Bell States
Two photons states
Consider two photons with two distinct external states, labeled \nu_1 and \nu_2. Each photon’s polarization resides in a two-dimensional space, spanned by basis states, for example, |\mathbf x\rangle and |\mathbf y\rangle.
The quantum state of these two photons exists in a four-dimensional space, with a complete orthonormal basis given by:
| \boldsymbol \Psi\left(\nu_1, \nu_2\right) \rangle \in \left\{ |\mathbf x_1, \mathbf x_2\rangle, |\mathbf x_1, \mathbf y_2\rangle, |\mathbf y_1, \mathbf x_2\rangle, |\mathbf y_1, \mathbf y_2\rangle \right\}
As in any vector space, we can form another basis by linear combination of the first basis vectors. A particularly important basis consists of the four so-called Bell states |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle, |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle, |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle, and |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle.
\begin{aligned} |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf y_2\rangle + |\mathbf y_1,\mathbf x_2\rangle \right) \\ |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf y_2\rangle - |\mathbf y_1, \mathbf x_2\rangle \right) \\ |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \end{aligned}
We can verify that these are mutually orthogonal, so they form a complete basis.
First Bell state
Let us verify that the Bell state |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle is normalized and orthogonal to the other Bell states.
Normalization of |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Psi^+(\nu_1, \nu_2) | \boldsymbol \Psi^+(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf y_2| + \langle \mathbf y_1, \mathbf x_2| \right) \left( |\mathbf x_1, \mathbf y_2\rangle + |\mathbf y_1, \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle + \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle \right. \\ & \left. + \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle + \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 \right) = 1 \end{aligned}
Orthogonality of |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle with |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Psi^+(\nu_1, \nu_2) | \boldsymbol \Psi^-(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf y_2| + \langle \mathbf y_1, \mathbf x_2| \right) \left( |\mathbf x_1, \mathbf y_2\rangle - |\mathbf y_1, \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle - \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle \right. \\ & \left. + \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle - \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 1 - 0 \cdot 0 + 0 \cdot 0 - 1 \cdot 1 \right) = 0 \end{aligned}
Orthogonality of |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle with |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Psi^+(\nu_1, \nu_2) | \boldsymbol \Phi^+(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf y_2| + \langle \mathbf y_1, \mathbf x_2| \right) \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle + \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle \right. \\ & \left. + \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle + \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 \right) = 0 \end{aligned}
Orthogonality of |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle with |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Psi^+(\nu_1, \nu_2) | \boldsymbol \Phi^-(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf y_2| + \langle \mathbf y_1, \mathbf x_2| \right) \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle - \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle \right. \\ & \left. + \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle - \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 0 - 0 \cdot 1 + 0 \cdot 0 - 1 \cdot 0 \right) = 0 \end{aligned}
Second Bell state
Let us verify that the Bell state |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle is normalized and orthogonal to the other Bell states.
Normalization of |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Psi^-(\nu_1, \nu_2) | \boldsymbol \Psi^-(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf y_2| - \langle \mathbf y_1, \mathbf x_2| \right) \left( |\mathbf x_1, \mathbf y_2\rangle - |\mathbf y_1, \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle - \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle \right. \\ & \left. - \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle + \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 1 - 0 \cdot 0 - 0 \cdot 0 + 1 \cdot 1 \right) = 1 \end{aligned}
Orthogonality of |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle with |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Psi^-(\nu_1, \nu_2) | \boldsymbol \Psi^+(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf y_2| - \langle \mathbf y_1, \mathbf x_2| \right) \left( |\mathbf x_1, \mathbf y_2\rangle + |\mathbf y_1, \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle + \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle \right. \\ & \left. - \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle - \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 1 + 0 \cdot 0 - 0 \cdot 0 - 1 \cdot 1 \right) = 0 \end{aligned}
Orthogonality of |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle with |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Psi^-(\nu_1, \nu_2) | \boldsymbol \Phi^+(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf y_2| - \langle \mathbf y_1, \mathbf x_2| \right) \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle + \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle \right. \\ & \left. - \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle - \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 0 + 0 \cdot 1 - 0 \cdot 0 - 1 \cdot 0 \right) = 0 \end{aligned}
Orthogonality of |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle with |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Psi^-(\nu_1, \nu_2) | \boldsymbol \Phi^-(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf y_2| - \langle \mathbf y_1, \mathbf x_2| \right) \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle - \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle \right. \\ & \left. - \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle + \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 0 - 0 \cdot 1 - 0 \cdot 0 + 1 \cdot 0 \right) = 0 \end{aligned}
Third Bell state
Let us verify that the Bell state |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle is normalized and orthogonal to the other Bell states.
