2
$\begingroup$

I'm reading this article about coherent exciton transport in photosynthetic light harvesting and the role of quantized vibrations. Along the way, I came across a section where the article claimed the exciton-phonon Hamiltonian (that is, the Hamiltonian representing exciton-phonon coupling) can be represented as follows:

$$H_{\text{elec-vib}}=\sum_nL_n\sum_\lambda\kappa_{n\lambda}(a_{n\lambda}^\dagger+a_{n\lambda}),$$where the system operators $L_n$ are the projectors $L_n=-|n\rangle\langle n|$. The coupling constants $\kappa_{n\lambda}$... describe the strength of the coupling of electronic excitation of pigment $n$ to the vibrational mode $\lambda$ of this pigment.

The article also notes that $a_{n\lambda}^\dagger$ is the phonon creation operator for the $\lambda^{\text{th}}$ vibrational mode of the $n^{\text{th}}$ pigment. My question is: where does the expression $(a_{n\lambda}^\dagger+a_{n\lambda})$ come from? Why not just use the operator $a_{n\lambda}^\dagger a_{n\lambda}$ to "count" the number of phonons on a given mode and then multiply that number by their coupling strength to the excitons? The operator $(a_{n\lambda}^\dagger+a_{n\lambda})$ seems very weird and unintuitive to me, and I can't even figure out what its eigenvalues would be.

$\endgroup$

2 Answers 2

2
$\begingroup$

Exciton (which can be though of as an electron bound to a hole) couples not to the number of photons (why would it indeed?) but to the electric field created by the phonon deformation (since the electron and the hole both carry electric charge.) This displacement field can be shown to have the form like the one given in the OP.

The situation is not unlike quantization of the electromagentic field, where the electric and magnetic field, as well as the potentials are linear combinations of creation and annihilation operators.

Electron-phonon is derived in many texts, e.g., the Introduction to Solid State Physics by Kittel.

$\endgroup$
1
$\begingroup$

 This is the standard coupling to phonons, for which we can have several intuitions / explanations:

  1. phonons are vibrations of the lattice, and the quantization of them is similar to quantum harmonic oscillator, with $a^{\dagger}(q)$ and $a(q)$ the ladder operators associated with a mode $q$. The combination $a^{\dagger} + a$ is therefore proportional to the displacement (as $a^{\dagger}+a \propto x$) which measures the effect of the phonon on the electronic transport (it creates a local potential that couples to the electronic density)
  2. phonons are not fermions, but bosons. So a coupling in terms of quadratic operators is not necessary and the simpler (i.e. lower order) terms are more significant, meaning the leading contribution will be linear in the phononic creation and annihilation operators
  3. phonons induce electric potential interaction with the electrons, so they follow the minimal coupling convention in which the EM field couples to electrons. The strength of the electric potential is proportional to the displacement (see 1)
  4. a coupling of the form $c^{\dagger}c a^{\dagger} a$ means that electron and phonon scatter off each other, but that doesn't allow for creation of phonons. A Hamiltonian quadratic in both electronic operators and phononic (bosonic) operators will have conservation of total number of phonons, which is not physical. Phonons are created when electrons scatter off nuclei in the lattice, and the way to describe such a creation is $c^{\dagger}_k c_q a^{\dagger}_{q-k}$
$\endgroup$

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.