Basic doubt superconducting qubits This is a basic doubt regarding the principle behind superconducting qubits:

What I understand is when we add a Josephson junction to a standard LC circuit, it adds anharmonic behaviour, but what I do not understand is what quantizes the energy levels? because of what do these eigen states come into existence, what are these eigen states
and can this be done to any classical system?
 A: What is a photon ? The quantised representation of the energy (exchange) of the electromagnetic field. So basically you take the Maxwell's equations, and you quantise them. Usually you start from the electromagnetic energy $\propto ED+BH$ ($E$ electric field, $D$ Maxwell's vector, $B$ and $H$ the magnetic field and/or induction). By linearity and superposition principle, you can represent the free electromagnetic field as a superposition of non-interacting harmonic oscillators (when $D\propto E$ and $B\propto H$ you recover the usual free energy $\propto E^{2}+B^{2} \equiv F_{\mu\nu}F^{\mu\nu}$), the internal degree of freedom of which you can quantise, and that's the photon. (really quickly stated)
What is a circuit ? A lumped element representation of Maxwell's equations, when wavelength are far bigger than the size of the circuit. It requires small elements in the quasi-static approximation, at least. As long as you have linear elements, you can use the same trick as for the Maxwell's equations. So you can quantise the internal degree of freedom of the associated harmonic oscillator, and get something formally similar to photons. What you basically quantise are current and voltage, or magnetic flux and local charge. The free energy is always $\propto J\Phi+VQ$ in circuit, $J$ current, $\Phi$ magnetic flux, $V$ voltage and $Q$ the charge. To remain linear, a circuit must contain only inductance ($\Phi = LJ$) and capacitance ($Q=CV$) for which you get a free energy $\propto \Phi^{2} + V^{2}$ or $\propto J^{2} + Q^{2}$ (remark that $\dot{Q}\propto J$ and $\dot{\Phi}\propto V$).
Now if you add matter to your free theory, you start making interactions between the free bosons (initially the photons -- I neglect vacuum-vacuum interaction here for commodity). In electromagnetism, matter would appear as some more or less complicated relations between $D$ and $H$ on the one side and $E$ and $B$ on the other side, called constitutive relations. In circuits matter appears as constitutive relations between $J$, $\Phi$, $V$ and $Q$. That's how superconductivity enters the stage of quantised circuits : it makes a non-linear relation between $J$ and $\Phi$ in the form of the Josephson's relation $J=J_{c}\sin\left(\Phi\right)$.
So, to answer your question directly : the complicated, collective internal degrees of freedom of a circuit can be quantised by virtue of the analogy between the free energy of a circuit $\propto J\Phi+VQ$ and the free energy of a classical electromagnetic field $\propto ED+BH$ in the linear approximation. Usually, people call photon the quantised version of the exchange of energy in circuit. Being intrinsically one dimensional, these photons have no polarisation, ...
What I forget to mention before is that Maxwell's equation are not complete by themselves, you need to add an equation of motion for the basic charge. In classical electromagnetism the equation of motion is the Newton's one. In quantum realm it is obviously the Schrödinger's equation. So to completely understand the problem you should resolve the full Schrödinger's equation of both matter and field problem. Above I just gave a classical argument about the quantisation of the energy. But I think you can got the idea, and think about refinement later.
More informations about quantisation of circuit can be found in

Devoret, M. H. (1997). Quantum fluctuations in electrical circuits. In S. Reynaud, E. Giacobino, & J. Zinn-Justin (Eds.), Quantum Fluctuations: Les Houches Session LXIII. Elsevier. PDF
Vool, U., & Devoret, M. H. (2016). Introduction to Quantum Electromagnetic Circuits.

