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The Dirac equation works well for predicting the spectrum of the hydrogen atom, famously incorporating relativistic effects like fine structure. Yet, there seems to be a sense in which this is accidental. Specifically, let me quote the following passage from Sidney Coleman's recently published lecture notes, where he argues against the usage of the Dirac equation as a single-particle wave-equation (as opposed to a quantum field theoretic description):

"As a general conclusion, the corrections of relativistic kinematics and corrections from multi-particle intermediate states are comparable; relativity forces you to consider many-body problems. There are however very special cases, due to the specific dynamics involved, where the kinematic effects of relativity are considerably larger than the effects of pair states. One of these is the hydrogen atom. That's why Dirac's theory gives excellent results to order $(v/c)^2$ for the hydrogen atom, even without considering pair production and multi-particle intermediate states. This is a fluke." - Quantum Field Theory, Lectures of Sidney Coleman (Page 1).

Of course the hydrogen atom is very special, being integrable, and being one of the few exactly solvable systems in quantum mechanics.

But physically, what exactly is it about the hydrogen atom which makes the single-particle Dirac equation work, where we expect it fails for generic relativistic systems? What is the fluke here?

More generally, are there generic physical conditions under which we can expect relativistic results to be faithfully represented by simplified single-body techniques?

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    $\begingroup$ Perhaps you should provide a link to these lecture notes. $\endgroup$ May 31, 2019 at 10:47
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    $\begingroup$ @StephenG They're not freely available, but here is the publisher's website. You can also hear him talking about this in his lectures posted on YouTube here around the 10 minute mark (beware the video quality isn't great). He really doesn't go into anymore detail than what I quote here. $\endgroup$
    – EuYu
    May 31, 2019 at 10:54
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    $\begingroup$ A "fluke" is not a physical qualification. The argument needs to be more detailed. Also what is true for hydrogen is true for all atoms, molecules, materials. That makes it a very common "fluke". $\endgroup$
    – my2cts
    May 31, 2019 at 11:07
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    $\begingroup$ Related: physics.stackexchange.com/q/65359. That question also asks about the "fluke" mentioned in Sidney Coleman's lectures on Quantum Field Theory. $\endgroup$ May 31, 2019 at 23:01
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    $\begingroup$ Assume the QED effects are negligible, and we're dealing with a free QFT of electrons. It lives on the fermionic Fock space built from the 1-particle space of solutions of the Dirac equation. In the presence of the static Coulomb potential of the nucleus, these solutions form a discrete spectrum, and the QFT states are given in terms of the occupation numbers. Solving the Dirac equation is fully equivalent to understanding the QFT. Are you asking why QED is negligible? QED corrections are computable and turn our small. But still measurable, i.e. the Lamb shift. $\endgroup$ Jun 9, 2019 at 16:10

2 Answers 2

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Several research groups aim to measure binding energies and excitation energies in highly charged ions with great precision. This is done with the aim to find physics beyond the standard model. The first level of theory to compare experiment with is the Dirac + Coulomb + Breit Hamiltonian. Dirac for the kinetic energy and potential of the nucleus, Coulomb for the interaction between the electrons and Breit to account approximately for the retardation in the Coulomb field. After this level one can have a look at QED corrections and corrections due to a finite nucleus with finite mass.

The Dirac equation works really well, even for highly charged ions, as can be seen in excitation energies for 187Re29+ https://www.nature.com/articles/s41586-020-2221-0 or the binding energies in 187Re https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.127.072502. The Dirac equation is also used to calculate several core level excitations by x-rays. (see for many examples www.quanty.org).

One can go even one step further, very often one does not need the full Dirac equation, let alone QED corrections. In many cases the Schrödinger equation with first order relativistic corrections due to the Dirac equation describe reality rather accurate.

What Coleman might have referred to is that the Dirac equation is only a good approximation at moderately low densities. If you want to describe matter at room temperature and ambient pressure the Dirac equation is fine. If you want to describe an experiment done at CERN where two particles collide with near the speed of light and form a complete zoo of new particles, then the Dirac equation is really not good enough. Furthermore, deriving the low energy limit from QED is not entirely trivial for bound-states. It is not obvious that the radiative corrections are as small as they are in the physical world around us. See also the answer by Thomas here What is the reason that relativistic corrections for hydrogen atom work?

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  • $\begingroup$ This doesn’t really answer the question. $\endgroup$ Apr 13 at 21:44
  • $\begingroup$ Maybe it doesn't answer the question, but it is "useful", IMHO. $\endgroup$ Apr 14 at 0:41
  • $\begingroup$ @ZeroTheHero: To answer the question: " are there generic physical conditions under which we can expect relativistic results to be faithfully represented by simplified single-body techniques?". There are not really hard proofs in which limits these series expansions work. You can even show that many series expansions diverge. That said at ambient pressure and energies small compared to $mc^2$, the electron hole pair production energy, the Dirac equation is, based on comparison to experiment, really accurate and does provide the leading relativistic corrections to the Schrödinger equation. $\endgroup$
    – mwh
    Apr 14 at 7:10
  • $\begingroup$ @mwh I reread your post and maybe the 2nd paragraph — which at least gives some examples where the Dirac equation works well — should come in first position. $\endgroup$ Apr 14 at 12:33
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The Dirac equation works well not only for the hydrogen atom, but for all atoms, molecules and materials in general. The use of the term "fluke" is imprecise and wholly unwarranted.

Edit (20221207): The Dirac equation is widely used in quantum chemistry. On top of it corrections for virtual pairs and radiative effects are often used. Anyone who wants to understand quantum mechanics may find it useful to take a deep dive into quantum chemistry.

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    $\begingroup$ This reads more like an opinion than an actual answer. Coleman gives at least a heuristic estimate of why the single-particle Dirac equation fails for generic relativistic systems. I am not aware of any work on non-hydrogenic atoms which can be obtained without including quantum field theoretic effects (which are often put in by hand as explicit corrections). In fact, it is known that these corrections are important for the structure of more complicated atoms and molecules. $\endgroup$
    – EuYu
    May 31, 2019 at 12:06
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    $\begingroup$ I am not arguing against the validity of the Dirac equation itself of course, merely its treatment as a single-particle wave equation (without the inclusion of corrections from pair creation). Can you provide any references to back up your claim that the Dirac equation (without second quantization) treats "all atoms, molcules and materials in general" without corrections from QED? Otherwise, this answer feels too dismissive without any real reason. $\endgroup$
    – EuYu
    May 31, 2019 at 12:07
  • $\begingroup$ @Euyu note the the qualification "fluke" is an opinion. The OP is close to being opinion based. $\endgroup$
    – my2cts
    May 31, 2019 at 12:11
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    $\begingroup$ @euyu . "I am not aware of any work on non-hydrogenic atoms which can be obtained without including quantum field theoretic effects" There is the large research field of relativistic quantum chemistry based on the many electron Dirac equation. Only small effects such a Lamb shifts require radiative and virtual pair formation corrections. The no pair approximation gives an error of order $(Z\alpha) ^3$. $\endgroup$
    – my2cts
    May 31, 2019 at 13:53
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    $\begingroup$ It may be of interest to note that as a metal, gold has its colour because of relativistic effects: link.springer.com/content/pdf/10.1007%2FBF03215471.pdf $\endgroup$
    – jim
    Jul 7, 2021 at 14:05

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