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We know a quantum state can be represented by a vector in a Hilbert space $\mathcal{H}$. For simpplicity let the dimension of $\mathcal{H}$ be two. The basis of the states are given by $\{|0 \rangle, |1 \rangle\}$, where $|0 \rangle = \begin{bmatrix} 1 \\0 \end{bmatrix}$ and $|1 \rangle = \begin{bmatrix} 0 \\1 \end{bmatrix}$. Then any state vector $|\psi \rangle \in \mathcal{H}$ can be expressed as $$|\psi \rangle = \alpha |0 \rangle + \beta |1 \rangle ,$$ where $\alpha, \beta \in \mathbb{C}$, the set of all complex numbers, and $|\alpha|^2 + |\beta|^2 = 1$. If the above statement is correct then the origine $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ is not s state vectors as it can not be expressed in the above form. Now my questions are:

  1. Is origin a state vector? If yes, how?

  2. If origin is not a state vector, then what is its physical interpretation? Particularly, in any neighbourhood of origin one may get uncountably infinite number of vectors which represents a quantum state after normalization.

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    $\begingroup$ The origin is not part of state space. The state space of quantum mechanics should properly be considered as a complex projective space. The origin is not a part of this space, thus does not have a neighborhood in this space. $\endgroup$ – Jahan Claes Jan 26 '18 at 20:02
  • $\begingroup$ @JahanClaes Would you write it as an answer? $\endgroup$ – Dutta Jan 27 '18 at 3:30

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