# Showing if scalar product of vectors is 1 then they are the same quantum state

How do you show that if $$|\langle \psi|\phi\rangle| = 1$$, then $$\phi$$ and $$\psi$$, both of dimension $$d$$, represent the same quantum state? (Same quantum state iff there exists a $$\theta$$ s.t. $$|\psi\rangle = e^{i\theta}|\phi\rangle$$)

I've tried doing given $$|\langle \psi|\phi\rangle| = 1$$

$$\Leftrightarrow \left|\sum_{k=0}^{d-1}{\psi_{k}\phi_{k}}\right| = 1$$

$$\Leftrightarrow \left|\exp(i\theta)\sum_{k=0}^{d-1}{\psi_{k}\phi_{k}}\right| = 1$$ for any $$\theta \in \mathbb{R}$$, but couldn't go much further.

• What have you tried? See How to ask a good question. Commented Jun 27, 2020 at 12:42
• By "same quantum state" do you mean $\psi = \alpha \phi$ with $|\alpha | = 1$ ? Commented Jun 27, 2020 at 17:45
• @KeithMcClary yes, thats what I mean. Thank you for pointing it out, I have edited the original post.
– M80
Commented Jun 27, 2020 at 18:20

As quantum states are normalized, we have $$|\langle \psi | \phi \rangle|^2 = 1 = \langle \psi | \psi \rangle \langle \phi | \phi \rangle,$$ so $$\phi$$ and $$\psi$$ are linearly dependent. Neither of them is zero, hence one must be a scalar multiple of the other. This scalar is in $$U(1)$$, as both states have magnitude $$1$$.
• How does $⟨ψ|ϕ⟩|^{2}=1=⟨ψ|ψ⟩⟨ϕ|ϕ⟩$ imply linear dependence?