Clarification on what we can consider a qubit to be

  • #1
eprparadox
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2
In a 2 level quantum system, should I consider the states

[tex] |0> [/tex]

and

[tex] |1|> [/tex]

to be qubits by themselves?

Or is only the SUPERPOSITION of these two states,

[tex] \alpha |0> + \beta |1> [/tex]

considered to be a qubit?
 
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  • #2
Nevermind, they have to be qubits as well. If we consider our superposition to be a qubit, then we can set ## \alpha = 1 ## and ## \beta = 0 ## and that should be an appropriate qubit state.
 
  • #3
eprparadox said:
In a 2 level quantum system, should I consider [...] to be qubits by themselves?
The qubit is the 2-state system, just like a classical bit is a binary variable. The states are not qubits, but the qauntum analogues of the classical values.
 
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  • #4
You should consider the pair to form a qubit. On their own they're not very useful.

Another useful way to think about what a qubit is is as an anti-commuting pair of observables, such as the observable ##|0\rangle##-vs-##|1\rangle## paired with the observable ##\frac{1}{\sqrt{2}} \left( |0\rangle + |1\rangle \right)##-vs-##\frac{1}{\sqrt{2}} \left( |0\rangle - |1\rangle \right)##.
 
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