Occurance and disappearance of degeneneracies in a periodic structure of (quantum) LC circuits

Introductory part

I'm currently studying an analytical model of coupled LC circuits, in preparation for actually performing measurements on such structures. While the final goal will struggle with a lot of experimental considerations, the purpose of this post is only about pure theory.

The idea is as follows. Lets use the example of what we call a circular tetra circuit. What we have is a piece of transmission line, capacitively coupled to four resonators. I depicted it below in the world's worst paint job; once I get to spending some time on doing it properly I'll go back and make it acceptable. In any case, we have a piece of transmission line which is capacitively coupled to RLC circuit one with $C_{\kappa}$. This circuit has some internal resistance $R_1$, capacitance $C_1$ and inductance $L_1$. You might guess that in a later step in the research the inductor is replaced by something non linear to get a qubit, but this is irrelevant to the problem. In any case, circuit one is capacitively coupled to circuit two with $C_{J1}$, which is connected to the third one, which is connected to the fourth one, and then the fourth one is again coupled to first one using $C_{J4}$. Of course the transmission line has some impedance associated with it and such, this we also take into account. So, what are we interested in in this system? In crude terms, we want to characterize the values of the individual components by looking at it's reflection coefficicent, which is the $S_{1,1}$ element of the scattering matrix. In essence it looks at which part of the signal is reflected by the circuit, and this is obviously a function of things like internal resistance and capacitance etc. But this is the experimental part; in this preparatory step we would like to find a model for this system so that we can gain some insight into what kind of spectrum this coefficient will have for certain parameter values. We'll mostly be interested in the case where are $L$ values are equal, as well as the $R$ and $C_{J..}$, and then see what the influence is of for example going from $C_1 = C_2 = C_3 = C_4$ to $C_1 = C_2 = C_3 \neq C_4$.

But why is this interesting? Our interest comes from studies done in the linear case (no $C_{J4}$) where even in the case where all $C$ values are equal our reflection coefficient shows 4 dips; so four different resonances of the circuit. But by introducing this periodic boundary conditions, the situation is more interesting: there is a certain symmetry. Before I get into how we actually find this, in the example where all $C$ are equal our reflection coefficient will only have three resonances instead of four, as apparently one of the modes of the system is degenerate.

Potentially superfluous theory

So how can we even analyze this coefficient? What we do is we use the input output formalism. For this one takes the Hamiltonian of the isolated system in the second quantization formalism, together with coupling and loss rates for each field in the system. The exact details are quite long, so I'll put a relatively summarized version here, but feel free to ignore this if you are familiar with it, it is not so crucial.

The idea is that we go from the lagrangian formalism to a hamiltonian of the form

$H = \frac{1}{2} Q C^{-1} Q - \frac{1}{2} \phi L^{-1} \phi$

where Q is the charge, $\dot{\phi} = V$ and $L$ and $C$ are matrices with the capacitance values of the system. Important might be that we take $C_{\kappa}$ to couple to the first circuit; as we later introduce the non unitary part we will treat it as if the driving is only on the first circuit, and that the losses to the transmission line also come from this circuit. In second quant the hamiltonian is then adjusted to

$H = \sum_i{\hbar\omega_i(\hat{a}^\dagger_i\hat{a}_i+\frac{1}{2})} + \sum_{i\neq j}{\hbar J_{ij}\hat{a}^\dagger_i\hat{a}_j}+hc$

The values of $\omega$ and $J$ are determined via the previously mentioned capacitance and inductance matrices. Now that this is set up we use the Langevin equations

$\dot{a}_i=\frac{-i}{\hbar}[a_i,H]-\frac{\kappa_i}{2}a_i-\frac{\gamma_i}{2}a_i+\sqrt{\kappa_i}a_{in,i}$

Combining this with the boundary condition $a_{out,i} = \sqrt{\kappa_i}a_{i}-a_{in,i}$ we can solve for the scattering matrix elements $S_{i,j}=\frac{a_{out,i}}{a_{in,i}}$.

Finally, after a lot of work, we get to where we want to be. We can now set up our equations, assume $a(t) = |a| e^{-i\omega t}$, $a_{in}(t) = |a_{in}| e^{-i\omega t}$, and choose which kappa's and gamma's to use. Now, like I said before, we treat it as if the only $\kappa$ part is for the first circuit, while each circuit will have some losses due to $\gamma$. How exactly we determine the values of these is a longer story, but safe to say kappa depends on the impedance of the transmission line, on the capactitance $C_{\kappa}$ and the other capacitances it is connected to, and the gamma's depend on the resistance and connected capacitances of each circuit.

