1
$\begingroup$

I've come across the term coupling strength $g/2\pi$ in various contexts (transmon, quantum dots, optomechanics etc)

Also, I've come across coupling term $J$ (exchange coupling) a lot.

I think in some sense, these coupling terms mean the same thing but I have a very cloudy view of all these coupling terms.

More specifically, I would like to know have intuitive understanding of $g/2\pi$ (inter-qubit capacitive coupling strength) in this paper.

Also, why are they all in the units of frequency and the higher the value, the stronger the coupling? (Intuitively, I suppose higher numbers mean higher frequency of interaction... but I still have lingering discomfort with my understanding)

$\endgroup$

1 Answer 1

3
$\begingroup$

Coupling strengths such as $g$ and $J$ are typically discussed in context of coupled oscillating systems, both quantum and classical. They are a measure of the rate at which excitations are exchanged between the coupled subsystems. Before explaining the meaning of $g/2\pi$ in the paper by Barends $\textit{et al.}$ (Nature 508, 500–503 (2014)), it is important to realize that similar concepts/quantities emerge in classical physics. For example, if one takes two pendulums with equal lengths coupled by a spring and sets off one pendulum while the other is initially at rest, then the two pendulums begin to exchange energy with one another. In this case, the coupling strength is really just a measure of how quickly energy moves from one pendulum to the other. In this example, if one replaces the coupling spring by a long piece of dental floss, this would correspond to the $g \rightarrow 0$ limit. On the other hand, if one replaces the springs by a rigid rod, this corresponds to the $g\rightarrow\infty$ limit.

Now to answer your question about the Barends paper, if one has two qubits that are coupled with strength $g$ (meaning that when the two qubits are made resonant with one another they repel by an angular frequency $2g$), then the time required for qubit #1 to give its excitation to qubit #2 is $$T_\text{swap} = \frac{1}{2}\frac{2\pi}{2g} = \frac{\pi}{2g} \, .$$ Since (1/frequency) = time, this explains why $g$ and $J$ are typically reported in units of angular frequency. Note that as $g\rightarrow0$ (corresponding to a capacitive coupling of 0), the time required to exchange energy between the two qubits $\rightarrow\infty$. On the other hand as $g\rightarrow\infty$ (the two qubits are coupled by a huge capacitance), the time required for a swap $\rightarrow0$.

$\endgroup$
10
  • $\begingroup$ Thank you so much for the answer. I have a question about two qubits repelling by an angular frequency of 2g. I suppose when you bring two qubits resonant, you get an avoided crossing. So the repulsion by 2g just means the separation distance in energy. Do i have this right? $\endgroup$
    – Blackwidow
    Sep 22, 2018 at 15:27
  • $\begingroup$ In the paper, the separation is $\sqrt(2)g/ \pi $though (figure 3a) $\endgroup$
    – Blackwidow
    Sep 22, 2018 at 15:34
  • $\begingroup$ In the paper here, arxiv.org/abs/1112.2458 (from the same group), they have the avoided crossing separation as $g/ \pi$. I guess it comes down to how they wrote the off diagonal terms in their hamiltonian. one has $\sqrt(2)$ in their off diagonal term and the other one doesnt. $\endgroup$
    – Blackwidow
    Sep 22, 2018 at 16:18
  • $\begingroup$ The $\sqrt{2}$ in the $\sqrt{2}g/\pi$ equation in Fig. 3a comes from the level repulsion with the $|2\rangle$. These things are really just a matter of definition, but is important to realize where the $\sqrt{2}$ comes from in that figure @Blackwidow. $\endgroup$ Sep 22, 2018 at 18:48
  • $\begingroup$ Thank you so much. I am still confused about it though. I thought it was just avoided crossing. Shouldn't it have the same factor as arxiv.org/pdf/1112.2458.pdf ? Is there a reference I can check out maybe..? $\endgroup$
    – Blackwidow
    Sep 22, 2018 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.