Hyperfine Structure in Hydrogen: the 21 Cm Line

Hyperﬁne structure in hydrogen: the 21 cm line

Author: Javier G´omezSubils Facultat de F´ısica, Universitat de Barcelona, Diagonal 645, 08028 Barcelona, Spain.∗

Advisor: Josep Taron Roca

Abstract: Quantum Mechanics are used to describe how hydrogen ground state is split in two when the interaction between electron and proton’s spins are considered. Using perturbation theory, the energetic diﬀerence between those levels is calculated. The corresponding wavelength is found to be close to λ = 21 cm. Lifetime of the decay is also calculated considering Fermi’s Golden Rule. Some applications are discussed.

I. INTRODUCTION where H0 is the sum of the hydrogen electrostatic Hamil- tonian and the Hamiltonian of the electromagnetic field Hydrogen ground state is split in two levels when the in the vacuum with no interaction between them (con- interaction between electron and proton magnetic mo- sidering the quantized electromagnetic field inside a box ments is considered. We expect that the difference in en- of size L with periodic boundary conditions, taking the quantized vector potential using linear polarization: ergy between this two levels being of the order of µpµe/ ∑ ∑ √ 3 −6 2 ~ 2 ⃗· † − ⃗· a ≈ 10 eV, where a0 is the Bohr radius. As we see, ⃗ 2π c ik ⃗r ik ⃗r 0 A(⃗r) = k λ=1 3 (a⃗ ⃗ϵ⃗ e + a ⃗ϵ⃗ e ) it is very small compared with hydrogen ground state ckL k,λ k,λ ⃗k,λ k,λ energy ≈ 10 eV. and knowing that L will go to infinity at the end of the H′ The energy between this two levels was calculated by calculations as it is usually made), is the interaction between the electron and the electromagnetic field and Fermi in [1]. In 1944 Hendrik Christoffel van de Hulst W suggested the possibility of measuring this radiation in is the interaction between the spin of the proton and the spectrum of galactic radiation in [2]. Few years the spin of the electron that will be given in next section. later, in 1951, Harold Ewen and Edward M. Purcel from Here we are considering for simplicity a static proton ≫ Harvard University measured for the first time this line. (mp me), so m will be the mass of the electron. gs Their results can be found in [3]. is the gyromagnetic factor of the electron and we take it What is the interest of this radiation? The answer is gs = 2 as QED corrections are not considered. simple: hydrogen is the most abundant element of common matter present in the Universe and some of it will be in atomic ground state. As we will see, this will have important consequences in cosmological and astrophysical A. Interaction between electron and proton spins observations among others. First of all some calculations are made in order to es- To study how the ground state is split, we take into tablish what the wavelength of this radiation is and de- account that both electron and proton behave as mag- termine the lifetime of the exited level. Then main ap- netic dipoles. So suppose that the proton with magnetic µN e~ plications are described. momentum ⃗µp = γpS⃗p (γp = gp where µN = ~ 2mpc is the nuclear magneton and gp = 5.586 is the gyromagnetic factor of the proton) is at ⃗r = ⃗0 and generates a II. HYPERFINE SPLITTING OF HYDROGEN magnetic field around it given by B⃗ (⃗r) = ∇×⃗ A⃗(⃗r), where GROUND STATE A⃗(⃗r) = −(⃗µ × ∇⃗ ) 1 [4]. Then, p ( r ) ( ( )) B⃗ (⃗r) = ∇×⃗ −⃗µ × ∇⃗ 1 = − ⃗µ ∇2 − ∇⃗ ⃗µ · ∇⃗ 1 For now on, we will work in the Coulomb gauge, so p r p p r ∇⃗ · A⃗ = 0. To begin with, we consider the hamiltonian Inside that field we have an electron with magnetic ′ µB e~ H = H0 + H + W: momentum ⃗µe = γeS⃗ (γe = −gs where µB = is ~ 2mec the Bohr magneton). The interaction between the two 2 2 ∑ ∑2 P e † dipoles is ( ( )( )) H0 = − + a a ~ωk ⃗k,λ ⃗k,λ 2 1 2m r W = −⃗µe · B⃗ = − ⃗µe · ⃗µp ∇ − ⃗µe · ∇⃗ ⃗µp · ∇⃗ ⃗k λ=1 (1) r ( ) e g e e2 In our case, the hydrogen atom will be in the ground H′ = A⃗ · P⃗ + s S⃗ · ∇⃗ × A⃗ + A2 mc 2mc 2mc2 state, and electron and proton will have their own spin, so we consider the uncoupled base |nlm⟩⊗|memp⟩ = |100⟩⊗ − r | ⟩ ⟨ | ⟩ √e a0 memp . In this case we have ψ100(⃗r) = ⃗r 100 = 3 πa0 ∗Electronic address: [email protected] and now we want to find the eigenvalues of W. Javier GómezSubils

