Expressing the vacuum projection operator in terms of number operator

Question

I've been reading this book, in which the author expresses the vacuum projection operator $\vert 0\rangle\langle 0\vert$ in terms of the number operator $\hat{N}=\hat{a}^{\dagger}\hat{a}$, where $\hat{a}^{\dagger}$ and $\hat{a}$ are the usual creation and annihilation operators, respectively. I can follow most of the derivation, however, I don't quite understand the following step: $$:\text{exp}\bigg\lbrace\hat{a}^{\dagger}\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg\rbrace\,\vert 0\rangle\langle 0\vert\,\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace :\bigg\vert_{Z^{\ast}=0} \ = \ :\text{exp}\big\lbrace\hat{a}^{\dagger}\hat{a}\big\rbrace\,\vert 0\rangle\langle 0\vert : \qquad (1)$$ How does one get from the left-hand side to the right-hand side of this equation? I'm assuming there are several steps that have been missed out, or am I missing something trivial?

Edit: Just in case the link is not viewable, let me elaborate on the details of the calculation a little further, in particular, how one arrives at eq. (1). Using the completeness relation for the basis of number-operator eigenstates, we have $$1\!\!1 \ = \ \sum_{n,m=0}^{\infty}\vert n\rangle\langle m\vert\,\delta_{n,m} \ = \ \sum_{n,m=0}^{\infty}\vert n\rangle\langle m\vert\,\frac{1}{\sqrt{n!m!}}\bigg(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg)^{n}\big(Z^{\ast}\big)^{m}\bigg\vert_{Z^{\ast}=0}\qquad (2)$$ where we make use of the identity $$\frac{1}{\sqrt{n!m!}}\bigg(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg)^{n}\big(Z^{\ast}\big)^{m}\bigg\vert_{Z^{\ast}=0} \ = \ \delta_{n,m}$$ Next, using that $\vert n\rangle =\frac{(\hat{a}^{\dagger})^{n}}{\sqrt{n!}}\,\vert 0\rangle$ we can rewrite (2) as $$\sum_{n,m=0}^{\infty}\frac{(\hat{a}^{\dagger})^{n}}{n!}\,\vert 0\rangle\langle 0\vert\,\frac{(\hat{a})^{m}}{m!}\bigg(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg)^{n}\big(Z^{\ast}\big)^{m}\bigg\vert_{Z^{\ast}=0} \ = \ \sum_{n,m=0}^{\infty}\frac{(\hat{a}^{\dagger})^{n}\big(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\big)^{n}}{n!}\,\vert 0\rangle\langle 0\vert\,\frac{(\hat{a})^{m}\big(Z^{\ast}\big)^{m}}{m!}\bigg\vert_{Z^{\ast}=0} \\[1cm] \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\, = \ \text{exp}\bigg\lbrace\hat{a}^{\dagger}\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg\rbrace\,\vert 0\rangle\langle 0\vert\,\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace\bigg\vert_{Z^{\ast}=0}$$ This final expression is already normal-ordered as all creation operators are placed to the left of all annihilation operators. As such we can express this last line as given in eq. (1).

I understand this derivation up to the left-hand side of eq. (1), I just don't understand how the author then gets to the right-hand side of eq. (1).

Answer 1

You can always expand your exponentials to get $$\exp(a^\dagger \frac{d}{dZ^*}) = \sum_n \frac{1}{n!} a^{\dagger n} \frac{d^n}{dZ^{*n}}$$

By expanding both exponentials, and since the vacuum state and the creation operators do not depend on $Z^*$, you end up with

$$ \exp(a^\dagger \frac{d}{dZ^*})|0⟩⟨0| \exp(a Z^*) = \sum_{n,m} \frac{1}{n!m!} a^{\dagger n} |0⟩⟨0| a^m \cdot \frac{d^n}{dZ^{*n}} Z^{*m}$$

Taking $Z^* = 0$, this last term is non-zero only when $n=m$, and equal to $n!$ when $n=m$, thus

$$ \exp(a^\dagger \frac{d}{dZ^*})|0⟩⟨0| \exp(a Z^*) |_{Z^* = 0}= \sum_{n} \frac{1}{n!} a^{\dagger n} |0⟩⟨0| a^n $$

Here, the author seems to assume that normal-ordering would imply $a^{\dagger n} |0⟩⟨0| a^n = (a^\dagger a)^n |0⟩⟨0|$, which might simply be notation. Reinjecting this into the expression above will give you the final result.