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 $vert 0ranglelangle 0vert$ 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: $$:text{exp}bigglbracehat{a}^{dagger}frac{mathrm{d}}{mathrm{d}Z^{ast}}biggrbrace,vert 0ranglelangle 0vert,text{exp}biglbracehat{a}Z^{ast}bigrbrace :biggvert_{Z^{ast}=0}  =  :text{exp}biglbracehat{a}^{dagger}hat{a}bigrbrace,vert 0ranglelangle 0vert : qquad (1)$$ 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 $$1!!1  =  sum_{n,m=0}^{infty}vert nranglelangle mvert,delta_{n,m}  =   sum_{n,m=0}^{infty}vert nranglelangle mvert,frac{1}{sqrt{n!m!}}bigg(frac{mathrm{d}}{mathrm{d}Z^{ast}}bigg)^{n}big(Z^{ast}big)^{m}biggvert_{Z^{ast}=0}qquad (2)$$ where we make use of the identity $$frac{1}{sqrt{n!m!}}bigg(frac{mathrm{d}}{mathrm{d}Z^{ast}}bigg)^{n}big(Z^{ast}big)^{m}biggvert_{Z^{ast}=0}  =  delta_{n,m}$$ Next, using that $vert nrangle =frac{(hat{a}^{dagger})^{n}}{sqrt{n!}},vert 0rangle$ we can rewrite (2) as $$sum_{n,m=0}^{infty}frac{(hat{a}^{dagger})^{n}}{n!},vert 0ranglelangle 0vert,frac{(hat{a})^{m}}{m!}bigg(frac{mathrm{d}}{mathrm{d}Z^{ast}}bigg)^{n}big(Z^{ast}big)^{m}biggvert_{Z^{ast}=0}  =  sum_{n,m=0}^{infty}frac{(hat{a}^{dagger})^{n}big(frac{mathrm{d}}{mathrm{d}Z^{ast}}big)^{n}}{n!},vert 0ranglelangle 0vert,frac{(hat{a})^{m}big(Z^{ast}big)^{m}}{m!}biggvert_{Z^{ast}=0} [1cm] qquadqquadqquadqquadqquadqquadqquadqquad,, =  text{exp}bigglbracehat{a}^{dagger}frac{mathrm{d}}{mathrm{d}Z^{ast}}biggrbrace,vert 0ranglelangle 0vert,text{exp}biglbracehat{a}Z^{ast}bigrbracebiggvert_{Z^{ast}=0}$$ 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).

Ronan · Answer

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.

Expressing the vacuum projection operator in terms of number operator

One Answer

Add your own answers!

Ask a Question