⟨100 m m | W |100 m m ⟩ = e p∫ e p The first we had observe is that the third term contains e2 so it is a second order correction. As we are interested ⟨ | W ∗ 3 | ⟩ = memp ψ100(⃗r) ψ100(⃗r)d ⃗r memp = in first order correction we will ignore this term. First R3 ∫ term will also disappear due to the fact that it does not 2 2 1 3 = − ⟨memp| ⃗µe · ⃗µp |memp⟩ |ψ100(⃗r)| ∇ d ⃗r + acts on the spin, as we will see in the next section. Notice 3 r ⟨ ∫ R ⟩ ⃗ that for A(⃗r) we consider the quantized vector potential i j | |2 1 3 given before. + memp µeµp ψ100(⃗r) ∂i∂j d ⃗r memp = R3 ∫ r 2 2 2 1 3 = − ⟨memp| ⃗µe · ⃗µp |memp⟩ |ψ100(⃗r)| ∇ d ⃗r = 3 R3 r A. Probability of the decay to a given final state −8π 1 ⟨ | ⃗ · ⃗ | ⟩ = γeγp 3 memp Se Sp memp = | ⟩ 3 πa0 Consider we have an initial state i and a final state 1 |f⟩ and we want to compute what is the probability of = A ⟨m m | S⃗ · S⃗ |m m ⟩ | ⟩ | ⟩ | ⟩ ~2 e p e p e p i going to f . By the one hand, i is the state of a A − 8 1 ~2 A hydrogen atom in the level with energy E+ (so with spin Taking = 3 a3 γeγp , and have units of energy. 0 W S = 1). Due to the fact that the energy of the exited As we are looking for the eigenvalues of we have to | ⟩ |↑ ↑ ⟩ | − ⟩ |↓ ↓ ⟩ diagonalize it. Fortunately, it will not be too difficult if state is the same for 11 = e p , 1 1 = e p and | ⟩ √1 |↑ ↓ ⟩ |↓ ↑ ⟩ ⃗ ⃗ ⃗ 10 = ( e p + e p ) we can arbitrarily choose one we define S =(Se + Sp and then) 2 ( ) of them to make the calculation. This is consequence of ⃗ · ⃗ 1 ⃗2 − ⃗2 − ⃗2 ~2 − 3 Se Sp = 2 S Se Sp = 2 S(S + 1) 2 the rotational symmetry of the problem and the fact that As S can take two different values, S = 0, 1 the ground we will integrate over all directions as can be seen during state is split in two levels: the triplet and the singlet, the calculations (we will comment this later). Then with an last having the lowest energy: appropriate transformation we can choose |↑e↑p⟩. ( ) | ⟩ | ⟩ ⊗ |↑ ↑ ⟩ ⊗ |∅⟩ A ~2 3 i = 100 e p ⟨W⟩ = S(S + 1) − = By the other hand, |f⟩ will be a hydrogen atom in ~2 2 2 { − A the energy level with energy E (so with spin S = 0, | ⟩ | ⟩ √1 E+ = E0 + 4 SMS = 1MS (|↓e↑p⟩ − |↑e↓p⟩)) and the electromagnetic field with = 3A 2 E− = E0 − |SMS⟩ = |00⟩ 4 a single photon with energy ε = ~ck0 with momentum ⃗ The difference between these two energies is the cause k0 and polarization( λ0: ) ⟩ of the radiation we were talking about, which energy is: √1 |f⟩ = |100⟩ ⊗ (|↓e↑p⟩ − |↑e↓p⟩) ⊗ γ⃗ − A − 8 ~2 4 2 me | | 2 k0,λ0 ε = E+ E− = = 3 γeγp = gsgpα E0 ′ 3a0 3 mp H − As we said before, 1 from (3) does not act on the where E0 = 13, 6 eV is the hydrogen ground state spin, so we will have energy, gs = 2 and gp = 5.586 as said before and mp = ⟨ |H′ | ⟩ ⟨ | ⊗ ⟨ | H′ | ⟩ ⊗ |∅⟩ · −6 f 1 i = ( 100 γ⃗k ) 1( 100 ) 1836me. Then we obtain: ε = A = 5.87 · 10 eV 0 ⟨S = 0 MS = 0| S = 1 MS = 1⟩ = 0 ⇒ λ = 21.1 cm ν = 1420 MHz (2) and it will not contribute to the probability, as said H′ before. So we will study 2. III. EXCITED STATE LIFETIME B. Fermi’s Golden Rule In the previous section we found the two levels in which hydrogen ground state is split when hyperfine structure is considered. As the atom is coupled to the electromag- The probability per unit of time of the decay is given by Fermi Golden’s rule: netic field, if it is in the E+ level (S = 1) it will decay to the E− level (S = 0) emitting a photon. In this section ∑ ∑2 we will answer what the lifetime of the S = 1 to S = 0 2π ′ 2 Γ = |⟨f |H | i⟩| δ (E − (~ω + E−)) (4) decay is. ~ 2 + ⃗ λ=1 To do this, we are going to take H′ from (1) and con- k H sider it as a perturbation of 0 in lowest order pertur- ⃗ K bation theory. Recovering H′ expression Considering√ the quantized A(⃗r) and taking (k) = ( ) e~ 2π~c2 H′ 2 3 , we have for 2: ′ e g e e 2mc ckL ( ) H = A⃗ · P⃗ + s S⃗ · ∇⃗ × A⃗ + A2 2 H′ e~ · ∇⃗ × ⃗ mc 2mc 2mc 2 = ⃗σ A = ′ ′ 2mc ( ) = H + H + O(e2) ∑ ∑ 1 2 2 K · ⃗ × i⃗k·⃗r − † −i⃗k·⃗r (3) = ⃗k λ=1 (k)i⃗σ (k ⃗ϵ⃗k,λ) a⃗k,λe a⃗ e H′ e ⃗ · ⃗ k,λ 1 = A P It is important to observe that, as H′ acts over the mc ( ) ( ) 2 ′ gse e~ vacuum, the annihilation operator will not contribute. H = S⃗ · ∇⃗ × A⃗ = ⃗σ · ∇⃗ × A⃗ | ⟩ 2 2mc 2mc If the emitted photon of the state f has momentum

