# Is there a convention for denoting $Y$ eigenstates?

Two common shorthands for eigenstates of the $$Z$$ operator are $$\{|0\rangle,|1\rangle\}$$ and $$\{|1\rangle,|-1\rangle\}$$, where in the first case we have $$Z|z\rangle=(-1)^z|z\rangle$$ and in the second case we have $$Z|z\rangle=z|z\rangle$$. The common shorthand for eigenstates of the $$X$$ operator is $$\{|+\rangle,|-\rangle\}$$, where $$X|\pm\rangle=\pm|\pm\rangle$$.

Is there any commonly used shorthand for eigenstates of the $$Y$$ operator?

There are several, these are the ones I have seen:

• $$|+\rangle_y,|-\rangle_y$$ a bit lazy but easy to remember
• $$|+i\rangle,|-i\rangle$$ the same as before but you replace the sub index with an imaginary unit $$i$$ inside the ket
• $$|\circlearrowleft\rangle,|\circlearrowright\rangle$$ this notation is borrowed from light polarization, as you can use photons for light too, circular polarization is the equivalent of $$y$$-axis for photons (It is easy to mess up which is right and which is left in this notation, but it is used by Qiskit, see section 2).
• $$|\mathrm R\rangle,|\mathrm L\rangle$$ for left and right, again based on the polarization of light (example: Quantum Inspire).

In all these notations the first state is $$(|0\rangle+i|1\rangle)/\sqrt{2}$$ and the second one is $$(|0\rangle-i|1\rangle)/\sqrt{2}$$.

What I like about the second notation $$|\pm i\rangle$$ is that it kind of tells you the coefficient (part of it) in front of the $$|1\rangle$$ state (analogous to the $$x$$ basis $$|\pm\rangle)$$.

• Just double checking. My brain equates counter-clockwise with left and clockwise with right. It's really the case that |↺⟩ is $|R\rangle$ ? Very confusing. Nov 16 at 17:56
• @FrankYellin Is it even rotating inwards or outwards? Anyway, I had to double check at the time (more than a year ago), but take your time to verify. The rotating notation is indeed the most confusing. Nov 16 at 22:51

As the other answer mentioned, they are often denoted as $$|+i\rangle= \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i\end{pmatrix} \ \ \ \textrm{and} \ \ \ |-i\rangle = \dfrac{|0\rangle - i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i\end{pmatrix}$$

but sometime you might also see them denoted as $$|R\rangle$$ and $$|L \rangle$$ correspond to the Right and Left circular polarization. This is because the matrix

$$U = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{pmatrix}$$

has the eigenvalues of $$e^{i\theta}$$ and $$e^{-i\theta}$$ with the corresponding eigenvectors

$$|R\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i\end{pmatrix} \hspace{1.2 cm} |L\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} i \\ 1\end{pmatrix} = \dfrac{i}{\sqrt{2}} \begin{pmatrix} 1 \\ -i\end{pmatrix} \equiv \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i\end{pmatrix} \ \ \textrm{since} \ e^{i \theta}|\psi \rangle \equiv |\psi \rangle$$

They are commonly denoted as

$$\left\{ |+i\rangle = \frac{|0\rangle + i|1\rangle}{\sqrt{2}}, |-i\rangle = \frac{|0\rangle - i|1\rangle}{\sqrt{2}}\right\}$$

So, you could use $$|\pm i\rangle = \frac{|0\rangle \pm i|1\rangle}{\sqrt{2}}$$, which would be very similar to the case of $$X$$, i.e., $$Y|\pm i\rangle = \pm|\pm i\rangle$$.