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Taking into account the fact that the theory of quantum gravity does not exist and the QED calculations are not possible for most realistic chemical systems, what levels of accuracy can we expect from a theoretical calculation on simple (small) materials? Examples that come to mind are:

  1. Simplest molecular materials: $^3\ce{He}$, $^4\ce{He}$, $\ce{H_2}$.
  2. Simplest periodic systems, i.e. metallic $\ce{Li}$.
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    $\begingroup$ I gave an answer, and then saw a close vote. Maybe it was because you used "etc.", which makes it "open ended". You have given 3 specific atomic/molecular systems, but for periodic systems there is no end. Specific examples would help. $\endgroup$
    – Nike Dattani - No Free Time
    Commented May 13, 2020 at 3:54
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    $\begingroup$ Thanks for the suggestion, I edited the question accordingly. There is still no answer for the periodic system though... $\endgroup$
    – Roman Korol
    Commented May 13, 2020 at 15:29

1 Answer 1

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Exhibit 1: Ground state hyperfine splitting of the H atom:

1420405751767(1) mHz (present most accurate experiment)
142045199        mHz (present most accurate theory)

The error in the theory is due to the difficulty in treating the nuclear structure (2 up quarks + 1 down).

Exhibit 2: Ground state hyperfine splitting of the muonium atom:

4463302780(050) Hz (present most accurate experiment)
4463302880(550) Hz (present most accurate theory)

Why does it agree so well? μ$^+$ is a fundamental particle and therefore has no nuclear structure. QED is the correct theory for describing the interaction between pure electric charges (e$^-$ and μ$^+$). The only QFD (quantum flavordynamics) needed is for the electro-weak interaction between the particles (not for interactions within sub-nuclear particles), and QFD calculations were done here in anticipation of more accurate experiments to come.

Exhibit 3: Ground state hyperfine splitting of the He atom:

6739701177(0016) Hz (present most accurate experiment)
6739699930(1700) Hz (present most accurate theory)

Notice how much harder it is when you add an electron.

Exhibit 5: $S\rightarrow P$ transition in the Li atom:

14903.632061014(5003) cm^-1 (present most accurate experiment)
14903.631765(0006670) cm^-1 (present most accurate theory)

Exhibit 6: Ionization energy of the Li atom:

43487.15940(18) cm^-1 (present most accurate experiment)
43487.1590(080) cm^-1 (present most accurate theory)

Exhibit 7: Ionization energy of the Be atom:

76192.64(0060) cm^-1 (present most accurate experiment)
76192.699(007) cm^-1 (present most accurate theory)

Notice that theory is 1 order of magnitude more accurate than experiment!!!

Exhibit 8: Atomization energy of the H$_2$ molecule:

35999.582834(11) cm^-1 (present most accurate experiment)
35999.582820(26) cm^-1 (present most accurate theory)

See here for more info.

Exhibit 9: Fundamental vibration of the H$_2$ molecule:

4161.16632(18) cm^-1 (present most accurate experiment)
4161.16612(90) cm^-1 (present most accurate theory)

See here for HD and D$_2$.

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    $\begingroup$ In "Ideas of Quantum Chemistry" by Lucjan Piela there is a table with the contributions of various physical effects (non-relativistic, Breit, QED, beyond QED) to the ionization energy and dipole polarizability of the He atom, compared with experiment. IP(th):5945204223(42) MHz, IP(exp): 5945204238(45) MHz. $\alpha$(th):1383760.79(23) a.u.$\times10^{-6}$, $\alpha$(exp):1383791(67) a.u.$\times10^{-6}$ $\endgroup$
    – Verktaj
    Commented May 13, 2020 at 4:07
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    $\begingroup$ Only 10 digits? Table II in my paper gives it to 11 digits: github.com/HPQC-LABS/Carbon/blob/master/C.pdf $\endgroup$
    – Nike Dattani - No Free Time
    Commented May 13, 2020 at 4:16
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    $\begingroup$ well, maybe the reference of the book is a bit old, from 2001 and 2004. $\endgroup$
    – Verktaj
    Commented May 13, 2020 at 4:51
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    $\begingroup$ I am curious how the theory can be more precise and more accurate than the experiment. How do we know that to be true? Precision I can believe, accuracy should be based on experiments should it not? $\endgroup$
    – Tristan Maxson
    Commented Sep 16, 2020 at 2:32
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    $\begingroup$ @NikeDattani Interesting approach to thinking about it. I think I am willing to believe that then based on that argument. $\endgroup$
    – Tristan Maxson
    Commented Sep 16, 2020 at 2:41

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