# Asymmetry in distributing phase change across components

The quantum computing text books and theory in general seems to have added an asymmetry in the distribution of change in phase across the components in the context of a qubit. Is there any reason for this? I know that global phase can be ignored and it allows us to assign all change to only one component. But I have observed that the impact would be different in multi-qubit contexts. For example, a controlled phase rotation between two qubits would no longer be symmetrical, if the phase distribution is symmetrical, in contradiction to what the textbook and many algorithms consider.

Is this asymmetry introduced due to any physics/hardware related problem which forbids distributing the phase change equally? Or is it only a software trick to reduce the operations (one instead of two with half change)?

I'm generally concerned about any loss of symmetry in such a beautiful and natural science. I don't think the nature itself pushes any changes entirely to one component. If it is a reference frame issue (we want to align our reference frame with one of the component for our convenience), then I'm afraid that this reference frame needs to a global one (single one for all qubits), but I see that this reference frame is local to each qubit, which appears quite artificial to me.

• Welcome to QCSE. I'm confused a little by your question. Are you asking about the difference between absolute and relative phase? – Mark S Aug 24 at 18:33
• Hi @Mark S - I'm asking about why the relative phase is added to ket_1 only, and not distributed equally? This might not matter for a single qubit, but I found it matter for multi-qubit context. – vrpbkp Aug 24 at 19:03
• You can set a phase for each state in superposition separately. For example, you can have a state (I did not add normalization constant) $|001\rangle + |010\rangle-|100\rangle$ or $|001\rangle - |010\rangle+|100\rangle$. As you can see the phase is not distributed equally. These states are of course different. If you distributed the phase equaly they would be the same which is in contradiction with observation. – Martin Vesely Aug 25 at 7:02

If I understand your question correctly, you're asking why we choose to write phase gates like $$\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{i\phi} \end{array}\right)$$ rather than $$\left(\begin{array}{cc} e^{-i\phi/2} & 0 \\ 0 & e^{i\phi/2} \end{array}\right).$$ As you rightly state, these are entirely equivalent due the the global phase, but controlled-versions of the gate are not. (Although the mapping is easily made by applying a phase gate on the control qubit.)
One might also argue that there's a far more fundamental break in the symmetry when you're considering two-qubit gates anyway. On the control qubit, there is a far more drastic distinction between $$|0\rangle$$ and $$|1\rangle$$. You might argue that we should be symmetrising that instead but, again, we'd find it much harder to think about and manipulate.