I was reading this article. A brief explanation: Here we have a circuit, the registers are a stepfunction state, an single photon state and a function state. The first two have position operators $X$ and $Y$, respectively. circuit

but I'm very confused on how can that the set

$$\{e^{i\frac{\pi}{2}(\hat{X}^2+\hat{P}^2)},e^{it_1\hat{X}}, e^{it_2\hat{X}^2},e^{it_3 \hat{X}^3},e^{i\tau \hat{X}_1\otimes \hat{X}_2}\}$$

be the universal set. I mean, where did the Y go?

In this case $\hat{A}=\lambda \mathbb{I}+\sum_{i=1}^n a_i \hat{X}_i + b_i \hat{P}_i + \alpha_i \hat{X}^2_i +\beta_i \hat{P}^2_i$

I just want a clue about how to start here, thanks everyone.

  • $\begingroup$ I think when they talk about their universal gates, they drop the indices, so on three wires , $e^{itX}$ acutally corresponds to having the three gates $e^{itX_1}, e^{itX_2},e^{itX_3}$. The same holds for the interaction operation where you have all possible pairs of wires. Finally, set $X = X_1, Y = X_2$ and then you're there. $\endgroup$ – Johannes Jakob Meyer Apr 14 '20 at 13:24
  • $\begingroup$ Thanks! I'll try that $\endgroup$ – tererecomchimarrao Apr 15 '20 at 15:17

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