Computing Scalar Product in Antisymmetric Fock Space w/ Creator Operators

[Boby37]

We use the antisymmetric Fock space ( "fermions"). We denote by $$c(h)$$ a creator operator.

I need to evaluate the following quantity:

$$< \Omega , \big(c(h_1)+c(h_1)^{*}\big)\big(c(h_2)+c(h_2)^{*}\big) \ldots \big(c(h_n)+c(h_n)^*\big)\Omega>$$

where $$\Omega$$ is the unit vector called vaccum, $$<\cdot\ ,\ \cdot>$$ the scalar product and $$h_1,\ldots,h_n$$ any vectors.

I need a reference or an explanation.

Thank you!

 

[strangerep]

I need a reference or an explanation.

You didn't really say what part of this is giving you trouble.

As a broad approach, I'd try explicit evaluation for some
low values on n, and try to see a pattern. Then try and
prove that the pattern holds for all n via induction.

 

[Boby37]

If n is odd, I understand that the quantity is 0: We can write
the quantity as a sum of monomials in which all
creators are to the right of all annihilators (anti-Wick ordered). A
such monomial is a product of an odd number of factors. Clearly
the vacuum state annihilates a such monomial. We deduce the result by
linearity.

If n=2k is even: if a creator is to the left of a annihilator note that we have the formula

$$<\xi,c(e)c(f)^*\eta>=0$$

proof:

$$<\xi,c(e)c(f)^*\eta>_{\mathcal{F}(H)} = <f,e>_H<\xi,\eta>_{\mathcal{F}(H)}-<\xi,c(f)^*c(e)\eta>_{\mathcal{F}(H)} = <f,e>_H<\xi,\eta>_{\mathcal{F}(H)}-<f\otimes \xi,e\otimes\eta>_{\mathcal{F}(H)} = 0$$

where in the first equality, I use $$c(f)^*c(e)+c(e)c(f)^* =\ <f,e>_{H} Id_{\mathcal{F}(H)}$$.

Now each term of the product is a sum of monomials. By the previous calculation, if a creator is
to the left of a annihilator then the vacumm state annihilates this monomial.
Then, we only must consider the monomial which all creators are to the right of
all annihilators (anti-Wick ordered). Moreover, it is clear that if
the number of creators and the number of annihilators is
different the vacuum state annihilates the (anti-Wick ordered) monomial.
Consequently, the quantity is equal to
$$<\Omega ,c(h_1)^*...c(h_k)^*c(h_{k+1})..c(h_{2k})\Omega>=<h_1,h_{2k}>...<h_k,h_{k-1}>$$
But this last quantity is not "symmetric" ($$c(h_1)^*$$ and $$c(h_2)^*$$ anticommute).

Then I have a BIG problem. Where is the mistake?

 

[strangerep]

If n=2k is even: if a creator is to the left of a annihilator note that we have the formula

$$<\xi,c(e)c(f)^*\eta>=0$$

proof:

$$<\xi,c(e)c(f)^*\eta>_{\mathcal{F}(H)} = <f,e>_H<\xi,\eta>_{\mathcal{F}(H)}-<\xi,c(f)^*c(e)\eta>_{\mathcal{F}(H)} = <f,e>_H<\xi,\eta>_{\mathcal{F}(H)}-<f\otimes \xi,e\otimes\eta>_{\mathcal{F}(H)} = 0$$

where in the first equality, I use $$c(f)^*c(e)+c(e)c(f)^* =\ <f,e>_{H} Id_{\mathcal{F}(H)}$$.

I'm not sure I understand your notation, I presume that

$$|\eta\rangle ~:=~ c^*(\eta) |\Omega\rangle ~~~~~~ ?$$

Hmm... permit me to simplify (i.e., abuse) the notation...

I'll write
$$c^*(f) ~\to~ f^* ~~~~~~\mbox{etc,}$$
and I'll use ordinary parentheses to denote the inner product in H, e.g., (f,g).
I'll also use "0" for the vacuum.

Then

$$\langle\xi,c(e)c(f)^*\eta\rangle ~\to~ \langle 0|\, \xi \, e \, f^* \, \eta^* \, |0\rangle ~=~ (f,e)\,(\eta,\xi) ~-~ (f,\xi) \, (\eta,e) ~\ne~ 0 ~~,$$

unless I've made a mistake, or misunderstood your notation.

 

[Boby37]

Thank you for your answer.
However, with my notations, we have
$$c(\eta) |\Omega\rangle ~:=~ |\eta\rangle$$
$$c(\eta)$$ is a creator and not a annhilator.

 

[strangerep]

[...] with my notations, we have
$$c(\eta) |\Omega\rangle ~:=~ |\eta\rangle$$
$$c(\eta)$$ is a creator and not a annhilator.

OK, but I still don't see how you get zero...


$$\langle\xi,c(e)c^*(f)\eta\rangle ~=~ \langle 0|\, c^*(\xi) \, c(e) \, c^*(f) \, c(\eta)\, |0\rangle ~=~ \langle 0| \big( (\xi, e) - c(e)c^*(\xi) \big) \big( (f,\eta) - c(\eta) c^*(f) \big) |0\rangle ~=~ (\xi, e)\,(f,\eta) ~\ne~ 0 ~~,$$

 

[Boby37]

Thank you very much!