On a view places, I've seen kets that look like this $\left|+\right>$ or this $\left|-\right>$ but I don't seem to find any explanation of this base online. Is it just a different notation for $0$ and $1$, or does it mean something else? Thanks for your help!

  • $\begingroup$ $|\pm \rangle = \frac{1}{\sqrt 2} (|0\rangle \pm |1\rangle)$ $\endgroup$ Feb 1, 2020 at 19:26
  • $\begingroup$ Thanks a lot! Does this Notation have any particular name, as I was not able to find it? $\endgroup$
    – Robinbux
    Feb 1, 2020 at 19:31

1 Answer 1


The set $\{ \left|+\right>, \left|-\right> \}$ is known as the polar basis. It easy to see that they are the result of applying the Hadamard transform $H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ to the standard basis vectors of a one-dimensional Hilbert space: $$ H\left|0\right> = \left| + \right>, $$ $$ H\left|1\right> = \left| - \right>. $$ You can read more about them Hadamard transform here.

  • $\begingroup$ I would just add that the basis is sometimes called Hadamard basis or x-basis. $\endgroup$ Feb 1, 2020 at 23:24
  • $\begingroup$ There is |0> and |1>. Why do we need different basis (in quantum computation), like polar basis and others? $\endgroup$
    – guest
    Feb 2, 2020 at 2:51

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