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My question is about the input output formalism. In the original paper, the authors, Gardiner & Collet, describe the interaction of a system with a continuum of modes via the Hamiltonian:

$$H=H_S+\int_{0}^{+\infty} d \omega \ \hbar \omega b^{\dagger}_{\omega} b_{\omega}+i \int_0^{+\infty} d \omega \ \hbar g(\omega) (b^{\dagger}(\omega) c - b(\omega) c^{\dagger})$$

Where the operators $c$ and $c^{\dagger}$ are annihilation/creation operators acting on the system S, the operators $b(\omega)$ satisfy the commutation relations:

$$[b(\omega),b^{\dagger}(\omega')]=\delta(\omega-\omega')$$ $$[b(\omega),b(\omega')]=0$$

In the paper, they say that they extend to $-\infty$ the integrals to make calculations easier (explained in the paragraphs right after 2.2).

My question is: how to properly get to this kind of Hamiltonian in a particular setup: qubit put at the end of a semi infinite waveguide.

The physical model I consider

I am following this paper. Basically a superconducting qubit is capacitively coupled to a waveguide as shown on this picture:

enter image description here

In the end, the Hamiltonian can be written as:

$$H=H_S+H_W+H_{int}$$

With the system Hamiltonian (I call $C_{\Sigma}=C_c+C_J$ and $\phi_0=\hbar/2e$)

$$H_S=\frac{p_J^2}{2 C_{\Sigma}}-E_J \cos(\Phi_J/\phi_0)$$ $$H_W=\frac{C^2_c}{2 C_{\sigma}} (\partial_t \phi(0,t))^2+\int_{-\infty}^{0} \frac{p(x,t)^2}{2C_0}+\frac{\partial_x \Phi(x,t)}{2 L_0}$$ $$H_{int}=\frac{C_c}{C_{\Sigma}} p_J \partial_t \Phi(0,t)$$

To simplify the discussion, we can assume that the system is approximatively an Harmonic oscillator, thus:

$$H_S \approx \frac{p_J^2}{2 C_{\Sigma}}+\frac{E_J}{\phi_0^2}\frac{\Phi_J^2}{2} =\frac{p_J^2}{2m}+\frac{1}{2}m \omega^2 \Phi_J^2$$

With $m=C_{\Sigma}$, and $\omega=\sqrt{E_J/C_{\Sigma}}*1/\phi_0$

The field $\Phi(x,t)$ and its conjugated momentum field: $p(x,t)$ verify for $x<0$ the equation of propagation of a Free field. It gives us:

$$\Phi(x,t)_{\leftrightarrows}=\sqrt{\frac{\hbar Z_0}{4 \pi}}\int_0^{+\infty} \frac{d \omega}{\sqrt{\omega}} a^{\leftrightarrows}(\omega) e^{-i\omega(t \mp x/c)}+h.c$$ $$p(x,t)_{\leftrightarrows}=-i\sqrt{\frac{\hbar Z_0}{4 \pi}}\int_0^{+\infty} d \omega \sqrt{\omega} a^{\leftrightarrows}(\omega) e^{-i\omega(t \mp x/c)}+h.c$$

With $Z_0$ the impedance of the ligne: $Z_0=\sqrt{\frac{L_0}{C_0}}$, and $c=\frac{1}{\sqrt{L_0 C_0}}$ the velocity of waves in this line.

In term of physics, $\dot{\Phi}$ represent the voltage on the associated point, while $p(x,t)$ is the charge density.

First question:

To get an Hamiltonian like the one I wrote, I would simply replace the expression of the free field $\Phi(0,t)$, and express $p_J$ in function of the annihilation/creation operators which would in principle give me an expression that looks like my interacting Hamiltonian $i \int_0^{+\infty} d \omega \ \hbar g(\omega) (b^{\dagger}(\omega) c - b(\omega) c^{\dagger})$ (after neglecting counter rotating terms). I skip the details.

However, I would have in this Hamiltonian two modes $a^{\leftarrow}(\omega)$ and $a^{\rightarrow}(\omega)$.

This might not be a problem in principle but what confuses me is the following.

If I imagine putting an atom not at $x=0$ but somewhat before, at $x=-x_0$ the boundary condition in $x=0$ would give me a relationship between $a^{\rightarrow}(\omega)$ and $a^{\leftarrow}(\omega)$ such that I would only have "half the modes" I am writing here. This is the situation I am more familiar with.

Here the boundary condition is given by some relationship between $a^{\leftarrow}(\omega)$,$a^{\rightarrow}(\omega)$ and the system operators which I don't really know how to interpret in order to get the "half of the modes" result.

The question is thus: how to cleanly write down an Hamiltonian for this system that is suited for the input/output formalism from Gardiner.

Second question

In the paper they proceed to solve the dynamic by calling:

$$\Phi_{in}(t)=\Phi_{\rightarrow}(0,t)$$ and they get some input/output relations. However the input-output formalism which I gave the ref is not simply taking the right moving waves and calling it the input field. It consists in taking the right moving waves and make them freely evolve, thus without the interaction. In their definition of input/output fields, their operators $a^{\rightarrow}(\omega)$ in $\Phi_{\rightarrow}(0,t)$ will not freely evolve because of the interaction with the qubit. For me what they call input and output is not the same as what is done with Gardiner method.