Normalization of |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Phi^+(\nu_1, \nu_2) | \boldsymbol \Phi^+(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf x_2| + \langle \mathbf y_1, \mathbf y_2| \right) \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle + \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle \right. \\ & \left. + \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle + \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 \right) = 1 \end{aligned}
Orthogonality of |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle with |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Phi^+(\nu_1, \nu_2) | \boldsymbol \Psi^+(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf x_2| + \langle \mathbf y_1, \mathbf y_2| \right) \left( |\mathbf x_1, \mathbf y_2\rangle + |\mathbf y_1, \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle + \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle \right. \\ & \left. + \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle + \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 \right) = 0 \end{aligned}
Orthogonality of |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle with |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Phi^+(\nu_1, \nu_2) | \boldsymbol \Psi^-(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf x_2| + \langle \mathbf y_1, \mathbf y_2| \right) \left( |\mathbf x_1, \mathbf y_2\rangle - |\mathbf y_1, \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle - \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle \right. \\ & \left. + \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle - \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 0 - 0 \cdot 0 + 0 \cdot 1 - 1 \cdot 0 \right) = 0 \end{aligned}
Orthogonality of |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle with |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Phi^+(\nu_1, \nu_2) | \boldsymbol \Phi^-(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf x_2| + \langle \mathbf y_1, \mathbf y_2| \right) \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle - \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle \right. \\ & \left. + \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle - \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 1 - 0 \cdot 0 + 0 \cdot 0 - 1 \cdot 1 \right) = 0 \end{aligned}
Fourth Bell state
Let us verify that the Bell state |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle is normalized and orthogonal to the other Bell states.
Normalization of |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Phi^-(\nu_1, \nu_2) | \boldsymbol \Phi^-(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf x_2| - \langle \mathbf y_1, \mathbf y_2| \right) \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle - \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle \right. \\ & \left. - \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle + \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 1 - 0 \cdot 0 - 0 \cdot 0 + 1 \cdot 1 \right) = 1 \end{aligned}
Orthogonality of |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle with |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Phi^-(\nu_1, \nu_2) | \boldsymbol \Psi^+(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf x_2| - \langle \mathbf y_1, \mathbf y_2| \right) \left( |\mathbf x_1, \mathbf y_2\rangle + |\mathbf y_1, \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle + \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle \right. \\ & \left. - \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle - \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 0 + 0 \cdot 0 - 0 \cdot 1 - 1 \cdot 0 \right) = 0 \end{aligned}
Orthogonality of |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle with |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Phi^-(\nu_1, \nu_2) | \boldsymbol \Psi^-(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf x_2| - \langle \mathbf y_1, \mathbf y_2| \right) \left( |\mathbf x_1, \mathbf y_2\rangle - |\mathbf y_1, \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle - \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle \right. \\ & \left. - \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle + \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 0 - 0 \cdot 0 - 0 \cdot 1 + 1 \cdot 0 \right) = 0 \end{aligned}
Orthogonality of |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle with |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle:
\begin{aligned} \langle \boldsymbol \Phi^-(\nu_1, \nu_2) | \boldsymbol \Phi^+(\nu_1, \nu_2) \rangle =& \frac{1}{2} \left( \langle \mathbf x_1, \mathbf x_2| - \langle \mathbf y_1, \mathbf y_2| \right) \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( \langle \mathbf x_1| \mathbf x_1\rangle \langle \mathbf x_2| \mathbf x_2\rangle + \langle \mathbf x_1| \mathbf y_1\rangle \langle \mathbf x_2| \mathbf y_2\rangle \right. \\ & \left. - \langle \mathbf y_1| \mathbf x_1\rangle \langle \mathbf y_2| \mathbf x_2\rangle - \langle \mathbf y_1| \mathbf y_1\rangle \langle \mathbf y_2| \mathbf y_2\rangle \right) \\ =& \frac{1}{2} \left( 1 \cdot 1 + 0 \cdot 0 - 0 \cdot 0 - 1 \cdot 1 \right) = 0 \end{aligned}
Single photon gates
In contrast to the initial basis vectors, which are factorized, the Bell states are entangled. These Bell states cannot be factorized as a product of a state of photon \nu_1 and a state of photon \nu_2.
While it is possible to produce all four Bell states, there is no practical method to implement a full measurement of the Bell observable of which these four Bell states are eigenstates, but it nevertheless possible to implement a partial measurement.
Birefringent plates are tools for manipulating the polarization of light beams. These plates enable any polarization transformation to be realized. For instance, they can change the direction of a linear polarization or transform a linear polarization into a circular polarization. This implies that the polarization state of a single photon can be changed at will.
