Is this proof that $a|0\rangle=0$ wrong (ladder operators and number operator)?

Question

I understand everything except when he says

"Then we get in particular $a|0\rangle=0$."

It seems the author substituted the formula with $n=0$, But this seems unsound, since we get $|n-1\rangle$ which is not defined

The book is Quantum Communications by Gianfranco Cariolaro

What makes me more suspicious is that in another reference (measuring the quantum state of light) its author explicitly states that we need to add the assumption $a|0\rangle=0$ which is more consistent with the question and answer I got here Why is there only one eigenket per eigenvalue of the number operator?

$\Vert\hat{a}|n\rangle\Vert=\langle n|\hat{N}|n\rangle=n$ works even when $n=0$, whence $\hat{a}|n\rangle$ is the zero vector. — J.G., Jul 22 '22 at 15:53
Indeed, you confirmed $|-1\rangle$ is not in the Hilbert space. — Cosmas Zachos, Jul 26 '22 at 13:41

score 3 · Answer 1 · edited Sep 11 '22 at 13:42

The proof is correct, just it is a proof by contradiction.

Suppose that $a|0\rangle\neq 0$, then we can normalise it and we call $|-1\rangle$ a unit vector parallel to it. Therefore $$a|0\rangle = c_0|-1\rangle$$ for some complex $c_0\neq 0$. The argument goes on proving that actually $c_0=0$ and thus the initial hypothesis $a|0\rangle\neq 0$ is untenable because it leads to a contradiction, that is $a|0\rangle =0$ necessarily.

Is this proof that $a|0\rangle=0$ wrong (ladder operators and number operator)?

1 Answers1

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