Actual question part

Okay, so now we have found a way to calculate the reflection coefficient for arbitrary parameter values. In Mathematica I set up a system to solve the Langevin equations and calculate the resulting coefficient. Of course I could have made a mistake somewhere, but drawing the circuit in Microwave Office gives a very similar spectrum, except you don't know how the actual calculations are done. So, there are a few cases, and understanding the difference in these cases is what my question is about. The first is if we take all $C_i$ to be equal. We get the spectrum shown below (I apologize, the y axis is the real part of the reflection coefficient, the x axis is the frequency in Ghz, and the fact that the y axis scale is so small is because I used a very small coupling $C_{kappa}$.) To me, this was very interesting! In contrast with the linear chain of 4, the spectrum for which was found in the exact same way, we have only three resonances instead of four. The symmetry of our system has to be the source of this; after all circuit 2 and circuit 4 are in the exact same position, surrounding wise. The spectrum is not symmetric because while $C_1$ and $C_3$ might be the same, circuit 1 is coupled to the transmission line with a capacitance, and the line has impedance. Now the question is, can we see 4 peaks if we break the capacitance symmetry? If we change $C_1$ or $C_3$ and leave the rest unchanged, our spectrum still has three dips. Intuitively this makes sense: $C_2$ and $C_4$ still have identical surroundings. What happens when we change $C_2$ or $C_4$? We see four dips, four frequencies! Again, this makes intuitive sense, the symmetry is broken.

Now, I tried thinking about why this degeneracy takes place. An idea was to look at the bare system hamiltonian, which in this case looks like $\left( \begin{array}{cccc} \text{$\omega $1} & \text{J12} & \text{J13} & \text{J14} \\ \text{J12} & \text{$\omega $2} & \text{J23} & \text{J24} \\ \text{J13} & \text{J23} & \text{$\omega $3} & \text{J34} \\ \text{J14} & \text{J24} & \text{J34} & \text{$\omega $4} \\ \end{array} \right)$

Obviously exact solutions of eigenvectors and eigenvalues will be terribly complicated, but we can look at them numerically. Lets analyse them for all $C$ equal. We get the results vizualized below: These are the eigenvectors and eigenvalues, also shown in a plot, in order of increasing frequency. Well, this is a nice intuitive picture. We have two eigenvectors with degenerate frequencies, and the eigenvectors themselves are also very similar. So this points towards this being our degenerate mode. What if we now change $C_2$, which previously broke our symmetry and gave us four dips? Okay, so the frequencies are no longer degenerate, and the eigenvectors also look quite different. Seems to make sense to me. But here comes the trouble.

Actual Actual question

Now lets look at the same picture for $C_1$ having a different value. Remember, in this case our spectrum still has only three dips, so the symmetry should still be present. But what do we see in the eigenvalues and vectors? The picture is essentially the same as with changing $C_2$! We see no degeneracy in the eigenvalues, and no clear equivalent eigenvectors. My question is this: how can I understand these eigenvalues and eigenvectors, which seem to indicate a broken symmetry, in combination with a reflection coefficient that clearly does show a symmetry being present in the system? What am I missing? Most general way of stating my question

In my system a certain symmetry is present. When all $C_i$ values are equal, the total systen exhibits only three resonances (the three dips in the coefficient) even though it has 4 resonators. Intuitively I would think one mode is degenerate, and this is also shown by numerical analysis. When $C_2$ or $C_4$ is changed, we get 4 resonances instead, and the degeneracy of the eigenvalues is also lifted. However, for changing $C_1$ or $C_3$ we still have only three resonances, even though none of the eigenvalues are still degenerate. Why do I see three resonances, but no apparent degeneracy or symmetry in the eigensystem?

Polar Plot

As CuriousOne suggested, I also looked at the polar plot for the different scenarios. As shown in the figure below, both the fully symmetric and the case for broken symmetry in C1 show three closed curves, while breaking the symmetry in C2 shows 4. I'm still not sure how to answer my question from this, however.. • I am having a hard time deciphering what your question is. Is this the smallest system that exhibits the problem that you are having? – CuriousOne Jun 5 '15 at 3:54
• Hm, I apologize. I was going to say yes, but in a system of three instead of four this also occurs. Here, if $C_1 = C_2 = C_3$, the reflection coefficient has two dips. If $C_2$ or $C_3$ is made different from the other two, you see three dips, and if you make $C_1$ different you still only get two. I could update the question with those pictures instead, but the problem is the same, when it comes to the eigenvalues/eigenvectors. So let me try to phrase my question in as general a way as possible, in an edit. – user129412 Jun 5 '15 at 8:19
• One thing I do notice is that both in the case of $C_i$ all being equal and $C_1$ being different, there is an eigenvector of which the first component (which is linked to the first resonater, which is coupled to the outside) is zero. However, in the case of changing $C_2$, this is no longer present. This pattern does seem to generalize to larger structures of 6 and 8. – user129412 Jun 5 '15 at 9:37
• So your problem comes down to the number of distinct eigenfrequencies/degeneracies of the system? – CuriousOne Jun 5 '15 at 14:42
• Yeah. The degeneracy is lifted when I change $C_1$, but I still get only three modes in the reflection coefficient. – user129412 Jun 5 '15 at 14:51