Considering the most general polarization of a photon in a linear polarization basis, expressed by two complex components \alpha and \beta:
|\boldsymbol \varphi(\nu)\rangle = \alpha | \mathbf x \rangle + \beta | \mathbf y \rangle
a birefringent plate can perform a transformation described by a unitary matrix \mathbf M:
\begin{bmatrix} \alpha^\prime \\ \beta^\prime \end{bmatrix} = \mathbf M \begin{bmatrix} \alpha \\ \beta \end{bmatrix}
This transformation is a qubit gate. With a half-wave plate aligned at 45^\circ from the \mathbf x and \mathbf y-axes, a gate can be implemented that transforms |\mathbf x\rangle into |\mathbf y\rangle, and vice versa. This gate is represented by the matrix:
\mathbf X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} and is called a Pauli \mathbf X-gate. A quarter-wave plate with axes along \mathbf x and \mathbf y can transform a linear polarization at 45^\circ from \mathbf x and \mathbf y into a circular polarization. This is represented by the matrix:
\mathbf S = \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}
and is called a phase gate. Applying these gates to one or both photons of an entangled pair allows for the transformation of one Bell state into another.
This capability allows the production of any of the Bell states from another.
We start from an entangled state |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle that can be produced.
If we apply a Pauli \mathbf X-gate to photon \nu_2, we obtain the state |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle.
\begin{aligned} (\mathbf I \otimes \mathbf X_2) |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle &= (\mathbf I \otimes \mathbf X_2) \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( (\mathbf I \otimes \mathbf X_2) |\mathbf x_1, \mathbf x_2\rangle + (\mathbf I \otimes \mathbf X_2) |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle \otimes (\mathbf X_2 |\mathbf x_2\rangle) + |\mathbf y_1\rangle \otimes (\mathbf X_2 |\mathbf y_2\rangle) \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf y_2\rangle + |\mathbf y_1, \mathbf x_2\rangle \right) \\ &= |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle \end{aligned}
Applying a phase gate twice, we obtain the state |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle in three ways.
We compute \mathbf S^2 first:
\mathbf S^2 = \mathbf S \cdot \mathbf S = \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & i^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
We can apply a phase gate twice to photon \nu_1:
\begin{aligned} (\mathbf S_1^2 \otimes \mathbf I_2) |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle &= (\mathbf S_1^2 \otimes \mathbf I_2) \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( (\mathbf S_1^2 \otimes \mathbf I_2) |\mathbf x_1, \mathbf x_2\rangle + (\mathbf S_1^2 \otimes \mathbf I_2) |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( (\mathbf S_1^2 |\mathbf x_1\rangle) \otimes |\mathbf x_2\rangle + (\mathbf S_1^2 |\mathbf y_1\rangle) \otimes |\mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle \otimes |\mathbf x_2\rangle + (- |\mathbf y_1\rangle) \otimes |\mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle \end{aligned}
We can apply a phase gate twice to photon \nu_2:
\begin{aligned} (\mathbf I_1 \otimes \mathbf S_2^2) |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle &= (\mathbf I_1 \otimes \mathbf S_2^2) \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( (\mathbf I_1 \otimes \mathbf S_2^2) |\mathbf x_1, \mathbf x_2\rangle + (\mathbf I_1 \otimes \mathbf S_2^2) |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle \otimes (\mathbf S_2^2 |\mathbf x_2\rangle) + |\mathbf y_1\rangle \otimes (\mathbf S_2^2 |\mathbf y_2\rangle) \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle \otimes |\mathbf x_2\rangle + |\mathbf y_1\rangle \otimes (- |\mathbf y_2\rangle) \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle \end{aligned}
We can apply a phase gate once to photon \nu_1 and once to photon \nu_2:
\begin{aligned} (\mathbf S_1 \otimes \mathbf S_2) |\boldsymbol \Phi^+(\nu_1, \nu_2)\rangle &= (\mathbf S_1 \otimes \mathbf S_2) \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( (\mathbf S_1 \otimes \mathbf S_2) |\mathbf x_1, \mathbf x_2\rangle + (\mathbf S_1 \otimes \mathbf S_2) |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( (\mathbf S_1 |\mathbf x_1\rangle) \otimes (\mathbf S_2 |\mathbf x_2\rangle) + (\mathbf S_1 |\mathbf y_1\rangle) \otimes (\mathbf S_2 |\mathbf y_2\rangle) \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle \otimes |\mathbf x_2\rangle + (i |\mathbf y_1\rangle) \otimes (i |\mathbf y_2\rangle) \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle + i^2 |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle \end{aligned}
Finally, we can find |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle in three ways.
