$|0⟩$ and $|1⟩$ are usually referred as the computational basis. $|+⟩$ and $|-⟩$, the polar basis.

What about $|i\rangle$ and $|\mbox{-}i\rangle$?

basis in Quirk

And collectively? Orthonormal states?

References are welcomed!


3 Answers 3


In my opinion the nature of these states becomes quite clear when we look at it from an optics angle. We can identify the computational basis states with the vertical and horizontal polarization directions: $$ |0\rangle \sim |\updownarrow\,\rangle \qquad |1\rangle \sim |\leftrightarrow\,\rangle $$ The superposition states then correspond to diagonally polarized light: $$ |+\rangle \sim |⤢\,\rangle \qquad |-\rangle \sim |⤡\,\rangle $$

Now, the superposition states that have an $i$ do actually correspond to circularly polarized light: $$ |+i\rangle \sim |\circlearrowright\,\rangle \qquad |-i\rangle \sim |\circlearrowleft\,\rangle $$ Which also explains the labels $R$ for right and $L$ for left in @Z..'s post.

This correspondence is explained by the fact that circularly polarized light is created by superposing vertical light with horizontal light that has a $\pi/2$ phase difference. This phase difference is exactly $\mathrm{e}^{i \pi/2}=i$.

  • $\begingroup$ So, circular basis? If so, that matches one of the comments from twitter (I also posted the question there). $\endgroup$
    – luciano
    Commented Aug 19, 2020 at 19:04
  • 1
    $\begingroup$ You could call it that way, but I think of this more as a tool for understanding. I'd refer to it as the Y basis. $\endgroup$ Commented Aug 20, 2020 at 10:50

Quirk refers to the $\frac{1}{\sqrt{2}}|0\rangle + \frac{i}{\sqrt{2}}|1\rangle$ state as $|i\rangle$ and to the $\frac{1}{\sqrt{2}}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle$ state as $|-i\rangle$:


When I implemented this it just seemed like a natural choice at the time. I didn't get it from a textbook or a paper.

  • 4
    $\begingroup$ This is not the only notation for these states, but it is one of the most common notations (if not actually the most common). I also think that this notation is often re-invented: we simply don't talk as much about the eigenstates of the Y operator, for what I would argue are somewhat superficial cultural reasons. $\endgroup$ Commented Aug 17, 2020 at 21:31
  • $\begingroup$ @NieldeBeaudrap any reference to this notation besides Quirk? I saw it in Quirk and I assumed it standard. $\endgroup$
    – luciano
    Commented Aug 18, 2020 at 0:46

This is another reference.

$|i\rangle$ and $|\mbox{-}i\rangle$ are two orthogonal y-basis states. In the above link they are called $|R\rangle$ and $|L\rangle$.

$$|i\rangle = \frac{1}{\sqrt{2}}\left[ \begin{array}{c} 1 \\ i \end{array} \right] \;\; , \;\; |\mbox{-}i\rangle = \frac{1}{\sqrt{2}}\left[ \begin{array}{c} 1 \\ -i \end{array} \right]$$

You can simply check the orthonormality by using the definition of inner product space $\mathbb{C}^2$, $\langle v | w\rangle =\sum(v_i^{*} w_i)$, and Kronecker delta function.

$$\langle i|i\rangle = [1.1 + (-i).i]/2 = 1$$

$$\langle i|\mbox{-}i\rangle = [1.1 + (-i).(-i)]/2 = 0$$


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