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I've been watching this video for the introduction to phase kickback. And here's a diagram:

enter image description here I got confused if we really need $|\psi_k\rangle$ to be an eigenstate to make the kickback work. It seems to me (from the math) that $|\psi_k\rangle$ could be any state.

My follow-up question is assume $|\psi_k\rangle$ has to be an eigenstate, then for this diagram:

enter image description here

Is it true only if the second qubit is set to $|1\rangle$ state?

I'm still a bit confused if there are any relations between the 2 diagrams. Why the top one requires the second register to be in the state $|\psi_k\rangle$, but the bottom one, it could be $|0\rangle$ or $|1\rangle$ state?

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2 Answers 2

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The first circuit equality fails when $|\psi_k\rangle$ is not an eigenstate of $U$.

A simple way to see this is to set the control qubit to $|1\rangle$. In this case, the RHS circuit is equivalent to $I\otimes I$ up to global phase while the LHS circuit is $I\otimes U$. These are equivalent up to global phase only when $|\psi_k\rangle$ is an eigenstate of $U$.

Alternatively, you may notice that the RHS circuit never creates any entanglement. On the other hand, if $|\psi_k\rangle$ is not an eigenstate of $U$ then the LHS circuit may produce entangled output. Therefore, LHS and RHS circuits cannot be equal.


The second circuit equality is always true. To see this, note that $\Delta$ is a scalar, so $e^{i\Delta}$ is a multiple of the identity, so every state is its eigenstate.

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  • $\begingroup$ Thanks so much for the answer!! Can I say the top diagram is the phase kickback for QPE, while the second (bottom) diagram is the phase kickback in general? $\endgroup$
    – IGY
    Commented Jun 19, 2022 at 21:55
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    $\begingroup$ Yeah, this sounds right, but note that the first equality is of course fairly general (as long as the assumptions about inputs are satisfied) and true outside the scope of QPE. In the second equality one should add that $\Delta$ is a phase angle (not a matrix), though that is implied if horizontal lines are understood to denote single qubits. $\endgroup$ Commented Jun 19, 2022 at 22:02
  • $\begingroup$ Thanks!! I'm still not super clear how the two equalities are connected. Can I say the controlled unitary has something more to consider, compared to just a controlled phase, so that an initial eigenstate is required for phase kickback? $\endgroup$
    – IGY
    Commented Jun 19, 2022 at 22:08
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    $\begingroup$ Updated the answer to clarify the connection between the two. In short, the second one is a special case with $U=e^{i\Delta}I$. In this case, all states are $U$'s eigenstates and so the equality is true for all inputs. $\endgroup$ Commented Jun 19, 2022 at 22:11
  • $\begingroup$ Thanks so much!! $\endgroup$
    – IGY
    Commented Jun 19, 2022 at 22:41
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Disclaimer: please do not consider this to be a definite answer as it is not fully rigorous. I am just giving you some evidence or clues from one practical example.

You can write the state $|\psi_k\rangle$ as a linear combination of eigenstates, given they form a basis - this is true for example if $U = \text{exp}(A)$, where $A$ is a Hermitian operator. Then the operator $U$ action on $|\psi_k\rangle$ is a linear combination of actions on the eigenstates.

You can find this arrangement in HHL algorithm, for example. In this case a right side $b$ of a linear system $Ax = b$ is on input to the phase estimation algorithm. The vector $b$ is expressed in basis composed of eigenvectors of $A$. As a result, you have a linear combination of all eigenvalues on output of the phase estimation.

Concerning, the second question. To show the equivalence of the two diagrams, you should write down matrix representation of both. The first one is $$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \mathrm{e}^{i\Delta} & 0 \\ 0 & 0 & 0 & \mathrm{e}^{i\Delta} \\ \end{pmatrix}. $$ This matrix can be writen as $$ \begin{pmatrix} 1 & 0 \\ 0 & \mathrm{e}^{i\Delta} \end{pmatrix} \otimes I, $$ which is the second diagram.

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  • $\begingroup$ Thanks for the answer! So does phase kickback dependent on the $|\psi_k\rangle$. I got confused about this. I thought I shouldn't. $\endgroup$
    – IGY
    Commented Jun 19, 2022 at 21:08
  • $\begingroup$ @IGY: I am afraid I do not understand your question....do you mean whether the phase changes based on the state $|\psi_k\rangle$? If so, the answer is yes. The phase estimation is used for getting a phase of the state,...so it is different for different input state. $\endgroup$ Commented Jun 19, 2022 at 21:18
  • $\begingroup$ Thanks, my question is for different initial states for the $e^{i\Delta}$ gate on my second graph, do we always have that property of phase kickback? (it's not about phase estimation) I think we always have $|0\rangle\langle 0|\otimes e^{i\Delta}|1\rangle\langle1| = $ that matrix $\otimes I$, no matter what $|\psi_k\rangle$ is $\endgroup$
    – IGY
    Commented Jun 19, 2022 at 21:20
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    $\begingroup$ @IGY: Concerning the second question, the equivalence between the two diagrams is valid for both $0\rangle$ and $1\rangle$ input on second qubit. This is called controlled global phase, - if the control qubit (the first one) is in state $1\rangle$, then phase $\delta$ is applied regardless the state of second qubit. $\endgroup$ Commented Jun 19, 2022 at 21:28
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    $\begingroup$ @IGY: I edited the question, hope it helps. $\endgroup$ Commented Jun 20, 2022 at 4:03

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