Abstract
In this chapter, we focus on hydrogen-like atoms and their spectral structure in great depth. We show that the spectrum consists of a gross structure resulting from the electrostatic interaction between an electron and the nucleus, a fine structure arising from the spin-orbit interaction, and a hyperfine structure stemming from the spin-spin interaction. These structures are not specific solely for the hydrogen-like atoms though, we encounter them in any system which we can describe in the first approximation within the framework of nonrelativistic quantum mechanics. As we will shortly see, the spin-orbit interaction is an effect of relativistic kinematics and the spin-spin interaction is nothing more than a quantum mechanical analogy of the interaction of two magnetic dipoles. In the case of such systems, the effects of relativistic kinematics are minor and likewise the magnetic interaction is substantially smaller in comparison to the electrostatic interaction. Furthermore, we focus on the problem of hydrogen-like atoms to illustrate several methods that we will then systematically develop in the next chapter.
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Notes
- 1.
A hydrogen-like atom is a bound state of two charged particles. The simplest example is hydrogen atom consisting of an electron and a proton. The other examples discussed extensively in this book are deuterium, where the proton is replaced by a deuteron, the bound state of a proton and a neutron, muonium, where the proton is replaced by an antimuon, an electron’s heavier cousin with the opposite charge, positronium, where the proton is replaced by a positron, electron’s antiparticle, see Chap. 7, and muonic hydrogen, where the electron is replaced by a muon. Under the term antiparticle we mean a particle with the same mass and spin as, but opposite charge to the “original” particle. Although this definition is too narrow, it suffices well in this book.
- 2.
The reason might probably be to stress that it is an electrostatic potential. Newton potential would be a more convenient name, though.
- 3.
Note that one is prone to make a mistake when finding \(\hat{\mathsf{p}}_{r}^{2}\):
$$\displaystyle{\left ({ \partial \over \partial r} +{ 1 \over r}\right )\left ({ \partial \over \partial r} +{ 1 \over r}\right ) ={ \partial ^{2} \over \partial r^{2}} +{ 1 \over r}{ \partial \over \partial r} +{ \partial \over \partial r}{ 1 \over r} +{ 1 \over r^{2}}\,;}$$now
$$\displaystyle{{ \partial \over \partial r}{ 1 \over r} ={ 1 \over r}{ \partial \over \partial r} + \left [{ \partial \over \partial r},{ 1 \over r}\right ] ={ 1 \over r}{ \partial \over \partial r} -{ 1 \over r^{2}}\,.}$$ - 4.
It might seem rather odd that we focus on the point r = 0 and yet we have kept only the leading term when expanding the Coulomb potential, see Eq. (3.41). The expression (3.46) should thus hold solely for large r, and we should take into consideration other terms of the expansion (3.41) to accurately capture the situation for small r. Experiments show, though, that particles we consider as elementary (such as electrons, muons, etc.) behave as electric monopoles and magnetic dipoles. That is, the remaining terms of the expansion (3.41) are zero. Although in the case of a proton the other terms are not null, the dipole term still prevails for r ≈ (mα)−1 ≈ 10−10 m and an electron practically does not approach the proton any closer. In other words, the probability of finding an electron in the space between 0 and 10−1 atomic units from the proton is ∫ 0 0. 1drr 2 | R 1s (r) |2 = 0. 11 × 10−2, see Eq. (3.35). We thus see that one needs to distinguish between a physical and mathematical zero!
- 5.
The easiest way to obtain this equation goes as follows. An object with two indices appears on the lhs, and the only plausible object with two indices on the rhs is the Kronecker delta, i.e., ∫n s n p dΩ = Kδ sp . Setting now s = p and keeping in mind that we are using the Einstein summation convention and \(\boldsymbol{\mathbf{n}}\) is a unity vector, we obtain 4π = 3K.
- 6.
The fact that the gyromagnetic ratios of a proton and a neutron do not equal one (or zero, respectively) was one of the first clues that these two particles are not elementary.
- 7.
\(\left <2\right \vert \left.\!3\right> = \left <e\!+\right \vert \left.\!e\!-\right>\left <p\!-\right \vert \left.\!p\!+\right> = 0\), see Eq. (1.9).
- 8.
