I'm revisiting my knowledge on phase kickback, and I realize that there are many holes in my understanding. I've come across the definition that phase kickback is the phenomenon that occurs when you apply a controlled unitary where the target qubit is in an eigenstate of the unitary thus kicking the phase to the ancilla qubit So here's my question:

Why is it that when applying a controlled operation where the target qubit is an eigenstate, that the phase of that state gets kicked up to the ancilla qubit?

  • 1
    $\begingroup$ Have you tried doing the math? I don't know a better explanation than just math :-) The first part of quantumcomputing.stackexchange.com/a/2568/2879 does it $\endgroup$ – Mariia Mykhailova Nov 19 '20 at 23:33
  • $\begingroup$ Seconded @MariiaMykhailova, the math is much more intuitive than trying to explain it with words $\endgroup$ – C. Kang Nov 19 '20 at 23:49
  • $\begingroup$ I haven't learned traces yet in linear alg unfortunately. $\endgroup$ – Sinestro 38 Nov 20 '20 at 20:37

Here is a basic example of a two system that might help you to see this better. Suppose I have these two circuits:

Circuit 1: Which put the "Controlled qubit" in the state $|1\rangle$ and the "Target qubit" state in $H \big(X|0\rangle \big) = \dfrac{|0\rangle - |1\rangle}{\sqrt{2}} $. Note that this state, $|psi \rangle = \dfrac{|0\rangle - |1\rangle}{\sqrt{2}} $ is an eigenvector of the NOT gate $X = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$ since $ X|\psi \rangle = X \bigg(\dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) = -1 \bigg(\dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) = -1|\psi \rangle $.

enter image description here

so here the state before the Controlled-NOT (CNOT) operation is $|1\rangle \bigg(\dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) = \dfrac{|10\rangle - |11\rangle}{\sqrt{2}}$.


$$ CNOT \bigg( \dfrac{|10\rangle - |11\rangle}{\sqrt{2}} \bigg) = \dfrac{|11\rangle - |10\rangle}{\sqrt{2}} = - \bigg( \dfrac{|10\rangle - |11\rangle}{\sqrt{2}}\bigg) = - \bigg( |1\rangle \otimes \dfrac{ |0\rangle - |1\rangle}{\sqrt{2}} \bigg) $$

But as you know, the state $|\psi \rangle = - \bigg( |1\rangle \otimes \dfrac{ |0\rangle - |1\rangle}{\sqrt{2}} \bigg) $ and the state $|\phi \rangle = |1\rangle \otimes \dfrac{ |0\rangle - |1\rangle}{\sqrt{2}}$ are equivalent.

Circuit 2: Which put the "Controlled qubit" in the state $\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$, and "Target qubit" still in $\dfrac{|0\rangle - |1\rangle}{\sqrt{2}} $

enter image description here

here the state before the Controlled-NOT (CNOT) operation is: $$\bigg(\dfrac{|0\rangle + |1\rangle}{\sqrt{2}} \bigg)\bigg(\dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) = \dfrac{|00\rangle - |01\rangle + |10\rangle - |11\rangle}{2}$$ Hence \begin{align} CNOT \bigg( \dfrac{|00\rangle - |01\rangle + |10\rangle - |11\rangle}{2} \bigg) &= \dfrac{|00\rangle - |01\rangle + |11\rangle - |10\rangle}{2}\\ &= \bigg( \dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) \otimes \bigg( \dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) \end{align}

From here we can see that the controlled qubit started in the state $\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ but ended in the state $\dfrac{|0\rangle - |1\rangle}{\sqrt{2}}$. Thus, it has picked up the overall phase of the state: $ -1 \bigg(\dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) $ as its relative phase.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.