I am not sure if $e^{-a^\dagger a}$ can be equal to $\left|0\right>\left<0\right|$. I found on my lecture notes that I can write it as $\sum_{jkn}\left|j\right>\left<j\right|a^{\dagger n} a^n \left|k\right>\left<k\right|$ using identity spectralization and do some quick algebra in order to get a newton binomial formula like $\sum_j \left|j\right>\left<j\right| (1-1)^j = \left|0\right>\left<0\right|$. But, if I write it in the Fock basis, shouldn't it be $\sum_{n=0}^\infty e^{-n} \left|n\right>\left<n\right|$, which can be also written as $\left|0\right>\left<0\right|+\sum_{n=1}^\infty e^{-n} \left|n\right>\left<n\right|$ or am I missing something?

I am not convinced because I couldn't find another source and also because this means, if it is true what I have written before, that $\sum_{n=1}^\infty e^{-n} \left|n\right>\left<n\right|$ is equal to zero operator, which seems odd to me.

I also suspect that $\sum_{jkn}\left|j\right>\left<j\right|a^{\dagger n} a^n \left|k\right>\left<k\right|$ is true ONLY for the normal ordered operator $:e^{-a\dagger a}:$

Thanks in advance, I am new to this forum and kinda new to physics in general.

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    $\begingroup$ Where did you get the idea that it is equal to |0><0|? $\endgroup$ – wnoise Mar 2 at 17:13
  • $\begingroup$ I edited my question in order to give you further information $\endgroup$ – Hub One Mar 2 at 17:31
  • $\begingroup$ I agree with the expression in your second sentence. It is not equal to $|0\rangle\langle0|$. $\endgroup$ – march Mar 2 at 17:35
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    $\begingroup$ Please make your question one cohesive post. There is an edit history available for those who need it. You do not need to explicitly mention edits in your post. $\endgroup$ – BioPhysicist Mar 2 at 19:49
  • $\begingroup$ Ok, done. Thank you for your comment $\endgroup$ – Hub One Mar 3 at 7:20

You are right, that $e^{-a^+a}$ is not equal to $|0\rangle\langle 0|$. But the following operator is equal: $$ :e^{-a^+a}:\ \equiv\ \sum_{n=1}^\infty \frac{(-1)^n}{n!} (a^+)^n a^n = |0\rangle\langle 0| $$

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    $\begingroup$ I suspected that normal ordering was involved... in fact the proof I found on my notes can be applied only to the normal ordered operator. Thank you very much! $\endgroup$ – Hub One Mar 2 at 17:50

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