Faithful description of a photonic setting with the circuit model

The above picture comes from this paper.

The circuit on the left and the one on the right are equivalent (up to the basis). However, there is an important difference: the circuit makes the input -- i.e. $$\alpha|\texttt{0}\rangle + \beta|\texttt{1}\rangle$$ -- interact with two qubits initialised to $$|\texttt{0}\rangle^{\otimes 2}$$, while the photonic settings achieves the same result with only one interaction (to an EPR pair).

Hence, I am looking for a circuit description which works as the photonic settings; in the sense that the input state interact with only one of the two entangled qubits.

Note: considering that Bell pairs are basis independent, I actually don't know how important is the fact that the circuit works on $$z$$-basis, while the photonic settings works on the $$y$$-basis.

• there's no need to mark edits as such. Keep in mind that posts should be readable by somehow stumbling upon the post in the future, which thus doesn't need/want to know the history of the post itself. That aside, could you define what you mean exactly with "implement and decode the photonic setting"? Also, what do you mean with "syntactically different", and "describes more faithfully the photonic evolution"? You also didn't define the symbols used in your circuit. I'm assuming it's a three-qubit circuit, and $H$ is the Hadamard. Are the double squares CNOTs or CZ? What is "Sdg"?
– glS
Jun 20 at 10:46
• @glS Thank you. I've been a bit cursory. Now I think the question is self-consistent. Jun 20 at 11:12
• that's much clearer, but I don't think you can directly compare photonic and logic circuits like that. To name one thing, photonic circuits actually operate on a much larger Hilbert space. You could describe the photonic circuit in the circuit model, but you'd get a nontrivial description. Each photon essentially amounts to two qubits, one qubit for its polarisation state and one for its position state (assuming the photon to only exist in one of two possible positions). Even worse, in these circuits you deal with multiple photons, which further enlarges the space
– glS
Jun 20 at 15:28
• also note that the two circuits in the picture use different input states. You could argue that the photonic circuit only uses "one interaction", the PBS, but (1) it's probabilistic, thus fundamentally different, and (2) it uses as input state a maximally entangled state. If these are allowed, you could make your logic circuit resulting in a GHZ with zero interactions: just start with $(\alpha|0\rangle+\beta|1\rangle)(|00\rangle+|11\rangle)$ and post-select on events of the form $|000\rangle$ or $|111\rangle$, and you'll get your GHZ.
– glS
Jun 20 at 15:33
• The PBS between the first and second qubits, is changing the states like 011 and 100 to states of (HV,vaccum,1) and (vaccum,HV,0) where HV menas that qubit has 1 photon in H and 1 in V (illegal). The 000 and 111 behave like identity - unchanged. States with 2 photons in one mode or vaccum, is not legal in circuit model, and therfore we can show equivalence only after remebering that after measuremnets you through to garbage all the states that had vaccum or 2 photons. Jun 20 at 16:52