I am aware of the fact that enantiomers have different energies, for example L-amino acids have different energy than D-amino acids. The difference is not significant and is most usually about $10^{-18}$ eV. (1) Recently I have read that antimatter mirror images of compounds have actually the same energy. So L-amino acids will actually have the same energy as antimatter D-amino acids.

Can someone explain in relatively simply terms (meaning not too much math) why enantiomers have different energies and why matter-antimatter enantiomers have the same energy? Also if L is the more stable enantiomer for normal matter, will D be the more stable enantiomer for antimatter?

(1) Amino Acids and the Asymmetry of Life: Caught in the Act of Formation - Uwe Meierhenrich

  • $\begingroup$ Aren't people trying to phase out the L/D notation because it's misleading? $\endgroup$
    – user191954
    Commented Oct 15, 2018 at 4:35
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    $\begingroup$ Not when it comes to biomolecules. All natural amino acids are L, while using R/S nomenclature not all of them are S. $\endgroup$
    – EJC
    Commented Oct 15, 2018 at 4:37
  • $\begingroup$ I suppose matter-antimatter enantiomers have the same energy from simple CPT symmetry. But it will be quite a while before we can verify that empirically, with actual antimatter compounds. ;) $\endgroup$
    – PM 2Ring
    Commented Oct 17, 2018 at 6:14
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    $\begingroup$ You say the energy difference between enantiomers is "most usually few eV". That's much much too high. If you had said 10^-20 eV, that would be more reasonable. For example: pnas.org/content/96/8/4252 $\endgroup$ Commented Oct 18, 2018 at 12:24
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    $\begingroup$ "Recently I have read that antimatter mirror images of compounds have actually the same energy" - read where? $\endgroup$ Commented Mar 7, 2019 at 10:08

2 Answers 2


Does anti-Alice take levo-glucose and dextro-fructose in her anti-tea?

The putative equality of the levo-dextro energy difference our world and the dextro-levo difference in an anti-world would follow from CP-invariance, but CP-invariance is subtly broken by the complex phase of the CKM matrix. The experimental evidence for CP-violation comes from ${{K}^{0}}\And {{B}^{0}}$ decays, but there is as yet no corresponding evidence about CP-violation in leptons. CP-violation is a necessary but probably insufficient condition for inequality, since it is hard to see how this known kind of CP-violation would lead to inequality.

Published articles have calculated tiny levo-dextro differences in ordinary matter from CP-conserving weak neutral current interactions mediated by ${{Z}^{0}}$. They finger electron-neutron interactions as the dominant effect, with the P-violating interaction ${{H}_{PV}}\propto {{(\bar{\psi }{{\gamma }_{0}}\psi )}_{N}}{{(\bar{\psi }{{\gamma }_{5}}{{\gamma }_{0}}\psi )}_{e}}$ yielding terms $\propto {{(\mathbf{p}\cdot \mathbf{s})}_{e}}{{\delta }^{3}}({{\mathbf{x}}_{e}}-{{\mathbf{x}}_{N}})$ for each electron near a nucleus. The $Z$’s vectorial coupling to protons is weaker than its coupling to neutrons, by a factor of $4{{\sin }^{2}}{{\theta }_{W}}-1=-0.11$. One may therefore sum $(N-0.11Z)(\mathbf{p}\cdot \mathbf{s})\rho ({{\mathbf{x}}_{N}})$ over nuclei, where $\rho (\mathbf{x})$ denotes local electron density.

Since $\left\langle ground|{{H}_{PV}}|ground \right\rangle =0$, these P-violating terms have no effect on energy in 1st-order perturbation theory, but they do admix excited states, notably triplet states with parallel spins, which result in bilocal $\mathbf{s{s}'}\And \mathbf{p{p}'}$ correlations.

The articles go on to argue that the P-violating term operates in tandem with a P-conserving spin-orbit term ${{H}_{SO}}\propto (\mathbf{E}\times \mathbf{p})\cdot \mathbf{s}=\mathbf{E}\cdot (\mathbf{p}\times \mathbf{s})$, where $\mathbf{E}$denotes the electric field from another nearby atom. They then calculate energies in the Born-Oppenheimer approximation, which assumes fixed nuclear positions. In an anisotropic environment, particular components of the $\mathbf{s{s}'}\And \mathbf{p{p}'}$ correlations may be dominant. Unless these dominant components are parallel, their cross-product will define a preferred direction for $\mathbf{E}$, hence the chiral preference. The levo-dextro energy difference is 1st-order in ${{H}_{PV}}$ after all.

Bakasov el al: Ab initio calculation of molecular energies including parity violating interactions, J Chem Phys 109 (1998) 7263

Quack & Stohner: How do Parity Violating Weak Nuclear Interactions Influence Rovibrational Frequencies in Chiral Molecules?, Zeitschrift für Physikalische Chemie, 214, 5, 6752703 (2000)


I will expand on this later, but there is a main difference between regular enantiomeres, in which the particles are the same but in a different configuration, versus an antimatter enantiomere, in which all particles reverse their properties. In the antimater case, chirality relationships for instance, remain the same, so no changes in energy; but in a regular enantiomere the particules are the same but in different configurations, and parity non-conserving energy differences can and have been calculated.

  • $\begingroup$ I suppose that antimatter enantiomers also have different energies? $\endgroup$
    – EJC
    Commented Oct 20, 2018 at 17:13
  • $\begingroup$ Also if L is the more stable enantiomer for normal matter, will D be the more stable enantiomer for antimatter? $\endgroup$
    – EJC
    Commented Oct 20, 2018 at 17:13
  • $\begingroup$ L and D will have the same energy difference either in matter than in antimatter, second question: yes. The reason is that the antimater enantiomeres are identical up to a reflection to the matter enantiomere, but the L and D versions in matter (or antimater) are not $\endgroup$
    – user65081
    Commented Oct 20, 2018 at 17:20

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