Is phase factor negligible in fidelity of quantum states?

One well-known fidelity is defined as $$(Tr\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}})^2$$. And for pure states, fidelity is always in the form $$|\langle\psi|\phi\rangle|^2$$. As we know, in the context of two-qubit quantum computation, we cannot tell the difference between $$|\psi\rangle$$ and $$e^{ia}|\psi\rangle$$($$a$$ is real and perhaps $$|\psi\rangle$$ need to be a pure state). And the definitions of fidelity above are all modes of some complex values. So does that means the phase factor of quantum states is negligible in calculating the fidelity?

And could we generalize it to quantum operation and say that $$U$$, a unitary, cannot be distinguished from $$e^{ia}U$$?

• I'm not sure what you mean, density matrices are hermitian and positive definite, you cannot multiply them by a complex phase, and $|\psi\rangle$ is always a pure state by definition, $\rho=|\psi\rangle\langle\psi|$. Your last point about the unitary is true. – user2723984 Jul 29 '19 at 9:41
• Ahhh..Thanks for reminding this... – raycosine Jul 29 '19 at 10:08

1 Answer

You are correct in both assumptions.

A total phase on a qubit state $$|{\psi}\rangle$$ is often referred to as the global phase. Any measurement of a quantum state is the expectation value $$\lambda_{M}$$ of some (Hermitian) observable $$M$$: $$\lambda_{M} = |\langle\psi|M|\psi\rangle|^{2}$$. Because this is invariant to the global phase, there is no physical meaning to adding said global phase $$e^{i\alpha}$$ to the state $$|\psi\rangle$$, and therefore it is often omitted.

This also means, as you correctly observed, that such a phase can be omitted from a Unitary $$e^{i\alpha}U$$ as well. Even more, we often restrict ourselves to only use elements from $$SU(2)$$; the group of unitary $$2\times2$$ matrices with $$\det(U) = 1$$; this is essentially the same as choosing a particular phase $$\alpha_{1}$$ for the unitary (although not necessarily $$\alpha_{1} = 0$$!).