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Number operator help 
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#1
Feb1613, 05:01 AM

P: 160

Hi can anyone tell me why in the fermionic number operator case:
[itex]<0N/V0>= \sum_{\pm r}\int d^3 k a^{\dagger}(t,r)a(t,r) [/itex] because if: [itex] N=a^{\dagger}(t,k)a(t,k) [/itex]then after fourier decomposition surely one gets: [itex] \int d^3 r d^3 r \frac{1}{(2Pi)^{3}} a^{\dagger}(t,r)a(t,rk) [/itex] 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 [itex] d^3k [/itex] term. This V term gives just a [itex] \frac{1}{V} [/itex] term in the final integral. 


#2
Feb1613, 07:55 PM

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P: 1,927

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 Fouriertransform each operator and then multiply them, you'll need different dummy integration symbols in the respective Fourier integral. 


#3
Feb1713, 05:06 AM

P: 160

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. help.pdf 


#4
Feb1713, 07:02 PM

Sci Advisor
P: 1,927

Number operator help
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.] 


#5
Feb1813, 01:46 AM

P: 1,020

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 d^{3}k/(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
Feb1813, 03:35 AM

P: 160

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 [itex] \hat{a}^{\dagger}\hat{a}[/itex], but im not sure what N is a function of, as in r or k.
I also understand that [itex] d^3k \rightarrow dk k^2 [/itex] 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 \ [itex] \pm r[/itex] but not sure). I also understand that the [itex]1/a^3[/itex] term arises from V. I also know that [itex] <0a^{\dagger} = a0>=0 [/itex], 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
Feb1813, 03:42 AM

P: 160

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



#8
Feb1813, 06:44 AM

Sci Advisor
P: 2,139




#9
Feb1813, 07:48 AM

P: 160

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



#10
Feb1813, 08:00 AM

Sci Advisor
P: 2,139




#11
Feb1813, 08:10 AM

Sci Advisor
P: 2,139

[itex]\psi_{+} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}[/itex] [itex]\psi_{} =\begin{pmatrix} 0 \\ 1 \end{pmatrix}[/itex] So the index [itex]r[/itex] means either [itex]+[/itex] or [itex][/itex]. 


#12
Feb1813, 11:22 AM

P: 160

Can you tell me then why it is [itex] \hat{a}(\eta,r) [/itex] or what they are in terms of the bogulobov transformations and operators, i.e. can i define them as i have done so.
Is [itex] \frac{N}{V} = \frac{1}{(2\pi)^3 a^3} \int d^3 k \hat{a}^{\dagger}(\eta, r)\hat{a}(\eta, r) [/itex] just an identity then, I can't see how they form it with only one integration variable d^3 k and why it contains [itex](2\pi)^3[/itex] not [itex] (2\pi)^{3/2} [/itex]? 


#13
Feb1813, 02:23 PM

Sci Advisor
P: 2,139

http://arxiv.org/pdf/hepph/0003045.pdf I'm almost positive that equation 2.20 contains a typo. Instead of [itex]\frac{1}{(2\pi)^3 a^3} \int d^3 k \hat{a}^{\dagger}(\eta, r)\hat{a}(\eta, r) [/itex] it should be [itex]\frac{1}{(2\pi)^3 a^3} \int d^3 k \hat{a}^{\dagger}(\eta, k)\hat{a}(\eta, k) [/itex] 


#14
Feb1813, 02:29 PM

P: 160

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


#15
Feb1913, 06:09 AM

P: 1,020

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 ∫d^{3}k/(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/m...et/node30.html 


#16
Feb1913, 06:18 AM

P: 160

Ok then is [itex] 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) [/itex]



#17
Feb1913, 06:44 AM

P: 1,020

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



#18
Feb1913, 07:34 AM

P: 160

I am sorry I don't know what you mean, is it like [itex] 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 [/itex]



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