- #1
JustinLevy
- 882
- 1
many body problem --> non-interacting particles
Okay, this is a question I brought up elsewhere, but I guess I was off topic from the original thread so I decided to post here. Any help would be appreciated.
It appears to me that the grand partition function indicates any problem can be rewritten in terms of non-interacting quasi-particles. Below is an example to "demonstrate" the method, and also to give specifics in which to discuss. I haven't been able to find a flaw or make physical sense of it yet, which is where my questions lie. Any help understanding this would be much appreciated.
Consider a potential with two sites for interacting fermions.
H = \epsilon c_1^\dagger c_1 + \epsilon c_2^\dagger c_2 + t (c_2^\dagger c_1 + c_1^\dagger c_2) + A (c_1^\dagger c_2^\dagger c_1 c_2 + c_2^\dagger c_1^\dagger c_2 c_1)
Where t is a tunnelling term coupling the two sites, and A is a repulsion term. t is real. A is real and greater than zero.
zero fermion eigen solution:
|0\rangle, energy is H |0\rangle = 0 |0\rangle
one fermion eigen solutions:
let |1+\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger + c_2^\dagger)|0\rangle, energy is H|1+\rangle = (\epsilon+t) |1+\rangle
let |1-\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger - c_2^\dagger)|0\rangle, energy is H|1-\rangle = (\epsilon-t) |1-\rangle
two fermion eigen solution:
let |2\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger c_2^\dagger - c_2^\dagger c_1^\dagger)|0\rangle, energy is H|2\rangle = (2 \epsilon+A) |2\rangle
The grand partition function is therefore:
\mathcal{Z} = 1 + e^{-\beta ((\epsilon+t)-\mu)} + e^{-\beta ((\epsilon-t)-\mu)} + e^{-\beta (2\epsilon+A-2\mu)}
let \lambda = e^{-\beta (\epsilon-\mu)}, then we can clearly see the polynomial:
\mathcal{Z} = 1 + e^{-\beta t}\lambda + e^{-\beta (-t)} \lambda + e^{-\beta A} \lambda^2}
\mathcal{Z} = 1 + (e^{-\beta t}+ e^{\beta t})\lambda + e^{-\beta A} \lambda^2
And of course we can factor this polynomial to rewrite the partition function as:
\mathcal{Z} = \prod_i Q_i
Where Q_i = (\lambda - \lambda_i) and \lambda_i are the zero's of the polynomial. Notice that for this example, \lambda_i are real, less than zero, and \prod_i \lambda_i = 1.
So I can rewrite this as:
\mathcal{Z} = \prod_i Z_i
Where Z_i = (1 + e^{-\beta (\epsilon_i-\mu)}). And now the partition functions are of non-interacting quasiparticle states!
So the Hamiltonian should be able to be rewritten as such:
H = \sum_i \epsilon_i b_i^\dagger b_i
Where b_i^\dagger is the creation operator for non-interacting state i with energy \epsilon_i.
This should probably work for any interacting system of particles. You can always rewrite the polynomial as a product of the zeros. So you can always rewrite the paritition function as a system of non-interacting quasi-particles. But what are these states physically? And how do they relate to the original creation operators c_1^\dagger and c_2^\dagger ?
Any insight people can provide on this would be appreciated.
Okay, this is a question I brought up elsewhere, but I guess I was off topic from the original thread so I decided to post here. Any help would be appreciated.
It appears to me that the grand partition function indicates any problem can be rewritten in terms of non-interacting quasi-particles. Below is an example to "demonstrate" the method, and also to give specifics in which to discuss. I haven't been able to find a flaw or make physical sense of it yet, which is where my questions lie. Any help understanding this would be much appreciated.
Consider a potential with two sites for interacting fermions.
H = \epsilon c_1^\dagger c_1 + \epsilon c_2^\dagger c_2 + t (c_2^\dagger c_1 + c_1^\dagger c_2) + A (c_1^\dagger c_2^\dagger c_1 c_2 + c_2^\dagger c_1^\dagger c_2 c_1)
Where t is a tunnelling term coupling the two sites, and A is a repulsion term. t is real. A is real and greater than zero.
zero fermion eigen solution:
|0\rangle, energy is H |0\rangle = 0 |0\rangle
one fermion eigen solutions:
let |1+\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger + c_2^\dagger)|0\rangle, energy is H|1+\rangle = (\epsilon+t) |1+\rangle
let |1-\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger - c_2^\dagger)|0\rangle, energy is H|1-\rangle = (\epsilon-t) |1-\rangle
two fermion eigen solution:
let |2\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger c_2^\dagger - c_2^\dagger c_1^\dagger)|0\rangle, energy is H|2\rangle = (2 \epsilon+A) |2\rangle
The grand partition function is therefore:
\mathcal{Z} = 1 + e^{-\beta ((\epsilon+t)-\mu)} + e^{-\beta ((\epsilon-t)-\mu)} + e^{-\beta (2\epsilon+A-2\mu)}
let \lambda = e^{-\beta (\epsilon-\mu)}, then we can clearly see the polynomial:
\mathcal{Z} = 1 + e^{-\beta t}\lambda + e^{-\beta (-t)} \lambda + e^{-\beta A} \lambda^2}
\mathcal{Z} = 1 + (e^{-\beta t}+ e^{\beta t})\lambda + e^{-\beta A} \lambda^2
And of course we can factor this polynomial to rewrite the partition function as:
\mathcal{Z} = \prod_i Q_i
Where Q_i = (\lambda - \lambda_i) and \lambda_i are the zero's of the polynomial. Notice that for this example, \lambda_i are real, less than zero, and \prod_i \lambda_i = 1.
So I can rewrite this as:
\mathcal{Z} = \prod_i Z_i
Where Z_i = (1 + e^{-\beta (\epsilon_i-\mu)}). And now the partition functions are of non-interacting quasiparticle states!
So the Hamiltonian should be able to be rewritten as such:
H = \sum_i \epsilon_i b_i^\dagger b_i
Where b_i^\dagger is the creation operator for non-interacting state i with energy \epsilon_i.
This should probably work for any interacting system of particles. You can always rewrite the polynomial as a product of the zeros. So you can always rewrite the paritition function as a system of non-interacting quasi-particles. But what are these states physically? And how do they relate to the original creation operators c_1^\dagger and c_2^\dagger ?
Any insight people can provide on this would be appreciated.