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In this PSE question If a basis set is complete, are the elements in it mutually orthonormal? it was argued that if a basis set is complete, then the elements need not be mutually orthogonal (or even orthonormal). What happens if we know that a set of vectors are mutually orthonormal? Is there a way of knowing whether or not this set is complete?

Can someone also give a physical example of orthonormal sets of vectors that are not complete? For instance, I am reading that an ensemble average of an operator is defined by $$\langle O\rangle=\sum_iw_i\langle\alpha_i|O|\alpha_i\rangle$$ where the number $w_i$ is the statistical weight, satisfying $\sum_iw_i=1$ and $w_i\ge0$, and the states $\{|\alpha_i\rangle\}$ need not be complete, but they are normalized. Does this mean that they need not be orthogonal as well?

Any help will be appreciated.

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Take as the Hilbert space $\mathbb{C}^2$ and as a set of vectors the set that consists of the vector $(1,0)$ only. This is a set of orthonormal vectors, but it is not complete. It misses a second basis vector, e.g. $(0,1)$.

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  • $\begingroup$ So, not being complete does not necessarily mean that they are not orthogonal, or am I wrong? $\endgroup$
    – schris38
    Commented Nov 17, 2022 at 12:58
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    $\begingroup$ @schris38: Setting aside the double negatives: A set of vectors can be (a) mutually orthogonal, (b) a basis for the space they're a part of, (c) both, or (d) neither. All options are possible. $\endgroup$
    – Michael Seifert
    Commented Nov 17, 2022 at 13:00
  • $\begingroup$ Okay thanks @MichaelSeifert $\endgroup$
    – schris38
    Commented Nov 17, 2022 at 13:20
  • $\begingroup$ +1, but a better example might have been $\mathbb{C}^3$ and the set $(1,0,0)$ and $(0,1,0)$. These are more obviously orthonormal but not complete. $\endgroup$
    – mmesser314
    Commented Nov 17, 2022 at 13:54

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