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Let $\mathcal{H}$ be a Hilbert space. We define the projective Hilbert space $\mathbb{P}\mathcal{H}$ as $\mathcal{H}\setminus \{ 0\}/\mathbb{C}^*$. Then $[\Psi]=\{ z\Psi :z\in \mathbb{C}^*\}$.

On the other hand, we know that $U(1)$ acts on $\mathcal{H}_1 :=\{ \Psi \in \mathcal{H}:\langle \Psi |\Psi \rangle=1\}$. So we can construct $\mathcal{H}_1/U(1)$ and so we have $[\Psi]=\{ e^{i\theta}:\theta \in \mathbb{R}\}$.

What is the difference between $(\mathcal{H}\setminus \{ 0\})/\mathbb{C}^*$ and $(\mathcal{H}_1/U(1)$? Are they the same?

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    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$
    – ACuriousMind
    Feb 7 at 16:14
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    $\begingroup$ Also, the text you quoted already answers your question: It doesn't say to take the quotient $(H\setminus\{0\})/\mathrm{U}(1)$, it says to take the unit sphere in $H$ and quotient it by $\mathrm{U}(1)$. $\endgroup$
    – ACuriousMind
    Feb 7 at 16:15
  • $\begingroup$ @ACuriousMind You are right. I've made a mistake. Assume that $\mathcal{H}_1 :=\{ \Psi \in \mathcal{H}:\langle \Psi |\Psi \rangle =1\}$. Are $(H\setminus \{ 0\})/\mathbb{C}^*$ and $\mathcal{H}_1 /U(1)$ the same? $\endgroup$
    – Mahtab
    Feb 7 at 16:19
  • $\begingroup$ Yes - the proof is not hard, try it. $\endgroup$
    – ACuriousMind
    Feb 7 at 16:20
  • $\begingroup$ @ACuriousMind Thanks for the comment. Could you please tell me you mean they are the same set? $\endgroup$
    – Mahtab
    Feb 7 at 16:30

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The text mentions two definitions of the space of physical states. One is $\mathcal{H} \setminus \{0\}$ modulo $\mathbb{C}^*$, and the other is the unit sphere in $\mathcal{H}$ modulo $U(1)$. They are equivalent, there's no distinction between using one or the other.

The unit sphere $\mathcal{H}_1$ is the subset of vectors with absolute value 1; topologically, it's equivalent to $\mathcal{H} \setminus \{0\}$ quotiented by $\mathbb{R}^*$. This is because in this latter set, the equivalence class of a vector $\Psi$ is the set of all the nonzero real multiples of $\Psi$, which contains exactly one element of absolute value 1; and conversely, every element of the unit sphere is in one equivalence class, so there's a one-to-one correspondence.

Now, $\mathbb{C}^* = \mathbb{R}^* \times U(1)$, because a nonzero complex number has an amplitude and a phase. We can think of $(\mathcal{H}\setminus \{0\}) / \mathbb{C}^*$ as being done in two steps: first we quotient by $\mathbb{R}^*$ and get the unit sphere, and then we quotient by $U(1)$ and get the space of rays.

This is a bit extra, but strictly speaking $(\mathcal{H}\setminus \{0\}) / \mathbb{C}^*$ and $\mathcal{H}_1 / U(1)$ are homeomorphic, but they are not the same set. Both are sets of equivalence classes; but each equivalence class $[\Psi]$ in the first quotient has elements with arbitrary amplitude, while the equivalence class in the second quotient only has unit vectors. Still, there's a one-to-one correspondence between the two quotients, and I'm sure that there are topological theorems assuring that everything is continuous.

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  • $\begingroup$ Dear @Javier thanks so much for the comment. I've put a part of a book concerning my question. Could you please tell my when we use $[\Psi]=\{ z\Psi :z\in \mathbb{C}^*\}$ and when we use $[\Psi]=\{ e^{i\theta}:\theta \in \mathbb{R}\}$ in quantum mechanics? I don't understand it. $\endgroup$
    – Mahtab
    Feb 7 at 16:16
  • $\begingroup$ Sorry. I've made a mistake in my question. I've edit it. $\endgroup$
    – Mahtab
    Feb 7 at 16:19
  • $\begingroup$ @Mahtab I've updated my answer. $\endgroup$
    – Javier
    Feb 8 at 2:01
  • $\begingroup$ Thank you very much for your nice explanation. I got it now. I appreciate you. $\endgroup$
    – Mahtab
    Feb 8 at 17:06

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