# The matrix element of a normal-ordered operator

Eq (1.137) in Negele and Orland gives the following identity for a normal-ordered operator $A(a_i^\dagger,a_i)$:

$$\langle \phi|A(a_i^\dagger,a_i)|\phi'\rangle=A(\phi_i^*,\phi'_i)e^{\sum \phi_i^*\phi'_i}$$

For the coherent (boson) states $|\phi\rangle=e^{\sum\phi_ia^\dagger_i}|0\rangle$. When I try to prove this I get a factor of 1/2 in the exponent. I'm going to take a simple example $A=a_i^\dagger a_i$ to try and find my mistake. First I define $\eta'=\sum \phi'_ia_i^\dagger$, and use the identity

$$a_i\eta'^n=n\phi'_i\eta'^{n-1}+\eta'^n a_i.$$

Then I can show

$$a_i|\phi'\rangle=(e^{\eta'}a_i+\phi'_ie^{\eta'})|0\rangle$$

One term of which kills the vacuum so I have shown

$$\langle \phi |a^\dagger_i a_i|\phi'\rangle=\langle 0| \phi^*_i\phi'_ie^{\eta^*}e^{\eta'}|0\rangle$$

Then I want to use the Baker-Campbell-Hausdorf formula. The commutator

$$[\eta^*,\eta']=[\sum\phi^*_ia_i,\sum\phi_ja^\dagger_j]=\sum[\phi^*_ia_i,\phi_i'a_i^\dagger]+\sum_{i\neq j}[\phi^*_ia_i,\phi_ja^\dagger_j]=\sum\phi_i^*\phi'_i,\qquad (1)$$ and when you commute everything through the second term and kill the vacuum. Since this is a scalar, the BCH formula gives

$$\ln(e^{\eta^*}e^{\eta'})=\eta^*+\eta'+\frac{1}{2}\sum\phi_i^*\phi_i'$$

Rearrange the sum, exponentiate, kill some more vacuum and I get the right expression but with a factor of 1/2!

What am I doing wrong? My first guess is in equation (1), maybe there is some dobule-counting I don't see but you can write things like

$$[\sum,\sum]=[1,\sum]+[2,\sum]+...=[1,1]+[1,\sum_{\neq 1}]+[2,2]+[2,\sum_{\neq 2}]+...$$ $$=\sum[i,i]+\sum_{i\neq j}[i,j].$$

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I) OP's factor $\frac{1}{2}$ comes from using the truncated version

$$\tag{1} e^Ae^B~=~e^{A+B+\frac{1}{2}[A,B]}$$

of the Baker-Campbell-Hausdorff formula. Formula (1) holds if the commutator $[A,B]$ commutes with both the operators $A$ and $B$.

II) What is actually needed in OP's calculation is rather this version

$$\tag{2} e^Ae^B~=~e^{[A,B]}e^Be^A$$

of the truncated Baker-Campbell-Hausdorff formula. Formula (2) contains no half. It can easily be derived by using formula (1) twice.

III) Concretely, with $\hbar=1$, the operators $A=\phi^{*}_i a_i$ and $B=\phi^{\prime}_i a^{\dagger}_i$ play the role of annihilation and creation parts, respectively. The commutator $[A,B]~=~\phi^{*}_i\phi^{\prime}_i {\bf 1}$ is a $c$-number. The normal ordered expression (which is needed in OP's calculation) is the rhs. of eq. (2) (as opposed to the rhs. of eq. (1), which is not normal ordered). In particular, the bracket of two coherent states reads $\langle\phi |\phi^{\prime} \rangle=e^{\phi^{*}_i\phi^{\prime}_i}$.

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But the commutator $[A,B]=[a,a^\dagger]=1$, a scalar here (I guess a scalar times the identity operator), and that DOES commute with $a$ and $a^\dagger$. I started with the full version of BCH, removed the higher terms that vanish (I think) and used the exponential map to get the thing I wanted. No? – levitopher May 3 '13 at 18:14
I updated the answer. – Qmechanic May 3 '13 at 23:43
I have to say, I am still surprised by this. How can you tell that (2) is normal-ordered if $A$ and $B$ are any old operators whose commutator is proportional to the identity? I see how you can't tell if (1) is normal-ordered, but how can you tell that (2) is? – levitopher May 4 '13 at 3:30