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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.

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

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Hint: use Cauchy–Schwarz inequality.

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 $1$.

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  • $\begingroup$ How does $⟨ψ|ϕ⟩|^{2}=1=⟨ψ|ψ⟩⟨ϕ|ϕ⟩ $ imply linear dependence? $\endgroup$
    – M80
    Commented Jun 27, 2020 at 21:24
  • $\begingroup$ This is a corollary of C–S inequality $\endgroup$ Commented Jun 28, 2020 at 7:55

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