I have seen that in many experimental papers dealing with some form of oscillator coupled to a bath, people model system as $H = H_s + H_b +H_{sb}$ with $$ H_s = \omega_s a^\dagger a, \quad H_b = \sum_n\omega_n b^\dagger_n b_n, \quad H_{sb}= −(a-a^\dagger) \sum_n h_n (b_n −b_n^\dagger). $$ I don't understand what $H_{sb}$ physically means or where it comes from.

I get that it can be rewritten in terms of $\hat{x}$ and $\hat{p}$ like $H_{sb}\sim \sum_n g_n \hat{p}^s \hat{p}_n^b$ but what does that physically mean?

I'm used to see terms in condensed matter as things like density-density $n_{i}n_{i+1}$ or just plain hopping $c_i^\dagger c_{i+1}$ and so on but nothing like that. So if I expand it is not very illuminating for me: $$ H_{sb}= −(a-a^\dagger) \sum_n h_n (b_n −b_n^\dagger)= -\sum_n h_n(ab_n-ab_n^\dagger-a^\dagger b_n+a^\dagger b^\dagger_n), $$ which kind of amounts to all possible things really, i.e. $ab_n^\dagger$ is like removing an excitation from the system and putting it in the bath, etc.

Can someone shine some light on this?

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    $\begingroup$ References? ... $\endgroup$ Jul 14, 2023 at 14:36
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    $\begingroup$ One possibility is a dipole Hamiltonian, where $\hat{V}=-\vec{E}\cdot\vec{d}$, where $\vec{E}$ is the electric field operator---roughly proportional to $\hat{a}-\hat{a}^{\dagger}$---and $\vec{d}$ is the dipole moment of the particle---proportional to the position $\hat{x}$, which roughly goes as $\hat{b}+\hat{b}^{\dagger}$ if the particle is an oscillator. The minus sign is just a phase that can be chosen independently, I believe. $\endgroup$
    – march
    Jul 14, 2023 at 15:29
  • $\begingroup$ @march I have been further reading inspired by your answer and indeed it seems like it might be something along those lines. I found the Jaynes-Cummings model, which looks like it contains an interaction term like $H_{sb}$ whose origin is some light-matter interaction as you suggested. It might be worth studying that model then! $\endgroup$ Jul 16, 2023 at 22:09

1 Answer 1


I don't understand what $H_{sb}$ physically means or where it comes from.

This, of course, depends on the specific problem at hand - the nature of the bath and the type of interaction. An often considered case is electron-phonon interaction, where an electron is coupled to the (polar) phonon field via something like $$ H_{int}=-e\mathbf{r}\cdot\mathbf{E}(\mathbf{r}). $$ Polar phonons correspond to the polarization (i.e., electric field/dipole moment) induced when two atoms in a unit cell of a crystal a displaced from their equilibrium positions. When quantizing the phonon field, the displacement is then treated as the momentum of the oscillator, that is $$ \mathbf{E}(\mathbf{r})\propto\sum_{\mathbf{k}}\lambda_\mathbf{k}\left(b_\mathbf{k}^\dagger e^{i\mathbf{k}\mathbf{r}} - b_{-\mathbf{k}}e^{-i\mathbf{k}\mathbf{r}}\right) $$ or something of the kind.

Likewise, if we consider an electron in a harmonic oscillator, its position operator is replace by something like $a^\dagger \pm a$.

There many different types of baths that can be modeled as collections of oscillators (at least in linear approximation) - phonons, plasmons, magnons, etc. Another key point to remark in the coupling is that it is assumed to be linear (dipole approximation), although more complex cases are possible.

Derivations can be found in many solid state and/or many-body texts: Kittel, Mahan, AGD, etc.

Coupling to EM field
Another commonly occurring case is the coupling of electrons (in atoms or solid state) to electromagnetic field. The minimal coupling Hamiltonian is $$ H=\frac{1}{2m}\left[\mathbf{p}+\frac{e}{c}\mathbf{A}(\mathbf{r},t)\right]^2-e\varphi(\mathbf{r},t).$$ One often chooses to work in Coulomb gauge, where $\nabla\varphi=0$, so that the Hamiltonian becomes $$ H=\frac{\mathbf{p}^2}{2m}+\frac{e}{2mc}\left[\mathbf{p}\mathbf{A}(\mathbf{r},t)+\mathbf{A}(\mathbf{r},t)\mathbf{p}\right]+\frac{e^2}{2mc^2}\mathbf{A}^2(\mathbf{r},t),$$ The last term is then neglected, as describing two-photon processes that are unlikely for the energies of interest, whereas $\mathbf{A}(\mathbf{r},t)$ can be assumed to be constant on the scales of atoms or the transition regions of interest in semiconductors (the wave length of might is hundreds of nanometers, whereas a size of atom is several angstroms.) We then have the coupling part $$ H_{int}\propto \mathbf{p}\sum_\mathbf{k}\left(b_\mathbf{k}+b_\mathbf{k}^\dagger\right) $$ (See Quantization of EM field.)