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⃗ ∗ ∗ † ⃗ ̸ ⃗ ∗ | ⟩ K 2 2 2 k and polarization λ , none of the a⃗ with k = k or (For 10 we obtain (k0) k0 sin θ∫but the equality raises k,λ 1 π 2 diﬀerent polarization will contribute because in that case ∫when the integral is made since 2 0 (1+cos θ) sin θdθ = ⟨ ⟩ ⟨ ⟩ π 3 0 sin θdθ). γ ∗ ∗ a ∅ = γ ∗ ∗ γ = 0 ⃗k ,λ ⃗k,λ ⃗k ,λ ⃗k,λ Now, going back∫ to (5) we have 2π L3 1 |⟨ |H′ | ⟩|2 2K2 2 2 So for each term f 2 i of the summation we only Γ = 2 3 k0 (k0) k0(1 + cos θ) sin θdθdφ c~ (2π) 2 2 have to consider one ⟨ mode: ⟩ S ∫ ∫ 2 3 2π π |⟨ |H′ | ⟩|2 2π L 4 2 1 2 f 2 i = f H ⃗ i = k K (k ) (1 + cos θ) sin θdθdφ 2,k,λ c~2 (2π)3 0 0 2 with √ ( )( ) 0 0 ~ ~ 2 † ⃗ 3 e 2π c · ⃗ × −ik·⃗r 2π L 4 2 1 8 Hs,⃗k,λ = 2mc ckL3 i⃗σ k ⃗ϵ⃗k,λ a⃗ e = k K (k ) 2π k,λ c~2 (2π)3 0 0 2 3 Taking this into account and in order to calculate Γ 2 3 K2 e ~ (2π) from (4), we make the volume go to inﬁnity and we obtain And now every thing is known, as (k0) = 2 3 [∑ ( ) ∫ ] 4m ck0L → L 3 3⃗ ⃗ ∀⃗ d k : k 2π k (2π)2 L3 e2~3(2π) 1 8 1 e2k2~ ∫ Γ = k4 = · 0 (7) 3 ∑2 ⟨ ⟩ ~2 3 0 2 3 2 2 2π d ⃗k 2 c (2π) 4m ck0L 2 3 3 m c − Γ = 2 3 f H2,⃗k,λ i δ(k0 k) = c~ ∀⃗ k (2π/L) λ=1 As this is the probability per unit of time, it is directly ∫ 3 ∑2 ⟨ ⟩ 2 related to the lifetime through the expression Γ = 1/τ. 2π L 2 2 2 = k sin θdθdφ f H ⃗ i 3m c ≈ × 14 ≈ 7 2 3 0 2,k,λ So, τ = 2~ 3 3.4 10 s 10 years. c~ (2π) S2 e k λ=1 0 (5) ⟨The next step⟩ is the calculation of the matrix elements 2 IV. HYPERFINE SPLITTING OF HYDROGEN

