Notes 3: Hydrogen Fine Structure
High-resolution spectra of atoms show lines that are not single but have closely spaced multiple components. This fine structure is from the splitting of energy levels due to effects that we have not included in the simple model of the hydrogen atom. A proper analysis of fine structure requires solving the relativistic Dirac equation. Here we consider an approach using three perturbations to the Schrödinger Hamiltonian. The first perturbation is a first-order relativistic correction to the kinetic energy operator. We then consider spin - orbit coupling arising from the interaction between the electron magnetic moment and the magnetic field generated by the motion of the proton relative to the electron. The third perturbation, called the Darwin term, only effects the l = 0 eigenstates. The Darwin term arises because the electron cannot be confined to a point and must be treated as fuzzy ball of radius equal to the electron Compton wavelength (i.e. the wavelength corresponding to the mass of the electron).
Relativistic correction The kinetic energy operator in the Schrödinger equation is derived by assuming that the motion of the particle(s) is non-relativistic, i.e. the relation between kinetic energy and momentum for a particle of mass M is1
1 p2 T = . (3.1) 2 M
The relativistic expression is
p2 T= c2 p 2 + M 24 c −= Mc 2 Mc 21 + − 1, (3.2) 22 Mc where c is the speed of light in vacuum. We can find the first order correction to the Hamiltonian by 2 expanding T as a power series in p2 ( Mc)
p2 1 p2 1 p 4 11 pp 24 T= Mc2211 + −=Mc − + = − +. (3.3) 22 22 44 32 Mc 2Mc 8 Mc 28 M Mc
The ratio of the first two terms in the series has magnitude p2(4 Mc 22) , which is of order the electron’s kinetic energy divided by its rest mass energy. Numerically, this ratio is ~10-5. Relativity effects are small and difficult to detect in light atoms but become sizable for the inner electrons of massive atoms.
1 I use M do avoid confusion with the magnetic quantum number m in application to the hydrogen atom. In many cases, M should be the reduced mass µ but GS do not make this distinction.
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Since p2 is a Hermitian operator and commutes with itself, p4 is also a Hermitian operator (see problem 2 2 3.4 c). It is easy to show that the operators p , L and Lz commute. Hence, by using the commutator identity from problem (3.14),
ˆˆˆ ˆ ˆ ˆ ˆˆ ˆ ABC, = A BC , + AC ,, B
ˆ ˆ 2 ˆ 2 4 2 2 with AB= = pand CL= or Lz , we can show that p commutes with L and Lz. Since L and Lz have distinct eigenvalues, the good state theorem allows us to use the n2 degenerate hydrogen wave functions
ψ nlm in non-degenerate perturbation theory to find the first-order energy corrections due to the perturbation
1 p4 H 1 = − . (3.4) 8 Mc32
The correction to the energy is
11 E1=−=ψψ0 p 40 −pp20ψψ 20, (3.5) 88Mc32 Mc32
where we has used the Hermitian property of p2. For the unperturbed eigenstates
p20ψψ=2, mE( 0 − V 0) 0 (3.6)
and so
1 110 0220 00 0 2 EEV=−−=−( ) (E) −2. EV +( V) (3.7) 22Mc22Mc
For the hydrogen atom
1 e2 Vr0 ( ) = − , (3.8) 4πε 0 r
and so equation (3.7) gives
222 121ee 11 E=−−+ EEnn2. (3.9) 22πε πε 2Mc 4400r r
An elegant way to evaluate the expectation values is to use the so-called Hellmann-Feynman theorem (Problems 7.38 and 7.42), which states that if the Hamiltonian H depends on a parameter β and has
(good state) orthonormal eigenfunctions ψβn ( ) and eigenvalues En (β ), then
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∂E ∂H n = ψψ. (3.10) ∂∂ββnn
The results are
11 1 1 = = 22,, (3.11) r na r 1 32 l+ na 2
where a is the Bohr radius. Hence the relativistic correction to the energy is
2 1ee22 11 12=−++ Erel 22 EEn2. n (3.12) 2Mc 44πε00n a πε 1 32 l+ na 2
The energy of the unperturbed state is
11e2 = − En 2 . (3.13) 24naπε 0
Using this to eliminate the Bohr radius from equation (3.12), we obtain
En2 4 E1 =−−n 3. (3.14) rel 2 1 2Mc l + 2
The degeneracy has been partially lifted as this correction depends on l but not on the magnetic quantum number m.
Problem 7.17 Find the lowest order relativistic correction to the energy levels of the one-dimensional harmonic oscillator.
Solution: The energy states of the one-dimensional harmonic oscillator are non-degenerate. The energy corrections are given by equation (3.7) but with
1 Vx0( ) = mxω 22. 2
Thus we need to calculate x2 and x4 . The operator x is in terms of the raising and lowering operators is
x=( aa + ), 2mω +−
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Hence
01 22 ωω2 V= mω x =( aa +) = aaaaaaaa +++ 24+− 4+++−−+−−
ωω21H =00 + +=n + . 4ω 22
Also
2 4 22 22 x= xψψnn x =(aa+− ++) ψn( aa +−) ψn. 2mω
Using
an+ψn=+=1, ψ n+−11 an ψψ nn −, we find
2 (aa+−+) ψψnn =( aaaaaaaa +++−−+−− +++)
=(n +2)( n + 1)ψψψnnn+−22 ++( 21 n) + nn( − 1) .
