# Phase shifter acting on double rail states

In Nielsen and Chuang, it is stated that the photonic phase shift gate acts on the single photon states as $$P|0\rangle \ = \ |0\rangle$$ and $$P|1\rangle \ = \ e^{i\Delta}|1\rangle$$, where $$\Delta \ = \ (n \ - \ n_0)L/c_0$$. $$n$$ is the index of refraction of the phase shifter, $$n_0$$ is the index of refraction of the surrounding space in which the photon is propagating, $$L$$ is the distance the photon travels through the phase shifter, and $$c_0$$ is the speed of light in a vacuum. It is then stated that the phase shifter acting on the dual-rail photon state is equal to: $$a_0e^{-i\Delta/2}|01\rangle \ + \ a_1e^{i\Delta/2}|10\rangle$$, where $$a_0$$ and $$a_1$$ are simply the coefficients on the different states. How exactly was this result obtained?

• Hint: multiply in a global phase of $e^{i \Delta / 2}$. "Equivalent" likely means observationaly equivalent; equivalent up to global phase. – Craig Gidney Jan 2 at 21:33

Start in

$$a_0 | 01 \rangle + a_1 | 10 \rangle$$

Then apply $$P \otimes I$$ to get

$$a_0 * 1 | 01 \rangle + a_1*e^{i \Delta} | 10 \rangle$$

But that is the same up to phase as

$$e^{-i \Delta /2} (a_0 * 1 | 01 \rangle + a_1*e^{i \Delta} | 10 \rangle)$$

which simplifies to

$$a_0 e^{-i \Delta /2} | 01 \rangle + a_1 e^{i \Delta /2} | 10 \rangle$$

For the former part: Even a simple slab of glass can act as a phase shift quantum gate. The difference in path covered is $$(n-n_0)L \equiv \delta L$$ and the time difference is $$\delta L/c_0$$ and then the phase shift is proportional to this time shift. Or just divide this time phase difference by some time $$T$$ taken by the light in a vacuum to travel the same length $$L$$, often taken to be any constant.

For the latter part: Simply apply the operator $$P\otimes I$$ on a dual rail $$a_0|01\rangle +a_1|10\rangle.$$ The fact that has been used here is that the global phase is not an observable and only the relative phase is what leads to observable quantities. It disappears when you calculate any inner product. We can safely drop the global phase as follows: $$$$a_1e^{i\theta}|\psi_1\rangle+a_2e^{i\phi}|\psi_2\rangle \equiv e^{i\theta}(a_1|\psi_1\rangle+a_2e^{i(\phi-\theta)}|\psi_2\rangle)$$$$ $$$$\equiv (a_1|\psi_1\rangle+a_2e^{i(\phi-\theta)}|\psi_2\rangle)$$$$ $$$$\equiv (a_1e^{-i(\phi-\theta)/2}|\psi_1\rangle+a_2e^{i(\phi-\theta)/2}|\psi_2\rangle)$$$$