f H2,⃗k,λ i were k = k0 is ﬁxed due to the energy GROUND STATE INSIDE HEAVY MAGNETIC conservation but the direction of ⃗k is not. FIELDS (ZEEMAN EFFECT) ⟨ ⟩

⟨ | | ⟩ ⟨ | −i⃗k·⃗r | ⟩· † ∅ · f H ⃗ i = 100 e 100 γ⃗ a ⃗ 2,k,λ ( k,λ ⃗k,λ) If there exists an external magnetic ﬁeld, B0 = B0zˆ the ( ) situation changes a little, just because the triplet state · ⟨↑ ↑ | · ⃗ × √1 |↓ ↑ ⟩ − |↑ ↓ ⟩ e p ⃗σ k ⃗ϵ⃗k,λ ( e p e p ) is split in three diﬀerent levels. This is because the new 2 interaction W∗ is As λ = 21∫ cm and r ≈ a0 (which means that the probability of |ψ (⃗r)|2d3⃗r ≈ 0 if we integrate for r > ∗ 100 W = W − ⃗µe · B⃗ 0 − ⃗µp · B⃗ 0 ⃗ −9 a0), we have that k · ⃗r ≈ 10 ≪ 1 and it is a good −i⃗k·⃗r −i⃗k·⃗r approximation e = 1 so ⟨100| e |100⟩ = 1 and as γp ≪ γe we will not consider the last term. So In order to perform the integral (5) it is convenient now we have to diagonalize: ⃗ to choose the axis in this way: k = k0(sin θ, 0, cos θ). ∗ A Then, ⃗ϵ⃗k,2 can be chosen as ⃗ϵ⃗k,2 = (0, 1, 0). Finally ⃗ϵ⃗k,1 = W = S⃗ · S⃗ − ⃗µ · B⃗ (8) ~2 e p e 0 1 ⃗ϵ × ⃗k = (cos θ, 0, sin θ). Hence, k0 ⃗k,2 ⃗ × where we can remember from (2) what A is. k ⃗ϵ⃗k,1 = k0⃗ϵ⃗k,2 = (0, k, 0) In the ordered coupled base of |SMS⟩, that is ⃗k × ⃗ϵ = −k ⃗ϵ = k (− cos θ, 0, sin θ) ⃗k,2 0 ⃗k,1 0 {|11⟩ , |1 − 1⟩ , |10⟩ , |00⟩}, the matrix for (8) becomes: ( Knowing) how do ⃗σ act on |↑e⟩ and |↓e⟩ we compute ⃗σ · 1 A 1 ~ 4 + 2 γeB0 0 0 0 ⃗ × | ⟩ |⟨ | ˆ| ⟩|2 |⟨ | ˆ| ⟩|2  1 A − 1 ~  k ⃗ϵ⃗k,i 00 (as f O i = i O f for an hermitian  0 4 2 γeB0 0 0   0 0 1 A 1 ~γ B  operator Oˆ): 4 2 e 0 0 0 1 ~γ B − 3 A √ 2 e 0 4 · ⃗ × |↓ ↑ ⟩ − |↑ ↓ ⟩ ⃗σ (k ⃗ϵ⃗k,1)1/ 2 ( e p e p ) = The four eigenvalues and eigenstates are (taking √ ~ tan 2θ = e B0 /A): = −1/ 2ik (|11⟩ + |1 − 1⟩) m 0 √ A 1 ⃗ ~ | ⟩ ⃗σ · (k × ⃗ϵ⃗ )1/ 2(|↓e↑p⟩ − |↑e↓p⟩) = + γeB0, ϖ1 = 11 √ k,2 4 2 A = 1/ 2k0 cos θ(|11⟩ − |1 − 1⟩) − sin θ |10⟩ 1 − ~γeB0, ϖ2 = |1 − 1⟩ Here we see that for |11⟩ and |1, −1⟩ we have the same 4 2 √ A A2 1 result: 2 − + + (~γeB0) , ϖ3 = cos θ |10⟩ + sin θ |00⟩ 2 ⟨ ⟩ 4 √ 4 4 ∑ 2 1 K2 2 2 A A2 1 f H2,⃗k,λ i = (k0) k0(1 + cos θ) (6) 2 2 − − + (~γeB0) , ϖ4 = − sin θ |10⟩ + cos θ |00⟩ λ=1 4 4 4

