# How to understand intuitively the quantum gate phase kickback?

I do understand the math behind phase kickback. The math makes sense. For more context, I find this document very helpful.

But I’m struggling a lot to intuitively understand, why the conditional phase change to the second qubit will end up change the phase of the first qubit (the control qubit)?

Can somebody help to explain this in an intuitive way?

• I think you will your answer here: quantumcomputing.stackexchange.com/a/2572/9858 Sep 11 '20 at 6:43
• Thanks! Yeah the explanation is helpful. "The very idea that there is a 'control' qubit is one centered on the standard basis, and embeds a prejudice about the states of the qubits that invites us to think of the operation as one-sided. But as a physicist, you should be deeply suspicious of one-sided operations. For every action there is an equal and opposite reaction; and here the apparent one-sidedness of the CNOT on standard basis states is belied by the fact that, for X eigenbasis states, it is the 'target' which unilaterally determines a possible change of state of the 'control'." Sep 12 '20 at 6:15

Part of the problem people usually have here is a sort-of-classical intuition. Because you're trying to describe the action of a gate such as controlled-$$U$$, we divide it up as "if the control qubit is something, do something on the target qubit". It makes it sound like the control qubit doesn't change, and it's only the target that changes. This is completely wrong. One semi-intuitive way to see this is to consider the specific example of controlled-phase. This is symmetric - it doesn't matter which qubit you identify as control and which as target. $$|0\rangle\langle 0|\otimes I+|1\rangle\langle 1|\otimes Z=I\otimes |0\rangle\langle 0|+Z\otimes|1\rangle\langle 1|$$ Given that the gate is not the identity, something must change, and hence both qubits must change! It is a genuinely two-qubit gate that changes the two-qubit state.