# 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|? Mar 2, 2020 at 17:13
• I edited my question in order to give you further information Mar 2, 2020 at 17:31
• I agree with the expression in your second sentence. It is not equal to $|0\rangle\langle0|$. Mar 2, 2020 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. Mar 2, 2020 at 19:49
• Ok, done. Thank you for your comment Mar 3, 2020 at 7:20

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