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⟨W⟩ A can not be explained with common matter alone. This is / 4 3 widely explained in [8].

ϖ1 Finally, we can add that the Zeeman splitting of the 2 ϖ3 levels we have discussed is also used to detect intergalactic magnetic fields. Here Ly-α line is split in different lines and the difference between them is directly related 1 with the magnetic field as we can see in Figure 1.

~γeB0 A

1 2 3 4 5 B. Cosmology: Epoch of Reionization

−1 Recently it has been discussed the importance of the 21 line in Cosmology, specially in the study of the Epoch of −2 ϖ2 Reionization (EoR). Using Statistical Mechanics, a spin ϖ4 temperature TS is associated to the transition S = 1 to −3 S = 0, being related to the proportion on hydrogen in S = 0 state and S = 1 state (kBΘS ≡ ε of equation (2); if TS ≫ ΘS almost all hydrogen will be in S = 1 level Figure 1: Splitting of hydrogen ground state inside an exter- whereas TS ≪ ΘS implies that it will be in S = 0 level). nal magnetic field When matter is decoupled from radiation and CMB is emitted, TS evolves as kinetic temperature of matter TK due to collisional coupling, which is below the tempera- V. RECENT APPLICATIONS ture of the radiation Tγ . Then 21 cm line is observed as an absorption line in the CMB between 30 < z < 150, Now we are going to outline few relevant applications of being z the redshift. 21 cm line in Physics nowadays. We have separated them This changes when density falls as the Universe ex- in three: astrophysical and cosmological applications and pands and less collisions take place, so now TS is coupled precision measurement of constants while testing theories with CMB. Then between 10 < z < 30 TS = Tγ and no (like QED of General Relativity). information can be obtained with the 21 cm line. For z < 10 radiation from stars that begin to form makes TS grater than Tγ , and from then on the line is detected A. Astrophysical applications as an emission line. This is an extremely recent field of research and it is very promising, although challenging Observation of 21 cm line allows us to obtain impor- measurement techniques and difficulties in the descrip- tant astrophysical data. This application is based in cal- tion of galaxy formation have to be faced and it is ex- culating the density of atomic hydrogen in galaxies just pected that it will give us constraints for EoR. A large measuring the intensity received of 21 cm line. We will explanation of the 21 cm line aplication to EoR is given explain three results: in [11]. First of all, this measure and the Doppler effect associated to the velocities of hydrogen clouds has given us an approximation of the structure of the Milky Way (radius, C. Precision Tests distance from the Sun to Galactic center,...); and also has helped to measure the interstellar medium density[5]. In second place, the 21 cm radiation is not only aplied The frequency corresponding to the 21 cm hyper- to study the Milky Way, but also other galaxies and their fine transition is one of the best measured quantities in kinematics. Thanks to this, we know that between 1 Physics, establishing that it is [12]: per cent and 10 per cent of the total mass galaxies is in atomic hydrogen form. The largest proportions are ν = 1 420 405.751 766 7(9) kHz 10 found in small galaxies ∼ 10 M⊙[6]. With respect to the kinematics, not only circular motion around galactic This has motivated using hyperfine transition between centers are considered, but also some noncircular motions atoms to define the second in the International Sistem of are distinguished and studied, such as motion associated Units and building atomic clocks: to spiral arms, large-scale symmetric deviations, large- scale asymmetries (interaction with other close galax- The second is the duration of 9 192 631 770 periods of ies,...) and small-scale asymmetries (see [7]). This is the radiation corresponding to the transition between the certainly one of the most important applications, just two hyperfine levels of the ground state of the caesium because it has meant one of the evidences of the exis- 133 atom. [...] This definition refers to a caesium atom tence of dark matter, as the rotation curves of galaxies at rest at a temperature of 0 K. [9]

