Is a qubit always in superposition?

Question

I am introduced to ancilla qubits which are usually initialized to $\vert 0 \rangle$. It seems that an ancilla qubit is equivalent to the $0$ bit in classical computing as it will evaluate to $\vert 0 \rangle$ 100% of the time. My question is that can I conclude any qubit will always be in superposition regardless of the probabilities of $\alpha$ and $\beta$ since $\vert 0 \rangle + 0\vert 1 \rangle = \vert 0 \rangle$ and $0\vert 0 \rangle + \vert 1 \rangle = \vert 1 \rangle$? How about physically, are qubits prepared according to some procedure that differ if it will be initialized to $\vert 0 \rangle$ or $\vert 1 \rangle$ or in superposition?

Edit: this is a related comment but it doesn't answer my question regarding if the process of physically preparing qubits different from non-entangled qubits vs entangled ones.

Does this answer your question? Represent qubit in a superposition — Martin Vesely, Dec 27 '19 at 10:31
Regarding your statement "It seems that an ancilla qubit is equivalent to the $0$ bit in classical computing as it will evaluate to $\vert 0\rangle$ 100$ of the time," you mean "it will initially evaluate to $\vert 0\rangle$...", right? — Mark Spinelli, Dec 27 '19 at 13:38
@Martin Vesely, thanks for the suggestion. I read the answers there but unfortunately, I still have doubts. — M. Al Jumaily, Dec 27 '19 at 13:50

score 8 · Answer 1 · edited Jun 24 '21 at 06:14

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Superposition is a basis dependent concept. Namely the $|0\rangle$ and $|1\rangle$ states are commonly said not to be superposition states exactly because one of the two coefficients in the $\{|0\rangle,|1\rangle\}$ basis expansion is zero. However, using the x-basis representation, $\{|+\rangle, |-\rangle\}$ one finds $|0\rangle = (|+\rangle + |-\rangle)/\sqrt{2}$ and $|1\rangle = (|+\rangle - |-\rangle)/\sqrt{2}$, which are superposition states with both coefficients non-zero.

Qubits naturally prepare themselves in the $|0\rangle$ state if one waits long enough (much longer than the T1 time) as it is the ground state of the system. To get any other state one must apply gates to the qubit that rotate the state to the desired position on the Bloch sphere.

edited Jun 24 '21 at 06:14

epelaez

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answered Dec 27 '19 at 09:52

Paul Nation

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There was a similar discussion here: https://quantumcomputing.stackexchange.com/questions/9267/represent-qubit-in-a-superposition/9269#9269 – Martin Vesely Dec 27 '19 at 10:30
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The $\vert 0\rangle$ state is often, but not always, the ground state of a qubit . This may depend on convention, as in by convention, classically we think high voltage as $0$ and low/ground voltage as $1$. More importantly this may depend on how the qubits are physically instantiated. For example what is the ground state of polarization states of a photon? – Mark Spinelli Dec 27 '19 at 13:34
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A good point. However without a concrete example, the second part of the question then just becomes “it depends on the physical implementation”. – Paul Nation Dec 27 '19 at 13:37

score 3 · Answer 2 · answered Jan 09 '20 at 00:24

Yes, any state can always be described as a coherent superposition of some other set of states. "Being in a superposition" is not a property of just a quantum state, but rather a property of a quantum state relative to some basis. It doesn't make sense to talk of a superposition if not relative to a basis. We often don't explicitly specify this basis simply because it's usually obvious from the context which one is it.

Is a qubit always in superposition?

2 Answers2