# How do you apply a CNOT on polarization qubits?

I read that a qubit can be encoded in a polarization state (horizontal or vertical polarization of a photon). How do you perform two-qubit operations on a polarization qubit?

If one qubit is encoded in the polarization degree of freedom of a single photon, and the second qubit in the path degree of freedom of the same photon, then a CNOT gate is trivially implemented by a polarizing beamsplitter. This is a kind of beamsplitter that only changes the path of the photon if its polarization is in some polarization state (say, $|V\rangle$), and leaves the photon in its path otherwise. This is therefore effectively a CNOT gate where the control qubit is the polarization and the target qubit is the path.
A very different story is the use of polarization qubits of many different photons. The main problem with this is that photons do not naturally interact with each other, so that two-qubit gates between such qubits are nontrivial. Indeed, it is easy to show that with linear optics alone it is impossible to implement arbitrary two-qubit gates in a deterministic way. For example, consider the case where one has two single photons, each one in a different spatial mode, and both in the initial polarization state $|H\rangle$. Using the standard second quantization notation, the set of transformations that can be implemented between these two photons within linear optics are given by $$a_H^\dagger b_H^\dagger \to\left(\sum_{k=H,V}\alpha_k c_k^\dagger +\beta_k d_k^\dagger\right)\left(\sum_{k=H,V}\gamma_k c_k^\dagger +\delta_k d_k^\dagger\right),$$ where $\alpha_k,\beta_k,\gamma_k,\delta_k$ are parameters characterising the linear transformation that is being implemented, $a_H^\dagger,b_H^\dagger$ are the creation operators of the input photons in the spatial modes $a$ and $b$ with polarization $H$, and $c$ and $d$ denote the two output modes of the photons. It can be seen that, for example, no set of values of $\alpha_k,\beta_k,\gamma_k,\delta_k$ can implement the transformation $$a_H^\dagger b_H^\dagger\to c_H^\dagger d_V^\dagger + c_V^\dagger d_H^\dagger,$$ meaning that it is not possible to generate deterministically, within linear optics, to transform $|00\rangle$ into the Bell state $|01\rangle+|10\rangle$.