Is qubit superposition a basis-dependent concept?

5

The qubit $a\left|0\right>+b\left|1\right>$ is a superposition w.r.t. the basis $\left\lbrace\left|0\right>,\, \left|1\right>\right\rbrace$ because it may collapse to $\left|0\right>$ or $\left|1\right>$ when measured w.r.t. the above basis. However the same qubit, when measured w.r.t the basis $\left\lbrace a\left|0\right>+b\left|1\right>, b^*\left|0\right>−a^*\left|1\right>\right\rbrace$, always collapses to the first vector.

Can I therefore conclude that superposition is a basis-dependent concept?

Mathlusiverse

Posted 2019-01-13T22:45:49.953

Reputation: 53

Answers

4

Yes. You already have the reasoning for why. Sometimes the basis is implied but not explicitly stated though. So caution when reading something like that.

AHusain

Posted 2019-01-13T22:45:49.953

Reputation: 3 383

Thanks! I thought it was a little bit strange and not sure if I misunderstand the concept. Thanks for the confirmation. – Mathlusiverse – 2019-01-14T05:34:43.173

2It might help to emphasise that in real physical systems, there is often a "natural" basis. For example, if we encode the levels 0 and 1 as two distinct energy levels of an atom, then there is typically an error mechanism called relaxation (aka amplitude damping) that performs an equivalent function to the measurement in the 0/1 basis. – DaftWullie – 2019-01-14T07:51:11.650