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).

  • $\begingroup$ Did you utilize the property of normal ordering that all commutators inside :: vanish? $\endgroup$ Commented Jul 25, 2021 at 20:21
  • $\begingroup$ Linked. $\endgroup$ Commented Mar 13, 2022 at 21:02

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.

  • $\begingroup$ I don't think the last identity works as you suggest. In general, $a^{\dagger n} | 0 \rangle \langle 0 | a^{n} = n! \, |n\rangle \langle n | \neq n! \, (a^{\dagger n} a^{n}) \,|0 \rangle \langle 0 |$. I think the ordering enforced by the colons $: \ :$ in the expression (I can't see the reference provided by OP so I can't tell which ordering is it) plays a role in the last step. In that case, the identity you provide will be somehow valid once each side is put into $: \ :$ . $\endgroup$
    – secavara
    Commented Mar 2, 2019 at 9:36
  • $\begingroup$ Yes, you are absolutely right. Without the colons, $(a^\dagger a) ^n|0⟩ = 0$ anyway, since we first act on the vacuum with a annhiliation operator. But I am guessing that this is more or less the approach the author took to derive his expression. $\endgroup$
    – Ronan
    Commented Mar 2, 2019 at 11:04
  • $\begingroup$ @Ronan What confuses me is that to get to the left-hand side of the equation in my OP, the author already uses the expansion that you suggest in your answer. They then use that argument to show that the expression is already normal-ordered, which allows them to rewrite it as I have put on the left-hand side of the equation in my OP. Why would they then immediately undo this again? How does the normal ordering allow one to re-express it as it is on the right-hand side of the equation? $\endgroup$
    – Will
    Commented Mar 2, 2019 at 11:32
  • $\begingroup$ Unfortunately, the link to your book has limited access, so I cannot see what the author wrote. But normal ordering is when the $a^\dagger$ operators are on the left of the $a$ operators. I don't see how in any way it would be an argument to get from the left-hand side to the right-hand side of the equation. $\endgroup$
    – Ronan
    Commented Mar 2, 2019 at 11:43
  • $\begingroup$ I have corrected the last part of my answer, as there was a mistake in the derivation. $\endgroup$
    – Ronan
    Commented Mar 2, 2019 at 11:57

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