Qubits 101: superposition, measurement, and what a “state” means
A practical mental model for qubits: vectors, probabilities, and why measurement isn’t just “looking.”
When people first meet quantum computing, “superposition” is usually presented as mystical. The reality is more boring—and more useful: a qubit is a 2D complex vector that you manipulate with unitary transforms, and you only ever get classical data out via measurement.
The smallest math you need
We label the computational basis as (|0\rangle) and (|1\rangle). A general pure qubit state is:
[ |\psi\rangle = \alpha|0\rangle + \beta|1\rangle ]
where (\alpha) and (\beta) are complex numbers and (|\alpha|^2 + |\beta|^2 = 1).
That normalization condition is the only thing keeping probabilities sane.
Measurement: probabilities, not “reading the value”
If you measure in the computational basis:
- You get outcome 0 with probability (|\alpha|^2)
- You get outcome 1 with probability (|\beta|^2)
After you measure, the state collapses to the basis state you observed. That’s why measurement is “destructive” in many algorithms: you don’t get to keep the coherent state you had before you asked for an answer.
So what is “superposition” in practice?
“Superposition” just means both basis amplitudes can be non-zero. But the important part is that amplitudes can be complex and can interfere.
If you only remember one sentence:
Quantum computing is about shaping amplitudes so the right answers interfere constructively and the wrong ones cancel out.
A concrete example: the Hadamard gate
The Hadamard gate (H) maps:
- (|0\rangle \rightarrow (|0\rangle + |1\rangle)/\sqrt{2})
- (|1\rangle \rightarrow (|0\rangle - |1\rangle)/\sqrt{2})
So if you start in (|0\rangle) and apply (H), you’ll measure 0 or 1 with 50/50 probability.
What to take away
- A qubit isn’t a probability distribution—it’s a vector of amplitudes.
- Measurement gives you classical bits sampled from those amplitudes.
- Quantum speedups come from interference + structure, not from “trying all answers at once.”
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