Two state vectors representing the same physical state

Question

I saw somewhere that if two state vectors $\lvert \psi \rangle$ and $\lvert \psi' \rangle$ represent the same physical state, then we have $$\lvert \psi \rangle = e^{i\alpha} \lvert \psi' \rangle$$ I first wonder what the phrase "same physical state" means. Does it mean if we have any operator $\Omega$ then $\langle \psi \lvert \Omega \rvert \psi \rangle = \langle \psi' \lvert \Omega \rvert \psi' \rangle$? And can we show that $\lvert \psi \rangle = e^{i\alpha} \lvert \psi' \rangle$ is the necessary/sufficient condition of $\lvert \psi \rangle$ and $\lvert \psi' \rangle$ representing the same physical state?

Thanks!!!

Answer 1

Yes, all expectation values of $|\psi \rangle$ and $|\psi' \rangle$ are identical, because $$\langle \psi' | \Omega | \psi' \rangle = e^{-i\alpha} e^{i\alpha} \langle \psi | \Omega | \psi \rangle = \langle \psi | \Omega | \psi \rangle$$ by the (anti)linearity of the inner product. So this is clearly a sufficient condition.

Answer 2

I first wonder what the phrase "same physical state" means.

It simply means that states are rays rather than vectors in Hilbert space. From Weinberg's "The Quantum Theory of Fields", pg 49 - 50:

Physical states are represented by rays in Hilbert space.

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A ray is a set of normalized vectors (i.e., $\langle \Psi | \Psi \rangle = 1$) with $\Psi$ and $\Psi'$ belonging to the same ray if $\Psi' = \zeta\;\Psi$. wjere $\zeta$ is an arbitrary complex number with $|\,\zeta\,| = 1$