalance ρ 0 K not affected by B – ρ Q K (Q ≠0) reduced and dephased 41

42. 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 ≈ 10 1 to 10 2 /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 ≈ 10 6 to 10 8 /sec B0 ~ 10 – 100 G, depending on the spectral line 42 BB A

43. In the case of He10830, 0-1 atom (red wing) –  B >> A 10 – Saturated B=10-100G: except for sunspots, in saturated. 1-0 atom (blue wing) –  B >> B 10 J 0 0 (  01  – B 10 J 0 0 (  01  A 10 ≈10 -4 – Saturated B=1mG: completely saturated in any case! 43

44. 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 ρ Q K (Ju)=1/(1+iQΓu) ρ Q K (Ju, B=0)

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 ρ 0 2 (Ju=1) This also means that the linear polarization signal does not have any sensitivity to the strength of magnetic fields. 45

46. In He-10830 saturated regime, Stokes profile determines only the cone angle θ B and azimuth Φ B (Casini and Landi Degl’Innocenti) More concretely, 46

47. 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 z line of sight direction B BB B     Chart taken from H. Lin presentation for SOLAR-C meeting

48. In 10803 red wing Hanle (saturated) regime (Casini and Landi Degl’Innocenti) cos2(Φ B +90)=-cos2 Φ B sin 2(Φ B +90)=-sin2 Φ B σ 0 2 (Ju=1) ≡ρ 0 2 (Ju=1) see section on multipole component Van-Vleck 90 degree ambiguity! 48

49. Chapter 6 Final comment: In my opinion, Population imbalance ρ 0 2 (Ju=1), in other word, cone angle θ B and azimuth Φ B 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

50. To continue 50