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In superconducting qubits, we use a circuit with a specific type of inductor and quantize the Hamiltonian. Because it's an anharmonic oscillator, we say that it has states -- $|0\rangle$ and $|1\rangle$.
This is perhaps a very basic question, but I don't understand how the circuit would exist in a superposition of these two states. How would we transform it into a superposition? I thought it exists only in these discrete energy eigenstates.
Any clarification would be appreciated.
First note that superposition is not a distinct type of quantum state. In quantum mechanics, every state may be cast as a superposition by appropriate choice of basis$^1$. For example, even though $|0\rangle$ is not a superposition in the computational basis, it is a superposition in the $|+\rangle$, $|-\rangle$ basis since $|0\rangle = \frac{1}{\sqrt2}|+\rangle+\frac{1}{\sqrt2}|-\rangle$. The basis in which quantum states are described is generally chosen to be the eigenbasis of some observable, often the Hamiltonian of the system. In this basis, only the states of definite energy are not superpositions.
In any case, superpositions are an inescapable feature of quantum mechanics. Consequently, if you choose to describe your LC oscillator using quantum mechanics, you will almost certainly encounter superpositions.
The greatest challenge in preparing and manipulating superpositions lies in ensuring the system is isolated. Uncontrolled interactions of the system with its environment act like measurement and collapse the system's state to an eigenstate of the associated observable.
In a closed quantum system, superpositions are very easily created. In fact, they are nearly impossible to avoid! This is a consequence of the fact that the basis states such as $|0\rangle$ and $|1\rangle$ are discrete and quantum evolution is continuous. Consider for example the Hamiltonian
$$ H = -\mu\sigma_x $$
where $\mu > 0$ is a real constant and $\sigma_x$ is the Pauli $X$ operator and assume that our system begins in the $|0\rangle$ state. According to the Schrödinger's equation, after time $t$ the system is in state
$$ |\psi(t)\rangle = e^{-iHt/\hbar}|0\rangle = e^{i\mu\sigma_xt/\hbar}|0\rangle = \cos\left(\frac{\mu t}{\hbar}\right)|0\rangle + i \sin\left(\frac{\mu t}{\hbar}\right)|1\rangle $$
and we see that for all except a few special values of $t$, $|\psi(t)\rangle$ is a superposition in the computational basis. More generally, if a system is in an eigenstate of an operator that does not commute with the Hamiltonian then the system's evolution very quickly takes it into a superposition of energy eigenstates.
The question
How would the circuit exist in a superposition?
seems philosophical and predicated on the assumption that the world is classical. Experimental evidence clearly indicates that the world is fundamentally quantum, so we might as well flip the question and ask instead
How does the circuit ordinarily appear to avoid existing in superposition?
A simple answer is that ordinarily on the quantum level the circuit interacts strongly with its environment and therefore is not a closed system.
$^1$ That said, superselection rules do impose certain limits on what superpositions are valid.
