Why can't we use a superconducting $LC$ circuit as a qubit? To construct a qubit, we need two quantum states, $|0\rangle$ and $|1\rangle$. Why can't we use the ground and first excited states of a quantum harmonic oscillator such as a superconducting $LC$ circuit?
 A: The key problem is a harmonic oscillator are the equally spaced energy levels $|0\rangle$, $|1\rangle$, $|2\rangle$, and so on.  
The issue is the following: Imagine you want to flip the state of the qubit, i.e., get $|0\rangle$ to $|1\rangle$ and vice versa. So you would drive the transition between $|0\rangle$ and $|1\rangle$ (e.g. using some microwave which matches the transition frequency).  However, the same microwave would also drive the transition from $|1\rangle$ to $|2\rangle$ -- that is, if your qubit was in the $|1\rangle$ state (or analogously for any superposition), it would now be brought into a superposition of the $|0\rangle$ and the $|2\rangle$ state.  Once you are in the $|2\rangle$ state, you would also get in the $|3\rangle$ state, and so on ...  and there goes your nice qubit: Your system constantly leaks into states which are part of your qubit space.
That's the key reason why one needs non-linearities (such as the Josephson junction in a SQUID): To get a non-linear oscillator (i.e. and LC circuit with a non-linearity), which does no longer have equally spaced energy levels, and thus it is possble to address the $|0\rangle\leftrightarrow|1\rangle$ transition without exciting other states as well.