For the very same reason, it was used as a unit of length and time on information plates for extra-terrestrial civilizations on the probes Pioneer 10, 11 and Voyager 1, 2, and also the program SETI (Search for Extra-Terrestrial Intelligence) operated on this frequency.
- 9.
In light of this finding, one may ask to which of the four 2s − 1s transitions the values in Table 3.1 correspond. The numbers are spin-averaged values, meaning the spin-spin interaction was subtracted. Due to the substantially smaller magnitude of the hyperfine splitting in comparison to the 2s − 1s transition, one can perform that with sufficient precision.
- 10.
Precisely speaking, we consider \(\hat{\mathsf{\boldsymbol{S}}} =\hat{ \mathsf{\boldsymbol{S}}}_{e} \otimes \mathbf{1}_{p} + \mathbf{1}_{e} \otimes \hat{\mathsf{\boldsymbol{S}}}_{p}\)
- 11.
A more accurate notation of these states would be \(\left \vert j,m,\left ({1 \over 2},{ 1 \over 2}\right )\right>\) where the halves stress that these states stem from two particles with spin 1∕2 each, see Sect. 4.2.
- 12.
One will be surely able to solve the sodium case after reading Chap. 5
- 13.
A kind reader surely manages to generalize these findings for the case when an external magnetic field is present.
- 14.
In this case
$$\displaystyle\begin{array}{rcl} \int { \mathrm{d}^{3}\boldsymbol{\mathbf{k}} \over (2\pi )^{3}\omega ^{2}}\mathrm{e}^{\mathrm{i}\omega r\cos \vartheta }\cos ^{2}\vartheta & =&{ 1 \over (2\pi )^{3}r}\int _{0}^{2\pi }\mathrm{d}\varphi \int _{ 0}^{\infty }\mathrm{d}\omega \int _{ 0}^{\pi }\mathrm{d}\vartheta \sin \vartheta \,\mathrm{e}^{\mathrm{i}\omega \cos \vartheta }\cos ^{2}\vartheta {}\\ & =&{ 1 \over (2\pi )^{2}r}\int _{0}^{\infty }\mathrm{d}\omega \left (-{\mathrm{d^{2}} \over \mathrm{d}\omega ^{2}}\right )\int _{0}^{\pi }\mathrm{d}\vartheta \sin \vartheta \,\mathrm{e}^{\mathrm{i}\omega \cos \vartheta } {}\\ & =&{ -2 \over (2\pi )^{2}r}\int _{0}^{\infty }\mathrm{d}\omega {\mathrm{d^{2}} \over \mathrm{d}\omega ^{2}}{ \sin \omega \over \omega } = 0\,. {}\\ \end{array}$$The last equality holds since
$$\displaystyle{\left [{\mathrm{d} \over \mathrm{d}\omega }{ \sin \omega \over \omega }\right ]_{0}^{\infty } = 0\,.}$$ - 15.
The placing of the operators \(\hat{\mathsf{\boldsymbol{p}}}_{1}\) and \(\hat{\mathsf{\boldsymbol{p}}}_{2}\) is arbitrary as both \((\hat{\mathsf{p}}_{1})_{i}\) and \((\hat{\mathsf{p}}_{2})_{j}\) commute with the expression (δ ij + (n 12) i (n 12) j )∕2r 12. The reason thereof is we use the gauge \(\nabla \cdot \boldsymbol{\mathbf{A}} = 0\).
- 16.
We will show later in Chap. 7 that even when the particles indeed are elementary, the gyromagnetic ratio g differs from 1 as a result of effects of quantum electrodynamics. The deviation is very small, though, as it is proportional to α, see Eq. (7.204). Within our current precision, we can surely neglect this minor deviation.
- 17.
The nucleus of helium is an α-particle, a bound-state of two protons and two neutrons.
References
R.P. Feynman, R.B. Leighton, M. Sands, Feynman Lectures on Physics 1, 2 (Addison-Wesley, Reading, 1977)
V. Meyer et al., Phys. Rev. Lett. 84, 1136 (2000)
P.J. Mohr, B.N. Taylor, D.B. Newell, Rev. Mod. Phys. 84, 1527 (2012)
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Zamastil, J., Benda, J. (2017). The Hydrogen Atom and Structure of Its Spectral Lines. In: Quantum Mechanics and Electrodynamics. Springer, Cham. https://doi.org/10.1007/978-3-319-65780-6_3
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