Should the local coupling constant g not be without a unit, as the Holstein Hamiltonian suggests?

$ H=\underbrace{\sum_{P, Q} h_{P Q}^{(0)} a_P^{\dagger} a_Q}_{\text{Fixed geometry}}+\underbrace{\sum_{P \in \Omega} \sum_\lambda \hbar \omega_\lambda g_{P P}^\lambda\left[b_\lambda^{\dagger}+b_\lambda\right] a_P^{\dagger} a_P}_{\text{Atom displacement}}+\underbrace{\sum_\lambda \hbar \omega_\lambda\left(b_\lambda^{\dagger} b_\lambda+\frac{1}{2}\right)}_{\text{Vibrations}}$

$H$ = Holstein-Peierls Hamiltonian .$P$ or $Q$ = electronic levels
λ = normal mode g = coupling constant h(0) = transfer integral
a† and a = creation and annihilation operator for fermions
b† and b = creation and annihilation operator for bosons ℏω = energy of phonon

To give more context, in Ref. 1 for example, the local coupling constant is given in eV.

(1) Shizu, K.; Sato, T.; Tanaka, K. Vibronic Coupling Density Analysis for α-Oligothiophene Cations: A New Insight for Polaronic Defects. Chemical Physics 2010, 369 (2–3), 108–121. https://doi.org/10.1016/j.chemphys.2010.03.014.


1 Answer 1


The local coupling constant is typically given in units of energy, such as electronvolts (eV), because it represents the strength of the coupling between the electronic and vibrational degrees of freedom in a system (which is thus an 'energy-related' quantity). The Holstein Hamiltonian, which is used to describe the interaction between electrons and phonons in a system, includes the atom displacement term, which represents the coupling between the electronic state $P$ and the vibrational state $\lambda$, with the coupling constant $g^\lambda$ being the coefficient in front of this term for unit equality in the equation.

  • 2
    $\begingroup$ Yes, the coupling should have units of energy, but a dimensional analysis of the Hamiltonian presented above tells me that g lambda itself should be dimensionless - the units of energy are provided by hbar omega in front of it. Is the language just lax here, and the quantity in eV really the product hbar * omega * g ? $\endgroup$
    – Ian Bush
    Jan 28 at 6:42
  • $\begingroup$ I believe g lambda should be dimensionless as it's a ratio between the strength of the coupling between the electronic and vibrational degrees of freedom to the energy of the vibrational mode. It's often reported in units of energy as well as the product, for consistency with the vibrational energy scale. $\endgroup$
    – Hendrix13
    Jan 29 at 14:18
  • $\begingroup$ OK - but could you say mathematically exactly what is being reported in units of energy? $\endgroup$
    – Ian Bush
    Jan 29 at 14:24

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