# Exponential of number operator

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 $$\sum_{jkn}\left|j\right>\left\left using identity spectralization and do some quick algebra in order to get a newton binomial formula like $$\sum_j \left|j\right>\left\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, which can be also written as $$\left|0\right>\left<0\right|+\sum_{n=1}^\infty e^{-n} \left|n\right>\left 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 is equal to zero operator, which seems odd to me.

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

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

• Where did you get the idea that it is equal to |0><0|? – wnoise Mar 2 at 17:13
• I edited my question in order to give you further information – Hub One Mar 2 at 17:31
• I agree with the expression in your second sentence. It is not equal to $|0\rangle\langle0|$. – march Mar 2 at 17:35
• 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. – BioPhysicist Mar 2 at 19:49
• Ok, done. Thank you for your comment – 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: $$:e^{-a^+a}:\ \equiv\ \sum_{n=1}^\infty \frac{(-1)^n}{n!} (a^+)^n a^n = |0\rangle\langle 0|$$