# Why does the phase of the eigenstate get kicked up to the ancilla qubit?

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?

• 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 – Mariia Mykhailova Nov 19 '20 at 23:33
• Seconded @MariiaMykhailova, the math is much more intuitive than trying to explain it with words – C. Kang Nov 19 '20 at 23:49
• I haven't learned traces yet in linear alg unfortunately. – 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 $$X|\psi \rangle = X \bigg(\dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) = -1 \bigg(\dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg) = -1|\psi \rangle$$.

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

hence

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

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 $$\dfrac{|0\rangle - |1\rangle}{\sqrt{2}}$$

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.