<s>[INST] Types of photon qubit encoding How many types of qubit encoding on photons exist nowadays? I know only two:


*

*Encoding on polarization:
$$ \lvert \Psi \rangle = \alpha \lvert H \rangle + \beta \lvert V \rangle $$
$$ \lvert H \rangle = \int_{-\infty}^{\infty} d\mathbf{k}\ f(\mathbf{k}) e^{-iw_k t} \hat{a}^\dagger_{H}(\mathbf{k}) \lvert 0 \rangle_\text{Vacuum} $$
$$ \lvert V \rangle = \int_{-\infty}^{\infty} d\mathbf{k}\ f(\mathbf{k}) e^{-iw_k t} \hat{a}^\dagger_{V}(\mathbf{k}) \lvert 0 \rangle_\text{Vacuum} $$

*Time-bin:
$$ \lvert \Psi \rangle = \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle $$
$$ \lvert 0 \rangle = \int_{-\infty}^{\infty} dz\ f\left(\frac{t -z/c}{\delta t_{ph}}\right) e^{-i w_0 (t-z/c)} \hat{a}^\dagger(z) \lvert 0 \rangle_\text{Vacuum} $$
$$ \lvert 1 \rangle = \int_{-\infty}^{\infty} dz\ f\left(\frac{t -z/c+\tau}{\delta t_{ph}}\right) e^{-i w_0 (t-z/c+\tau)} \hat{a}^\dagger(z) \lvert 0 \rangle_\text{Vacuum} $$
Is there anything else?
 [/INST] Yeah there is a couple of 'em - off the top of my head I can think of:
\begin{align}
|\uparrow\;\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|\downarrow\;\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|g\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|e\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|L\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|R\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|H\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|V\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|0\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|1\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|+\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|-\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}
\end{align}
But these are only the naming conventions in the northern hemisphere. 
In the southern hemisphere they are labeled;
\begin{align}
|\downarrow\;\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|\uparrow\;\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|e\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|g\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|R\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|L\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|V\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|H\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|1\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|0\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}\\
|-\rangle\;=\;
\begin{pmatrix}
1 \\ 0
\end{pmatrix}\qquad&\qquad
|+\rangle\;=\;
\begin{pmatrix}
0 \\ 1
\end{pmatrix}
\end{align}
So as you can see you have to be careful about who you are talking to, otherwise you could get your calculations all backwards.
</s>