# What is the quantum circuit equivalent of a (delayed choice) quantum eraser?

Quantum computers are efficiently able to simulate any other quantum system. Hence there must be some sort of equivalent of a (possibly simulated) quantum eraser setup. I would like to see such an equivalent drawn as a quantum circuit, ideally in the variant of a delayed choice quantum eraser.

One (quantum) experimental realization of a quantum eraser is this: You create a double slit interference experiment where you obtain which-way information by "doubling" photons in front of each slit using spontaneous parametric down conversion (the physics of which are not important for my argument, the point being that we have a new photon we can measure to obtain which-way information). The interference pattern naturally disappears, unless we build a quantum eraser: If the two "doubled" photons carrying the which-way information are superimposed via a 50-50 beamsplitter in such a manner that the which-way information can no longer be measured, the interference pattern reappears. Curiously, this is experimentally the case even if this superimposing happens after the interference pattern is recorded.

I seem to be unable to find a convincing equivalence for the interference pattern and for the quantum eraser in simple qubit gates. But I would love to make the thought (and ideally, the real) experiment on a quantum computer, too. What program (quantum circuit) would I need to run on a quantum computer to do that?

We associate the blue path with $|0\rangle$ and the red with $|1\rangle$. The double slit can be described by a Hadamard gate. The BBO corresponds to a CNOT-gate. The state after the BBO is $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. There is a phase $\varphi$ depending on the position $x$ of detector $D_0$, which corresponds to a phase gate $R_\varphi=\operatorname{diag}(1,e^{i\varphi})$. Finally superposing the beams on $D_0$ corresponds to another Hadamard gate and the measurement of $D_0$ can be seen to project onto $|0\rangle$. The complete circuit looks like this:
The state before the measurement is: $$\frac{1}{2}(|00\rangle+|10\rangle + e^{i\varphi}|01\rangle - e^{i\varphi}|11\rangle)=\frac{1}{2}(((1+e^{i\varphi})|0\rangle+(1-e^{i\varphi})|1\rangle)|+\rangle + ((1-e^{i\varphi})|0\rangle +(1+e^{i\varphi})|1\rangle)|-\rangle)$$
Let's look at the probability to measure the first photon in $D_0$ ($|0\rangle\langle 0|$).
If we measure the second in z-basis ($D_3$ and $D_4$), the probability for a click in $D_0$ is $\frac{1}{2}$ (the post-measurement state is $|\pm\rangle$). This is independent of the phase: no interference here. For the x-basis ($D_1$ and $D_2$) the probability for a click at $D_0$ is $\frac{1}{2}(1\mp \cos \varphi)$, so here we see the interference. Whether we see interference or not depends on the basis choice on the second system, which can be delayed. Of course we need to know the outcome, so faster than light communication is not possible with this setup.