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Hyperfine Structure in Hydrogen: the 21 Cm Line

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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= P P q 3 −6 2 ~ 2 ~· y − ~· 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 (a~ ~ϵ~ e + a ~ϵ~ 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 H0 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 com- mon matter present in the Universe and some of it will be in atomic ground state. As we will see, this will have im- portant 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 = γpS~p (γp = gp where µN = ~ 2mpc is the nuclear magneton and gp = 5:586 is the gyromag- netic factor of the proton) is at ~r = ~0 and generates a II. HYPERFINE SPLITTING OF HYDROGEN magnetic field around it given by B~ (~r) = r×~ A~(~r), where GROUND STATE A~(~r) = −(~µ × r~ ) 1 [4]. Then, p ( r ) ( ( )) B~ (~r) = r×~ −~µ × r~ 1 = − ~µ r2 − r~ ~µ · r~ 1 For now on, we will work in the Coulomb gauge, so p r p p r r~ · A~ = 0. To begin with, we consider the hamiltonian Inside that field we have an electron with magnetic 0 µB e~ H = H0 + H + W: momentum ~µe = γeS~ (γe = −gs where µB = is ~ 2mec the Bohr magneton). The interaction between the two 2 2 X X2 P e y dipoles is ( ( )( )) H0 = − + a a ~!k ~k,λ ~k,λ 2 1 2m r W = −~µe · B~ = − ~µe · ~µp r − ~µe · r~ ~µp · r~ ~k λ=1 (1) r ( ) e g e e2 In our case, the hydrogen atom will be in the ground H0 = A~ · P~ + s S~ · r~ × A~ + A2 mc 2mc 2mc2 state, and electron and proton will have their own spin, so we consider the uncoupled base jnlmi⊗jmempi = j100i⊗ − r j i h j i pe a0 memp . In this case we have 100(~r) = ~r 100 = 3 πa0 ∗Electronic address: [email protected] and now we want to find the eigenvalues of W. Javier G´omezSubils h100 m m j W j100 m m i = e pZ 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 h j W ∗ 3 j i = memp 100(~r) 100(~r)d ~r memp = in first order correction we will ignore this term. First R3 Z term will also disappear due to the fact that it does not 2 2 1 3 = − hmempj ~µe · ~µp jmempi j 100(~r)j r d ~r + acts on the spin, as we will see in the next section. Notice 3 r ⟨ Z R ⟩ ~ that for A(~r) we consider the quantized vector potential i j j j2 1 3 given before. + memp µeµp 100(~r) @i@j d ~r memp = R3 Z r 2 2 2 1 3 = − hmempj ~µe · ~µp jmempi j 100(~r)j r d ~r = 3 R3 r A. Probability of the decay to a given final state −8π 1 h j ~ · ~ j i = γeγp 3 memp Se Sp memp = j i 3 πa0 Consider we have an initial state i and a final state 1 jfi and we want to compute what is the probability of = A hm m j S~ · S~ jm m i j i j i j i ~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 j i j" " i j − i j# # i diagonalize it. Fortunately, it will not be too difficult if state is the same for 11 = e p , 1 1 = e p and j i p1 j" # i j# " i ~ ~ ~ 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 j"e"pi. ( ) j i j i ⊗ j" " i ⊗ j;i A ~2 3 i = 100 e p hWi = S(S + 1) − = By the other hand, jfi will be a hydrogen atom in ~2 2 2 { − A the energy level with energy E (so with spin S = 0, j i j i p1 E+ = E0 + 4 SMS = 1MS (j#e"pi − j"e#pi)) and the electromagnetic field with = 3A 2 E− = E0 − jSMSi = j00i 4 a single photon with energy " = ~ck0 with momentum ~ The difference between these two energies is the cause k0 and polarization( λ0: ) E of the radiation we were talking about, which energy is: p1 jfi = j100i ⊗ (j#e"pi − j"e#pi) ⊗ γ~ − A − 8 ~2 4 2 me j j 2 k0,λ0 " = E+ E− = = 3 γeγp = gsgpα E0 0 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 jH0 j i h j ⊗ h j H0 j i ⊗ j;i · −6 f 1 i = ( 100 γ~k ) 1( 100 ) 1836me. Then we obtain: " = A = 5:87 · 10 eV 0 hS = 0 MS = 0j S = 1 MS = 1i = 0 ) λ = 21:1 cm ν = 1420 MHz (2) and it will not contribute to the probability, as said H0 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 X X2 we will answer what the lifetime of the S = 1 to S = 0 2π 0 2 Γ = jhf jH j iij δ (E − (~! + E−)) (4) decay is. ~ 2 + ~ λ=1 To do this, we are going to take H0 from (1) and con- k H sider it as a perturbation of 0 in lowest order pertur- ~ K bation theory. Recovering H0 expression Consideringq the quantized A(~r) and taking (k) = ( ) e~ 2π~c2 H0 2 3 , we have for 2: 0 e g e e 2mc ckL ( ) H = A~ · P~ + s S~ · r~ × A~ + A2 2 H0 e~ · r~ × ~ mc 2mc 2mc 2 = ~σ A = 0 0 2mc ( ) = H + H + O(e2) P P 1 2 2 K · ~ × i~k·~r − y −i~k·~r (3) = ~k λ=1 (k)i~σ (k ~ϵ~k,λ) a~k,λe a~ e H0 e ~ · ~ k,λ 1 = A P It is important to observe that, as H0 acts over the mc ( ) ( ) 2 0 gse e~ vacuum, the annihilation operator will not contribute.
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