# Number operator help

1. Feb 16, 2013

Hi can anyone tell me why in the fermionic number operator case:
$<0|N/V|0>= \sum_{\pm r}\int d^3 k a^{\dagger}(t,r)a(t,r)$
because if:
$N=a^{\dagger}(t,k)a(t,k)$then after fourier decomposition surely one gets:

$\int d^3 r d^3 r \frac{1}{(2Pi)^{3}} a^{\dagger}(t,r)a(t,rk)$
and when fourier decomposing back i dont see how one can get the creation/annhilation operators as a function of r or how to get this sum term or the $d^3k$ term. This V term gives just a $\frac{1}{V}$ term in the final integral.

2. Feb 16, 2013

### strangerep

There's seem to be some typos in your integrals, e.g., integration variables not matching the arguments in the integrands. (?)

But -- and I'm just guessing here -- if you intended to Fourier-transform each operator and then multiply them, you'll need different dummy integration symbols in the respective Fourier integral.

3. Feb 17, 2013

yeah the first line was something i read in a paper and I don't understand how they get there.

I have attached the pdf copy of the issue where the important things are marked with a blue blob. I simply dont understand how they can get an answer in the form that they do, for the number operator.

View attachment help.pdf

4. Feb 17, 2013

### strangerep

OK, first things first...

Do you understand how they get eq(2.17) ? Do you understand that $\hat a(\eta,k)$ is a different operator than $a(k)$, and that the latter annihilates the original (time=0) vacuum state?

[For the benefit of other readers: this is a standard calculation of a Bogoliubov transformation on fermionic operators, and calculation of the vev of the transformed number operator wrt an initial vacuum.]

Last edited: Feb 17, 2013
5. Feb 18, 2013

### andrien

is not that simply the sum over all modes.In case of summation over infinite modes the usual prescription is to replace the sum by an integral containing the factor d3k/(2∏)3,that is what is usually done in summing over all the modes of electromagnetic field when quantized,also with a sum over r,just like sum over polarization.

6. Feb 18, 2013

I understand that they are different because they dependent on the things in the bracket and to get between them one fourier transforms. I no that N is of the form $\hat{a}^{\dagger}\hat{a}$, but im not sure what N is a function of, as in r or k.

I also understand that $d^3k \rightarrow dk k^2$ by using the solid angle transformation, but im unsure why it would be a dk and what happens to the sum (i guessed this goes to 2 as it is \ $\pm r$ but not sure). I also understand that the $1/a^3$ term arises from V.

I also know that $<0|a^{\dagger} = a|0>=0$, so one can cancel away the operators.

I am really just unsure initially as to what N is and then why the integration variables (dk's etc) are present in the way that they are.

7. Feb 18, 2013

also can one just decide that $\hat{a}(\eta,k) = \alpha(\eta) a(k) + \beta(\eta) b^{\dagger}(-k)$ can then be rewritten as $\hat{a}(\eta,r) = \alpha(\eta) a(r) + \beta(\eta) b^{\dagger}(-r)$ or does one need to fourier transofrm between the two?

8. Feb 18, 2013

### stevendaryl

Staff Emeritus
I'm not sure if you're making this mistake, or not, but the $r$ in that section does not refer to position; note that in the text, r is summed over, not integrated over.

9. Feb 18, 2013

Dince r is in the integral then can one do as i decided and claim that $\hat{a}(\eta,r) = \alpha(\eta)a(r) + \beta(\eta)b^{\dagger}(-r)$. and then just sub this into the integral?

10. Feb 18, 2013

### stevendaryl

Staff Emeritus
As I said, in the paper you showed an excerpt from, $r$ is not position. The paper sums over all possibly values of $r$ and it says that it has values $+/- 1$. Maybe it's a spin index?

11. Feb 18, 2013

### stevendaryl

Staff Emeritus
In the full paper, they define
$\psi_{+} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$
$\psi_{-} =\begin{pmatrix} 0 \\ 1 \end{pmatrix}$

So the index $r$ means either $+$ or $-$.

12. Feb 18, 2013

Can you tell me then why it is $\hat{a}(\eta,r)$ or what they are in terms of the bogulobov transformations and operators, i.e. can i define them as i have done so.

Is $\frac{N}{V} = \frac{1}{(2\pi)^3 a^3} \int d^3 k \hat{a}^{\dagger}(\eta, r)\hat{a}(\eta, r)$ just an identity then, I can't see how they form it with only one integration variable d^3 k and why it contains $(2\pi)^3$ not $(2\pi)^{3/2}$?

13. Feb 18, 2013

### stevendaryl

Staff Emeritus
Looking at the paper here
http://arxiv.org/pdf/hep-ph/0003045.pdf
I'm almost positive that equation 2.20 contains a typo. Instead of
$\frac{1}{(2\pi)^3 a^3} \int d^3 k \hat{a}^{\dagger}(\eta, r)\hat{a}(\eta, r)$

it should be

$\frac{1}{(2\pi)^3 a^3} \int d^3 k \hat{a}^{\dagger}(\eta, k)\hat{a}(\eta, k)$

14. Feb 18, 2013

Yeah unfortunately it has a lot of typos, how is it formed by using:
$N=\hat{a}^{\dagger}(\eta , x) \hat{a}(\eta , x)$ and fourier expanding it? with k hence the $d^3 k$ and $\frac{1}{(2\pi)^3}$, although shouldn't there be $d^3 k d^3 k$,or is it just a fixed identity?

15. Feb 19, 2013

### andrien

are you aware how to go from fourier summation to fourier integration.In one dimension case,
when one converts a Ʃ into ∫,it is accompanied by a factor of ∫dk/2∏ per unit length.Similarly in three dimensional case,the summation is replaced by ∫d3k/(2∏)3 per unit volume.it is more or less like identity.it is used exhaustively at many places.I think this reference will be some useful,however it is at best suggestive.the result used is much more general.
http://www.math.osu.edu/~gerlach.1/math/BVtypset/node30.html

16. Feb 19, 2013

Ok then is $N=\hat{a}^{\dagger}(\eta , x)\hat{a}(\eta , x) = \int d^3k \frac{1}{(2 \pi)^3}\hat{a}^{\dagger}(\eta , k)\hat{a}(\eta , k)$

17. Feb 19, 2013

### andrien

there is a summation in first over k which is converted to integral in second(per unit volume)

18. Feb 19, 2013

I am sorry I don't know what you mean, is it like $N/V = \int (\frac{1}{(2\pi)^3}) d^3 k d^3 k' a^\{dagger} a = \int (\frac{1}{(a2\pi)^3}) d^3 k a^\{dagger} a$

19. Feb 19, 2013