Annihilation operators of two different types of Fermions

Tanmoy
Messages
3
Reaction score
0
Moved from a technical form
IfA=cd, where c and d are
annihilation operators of two different types of Fermions, then {A,A°}is?
A.1+n1+n2
B.1-n1+n2
C.1-n2+n1
D.1-n1-n2
Where,n1 and n2 are corresponding number operator,
A° means A dagger or creation operator,as the particles are fermions they will obey anti-commutation I think
 
Last edited by a moderator:
Physics news on Phys.org
Welcome to PF. We ask posters to show some work first.
What have you tried?
By the way, what is n1 and h2 in terms of creation/annihilation operators?
 
nrqed said:
Welcome to PF. We ask posters to show some work first.
What have you tried?
By the way, what is n1 and h2 in terms of creation/annihilation operators?
n1=c°c
n2=d°d
° means dagger
 

Attachments

  • IMG_20200415_001441.jpg
    IMG_20200415_001441.jpg
    68.5 KB · Views: 227
Tanmoy said:
n1=c°c
n2=d°d
° means dagger
Ok, good. Now can you show the first few steps in the calculation of ##\{ A, A^\dagger \} ##?
(the very first step is to write what ##A^\dagger## is equal to).
 

Thread 'Please tell me where I am going wrong in this integral'

##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
View full post »

Thread 'Direction of the magnetic field of an oscillating charge'

Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
View full post »

Thread 'Initial conditions for orbits around a wormhole'

Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
View full post »

Similar threads

Fourier transform of t-V model for t=0 case
Replies
0
Views
1K
Second Quantization - Quasiparticles
Replies
3
Views
2K
Electromagnetic waves incident on an anti-reflective coating
Replies
3
Views
2K
I What causes the unexpected annihilation point in the Magic-Tee configuration?
Replies
15
Views
2K
Diagonalization of a Hamiltonian for two fermions
Replies
10
Views
4K
Exercise involving Dirac fields and Fermionic commutation relations
Replies
1
Views
2K
Why are fermion states anti-symmetric under exchange operator?
Replies
1
Views
1K
Fundamental noise limit for an ideal photodetector
Replies
1
Views
1K
Anticommutation relations Fermion creation and annihilation
Replies
18
Views
3K
The average distance between two particles in a 1d box
Replies
1
Views
4K

Hot Threads

Recent Insights

Back
Top