(This question is a kind of sequel to a prior question at Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm?)

This question asks for someone to identify the error in what seems to me to be a reasonable interpretation of the phase kickback math. Here's the setup for the question. Suppose we have a unitary gate $U$ with eigenvector $|u\rangle$ and eigenvalue $e^{i\phi}$, so $U|u\rangle = e^{i\phi}|u\rangle$. If we use $|+\rangle$ as the control of a controlled-$U$ gate receiving $|u\rangle$, then the input system is $|+\rangle|u\rangle$, and the output system can be written (excluding the normalization factor $1/\sqrt{2}$) as $|0\rangle|u\rangle + e^{i\phi}|1\rangle|u\rangle$. The $e^{i\phi}$ is a term we can't directly measure, but it is part of the physical system, and can affect computations after this step. We can rewrite this output of the system in two forms:

$$ \begin{align} \begin{split} |0\rangle|u\rangle + e^{i\phi}|1\rangle|u\rangle &= |0\rangle|u\rangle + |1\rangle\left(e^{i\phi}|u\rangle\right) \,\,\,\,\, & \text{Form 1} \\ &= \left(|0\rangle + (e^{i\phi}|1\rangle\right) \, |u\rangle \,\,\,\,\, & \text{Form 2} \\ \end{split} \end{align} $$

Interpretation of Form 1: The left qubit is either $|0\rangle$ or $|1\rangle$. If we measure the left qubit and get $|0\rangle$, then we know that the right qubit is $|u\rangle$, and if we measure $|1\rangle$ on the left qubit, then we know that the right is $e^{i\phi}|u\rangle$. We can't measure that phase directly, but applying something like QPE to the right qubit we can turn the phase into something we can measure, so we could detect that the phase shift is on the right qubit, not the left. In short, the left qubit is unchanged, and the right changes.

Interpretation of Form 2: This the typical phase kickback formulation, where the phase is associated with the control qubit, here the left one. Now we have the opposite situation: the left qubit changes, and the right does not.

These interpretations appear to describe different physical states, which can result in different measurements after additional circuitry, yet only algebra distinguishes these two forms. Since the second interpretation is widely accepted and used in analysis, this suggests that the first interpretation is wrong. But specifically why?

That is, rather than a walkthrough of a "right" interpretation, I hope someone can answer identify and correct the specific error in logic or understanding in Interpretation 1.

  • $\begingroup$ Why are you applying the QPE to the right (second) register in the first case? $\endgroup$ May 29, 2022 at 4:40
  • $\begingroup$ The QPE is not integral to the question; it's only an example of a computation that could depend on the phase of the right qubit. $\endgroup$
    – Eleeza
    May 29, 2022 at 5:45

2 Answers 2


Physically, there is no difference between measuring the left qubit in the state $|0\rangle$ or $|1\rangle$. As stated in a previous answer, if the left qubit is measured in $|1\rangle$ the phase $e^{i\phi}$ becomes a global phase to the state $|u\rangle$, so you can not measure it.

Regarding your interpretations. I think an easier way to interpret what's happening is to consider the state

$$|0⟩|u⟩+e^{iϕ}|1⟩|u⟩ \qquad (1)$$

in the computational basis. Let's suppose that the state $|u\rangle$ can be written as:

$$|u\rangle = u_0 |0\rangle + u_1 |1\rangle $$

the state in (1) becomes:

$$ |0⟩(u_0 |0\rangle + u_1 |1\rangle) + e^{iϕ}|1⟩(u_0 |0\rangle + u_1 |1\rangle) $$

$$ = u_0 |00\rangle + u_1 |01\rangle + e^{iϕ}u_0 |10\rangle + e^{iϕ} u_1 |11\rangle $$

Using this form you can see that the states $|10\rangle$ and $|11\rangle$ are the ones being 'changed' with a phase.

You can find more information here:


A related question is:

Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm?


Regarding the first interpretation it appears that you wish to measure the left qubit and, conditioned on measuring $|1\rangle$ on the left qubit, "apply something like QPE to the right qubit".

But, once the left qubit is measured as $|1\rangle$, the phase of the right qubit then becomes global, and is not able to be determined. This phase cannot be determined by any computation once the left qubit is measured.

Furthermore measuring the left qubit in the Hadamard basis, or equivalently performing a Hadamard gate on the left qubit and measuring the left qubit in the computational basis, is what's meant by "applying the QPE" - but that is done on the left qubit and not the right qubit.

In short, I think the statement that "applying something like QPE to the right qubit" may be the part that's confused - we apply something like QPE to the left qubit, not the right.


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