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Such accuracy made possible one of the experiments VI. CONCLUSIONS that gave support to Einstein’s General Relativity in [10]. Placing one hydrogen maser in a rocket that goes up to The hyperfine splitting of hydrogen ground state is one 10.000 km and measuring the frequency received from of the most precisely measured quantities of Physics. In- the maser by an observer on the ground (and correcting teracion between proton and electron magnetic momenta the Doppler effect due to rocket velocity), gravitational split the ground state in two levels; the wavelength of the redshift was tested with great precision. The experiment transition between them is around 21 cm and the transi- · −5 was in agreement with theory with an accuracy of 7 10 . tion has a lifetime of about 107 years that can be analyti- Even more, as we have great precision in the measure- cally calculated. Measurements of this radiation gives us ment of the ground state hydrogen hyperfine splitting, a lot of information of Universe structure and evolution this has given a precision test of standard model and due to the high quantity of atomic hydrogen we can find. the extraction of the values of fundamental physical con- In this sense, information is got from the Epoch of stants (fine structure constant, the electron and muon Reionization and from structure and motion of galax- masses, proton charge radius,...); see for example [13]. ies. Furthermore, from Zeeman effect on hyperfine lev- Finally we mention that theoretical expression for the els we obtain information about intergalactic magnetic hyperfine splitting in the hydrogen can be corrected: fields. In addition to this, the great precision we have ( ) measuring hyperfine splitting in hydrogen ground state ∆EHFS = ε∗ 1 + δQED + δp, str + δp, pol + δHVP gives us precision tests of General Relativity and Stan- where ϵ∗ is ground state splitting taking into account fi- dard Model. Even more, 21 cm line has supported the nite mass of the proton, δQED represents the quantum existence of dark matter, a requirement of measurements electrodynamic contribution, δp, str and δp, pol are associ- that nowadays is still theoretically unexplained. ated to the fact that proton is not an elementary particle It is amazing how one of the less energetic transitions and they correspond to proton structure and polarizabil- and with such a long lifetime opens such a wide window ity respectively and δHVP is the contribution of hadronic to the understanding of the Universe. vacuum polarization. In order to measure this quantities, hyperfine splitting in 2s and 1s levels are considered due to the fact that Acknowledgments some corrections in both levels are equal and vanish when we make the difference, so the constants stated above can I would like to thank my advisor Josep Taron for all his easier be determined. This higher order corrections are patience and implication. I specially thank my parents studied for example in [14], were theoretical predictions and brothers, that have been always there whenever I had are made not only for hydrogen hyperfine structure but needed them. I also thank my friends, specially Marcos, also for deuterium and helium-3 ion. for being by my side in the good and the bad moments.

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