Let me first tell my problem statement. Suppose I have a uniform superposition of states $$|A\rangle=\dfrac{1}{2^{9}}\sum_{i,j,k=0}^{2^6-1}|0\rangle^{\otimes 8}|i\rangle|j\rangle|k\rangle,$$ where $|0\rangle^{\otimes 8}$ is pre-defined for each $i,j,k$. So suppose for a particular instance in this superposition state we have the $|0\rangle^{\otimes 8}$ register as $|b_7b_6b_5b_4b_3b_2b_1b_0\rangle$ where each $b_i\in\{0,1\}$. Now I want to transform this register by a certain rule and that rule is: $$|b_7b_6b_5b_4b_3b_2b_1b_0\rangle\to |b_6~ b_5 b_4 (b_3\oplus b_7) (b_2\oplus b_7)~b_1~(b_1\oplus b_7) ~b_7\rangle$$ so for this transform I initialize a register $|0\rangle^{\otimes 8}$. So to do this computation I first

  1. Modify my original superposition state as $\displaystyle |\tilde{A}\rangle=\dfrac{1}{2^{9}}\sum_{i,j,k=0}^{2^6-1}|0\rangle^{\otimes 8}|0\rangle_B^{\otimes 8}|i\rangle|j\rangle|k\rangle$, then

  2. I start the transformation process by first applying the controlled NOT gate operation between the working register and the output register i.e $|b_7b_6b_5b_4b_3b_2b_1b_0\rangle$ and $0\rangle_B^{\otimes 8}$. This C-NOT gate changes the 3 qubit of $0\rangle_B^{\otimes 8}$ from the right to $b_7$. To do this I apply the transform $$U_1= P_1\otimes I^{\otimes 12}\otimes X\otimes I^{\otimes 2}\otimes I^{\otimes 3l}+P_0\otimes I^{\otimes 15+3l}$$ So after this step i have the superposition state as $$\displaystyle\dfrac{1}{2^{9}}\sum_{i,j,k=0}^{2^6-1}|b_7b_6b_5b_4b_3b_2b_1b_0\rangle|00000b_700\rangle|i\rangle|j\rangle|k\rangle.$$

  3. Next, I apply controlled not operation between the working and output registers to change the state of the output register as $ |000b_7b_70b_7b_7\rangle $, the operator I define is $$ U_2=\prod_{i=0}^{4}(P_1\otimes I^{\otimes 10+i}\otimes X\otimes I^{\otimes 4-i}\otimes I^{\otimes 3l}+ P_0\otimes I^{\otimes 10+i}\otimes I\otimes I^{\otimes 4-i}\otimes I^{\otimes 3l})$$ At the end of this step I have the working register as $ |b_7b_6b_5b_4b_3b_2b_1b_0\rangle$ and the output register as $|000b_7b_70b_7b_7\rangle$.

The final step is I apply the C-NOT between these two registers with control starting from $b_6$. So I define the operator $$U_3=\prod_{i=1}^{8}I^{\otimes i}\otimes P_1\otimes I^{\otimes 6}\otimes X\otimes I^{\otimes 8-i}\otimes I^{\otimes 3l}$$ after this step I have the state as $$\dfrac{1}{2^{9}}\sum_{i,j,k=0}^{2^6-1}|b_7b_6b_5b_4b_3b_2b_1b_0\rangle~ |b_6~ b_5 b_4 (b_3\oplus b_7) (b_2\oplus b_7)~b_1~(b_1\oplus b_7) ~b_7\rangle~\rangle|i\rangle~|j\rangle~|k\rangle$$ So the overall operation need to reach the desired state would be $\mathbf{U}=U_3U_2U_1$ operated on $|\tilde{A}\rangle$. Now I have a few question about this whole transformation process:

Q1. Are the transformation that I have written correct? Can somebody check?

Q2. If it is correct, then am I able to convey the operation that I am trying to do, by the explanations including the quantum computing terms like qubit, register, working, output register?

Q3. As we can see, the classical transformation is itself invertible, since given $$|b_6~ b_5 b_4 (b_3\oplus b_7) (b_2\oplus b_7)~b_1~(b_1\oplus b_7) ~b_7\rangle$$ we can get $|b_7b_6b_5b_4b_3b_2b_1b_0\rangle$ but for this to be done via quantum computation we had to introduce the ancillary register $|0\rangle_B$ to store the output, but given the output since I can always get back the input, why do I keep that working register now so is there some unitary/non unitary transform that does resets the first register to the state $|00000000\rangle$ and after that to get the superposition as $$ |B\rangle=\dfrac{1}{2^{9}}\sum_{i,j,k=0}^{2^6-1}|0\rangle^{\otimes 8}|i\rangle|j\rangle|k\rangle$$ where the register $|0\rangle^8$ here has the modified output from the transform $\mathbf{U}$.

Edit: Do i just measure the first register and then throw them away ?

  • $\begingroup$ Did you figure this out? For the last point, are you familiar with the partial trace operation? Something looks fishy to me that your final expression is a product state in the first 2 registers (i.e. they're not entangled) $\endgroup$ – bRost03 Jun 10 '19 at 3:53
  • $\begingroup$ I wrote what i wanted to implement, where does your doubt come can you explain? $\endgroup$ – Upstart Jun 10 '19 at 20:46

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