Therefore
22 2 02 1 24 ωω2 2 (V) = mω x = (n +2)( n ++ 1) ( 2 n + 1) + nn( − 1) =( 6n + 6 n + 3.) 24 4
The energy correction is
22 11 1 1 1=−+−++++22ω22 ωω 22 2 E2 n n (6nn 63) 2mc 2 2 16 2 (ω) 6nn2 ++ 63 = − . mc2 32
For both the hydrogen atom and the quantum harmonic oscillator the relative energy correction is
EE10 . E02 Mc
For the hydrogen atom
2 Ee02 = α 2 2 , Mc4πε 0 c
4 where α is the fine structure constant,
e2 1 α = ≈ . 4πε 0c 137.036
The fine structure constant is a fundamental dimensionless quantity and has many physical interpretations. One is that it is a coupling constant for the strength of the electron - photon interaction.
In terms of α, the zeroth-order hydrogen energies are
1 E= − α 22 Mc , (3.15) n 2n2
and the Bohr radius is
1 a = . (3.16) α Mc
The first order relativistic corrections are from equation (3.14)
En1 14 rel =−−α 2 3. (3.17) 0 2 1 E 4n l + 2
Spin – Orbit Coupling The electron has an intrinsic magnetic moment that contributes to the total energy of the hydrogen atom due to its interaction with the magnetic field from the orbital motion of the nucleus in the electron rest frame.
The magnetic field from the proton’s motion is estimated by treating it as a current loop of radius r with current equal to the charge divided by orbital period
e I = Ω, (3.18) 2π where Ω is the orbital angular velocity. The magnetic field at the center of the loop (i.e. at the electron’s location) is obtained from the Biot-Savart Law, which gives
µµIeΩ Bz=00ˆˆ = z. (3.19) 24rrπ
We now eliminate the orbital angular velocity by using the orbital angular momentum of the electron (since the angular velocity is the same for the proton and electron)
Lz=Mr 2 Ωˆ. (3.20)
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The magnetic field is
µ ee1 =0 = BL3 23 L. (3.21) 44πMr πε 0 Mc r
We estimate the magnetic moment of the electron in a similar manner. The magnetic moment of a current loop is equal to its area times the current
eΩ μz= −π r 2 ˆ. (3.22) 2π
The angular momentum of the charge is
Sz=Mr 2 Ω.ˆ (3.23)
Hence
e μS= − . (3.24) 2M
The ratio of a particle’s magnetic moment to its spin is called the gyromagnetic ratio.
The electron is a structureless point particle and treating it as a current loop is hard to justify. The deviation from the ‘classical’ result for the magnetic moment is accounted for by introducing a gyromagnetic ratio factor (usually called a g-factor) so that
e μS= −g . (3.25) 2M
For the electron, g is very close to 2.
Also in estimating the magnetic field, the electron was taken to be at rest. However, the electron rest frame is a non-inertial frame and rotates relative to an inertial frame. This deficiency is accounted for by including what is called Thomas precession.
When these additional considerations are included, the perturbation to the Hamiltonian due to spin – orbit coupling is
ee2211 1 =−⋅= − ⋅≈ ⋅ HgSOC μB ( 1.) 223SL 223SL (3.26) 88πε 00Mcr πε Mcr
This perturbation does not commute with L nor with S. Hence, the energy and Lz or Sz are not 1 2 2 simultaneously observable. However, H SOC does commute with L , S and the total angular momentum J
= L + S. In particular, the Hamiltonian commutes with Jz, which has distinct eigenvalues. Thus non- degenerate perturbation theory can be used.
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Problem 7.19 Evaluate the following commutators: (a) [LSL⋅ ,,] (b) [LSS⋅ ,,] (c) [LSJ⋅ ,,] (d)
2 2 2 LS⋅ ,,L (e) LS⋅ ,,S (f) LS⋅ ,.J
= Solution: We make use of the commutator relation Lxy, L iL z and its permutations, and that L and L2 commute.
(a) We have
⋅ = + + = + =−+ = × [LS,Lx] LS xx LS yy LS zzx, L S y L yx ,, L S z[ L zx L] i( S yz L SL zy) i (LS)x .
By also considering the other two components of the commutator, we conclude
[LSL⋅=×,.] i( L S)
(b) By exchanging L and S in the result of part (a), we obtain
[LSS⋅=−×,.] i( L S)
(c) Since J = L + S, from parts (a) and (b), we see that [LSJ⋅=,] 0.
(d) Now
2 22 2 2 2 LS⋅, L =( LS ⋅) LL −( LS ⋅) =( L L) ⋅− S( L L) ⋅= S L, L ⋅= S 0.
2 (e) Applying the same method as in part (d), we obtain LS⋅=,S 0.
