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Jul 26, 2023 at 9:05 history bounty ended CommunityBot
Jul 20, 2023 at 8:11 history edited Roger V. CC BY-SA 4.0
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Jul 19, 2023 at 13:27 comment added Roger V. @FriendlyLagrangian on the other side, you could work in a different gauge and have EM field coupled directly to momentum - e.g., as one does when deriving Kubo formula.
Jul 19, 2023 at 12:43 comment added Roger V. @FriendlyLagrangian what I mean is that we can always choose $p=a+a^\dagger, x=a-a^\dagger$ - it is a matter of how we define $a$: $a=x+ip$ or $a=p+ix$ (the Hamiltonian is $H=p^2+x^2$ - position and momentum enter on equal footing )
Jul 19, 2023 at 10:36 comment added FriendlyLagrangian Regardless of this choice, there is a crucial sign difference between $F(a+a^\dagger,b-b^\dagger)=F(x,p)$ and $F(a-a^\dagger,b-b^\dagger)=F(p,p)$ which won't go away even if $x \mapsto p$ and $p \mapsto x$.
Jul 19, 2023 at 10:33 comment added FriendlyLagrangian Sure I understand that freedom in those definitions. However, my point is that $\hat{x}$ and $\hat{p}$ have already been chosen as $\hat{x}\propto a+a^\dagger$ and $\hat{p}\propto a-a^\dagger$ as the hamiltonian is an oscillator with $H_s= \omega a^\dagger a + \sum g_n (a+a^\dagger)^n \sim p^2+x^2+\sum_n x^n$. So $\hat{x}$ really is $\propto a+a^\dagger$ and not otherwise. Deciding now to use $\hat{x}\propto a-a^\dagger$ amounts to having $H_s \sim x^2+p^2+\sum_n p^n$ which is not the physical case at hand as far as I understand.
Jul 19, 2023 at 10:20 comment added Roger V. @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.
Jul 19, 2023 at 9:45 comment added FriendlyLagrangian 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?
Jul 18, 2023 at 14:43 comment added Roger V. @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.
Jul 18, 2023 at 14:33 comment added FriendlyLagrangian Maybe it might be worth mentioning they consider a matter-light interaction. I.e. they have an oscillator coupled with light (lasers I take).
Jul 18, 2023 at 14:29 comment added FriendlyLagrangian 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).
Jul 18, 2023 at 10:11 history answered Roger V. CC BY-SA 4.0