We first get |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle, applying a phase gate twice to photon \nu_1, or twice to photon \nu_2, or once to both, then we apply a Pauli \mathbf X-gate to photon \nu_2:
\begin{aligned} (\mathbf I_1 \otimes \mathbf X_2) |\boldsymbol \Phi^-(\nu_1, \nu_2)\rangle &= (\mathbf I_1 \otimes \mathbf X_2) \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf x_2\rangle - |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( (\mathbf I_1 \otimes \mathbf X_2) |\mathbf x_1, \mathbf x_2\rangle - (\mathbf I_1 \otimes \mathbf X_2) |\mathbf y_1, \mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle \otimes (\mathbf X_2 |\mathbf x_2\rangle) - |\mathbf y_1\rangle \otimes (\mathbf X_2 |\mathbf y_2\rangle) \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle \otimes |\mathbf y_2\rangle - |\mathbf y_1\rangle \otimes |\mathbf x_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf y_2\rangle - |\mathbf y_1, \mathbf x_2\rangle \right) \\ &= |\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle \end{aligned}
Measurements
Surprisingly, although the four Bell states form a complete orthogonal basis for the state space of two polarization-entangled photons, and as a consequence a mathematical observable is associated with them, we do not know how to implement an apparatus for a complete measurement of that observable.
Such an apparatus would ideally have four outputs, each corresponding to one of the four Bell states of the two photons. However, we can implement measuring setups that perform partial Bell measurements, meaning they have outputs associated with at least one of the four Bell states.
Second Bell state
Let us consider a pair of photons \nu_\alpha and \nu_\beta entering the two input ports \alpha and \beta of a beam splitter:
|\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle
Their polarization state is in the four-dimensional space previously discussed. We will see that the device shown here allows for a measurement associated with the projection onto the Bell state |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle:
|\boldsymbol \Psi^-(\nu_1, \nu_2)\rangle = \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf y_2\rangle - |\mathbf y_1, \mathbf x_2\rangle \right))
This means that the apparatus will sometimes produce a signal if the input state has a non-zero component along |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle. The probability of obtaining this signal is given by the squared modulus of that component. Furthermore, when such a signal is detected, the initial state is projected onto the eigenspace associated with |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle. The corresponding projector is \mathbf P_{\boldsymbol \Psi^-}:
\mathbf P_{\boldsymbol \Psi^-} = |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle \langle \boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)|
Let us find the expression of |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle in the output space. We express it in the input space with the creation operators and we transform them. Let us do it for the first term |\mathbf x_\alpha, \mathbf y_\beta\rangle expressed with creation operators applied to the vacuum and transformed.
The first term of |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle is |\mathbf x_\alpha, \mathbf y_\beta\rangle, which in terms of creation operators is \mathbf a^\dagger_{\alpha, \mathbf x} \mathbf a^\dagger_{\beta, \mathbf y} |\mathbf 0\rangle.
We use the beam splitter relations for creation operators:
\begin{aligned} \mathbf a^\dagger_{\alpha} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma} + \mathbf a^\dagger_{\delta})\\ \mathbf a^\dagger_{\beta} &= \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma} - \mathbf a^\dagger_{\delta}) \end{aligned}
These expressions are valid for each polarization component. Therefore, we have:
\begin{aligned} \mathbf a^\dagger_{\alpha, \mathbf x} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x}) \\ \mathbf a^\dagger_{\beta, \mathbf y} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y}) \end{aligned}
So the first term is:
\begin{aligned} |\mathbf x_\alpha, \mathbf y_\beta\rangle & = \mathbf a^\dagger_{\alpha, \mathbf x} \mathbf a^\dagger_{\beta, \mathbf y}| \mathbf 0 \rangle = \frac{1}{2} (\mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x}) (\mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y}) | \mathbf 0 \rangle \\ = & \frac{1}{2} \left( \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} \right) | \mathbf 0 \rangle \end{aligned}
The second term is:
\begin{aligned} |\mathbf y_\alpha, \mathbf x_\beta\rangle & = \mathbf a^\dagger_{\alpha, \mathbf y} \mathbf a^\dagger_{\beta, \mathbf x} | \mathbf 0 \rangle = \frac{1}{2} (\mathbf a^\dagger_{\gamma, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y}) (\mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf x}) | \mathbf 0 \rangle \\ = & - \frac{1}{2} \left( - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right) | \mathbf 0 \rangle \\ = & - \frac{1}{2} \left( - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right) | \mathbf 0 \rangle \\ = & \frac{1}{2} \left( \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right) | \mathbf 0 \rangle \end{aligned}
We can now compute |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle:
\begin{aligned} |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle = \frac{1}{\sqrt{2}} & \left( |\mathbf x_\alpha, \mathbf y_\beta\rangle - |\mathbf y_\alpha, \mathbf x_\beta\rangle \right) \\ = \frac{1}{\sqrt{2}} & \left[ \frac{1}{2} \left( \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} \right) \right. \\ & \left. - \frac{1}{2} \left( \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right) \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y}\right. \\ & \left. - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x}) - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right. \\ & \left. + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right] | \mathbf 0 \rangle \\ \end{aligned}
Since \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} = \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} and \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} = \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x}, the terms in the first parenthesis and the last two terms in the bracket cancel out.
