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This question already has an answer here:

In Sakurai's Modern Quantum Mechanics, it is stated that

One of the physics postulates is that $|\alpha\rangle$ and $c|\alpha\rangle$, with $c\neq 0$, represent the same physical state. In other words, only the "direction" in vector space is of significance.

Apart from the complex number $c\neq 0$, shouldn't we also require $|c|=1$? Shouldn't the normalizability of physical states force $|c|=1$?

As far as I know, physical states are represented by rays which contains $|\psi\rangle$ and $e^{i\theta}|\psi\rangle$, $\forall\theta\in\mathbb{R}$. This is how Weinberg defines a ray, in his QFT text (Vol. 1, section 2.1). It says that

A ray is a set of normalized vectors (i.e., $(\psi,\psi)=1$) $|\psi$ and $\psi^\prime$ belonging to the same ray if $\psi^\prime=\psi$, where $\zeta$ is an arbitrary complex number with $|\zeta|=1$.

  1. So which definition of a ray is the correct one? Weinberg's or Sakurai's? I read this post. Does it mean Weinberg's definition is not the general one?

  2. More displeasingly, if the criterion of $|c|=1$ (or $|\zeta|=1$, in Weinberg's notation) is relaxed, how does the Born's probability interpretation work? For example, how would one interpret a state $|\psi\rangle$ whose norm is $\langle\psi|\psi\rangle>1$ or $<1$. Shouldn't the probability amplitude for a particle to be in a state $|\psi\rangle$ when it is known to be in a state $|\psi\rangle$ must be unity? This is also not addressed here.

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marked as duplicate by Mark Mitchison, Kyle Kanos, AccidentalFourierTransform, John Rennie quantum-mechanics Feb 5 '17 at 8:32

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    $\begingroup$ There's nothing problematic about considering states that don't have unit norm. $\endgroup$ – gj255 Feb 3 '17 at 12:47
  • $\begingroup$ What you wrote down there is not a ray, a ray by definition is the set of all $c\lvert\psi\rangle,c\in\mathbb{C}$. $\endgroup$ – ACuriousMind Feb 3 '17 at 13:57
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    $\begingroup$ @gj255 - Except when they have infinite norm. Then you do have to be careful. $\endgroup$ – Prahar Feb 3 '17 at 20:11
  • $\begingroup$ See also this question for a discussion of the necessity of normalization. $\endgroup$ – ACuriousMind Jul 18 '17 at 14:13
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The space of states of a quantum system is the space of rays of some Hilbert space $\mathcal{H}$. That is, given the equivalence relation $\psi\sim z\psi$ for any $z\in\mathbb{C}-\{0\}$ and $\psi\in\mathcal{H}$, the states are elements of the quotient $\mathcal{H}/\sim$. Then, by definition, the states represented by $\psi$ and by $z\psi$ are the same.

Normalization of the vectors representing a state is convenient for calculations in quantum mechanics, but it is not necessary. For any representative $\psi$ of a ray representing the current state of the system the probability of finding the state in a ray represented by $\phi$ is

\begin{equation} \frac{\left|\left<\phi | \psi\right>\right|^2} {\left<\phi|\phi\right>\left<\psi|\psi\right>} \end{equation}

This simplifies to the usual $\left|\left<\phi | \psi\right>\right|^2$when the states are normalised.


About the space $\mathcal{H}/\sim$

The space of states $\mathcal{H}/\sim$ is what is known as a projective space. We can form a mental image of it by taking a representative of each equivalence class.

To understand a simple example, consider $\mathcal{H}=\mathbb{C}^2$. We can partly fix the representatives by imposing that their norm is one. That is, we reduce $\mathcal{H}$ to $\mathcal{H}'=\{(z,w)\in\mathbb{C}^2 : |z|^2+|w|^2=1\}$. The remaining redundancy is a global phase for $(z,w)$, so we can take $z$ to be real, for example. Defining $x=\operatorname{Re}w$ and $y=\operatorname{Im}w$ we end up with

\begin{equation} (\mathcal{H}/\sim) \cong \{(x,y,z)\in\mathbb{R}^3 : x^2 + y^2 + z^2 = 1\} = S^2, \end{equation}

the 2-dimensional sphere (which we can embed in $\mathbb{R}^3$). This is the space of states of our system described (with redundancy) by the original $\mathcal{H}$.

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  • $\begingroup$ I have thoroughly revised the question to be clear about my confusion. You said that the normalization is not necessary. But do you give up Born's probability interpretation for those states? Moreover, how does space $\mathcal{H}/\sim$ differ from the Hilbert space $\mathcal{H}$? @coconut $\endgroup$ – SRS Jul 18 '17 at 13:45
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The quantum state associated to a Hilbert space vector $\psi $ is isomorphic to the orthogonal projector on the subspace spanned by $\psi $. Since $\psi $ and $c\psi $ span the same subspace for any $c\neq 0$, the state associated to either one is the same.

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  • $\begingroup$ I've thoroughly edited my question. You might want to have a look for modifying your answer or making additional remarks. Moreover, can you explain easily what you mean by "isomorphic to the orthogonal projector"? It sounds terrifying :-) @yuggib $\endgroup$ – SRS Jul 18 '17 at 13:41

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