4
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ Aren't people trying to phase out the L/D notation because it's misleading? $\endgroup$
    – user191954
    Commented Oct 15, 2018 at 4:35
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ "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

1
$\begingroup$

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.

References:
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)

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.