\begin{aligned} |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle &= \frac{1}{2\sqrt{2}} \left[ - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} \right] | \mathbf 0 \rangle \\ &= \frac{1}{2\sqrt{2}} \left[ (\mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y}) + (\mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x}) \right] | \mathbf 0 \rangle \\ &= \frac{1}{\sqrt{2}} \left[ (\mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y}) \right] | \mathbf 0 \rangle \\ &= \frac{1}{\sqrt{2}} \left( - |\mathbf x_\gamma, \mathbf y_\delta\rangle + |\mathbf y_\gamma, \mathbf x_\delta\rangle \right) \end{aligned}
Based on the expression of |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle in the output space, |\boldsymbol \Psi^-(\nu_\gamma, \nu_\delta)\rangle = \frac{1}{\sqrt{2}} \left( - |\mathbf x_\gamma, \mathbf y_\delta\rangle + |\mathbf y_\gamma, \mathbf x_\delta\rangle \right), we can determine the detector signals. There are two possible outcomes, each with a probability of one-half.
The first possibility is joint detection of \mathbf x polarization in output mode \delta and \mathbf y polarization in output mode \gamma. This corresponds to detecting |\mathbf x\rangle in detector for mode \delta and |\mathbf y\rangle in detector for mode \gamma.
The second possibility is joint detection of \mathbf y polarization in output mode \delta and \mathbf x polarization in output mode \gamma. This corresponds to detecting |\mathbf y\rangle in detector for mode \delta and |\mathbf x\rangle in detector for mode \gamma.
In both scenarios, one photon is detected in channel \gamma and one photon is detected in channel \delta. If we were to remove the polarization splitters and simply place one detector in channel \gamma and another in channel \delta, we would still observe a joint detection event in channels \gamma and \delta. Among all four Bell states, |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle is unique in leading to such a joint detection in output channels \gamma and \delta.
First Bell state
Let us consider a pair of photons \nu_\alpha and \nu_\beta entering the two input ports \alpha and \beta of a beam splitter in the first bell state:
|\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle = \frac{1}{\sqrt{2}} \left( |\mathbf x_1, \mathbf y_2\rangle + |\mathbf y_1,\mathbf x_2\rangle \right)
With the expression previously computed for |\mathbf x_1, \mathbf y_2\rangle and |\mathbf y_1, \mathbf y_2\rangle we have:
\begin{aligned} |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle = \frac{1}{\sqrt{2}} & \left( |\mathbf x_1, \mathbf y_2\rangle + |\mathbf y_1,\mathbf x_2\rangle \right) \\ = \frac{1}{\sqrt{2}} & \left[ \frac{1}{2} \left( \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} \right) | \mathbf 0 \rangle \right. \\ & \left. + \frac{1}{2} \left( \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right) | \mathbf 0 \rangle \right] \\ = \frac{1}{2\sqrt{2}} & \left[ 2 \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right. \\ & \left. + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} - 2 \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} \right] | \mathbf 0 \rangle \end{aligned} Since \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} = \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} and \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} = \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x}, we can further simplify:
\begin{aligned} |\boldsymbol \Psi^+(\nu_1, \nu_2)\rangle = \frac{1}{2\sqrt{2}} & \left[ 2 \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} \right.\\ & \left. + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} - 2 \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ 2 \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - 2 \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} +\right.\\ & \left. \left( - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf x} \right) \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ 2 \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - 2 \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} + \right.\\ & \left. \left( - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} \right) \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ 2 \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - 2 \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} \right] | \mathbf 0 \rangle \\ = \frac{1}{\sqrt{2}} & \left[ \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf y} \right] | \mathbf 0 \rangle \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_\gamma, \mathbf y_\gamma\rangle - |\mathbf x_\delta, \mathbf y_\delta\rangle \right) \end{aligned}
The output state can be interpreted in terms of joint detections in either channel \gamma or channel \delta, but not both simultaneously. The probability is one half.
The apparatus will sometimes return a signal corresponding to a joint detection of both polarizations (\mathbf x and \mathbf y) in channel \gamma, or alternatively, a joint detection of both polarizations (\mathbf x and \mathbf y) in channel \delta. These two possibilities are mutually exclusive. This is in contrast to the |\boldsymbol \Psi^-(\nu_\alpha, \nu_\beta)\rangle state, where the joint detection occurs in channels \gamma and \delta with different polarizations in each channel.