(f) Since J2=(LS +) ⋅( LS +) = LS22 +2, SL ⋅+ using the results from parts (d) and (e) gives
2 LS⋅=,J 0.
2 2 2 To summarize, LS⋅ commutes with L , S , J and J but not with L or S. Hence ml and ms are no longer ‘good’ quantum numbers but n, l, s, j and mj are.
Note that there is no spin – orbit coupling for spherically symmetric states, i.e. those for which l = 0.
To evaluate the first-order correction to the energy, we need to evaluate SL⋅ and r −3 . The easiest way to get the former is to use
JL22= +2,SL ⋅+ S2 (3.27)
so that
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J22−− LS 2 SL⋅= . (3.28) 2
Then
2 SL⋅ =j( j +−1) ll( +− 1) ss( + 1.) (3.29) 2
This is zero for l = 0, since then j = s.
The results of problem 7.43 can be used to find that, for l > 0,
2 r −3 = . (3.30) nl33( l++12)( l 1) a
Putting the pieces together, we obtain
ee21122 j( j+−11) ll( +−) ss( + 1) 11= ≈ −3⋅= EHSOC SOC 22 rSL 22 3 3 8πε 00Mc 8πε Mc a nll( ++12)( l 1) (3.31) E 2 j( j+−11) ll( +−) ss( + 1) = n n . Mc2 l( l++12)( l 1)
and so
E1 j( j+−11) ll( +−) ss( + 1) SOC = α 2 . (3.32) E 0 2nl( l++ 12)( l 1)
The relativistic and spin-orbit corrections together are called fine structure. By considering separate cases (problem 7.20), a concise expression for the fine structure energy correction can be obtained:
α 2 3 n EE10= − . (3.33) fs 2 1 n 4 j + 2
The Darwin term Strictly speaking, the semi-classical derivation of this result is valid only for l > 0. However, we have yet to include the Darwin term, which has Hamiltonian
π 22e 1 = δ3 = πα22 33 δ HDar 22 (rr) nEan ( ). (3.34) 24Mc πε 0
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Because the l > 0 eigenfunctions are zero at the origin, they do not ‘see’ the delta function. Hence, the Darwin term only affects the l = 0 states.
The energy corrections due to the Darwin term are
1 22 3 HDar = πα nEan ψ (0.) (3.35)
−32 For l = 0, ψπ(01) = ( )(na) . Hence
1 HE12= α . (3.36) Dar n n
Adding this correction to the relativistic correction leads to the same result as given by equation (3.33).
Splitting of Lyman alpha and Balmer alpha lines The Lyman alpha spectral line is produced in the transition from the n = 2, l = 1 state to the n = 1, l = 0 state. The table shows the energy of these states with and without the fine structure correction. The energy unit is the ionization potential of hydrogen.
n l j En Enj Enj-En 1 0 1/2 -1 -1.000013313 -0.000013313 2 1 3/2 -0.25 -0.250000832 -8.321 10-7 2 1 1/2 -0.25 -0.250004160 -4.160 10-6
The fine structure correction splits the Lyman alpha line into a doublet of wavelengths 121.5668 and 121.5674 nm. The separation between components is only 0.0006 nm.
The Balmer alpha spectral line is produced in the transition from the n = 3 state to the n = 2 state. Since the orbital angular momentum can change by ±1, the allowed transitions are 3s to 2p, 3p to 2s and 3d to 2p. The table shows the energy of the states with and without the fine structure correction.
n l j En Enj Enj-En 2 0 1/2 -0.25 -0.250004160 -4.160 10-6 2 1 3/2 -0.25 -0.250000832 -8.321 10-7 2 1 1/2 -0.25 -0.250004160 -4.160 10-6 3 0 1/2 -0.111111111 -0.111112590 -1.479 10-6 3 1 1/2 -0.111111111 -0.111112590 -1.479 10-6 3 1 3/2 -0.111111111 -0.111111604 -4.931 10-7 3 2 3/2 -0.111111111 -0.111111604 -4.931 10-7 3 2 5/2 -0.111111111 -0.111111216 -1.644 10-7
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The next table shows the possible photon wavelengths (in vacuum) for a given transition. The wavelengths are in nm.
Transition 3s to 2p 656.4726663 656.4569364 3p to 2s 656.4569364 656.4522763 3d to 2p 656.4680060 656.4661719 656.4522763 656.4504427
The wavelength shown in red requires an electron spin flip. The wavelengths in italics are the same as other wavelengths in the table. We see that we expect there to be five distinct peaks in a high-resolution spectrum, spread over a wavelength interval of ~ 0.02 nm. When an additional effect called the Lamb shift is included the number of expected peaks increases to seven. In atomic hydrogen gases, because of the Doppler effect, usually only two peaks are observed. When saturation spectroscopy of gas cells or collimated beams of hydrogen atoms are used, four peaks can be resolved. For more complicated atoms, fine structure can be more easily visible. For example, the sodium D lines have a separation of 0.6 nm
The Lamb shift has some similarity to the Darwin term in that it affects only the l = 0, i.e. S states. Due to the quantum fuzziness of the electron it has a slightly weaker interaction with the nucleus in the l = 0 state than in l > 0 states. Hence, the 2S1/2 state has a higher energy than the 2P1/2 state.
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