I have a state $$ |\tilde{\Phi_2}\rangle =\dfrac{1}{2^{3l/2}}\sum_{x=0}^{2^l-1}\sum_{y=0}^{2^l-1}\sum_{z=0}^{2^l-1}|0\rangle^{\otimes q}\otimes |x\rangle^{\otimes l}\otimes |y\rangle^{\otimes l}\otimes |z\rangle^{\otimes l} $$ Now I want to change the state $|z\rangle$ using the states $|x\rangle$ and $|y\rangle$. The controls originally $|x\rangle $ and $|y\rangle$ are originally in states $|x_0x_1...x_{l-1}\rangle$ and $|y_0x_1...y_{l-1}\rangle$, then if $$ |x_0\rangle=|y_0\rangle=|1\rangle$$ then the state $|z\rangle$ has to be changed. And the state $|z\rangle$ is changed to $|00....z_{l-3}\rangle$.

Since the $|0\rangle^{\otimes q}$ is unaltered we just operate an identity operator $I^{\otimes q}$.

The next thing I understand is that we have to break the summation into two parts where in the first part the summation indexes of $i,j$ go from $0$ to $\lfloor \dfrac{2^l-1}{2}\rfloor$.

But after that I am unable to write deduce further. Can somebody help me in writing the Dirac notation for this operator?

Edit: The transformation on the $z$ register is the right shift by $2$ operator $|z,0\rangle \to |z, \frac{z}{4}\rangle$


As you say, any register on which you do nothing, use the identity, $I$. This is also going to be the case on $x_1,x_2,\ldots$ and $y_1,y_2,\ldots$.

When you want to control something, use the projectors $P_0=|0\rangle\langle 0|$ and $P_1=|1\rangle\langle 1|$. So, controlled-$U$ looks like $P_0\otimes I+P_1\otimes U$. For controlled-controlled-$U$, you can make this notation a bit more concise: $$ I\otimes I\otimes I+P_1\otimes P_1\otimes(U-I). $$ This is essentially what you're going to be doing for your operation, where the two controls are $x_0$ and $y_0$.

So, the next question is what the $U$ looks like for transforming the register $z$. If I understand correctly, you basically want to introduce an additional register $|0\rangle^{\otimes(l-2)}$ and to perform the joint transformation $$ |z_0z_1z_2\ldots z_{l-1}\rangle|0\rangle^{\otimes(l-2)}\rightarrow|z_0z_1z_2\ldots z_{l-1}\rangle|z_0z_1z_2\ldots z_{l-3}\rangle $$ This is achieved in a straightforward manner. Let $C_i$ denote the controlled-not gate controlled off qubit $i$ of the first register, and targetting qubit $i$ on the second register (and identity on everything else). Then, $$ U=\prod_{i=0}^{l-3}C_i. $$ I remain unclear if that target register is supposed to be the one that starts $|0\rangle^{\otimes q}$, or not. I'm assuming not.

Overall, you would thus be left with $$ \prod_{i=0}^{l-3}I^{\otimes q}\otimes(I^{\otimes (4l-2)}+P_1\otimes I^{\otimes(l-1)}\otimes P_1\otimes I^{\otimes(l+i-1)}\otimes P_1\otimes I^{\otimes (l-1)}\otimes(X-I)\otimes I^{\otimes(l-3-i)}). $$ Note, the first two $P_1$s are the controls off $x_0$ and $y_0$. The third is the control off $z_i$, and the $(X-I)$ term should be acting on the target, $i$ of the extra register. Now you can express each of these operators in Dirac notation if you really want to.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – heather May 23 '19 at 0:14
  • $\begingroup$ Yes sir now i think it is what i was looking for. Let me explain to you what i understood $\endgroup$ – Upstart May 23 '19 at 16:43
  • $\begingroup$ The operator can be $$ \prod_{i=0}^{l-3}I^{\otimes q}\otimes(I^{\otimes (4l-2)}+P_1\otimes I^{\otimes(l-1)}\otimes P_1\otimes I^{\otimes(l+i-1)}\otimes P_1\otimes I^{\otimes (l-1)} \otimes I^{\otimes 2}\otimes(X-I)\otimes I^{\otimes(l-3-i)})$$If i add the register $|0\rangle^{\otimes l}$ instead of $|0\rangle^{\otimes l-2}$ The change would just be that the control will still be $i$-th bit of $z$ but the target would be $i+3$-th qubit of the register $|0\rangle^{\otimes l}$ $\endgroup$ – Upstart May 23 '19 at 22:31
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    $\begingroup$ Yes, that would do. $\endgroup$ – DaftWullie May 24 '19 at 6:53
  • $\begingroup$ Sir i really appreciate the way you explained the answer inspite of my regular petty doubts. $\endgroup$ – Upstart May 24 '19 at 10:04

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