Quality Factor in a Parallel LC Circuit I was wondering if there is a notion of a quality factor in a parallel LC circuit, since there is no resistance.
One can show that this circuit has a resonance frequency as follows:
Impedance:
\begin{equation}
\frac{1}{Z_{parallel}}=\frac{1}{Z_{C}}+\frac{1}{Z_{L}}=\frac{1}{\frac{1}{i\omega C}}+\frac{1}{i\omega L}=i\omega C+\frac{1}{i\omega L}
\end{equation}
\begin{equation}
Z_{parallel}=\frac{1}{i\omega C+\frac{1}{i\omega L}}=\frac{i\omega L}{1-\omega^{2}LC}
\end{equation}
Resonance frequency:
\begin{equation}
\omega_{0}=\frac{1}{\sqrt{LC}}
\end{equation}
How about the quality factor? For a RLC circuit it would be:
\begin{equation}
Q=\omega_{0}RC=\frac{RC}{\sqrt{LC}}=\frac{R}{\sqrt{\frac{L}{C}}}
\end{equation}
\begin{equation}
Z_{0}=\sqrt{\frac{L}{C}}\Rightarrow Q=\frac{R}{Z_{0}}
\end{equation}
But since there is no resistance, how does one think about the quality factor in this circuit?
 A: 
since there is no resistance.

That's not quite correct; in fact, there is infinite parallel resistance or, better, zero parallel conductance.
Recall that, for a parallel RLC circuit, the circuit elements are parallel connected.  If the parallel resistance were zero, the Q would be zero since the resistance is effectively an ideal wire shunt across the L and C.
For the parallel RLC, the greater the (parallel) resistance, the greater the Q.  In the limit as the resistance goes to infinity, there is simply a parallel LC circuit for which the Q is 'infinite'.
In other words, there is no dissipation and, at the resonance frequency, the parallel LC appears as an 'infinite' impedance (open circuit).
(The above assumes ideal circuit elements - any physical LC circuit has finite Q).
A: One way of thinking about a $Q$ factor is as a measure of how many periods of oscillation it takes for the amplitude to dissipate (for concreteness, lets say the number of periods required for the amplitude to decrease by a factor of 2). Then an LC circuit (that is, a series RLC circuit with $R=0$) has no dissipation and therefore will undergo infinitely many oscillations without the amplitude decreasing. Therefore, we would think of the $Q$ factor as being infinite.
