Fermionic Number Operator Help

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In summary, the conversation discusses the calculation of the fermionic number operator in the case of a Fourier decomposition. There is confusion about the integration variables and the presence of a V term in the final integral. The conversation also touches on the definition of operators and their relationship to the Bogoliubov transformation. Some typos in the equations are pointed out and corrected.
  • #1
pleasehelpmeno
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Hi can anyone tell me why in the fermionic number operator case:
[itex]<0|N/V|0>= \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 don't 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.
 
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  • #2
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
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 don't understand how they can get an answer in the form that they do, for the number operator.

View attachment help.pdf
 
  • #4
pleasehelpmeno said:
I have attached the pdf copy of the issue where the important things are marked with a blue blob. I simply don't understand how they can get an answer in the form that they do, for the number operator.
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.]
 
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  • #5
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
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 I am 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 I am 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] <0|a^{\dagger} = a|0>=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
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
pleasehelpmeno said:
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?

I'm not sure if you're making this mistake, or not, but the [itex]r[/itex] in that section does not refer to position; note that in the text, r is summed over, not integrated over.
 
  • #9
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
pleasehelpmeno said:
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?

As I said, in the paper you showed an excerpt from, [itex]r[/itex] is not position. The paper sums over all possibly values of [itex]r[/itex] and it says that it has values [itex]+/- 1[/itex]. Maybe it's a spin index?
 
  • #11
stevendaryl said:
As I said, in the paper you showed an excerpt from, [itex]r[/itex] is not position. The paper sums over all possibly values of [itex]r[/itex] and it says that it has values [itex]+/- 1[/itex]. Maybe it's a spin index?

In the full paper, they define
[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
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
pleasehelpmeno said:
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]?

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
[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
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
pleasehelpmeno said:
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?
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
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
there is a summation in first over k which is converted to integral in second(per unit volume)
 
  • #18
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]
 
  • #19
so as in the Volume cancels with the second d^3 k' term?
 
  • #20
N=Ʃk a+a=V/(2∏)3∫d3k a+a,V is volume.
 

What is the Fermionic Number Operator?

The Fermionic Number Operator is a mathematical operator used to describe the number of fermions (particles with half-integer spin) in a quantum system. It is denoted by N and is defined as the sum of all the occupation numbers for each single-particle state.

How is the Fermionic Number Operator applied in quantum mechanics?

In quantum mechanics, the Fermionic Number Operator is used to determine the number of particles in a given quantum state. It is also used to calculate the total energy of a fermionic system and is an important tool in studying many-body systems.

What is the commutation relation for the Fermionic Number Operator?

The commutation relation for the Fermionic Number Operator is given by [Ni, Nj] = 0 for i ≠ j and [Ni, Nj] = Ni for i = j. This means that the operator commutes with itself and with all other operators corresponding to different single-particle states.

How does the Fermionic Number Operator relate to the Pauli exclusion principle?

The Pauli exclusion principle states that no two fermions can occupy the same quantum state simultaneously. The Fermionic Number Operator helps to enforce this principle by providing a way to count the number of particles in a given state, ensuring that only one particle can occupy each state at a time.

What are some applications of the Fermionic Number Operator?

The Fermionic Number Operator is used in various areas of quantum physics, including in the study of atoms, molecules, and condensed matter systems. It is also essential in understanding many-body phenomena and is a fundamental tool in quantum field theory.

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