Clashing definition of rays in Weinberg and Sakurai and Born interpretation without normalizability [duplicate]

Question

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., $(\psi,\psi)=1$) with $\psi$ and $\psi^\prime$ belonging to the same ray if $\psi^\prime=\xi\psi$, where $\xi$ is an arbitrary complex number with $|\xi|=1$

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.

Answer 1

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.

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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 $\mathcal{H}$.

Answer 2

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.