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I have a secret say $s$. I have a dealer $D$ and three participants $A, B, C$. I want to share this secret $s$ in such a way that the participation of all $3$ is essential to reconstruct the secret. Now for creating the shares, I use some classical sharing algorithms and create shares $s_A,s_B,s_C$. Now how do I distribute these shares among the participants quantum mechanically using qudits? What I thought is the following steps.

First, let the basis be $\{|0\rangle, |1\rangle,.....,|d-1\rangle\}.$ Now since each of the participant $A, B, C$ has his/her share, one of them starts the reconstruction process by first preparing a $|0\rangle$ and taking its Fourier transform, so I get $$|\phi\rangle_1=\sum_{y=0}^{d-1}|y\rangle_1$$Now the next step is to initialize two $|0\rangle$ states and perform the CNOT gate on them with the first qudit as the control, so to get $$|\phi\rangle_2=\sum_{y=0}^{d-1}|y\rangle_1|y\rangle_2|y\rangle_3$$After this step we perform the Quantum Fourier transformation on all the particles to get $$|\phi\rangle_3=\sum_{y=0}^{d-1}\sum_{k_1=0}^{d-1}\sum_{k_2=0}^{d-1}\sum_{k_3=0}^{d-1}\omega^{(k_1+k_2+k_3)y}|k_1\rangle_1|k_2\rangle_2|k_3\rangle_3$$ Now since the summation is finite i rearrange the terms to get $$|\phi\rangle_3=\sum_{k_1=0}^{d-1}\sum_{k_2=0}^{d-1}\sum_{k_3=0}^{d-1}\sum_{y=0}^{d-1}\omega^{(k_1+k_2+k_3)y}|k_1\rangle_1|k_2\rangle_2|k_3\rangle_3$$ With $\sum_{i=0}^{d-1}\omega^i=0$, we have the condition that the state left after this operation will be subject to the condition that $k_1+k_2+k_3=0\;mod\;d$ , we will have $$|\phi\rangle_3=\sum_{k_1=0}^{d-1}\sum_{k_2=0}^{d-1}\sum_{k_3=0}^{d-1}|k_1\rangle_1|k_2\rangle_2|k_3\rangle_3$$

now after preparing this state each participant $A,B,C$ applies a transformation $U_{s_B},U_{s_A},U_{s_C}$ which gives the state as $$|\phi\rangle_3=\sum_{k_1=0}^{d-1}\sum_{k_2=0}^{d-1}\sum_{k_3=0}^{d-1}|k_1+s_A\rangle_1|k_2+s_B\rangle_2|k_3+ s_C\rangle_3$$ After peparing this state the state is returned by the participants to the dealer who measures state for the shares and if it is right then announces the result/secret. Now my questions are:

(i) Even though this is a very preliminary effort, can somebody tell me whether can we can actually do this?

(ii) My second question is if this is possible then can we improve this scheme to achieve the condition for the detection of a fraudulent participant? Can somebody help??