Oscillator raising and lowering operators
Note that variables $x$ and $p$ in an oscillator Hamiltonian enter in symmetric way (up to coefficients): $$ H=\frac{\hat{p}^2}{2m}+\frac{m\omega^2\hat{x}^2}{2}. $$ E.g., in momentum space $\hat{x}=x,\hat{p}=-i\hbar\partial_x$, and the Schrödinger equation takes form $$\left[\frac{1}{2m}\frac{d^2}{dx^2}+\frac{m\omega^2x^2}{2}\right]\psi(x)=E\psi(x),$$ with the well-known solutions expressed in terms of Hermit polinomials.
In momentum space we have $\hat{x}=i\hbar\partial_p,\hat{p}=p$ and the SE becomes $$ \left[\frac{m\omega^2}{2}\frac{d^2}{dp^2}+\frac{p^2}{2m}\right]\phi(p)=E\phi(p),$$ which has the solutions expressed in terms of Hermit polinomials of variable $p$.

Likewise, the definitions of raising and lowering operators are interchangeable in terms of momentum and position: traditionally one defines them in such a way that $$x\propto a+a^\dagger, p\propto a-a^\dagger,$$ but it could as well be $$x\propto a-a^\dagger, p\propto a+a^\dagger,$$ the difference being the phase factors of the wave functions.

  • $\begingroup$ Thank you for your insight! I'm still wondering why $\hat{x} \sim a^\dagger \pm a$, I get that there is a freedom in the labeling of what $\hat{x}$ and $\hat{p}$ are, after all they are just a basis of sorts. However, in these papers they talk about a non-linear oscillator described by $H_s= \omega a^\dagger a + \sum g_n (a+a^\dagger)^n$ which makes me think that $\hat{x} \propto a+a^\dagger$ and $\hat{p} \propto i (a-a^\dagger)$. This brings us back to my question: it looks like it is a $\hat{p}^s \hat{p}^b$ type of interaction rather than what you described (at least superficially). $\endgroup$ Jul 18, 2023 at 14:29
  • $\begingroup$ Maybe it might be worth mentioning they consider a matter-light interaction. I.e. they have an oscillator coupled with light (lasers I take). $\endgroup$ Jul 18, 2023 at 14:33
  • $\begingroup$ @FriendlyLagrangian The Hamiltonian in your question is linear - unless you mean that $h_n$ is not a coefficient but a function? In any case, the generalization is pretty straightforward: if your systems and your bath are oscillators, while the interaction is a non-linear function of their variables, you simply plug the variables into this function: $H_{int}=F(x,E)=F(a+a^\dagger, b-b^\dagger).$ Note that this is already the case in the phonon Hamiltonian as I wrote it (it is non-linear in $\mathbf{r}$), although one often linearizes the exponents. $\endgroup$
    – Roger V.
    Jul 18, 2023 at 14:43
  • $\begingroup$ I omitted the non-linear interaction for the sake of clarity as it just complicates the question unnecessarily. I only mentioned this in the comment to illustrate that $\hat{x}\propto a+a^\dagger$ and not $\hat{x} \propto a-a^\dagger$. This would mean that your $H_{int}=F(x,E)=F(a+a^\dagger,b-b^\dagger)$ is not of the form of my $H_{sb}$ above as the latter is an interaction like $H_{int}=F(a-a^\dagger,b-b^\dagger)=F(p,p)$, coupling momentum of the system with the momentum of the bath. Thus, I am not so convinced this originates from $H_{int}=-e \vec{r}\cdot \vec{E}$. What am I missing? $\endgroup$ Jul 19, 2023 at 9:45
  • $\begingroup$ @FriendlyLagrangian I don't quite see the difference between using $a+a^\dagger$ vs. $a-a^\dagger$ - it is a matter of how one defines the operators, since oscilator Hamiltonian is symmetric in interchanging momentum and position. $\endgroup$
    – Roger V.
    Jul 19, 2023 at 10:20

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