It is not necessarily a problem but I would be interested to confirm if indeed the notion of input/output field they use here is different from the one from Gardiner.

And if it is the case, how to make the connection between the different input-output relations: one I would obtain from Gardiner, and the one they obtain here.

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  • $\begingroup$ What the bath operators in input-output theory correspond to physically is largely setup dependent. The original treatment by Gardiner-Collett was for a cavity, not for a waveguide. You may have to adjust the precise Hamiltonian depending on your setup. With that in mind, I think your question would profit from a concrete example/setup, e.g. of the waveguide that you are studying. $\endgroup$ – Wolpertinger Apr 24 at 12:28
  • $\begingroup$ With regards to the frequency extension to negative infinity: this approximation can be removed, as has been shown more recently: doi.org/10.1103/PhysRevX.10.011008. $\endgroup$ – Wolpertinger Apr 24 at 12:29
  • $\begingroup$ @Wolpertinger thank you for your comment. Why do you say it corresponds to a cavity ? In a cavity wouldn't he have only discrete set of modes ? My question can actually be rephrased as: what motivates him to only consider positive frequencies. What are the assumption behind his model. $\endgroup$ – StarBucK Apr 24 at 14:10
  • $\begingroup$ @Wolpertinger Actually I have some related questions on this topic: applications in circuit QED in order to drive qubits. But they deserve a post in themselves. I think we can understand my post as: to which physical system would his description be valid for. And then, for this system how to interpret the relationship between input/output operator and their relationship with left and right moving waves. Thank you very much for your time. $\endgroup$ – StarBucK Apr 24 at 14:12
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    $\begingroup$ @Wolpertinger I just edited to make the situation more focused $\endgroup$ – StarBucK Apr 25 at 11:09
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The question is thus: how to cleanly write down an Hamiltonian for this system that is suited for the input/output formalism from Gardiner.

I would suggest to do it the other way around! Instead of approximating your Hamiltonian and squeezing it into Gardiner-Collett form, try to apply the idea of their input-output approach directly to your Hamiltonian.

Specifically, set up the equations of motion for your circuit degrees of freedom. You will then find that the left/right field operators satisfy a certain equation which is very similar to the one in the Gardiner-Collett paper. The reason is that the approach is rather general; it mainly relies on bilinear coupling in some form, possibly on a Markov approximation if you want to go to the time domain and further approximations if your system is complicated. So the main task is to understand how to do this.

Indeed, this seems to be precisely what is done in the linked circuit QED paper. Eq. (14) seems to be the relevant input-output relation and it appears to be more complicated than the Gardiner-Collett case due to additional time derivatives. Trying to map to the Gardiner-Collett Hamiltonian is probably not a good approach in that case.

It is not necesseraly a problem but I would be interested to confirm if indeed the notion of input/output field they use here is different from the one from Gardiner.

The concept is very similar, but the underlying Hamiltonian and field degrees of freedom differ.

With regards to identifying what to call an input and an output: This is a physical question, namely which observables you are interested in/where you place your detectors. In the linked paper, they seem to measure something (voltages? excitation clicks? I don't know enough about circuit QED here...) along the transmission lines. So they identify their input/output channels as the corresponding asymptotic degrees of freedom.

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  • $\begingroup$ For your first point. I roughly see what you mean but what confuses me is that their input and output field are physically different from Gardiner because they don't have the "free evolution" around, they are Heisenberg operators. Because of that how can I be sure to not being wrong when I map this situation to Gardiner. Isn't it like associating carrot to apples ? $\endgroup$ – StarBucK Apr 25 at 11:43
  • $\begingroup$ @StarBucK In the paper they don't seem to map to Gardiner, do they? They just write down equations of motion and then they get the relation they need. If you do that consistently, it should not matter much whether it is Heisenberg operators or something else. It also seems that the free evolution is accounted for by the $e^{i\omega t}$-terms in Eq. (8), which seems to be the definition of these operators in their approach. $\endgroup$ – Wolpertinger Apr 25 at 11:51
  • $\begingroup$ Indeed they do not map to Gardiner. I want to find the mapping to Gardiner because I am more familiar with its approach. My issue is that, fine, I could use the relation they get. But then I would be a little bit lost in the intepretation. For instance I know that in input/output the injected power is $\hbar \omega_0 \langle b^{\dagger}_{in} b_{in} \rangle$. In a waveguide the power is proportional to the square of the voltage (thus square of $\dot{\Phi}$). Here I could then use $\hbar \omega_0 \langle \phi_{\rightarrow}^{\dagger} \phi_{\rightarrow} \rangle$ $\endgroup$ – StarBucK Apr 25 at 11:57
  • $\begingroup$ But as their operator are in Heisenberg picture I don't know if it would really physically mean the injected power. The free evolution is not really taken in account because their $a(\omega)$ can have a time-dependance in themselves. In Gardiner, we do $a(\omega_{t_0})e^{-i \omega (t-t_0)}$. This is different. $\endgroup$ – StarBucK Apr 25 at 11:58
  • $\begingroup$ The injected power should not be different in the Heisenberg picture though, should it? The expectation value should take care of that. I'm afraid that I don't really see a problem here, but maybe someone else has a better perspective. $\endgroup$ – Wolpertinger Apr 25 at 12:04

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