 #1
dRic2
Gold Member
 715
 165
Summary:

Notation used:
With ##n_m( \mathbf x) = \hat \psi_m^{†}( \mathbf x) \hat \psi_m( \mathbf x) ## I mean the density of the particle with spin ##m## and ##\psi_m^{†}## ##\hat \psi_m## are the field operator
##F>## = (ground) state of a ideal fermi gas (no interaction)
Main Question or Discussion Point
Hi,
some time ago our professor told us (en passant) to evaluate this quantity:
$$<Fn_m( \mathbf x) n_{m'}(\mathbf x) F>  <Fn_m( \mathbf x)F><Fn_{m'}(\mathbf x) F>$$
And then he said: "you'll find that this quantity may not be zero. In particular when the electron are correlated it will not be zero." Then he said something about the spin of the electrons being "parallel". I'm really sorry if I'm being vague but I really don't remember because it was some time ago. Anyway... I calculated the first part using Wick's theorem and I got:
$$<Fn_m( \mathbf x) n_{m'}(\mathbf x) F> = <F\hat \psi_m^{†}( \mathbf x) \hat \psi_m( \mathbf x) \hat \psi_{m'}^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x) F>=$$
$$ =<F\hat \psi_m^{†}( \mathbf x) \hat \psi_m( \mathbf x)F><F\hat \psi_{m'}^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x) F> + <F\psi_m^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x)F><F\psi_m( \mathbf x) \hat \psi_{m'}^{†}( \mathbf x)F> =$$
$$= <Fn_m( \mathbf x)F><Fn_{m'}(\mathbf x) F> + <F\psi_m^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x)F><F\psi_m( \mathbf x) \hat \psi_{m'}^{†}( \mathbf x)F>$$
So the interesting quantity is:
$$<F\psi_m^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x)F><F\psi_m( \mathbf x) \hat \psi_{m'}^{†}( \mathbf x)F>$$
which I can rewrite in momentum base as (I didn't use vector notation here sorry):
$$<F \sum_{kk'} e^{(kk')x} c_{k,m}^{†} c_{k',m'}F><F\sum_{qq'} e^{(qq')x} c_{q,m} \hat c_{q',m'}^{†}F>$$
Is it okay ?
Because from here I don't really see what kind of conclusion I could draw... The only thing I could think of is the following:
in order for that term not to vanish I should have (correct me if I'm wrong) a state with the energy levels filled like this:
Does it make sense ? Is there something more one can say about this ?
I'm again very sorry if it all seems vague but I can't remember and I hope that maybe some of you can recognize it as a well know example or something...
Thanks
Ric
some time ago our professor told us (en passant) to evaluate this quantity:
$$<Fn_m( \mathbf x) n_{m'}(\mathbf x) F>  <Fn_m( \mathbf x)F><Fn_{m'}(\mathbf x) F>$$
And then he said: "you'll find that this quantity may not be zero. In particular when the electron are correlated it will not be zero." Then he said something about the spin of the electrons being "parallel". I'm really sorry if I'm being vague but I really don't remember because it was some time ago. Anyway... I calculated the first part using Wick's theorem and I got:
$$<Fn_m( \mathbf x) n_{m'}(\mathbf x) F> = <F\hat \psi_m^{†}( \mathbf x) \hat \psi_m( \mathbf x) \hat \psi_{m'}^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x) F>=$$
$$ =<F\hat \psi_m^{†}( \mathbf x) \hat \psi_m( \mathbf x)F><F\hat \psi_{m'}^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x) F> + <F\psi_m^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x)F><F\psi_m( \mathbf x) \hat \psi_{m'}^{†}( \mathbf x)F> =$$
$$= <Fn_m( \mathbf x)F><Fn_{m'}(\mathbf x) F> + <F\psi_m^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x)F><F\psi_m( \mathbf x) \hat \psi_{m'}^{†}( \mathbf x)F>$$
So the interesting quantity is:
$$<F\psi_m^{†}( \mathbf x) \hat \psi_{m'}( \mathbf x)F><F\psi_m( \mathbf x) \hat \psi_{m'}^{†}( \mathbf x)F>$$
which I can rewrite in momentum base as (I didn't use vector notation here sorry):
$$<F \sum_{kk'} e^{(kk')x} c_{k,m}^{†} c_{k',m'}F><F\sum_{qq'} e^{(qq')x} c_{q,m} \hat c_{q',m'}^{†}F>$$
Is it okay ?
Because from here I don't really see what kind of conclusion I could draw... The only thing I could think of is the following:
in order for that term not to vanish I should have (correct me if I'm wrong) a state with the energy levels filled like this:
Does it make sense ? Is there something more one can say about this ?
I'm again very sorry if it all seems vague but I can't remember and I hope that maybe some of you can recognize it as a well know example or something...
Thanks
Ric