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When giving examples of universal gate sets in the paper Qudits and High-Dimensional Quantum Computing, the authors first define the transformation that maps any given qudit state to $|d-1\rangle$: $$ U_d(\alpha): \sum_{l=0}^{d-1}\alpha_l|l\rangle \rightarrow |d-1\rangle, \ \alpha = (\alpha_0, \alpha_1,\dots,\alpha_{d-1}) $$ This can be decomposed into $d-1$ unitary transformations $$ U_d = X_d^{(d-1)}(a_{d-1},b_{d-1}),\dots,X_d^{(1)}(a_1,b_1), \ a_l = \alpha_l,\ b_l = \sqrt{\sum_{l=0}^{l-1}\alpha_i^2} $$ with $$ X_d^{(l)}(x,y)= \begin{bmatrix} \mathbb{I}_{l-1} & & & \\ & \frac{x}{\sqrt{|x|^2 + |y|^2}} & \frac{-y}{\sqrt{|x|^2 + |y|^2}} & \\ & \frac{y^*}{\sqrt{|x|^2 + |y|^2}} & \frac{x^*}{\sqrt{|x|^2 + |y|^2}} & \\ & & & \mathbb{I}_{d-l-1} \\ \end{bmatrix} $$ So in the qutrit case, the unitary transformations are $$ U_3 = X_3^{(2)}(a_2,b_2),X_3^{(1)}(a_1,b_1) $$

where each is transformation is a $3\times3$ matrix.

Considering the example of mapping the state $|0\rangle \rightarrow |2\rangle$, we have $\alpha_0 = 1, \ \alpha_1 = \alpha_2 = 0 = a_1 = a_2$ and $b_1 = b_2 = 1$. Hence we see $$ X_3^{(2)}(0,1)X_3^{(1)}(0,1)|0\rangle = |2\rangle $$ However, when mapping $|1\rangle \rightarrow |2\rangle$ we see $$ X_3^{(1)}(1,1)X_3^{(2)}(0,1)|1\rangle = |2\rangle \\ $$

As it is not intuitively clear (to me), why is it necessary to switch the order of the gates (or where am I going wrong)? More importantly, given $U_d$, is there any method to determine the appropriate sequence for applying the gates $X_d^{(d-1)}(a_{d-1},b_{d-1}),\dots,X_d^{(1)}(a_1,b_1)$?

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  • $\begingroup$ Well, in the first case, you swap |0> and |1> and then |1> and |2>. In the second case you first swap |1> and |2>. The second operation is not really necessary. $\endgroup$
    – EvgeniyZh
    Commented Sep 3, 2023 at 13:59
  • $\begingroup$ Thanks @EvgeniyZh, however I had already noticed this and, unless I am missing something, this doesn't really answer how to determine the gate sequence when given $U_d$. $\endgroup$
    – banercat
    Commented Sep 3, 2023 at 16:04
  • $\begingroup$ If you are guaranteed that $U_d$ is created via this construction, then you can invert it and look at what $|d-1 \rangle$ maps to; then repeat the construction. $\endgroup$
    – EvgeniyZh
    Commented Sep 4, 2023 at 7:37

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