Elementary understanding on Hanle effect No.1 atomic polarization Rev. -2: 6 March 2009 Saku Tsuneta (NAOJ)

Table of contents 1.Atomic polarization (this handout) 2.Hanle effect (atomic polarization with B) 3.Van Vleck effect 4.More formal treatment 1.Density matrix approach

Hanle effect in general 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 the following excellent text: Javier Trujillo Bueno, Atomic Polarization and the Hanle Effect, AIP conference series volume 236, , Please point out any incorrect description for better understanding! The outstanding textbooks for basic quantum mechanics are R. P. Fynman, Lectures on physics: Quantum Mechanics J. J. Sakurai, Modern Quantum Mechanics

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 quantanization 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

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 quantanization 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

If unpolarized light comes in from horizontal direction, Take quantanization 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’>.

What is the relation between |Jm> and |J’m’> base states? |1>,|0>,|-1> |1’>,|0’>,|1’> |1’>|0’>|-1’> <1|(1+cosθ)/2sinθ/√2(1-cosθ)/2 <0|-sinθ/√2cosθsinθ/√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’> ) =1/4 |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 90 degree, |1> = 1/ √2 =1/2 |0> = 0 =0 |-1> = 1/ √2 =1/2 Thus, illumination from side provides different atomic polarization! Rotation matrix for spin 1 (any text book in quantum mechanics) Normal to stellar surface

This mean 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 Quantanization axis

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!

He Blue A – J(low)=1 J(up)=0 Red A – J(low)=1 J(up)=1 Red A – J(low)=1 J(up)=2

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 (LP exists with horizontal B) Hanle effect! Can not exist without B due to symmetry Prominence No Stokes Stokes LP even with zero B 1/3 Incoherent states

To understand LP 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>)

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

He with horizontal B prominence filament Blue A no LP LP – J(low)=1 J(up)=0 Red A LP LP – J(low)=1 J(up)=1 Red A LP LP – J(low)=1 J(up)=2

If with magnetic field, the story becomes different, namely to be continued to No.2 memo