Third Bell state
To express the normalized state |\boldsymbol \Phi^-(\nu_\alpha, \nu_\beta)\rangle in terms of output modes \gamma and \delta, we begin with the definition of the \Phi^- Bell state:
|\boldsymbol \Phi^-(\nu_\alpha, \nu_\beta)\rangle = \frac{1}{\sqrt{2}} \left( |\mathbf x_\alpha, \mathbf x_\beta\rangle - |\mathbf y_\alpha, \mathbf y_\beta\rangle \right)
The first term of |\boldsymbol \Phi^-(\nu_\alpha, \nu_\beta)\rangle is |\mathbf x_\alpha, \mathbf x_\beta\rangle, which in terms of creation operators is \mathbf a^\dagger_{\alpha, \mathbf x} \mathbf a^\dagger_{\beta, \mathbf x} |\mathbf 0\rangle.
We use the beam splitter relations for creation operators:
\begin{aligned} \mathbf a^\dagger_{\alpha} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma} + \mathbf a^\dagger_{\delta})\\ \mathbf a^\dagger_{\beta} &= \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma} - \mathbf a^\dagger_{\delta}) \end{aligned}
These expressions are valid for each polarization component. The first term of |\boldsymbol \Phi^-(\nu_\alpha, \nu_\beta)\rangle is |\mathbf x_\alpha, \mathbf x_\beta\rangle, which in terms of creation operators is \mathbf a^\dagger_{\alpha, \mathbf x} \mathbf a^\dagger_{\beta, \mathbf x} |\mathbf 0\rangle.
We use the beam splitter relations for creation operators:
\begin{aligned} \mathbf a^\dagger_{\alpha} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma} + \mathbf a^\dagger_{\delta})\\ \mathbf a^\dagger_{\beta} &= \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma} - \mathbf a^\dagger_{\delta}) \end{aligned}
These expressions are valid for each polarization component. Therefore, we have:
\begin{aligned} \mathbf a^\dagger_{\alpha, \mathbf x} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x}) \\ \mathbf a^\dagger_{\beta, \mathbf x} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf x}) \end{aligned}
So the first term is:
\begin{aligned} |\mathbf x_\alpha, \mathbf x_\beta\rangle &= \mathbf a^\dagger_{\alpha, \mathbf x} \mathbf a^\dagger_{\beta, \mathbf x} | \mathbf 0 \rangle \\ &= \left[ \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x}) \right] \left[ \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf x}) \right] | \mathbf 0 \rangle \\ &= \frac{1}{2} (\mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x}) (\mathbf a^\dagger_{\gamma, \mathbf x} - \mathbf a^\dagger_{\delta, \mathbf x}) | \mathbf 0 \rangle \\ &= \frac{1}{2} \left( (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right) | \mathbf 0 \rangle \end{aligned}
Similarly, we also have:
\begin{aligned} \mathbf a^\dagger_{\alpha, \mathbf y} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y}) \\ \mathbf a^\dagger_{\beta, \mathbf y} & = \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y}) \end{aligned}
So the second term is:
\begin{aligned} |\mathbf y_\alpha, \mathbf y_\beta\rangle &= \mathbf a^\dagger_{\alpha, \mathbf y} \mathbf a^\dagger_{\beta, \mathbf y} | \mathbf 0 \rangle \\ &= \left[ \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y}) \right] \left[ \frac{1}{\sqrt{2}} (\mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y}) \right] | \mathbf 0 \rangle \\ &= \frac{1}{2} (\mathbf a^\dagger_{\gamma, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y}) (\mathbf a^\dagger_{\gamma, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y}) | \mathbf 0 \rangle \\ &= \frac{1}{2} \left( (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right) | \mathbf 0 \rangle \end{aligned}
We can now compute |\boldsymbol \Phi^-(\nu_\alpha, \nu_\beta)\rangle:
\begin{aligned} |\boldsymbol \Phi^-(\nu_\alpha, \nu_\beta)\rangle = \frac{1}{\sqrt{2}} & \left( |\mathbf x_\alpha, \mathbf x_\beta\rangle - |\mathbf y_\alpha, \mathbf y_\beta\rangle \right) \\ = \frac{1}{\sqrt{2}} & \left[ \frac{1}{2} \left( (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right) \right. \\ & \left. - \frac{1}{2} \left( (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right) \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ \left( (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right) \right. \\ & \left. - \left( (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right) \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right. \\ & - \left. (\mathbf a^\dagger_{\gamma, \mathbf y})^2 + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} + (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \end{aligned}
Using the commutation relation for creation operators \mathbf a^\dagger_i \mathbf a^\dagger_j = \mathbf a^\dagger_j \mathbf a^\dagger_i, we can simplify the expression:
\begin{aligned} |\boldsymbol {\Phi^-}^\prime(\nu_\alpha, \nu_\beta)\rangle = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right. \\ & \left. - (\mathbf a^\dagger_{\gamma, \mathbf y})^2 + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} + (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right. \\ & \left. - (\mathbf a^\dagger_{\gamma, \mathbf y})^2 + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - (\mathbf a^\dagger_{\delta, \mathbf x})^2 - (\mathbf a^\dagger_{\gamma, \mathbf y})^2 + (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} | \mathbf 0 \rangle - \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} | \mathbf 0 \rangle - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} | \mathbf 0 \rangle + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} | \mathbf 0 \rangle \right] \\ = \frac{1}{2\sqrt{2}} & \left[ |\mathbf x_\gamma, \mathbf x_\gamma\rangle - |\mathbf x_\delta, \mathbf x_\delta\rangle - |\mathbf y_\gamma, \mathbf y_\gamma\rangle + |\mathbf y_\delta, \mathbf y_\delta\rangle \right] \end{aligned}
This state is not normalized. The norm squared is given by:
\begin{aligned} \left|\boldsymbol {\Phi^-}^\prime(\nu_\alpha, \nu_\beta)\right|^2 = & \langle \boldsymbol {\Phi^-}^\prime | \boldsymbol {\Phi^-}^\prime \rangle = \left( \frac{1}{2\sqrt{2}} \right)^2 \langle \left( |\mathbf x_\gamma, \mathbf x_\gamma\rangle - |\mathbf x_\delta, \mathbf x_\delta\rangle - |\mathbf y_\gamma, \mathbf y_\gamma\rangle \right. \\ & \left. + |\mathbf y_\delta, \mathbf y_\delta\rangle \right) | \left( |\mathbf x_\gamma, \mathbf x_\gamma\rangle - |\mathbf x_\delta, \mathbf x_\delta\rangle - |\mathbf y_\gamma, \mathbf y_\gamma\rangle + |\mathbf y_\delta, \mathbf y_\delta\rangle \right) \rangle \\ = & \frac{1}{8} \left[ \langle \mathbf x_\gamma, \mathbf x_\gamma | \mathbf x_\gamma, \mathbf x_\gamma \rangle + \langle \mathbf x_\delta, \mathbf x_\delta | \mathbf x_\delta, \mathbf x_\delta \rangle \right. \\ & \left. + \langle \mathbf y_\gamma, \mathbf y_\gamma | \mathbf y_\gamma, \mathbf y_\gamma \rangle + \langle \mathbf y_\delta, \mathbf y_\delta | \mathbf y_\delta, \mathbf y_\delta \rangle \right] \\ & = \frac{1}{8} \left[ 1 + 1 + 1 + 1 \right] = \frac{4}{8} = \frac{1}{2} \end{aligned}
we can then normalize the status:
|\boldsymbol {\Phi^-}(\nu_\alpha, \nu_\beta)\rangle = \frac{|\boldsymbol {\Phi^-}^\prime(\nu_\alpha, \nu_\beta)\rangle}{\left|\boldsymbol {\Phi^-}^\prime(\nu_\alpha, \nu_\beta)\right| } = \frac{1}{2} \left[ |\mathbf x_\gamma, \mathbf x_\gamma\rangle - |\mathbf x_\delta, \mathbf x_\delta\rangle - |\mathbf y_\gamma, \mathbf y_\gamma\rangle + |\mathbf y_\delta, \mathbf y_\delta\rangle \right]
The output state can be interpreted in terms of detection of two photons in either of the four channel with a probability is one fourth.
