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This is a followup to my previous question Is this proof that $a|0\rangle=0$ wrong (ladder oeprators and number operator)?

According to the comment, we should always have $a|0\rangle= 0$, because the norm of $a|0\rangle$ is 0 which I agree with. By definition of a Hilbert space, a norm of 0 is only possible for the null vector o.

However in "ulf leonhardt measuring the quantum state of light" which a copy is available here, in page 22, proposition 2.31, the author talks about the possibilty that $a|0\rangle\neq 0$ and explores it at the end of page 24. How can we even make this assumption if the norm of $a|0\rangle$ is always 0 ??

Elio Fabri
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newtophysics
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  • ? The author tells you, quite clearly and emphatically, that the wavefunction of $a|0\rangle\neq 0$, so $\varphi_{-1}(q)= \langle q|a|0\rangle$ is not normalizable, and so unacceptable, as not being in the Hilbert space. He demonstrates that. What is the problem? – Cosmas Zachos Jul 25 '22 at 19:13
  • So basically it's just an alternative proof ? – newtophysics Jul 25 '22 at 19:20
  • I am troubled by your logical path network. It is, of course, a confirming restatement that vacua not annihilated by a are unacceptable as not in the Hilbert space. – Cosmas Zachos Jul 25 '22 at 19:22
  • Thank you maybe I was just overthinking it. – newtophysics Jul 25 '22 at 20:29

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