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Many body problem -> non-interacting particles

  1. Apr 19, 2008 #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.
    [tex]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) [/tex]
    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:
    [tex]|0\rangle[/tex], energy is [tex]H |0\rangle = 0 |0\rangle[/tex]

    one fermion eigen solutions:
    let [tex] |1+\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger + c_2^\dagger)|0\rangle[/tex], energy is [tex] H|1+\rangle = (\epsilon+t) |1+\rangle [/tex]
    let [tex] |1-\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger - c_2^\dagger)|0\rangle[/tex], energy is [tex] H|1-\rangle = (\epsilon-t) |1-\rangle [/tex]

    two fermion eigen solution:
    let [tex] |2\rangle = \frac{1}{\sqrt{2}}(c_1^\dagger c_2^\dagger - c_2^\dagger c_1^\dagger)|0\rangle[/tex], energy is [tex] H|2\rangle = (2 \epsilon+A) |2\rangle [/tex]

    The grand partition function is therefore:
    [tex]\mathcal{Z} = 1 + e^{-\beta ((\epsilon+t)-\mu)} + e^{-\beta ((\epsilon-t)-\mu)} + e^{-\beta (2\epsilon+A-2\mu)}[/tex]

    let [tex]\lambda = e^{-\beta (\epsilon-\mu)}[/tex], then we can clearly see the polynomial:
    [tex]\mathcal{Z} = 1 + e^{-\beta t}\lambda + e^{-\beta (-t)} \lambda + e^{-\beta A} \lambda^2}[/tex]
    [tex]\mathcal{Z} = 1 + (e^{-\beta t}+ e^{\beta t})\lambda + e^{-\beta A} \lambda^2[/tex]

    And of course we can factor this polynomial to rewrite the partition function as:
    [tex]\mathcal{Z} = \prod_i Q_i [/tex]
    Where [tex]Q_i = (\lambda - \lambda_i)[/tex] and [tex]\lambda_i[/tex] are the zero's of the polynomial. Notice that for this example, [itex]\lambda_i[/itex] are real, less than zero, and [tex]\prod_i \lambda_i = 1[/tex].

    So I can rewrite this as:
    [tex]\mathcal{Z} = \prod_i Z_i [/tex]
    Where [tex]Z_i = (1 + e^{-\beta (\epsilon_i-\mu)})[/tex]. And now the partition functions are of non-interacting quasiparticle states!

    So the Hamiltonian should be able to be rewritten as such:
    [tex]H = \sum_i \epsilon_i b_i^\dagger b_i [/tex]
    Where [itex]b_i^\dagger[/itex] is the creation operator for non-interacting state [itex]i[/itex] with energy [itex]\epsilon_i[/itex].

    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 [tex]c_1^\dagger[/tex] and [tex]c_2^\dagger[/tex] ?

    Any insight people can provide on this would be appreciated.
  2. jcsd
  3. Apr 21, 2008 #2
    I've thought about this some more, and started wondering: what is the commutation relation for these new creation/annihilation operators? Are they still fermions? I'm not sure how to check. Basically I want to someone prove whether or not the Hamiltonian is still Hermitian. If not, then clearly there is a mistake and using this can help me track down the mistake (hopefully).

    Any ideas anyone?
  4. Apr 22, 2008 #3
    Those operators will probably not diagonalize the Hamiltonian in the interacting case, so calling them "particles" isn't a good idea. In other words, particle number will not be conserved (doesn't commute with H)
  5. Apr 22, 2008 #4


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    hmmm... phys forums isn't letting me quote and post... so maybe this will work.

    anyway, you have to factor out the e^{\beta A} in order to write your polynomial in the way yuo wrote it. and then I don't see how you rewrite it in the "quasiparticle" form. the algebra doesnt work.
  6. Apr 22, 2008 #5


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    you have to factor our the coefficient of the lambda^2 term.
    Z=e^{-\beta A}(\lambda-\lambda_+)(\lambda-\lambda_-)
    \lambda_{\pm}=\frac{-\cosh(\beta t)e^{\beta A}\pm\sqrt{\cosh^2(\beta t)e^{2\beta A}-e^{\beta A}}}{}
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