Fourth Bell state
Let us consider a pair of photons \nu_\alpha and \nu_\beta entering the two input ports \alpha and \beta of a beam splitter in the fourth bell state:
|\boldsymbol \Phi^+(\nu_\alpha, \nu_\beta)\rangle = \frac{1}{\sqrt{2}} \left( |\mathbf x_\alpha, \mathbf x_\beta\rangle + |\mathbf y_\alpha, \mathbf y_\beta\rangle \right)
With the expression previously computed for |\mathbf x_\alpha, \mathbf x_\beta\rangle and |\mathbf y_\alpha, \mathbf y_\beta\rangle we have:
\begin{aligned} |\boldsymbol \Phi^+(\nu_\alpha, \nu_\beta)\rangle = \frac{1}{\sqrt{2}} & \left( |\mathbf x_\alpha, \mathbf x_\beta\rangle + |\mathbf y_\alpha, \mathbf y_\beta\rangle \right) \\ = \frac{1}{\sqrt{2}} & \left[ \frac{1}{2} \left( (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right) \right. \\ & \left. + \frac{1}{2} \left( (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right) \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ \left( (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right) \right. \\ & \left. + \left( (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right) \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right. \\ & + \left. (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \end{aligned}
Using the commutation relation for creation operators \mathbf a^\dagger_i \mathbf a^\dagger_j = \mathbf a^\dagger_j \mathbf a^\dagger_i, we can simplify the expression:
\begin{aligned} |\boldsymbol {\Phi^+}^\prime(\nu_\alpha, \nu_\beta)\rangle = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right. \\ & \left. + (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 \right. \\ & \left. + (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 + (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} +\right. \\ & \left. \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} + \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 + (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} - \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} \right. \\ & \left.+ \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} - (\mathbf a^\dagger_{\delta, \mathbf x})^2 - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ (\mathbf a^\dagger_{\gamma, \mathbf x})^2 + (\mathbf a^\dagger_{\gamma, \mathbf y})^2 - (\mathbf a^\dagger_{\delta, \mathbf x})^2 - (\mathbf a^\dagger_{\delta, \mathbf y})^2 \right] | \mathbf 0 \rangle \\ = \frac{1}{2\sqrt{2}} & \left[ \mathbf a^\dagger_{\gamma, \mathbf x} \mathbf a^\dagger_{\gamma, \mathbf x} | \mathbf 0 \rangle + \mathbf a^\dagger_{\gamma, \mathbf y} \mathbf a^\dagger_{\gamma, \mathbf y} | \mathbf 0 \rangle - \mathbf a^\dagger_{\delta, \mathbf x} \mathbf a^\dagger_{\delta, \mathbf x} | \mathbf 0 \rangle - \mathbf a^\dagger_{\delta, \mathbf y} \mathbf a^\dagger_{\delta, \mathbf y} | \mathbf 0 \rangle \right] \\ = \frac{1}{2\sqrt{2}} & \left[ |\mathbf x_\gamma, \mathbf x_\gamma\rangle + |\mathbf y_\gamma, \mathbf y_\gamma\rangle - |\mathbf x_\delta, \mathbf x_\delta\rangle - |\mathbf y_\delta, \mathbf y_\delta\rangle \right] \end{aligned}
This state is not normalized. The norm squared is given by:
\begin{aligned} \left|\boldsymbol {\Phi^+}^\prime(\nu_\alpha, \nu_\beta)\right|^2 = & \langle \boldsymbol {\Phi^+}^\prime | \boldsymbol {\Phi^+}^\prime \rangle = \left( \frac{1}{2\sqrt{2}} \right)^2 \langle \left( |\mathbf x_\gamma, \mathbf x_\gamma\rangle + |\mathbf y_\gamma, \mathbf y_\gamma\rangle - |\mathbf x_\delta, \mathbf x_\delta\rangle \right. \\ & \left. - |\mathbf y_\delta, \mathbf y_\delta\rangle \right) | \left( |\mathbf x_\gamma, \mathbf x_\gamma\rangle + |\mathbf y_\gamma, \mathbf y_\gamma\rangle - |\mathbf x_\delta, \mathbf x_\delta\rangle - |\mathbf y_\delta, \mathbf y_\delta\rangle \right) \rangle \\ = & \frac{1}{8} \left[ \langle \mathbf x_\gamma, \mathbf x_\gamma | \mathbf x_\gamma, \mathbf x_\gamma \rangle + \langle \mathbf y_\gamma, \mathbf y_\gamma | \mathbf y_\gamma, \mathbf y_\gamma \rangle \right. \\ & \left. + \langle \mathbf x_\delta, \mathbf x_\delta | \mathbf x_\delta, \mathbf x_\delta \rangle + \langle \mathbf y_\delta, \mathbf y_\delta | \mathbf y_\delta, \mathbf y_\delta \rangle \right] \\ & = \frac{1}{8} \left[ 1 + 1 + 1 + 1 \right] = \frac{4}{8} = \frac{1}{2} \end{aligned}
we can then normalize the status:
|\boldsymbol {\Phi^+}(\nu_\alpha, \nu_\beta)\rangle = \frac{|\boldsymbol {\Phi^+}^\prime(\nu_\alpha, \nu_\beta)\rangle}{\left|\boldsymbol {\Phi^+}^\prime(\nu_\alpha, \nu_\beta)\right| } = \frac{1}{2} \left[ |\mathbf x_\gamma, \mathbf x_\gamma\rangle + |\mathbf y_\gamma, \mathbf y_\gamma\rangle - |\mathbf x_\delta, \mathbf x_\delta\rangle - |\mathbf y_\delta, \mathbf y_\delta\rangle \right]
The output state can be interpreted in terms of detection of two photons in either of the four channel with a probability is one fourth.