3
$\begingroup$

I have a question about a quantum field theory (QFT) partition function in the following example. Consider a single-site Hubbard model with the Hamiltonian given by:

$$H = \varepsilon_I \left(a_\uparrow^\dagger a_\uparrow + a_\downarrow^\dagger a_\downarrow \right) + U a_\uparrow^\dagger a_\uparrow a_\downarrow^\dagger a_\downarrow.$$

Typically, if we want to write down the partition function, we sum the Boltzmann factors over the Fock space. In this case, we have four states: one empty, two single-filled, and one double-filled. The partition function is:

$$\mathcal{Z} = 1 + 2e^{-\beta\varepsilon_I} + e^{-\beta(2\varepsilon_I + U)}.$$

Now, I would like to obtain the same using QFT.

The first step is writing down the action:

$$ S\left[\bar\phi,\phi,\bar\psi,\psi\right] = \int_0^\beta d\tau \left[\bar\psi(\tau)\left(\partial_\tau+\varepsilon\right)\psi(\tau) + \bar\phi(\tau)\left(\partial_\tau+\varepsilon\right)\phi(\tau) + U\bar\phi(\tau)\phi(\tau)\bar\psi(\tau)\psi(\tau)\right]\,, $$ where $\psi$ and $\phi$ correspond to spin-up and spin-down states.

Defining the Fourier transforms of the fields using Matsubara frequencies gives $$\psi(\tau) = \frac{1}{\sqrt{\beta}}\sum_{i}\psi_ie^{-i\omega_i\tau}\,,\quad \bar\psi(\tau) = \frac{1}{\sqrt{\beta}}\sum_{j}\bar\psi_je^{i\omega_j\tau}\,,\quad \phi(\tau) = \frac{1}{\sqrt{\beta}}\sum_{k}\phi_ke^{-i\omega_k\tau}\,, \quad \bar\phi(\tau) = \frac{1}{\sqrt{\beta}}\sum_{l}\bar\phi_le^{i\omega_l\tau}\,.$$

This results in $$ S\left[\bar\phi,\phi,\bar\psi,\psi\right] = \sum_{j}\bar\psi_j\left(-i\omega_j+\varepsilon\right)\psi_j + \sum_{l}\bar\phi_l \left(-i\omega_l+\varepsilon\right)\phi_l +\frac{U}{\beta}\sum_{l,k,j,i} \bar\phi_l\phi_k\bar\psi_j\psi_i\delta_{\omega_l+\omega_j,\omega_i+\omega_k} \\ = \sum_{j}\bar\psi_j\left(-i\omega_j+\varepsilon\right)\psi_j + \sum_{l}\bar\phi_l \left(-i\omega_l+\varepsilon\right)\phi_l +\frac{U}{\beta}\sum_m\left[\sum_{l}\bar\phi_l\phi_{l-m}\right]\left[\sum_j\bar\psi_j\psi_{j+m}\right] \\ = \sum_{j}\bar\psi_j\left(-i\omega_j+\varepsilon\right)\psi_j + \sum_{l}\bar\phi_l \left(-i\omega_l+\varepsilon\right)\phi_l +\frac{U}{\beta}\sum_{m}\Phi_{-m}\Psi_m \nonumber \\ = \sum_{j,n}\bar\psi_j\left[\delta_{j,n}\left(-i\omega_j+\varepsilon\right)+\frac{U}{\beta}\Phi_{j-n}\right]\psi_{n} + \sum_{l}\bar\phi_l \left(-i\omega_l+\varepsilon\right)\phi_l \,. $$

The partition function is

$$ \mathcal{Z} = \int D(\bar\psi,\psi)D(\bar\phi,\phi)e^{-S} = \int D(\bar\phi,\phi) \det\left[-i \hat\omega + \varepsilon + \frac{U}{\beta}\hat\Phi\right]\exp\left[-\sum_l\bar\phi_l \left(-i\omega_l+\varepsilon\right)\phi_l\right]\,. $$

We can put the determinant inside the exponential:

$$ \mathcal{Z} = \int D(\bar\phi,\phi) \exp\left[-\sum_l\bar\phi_l \left(-i\omega_l+\varepsilon\right)\phi_l + \mathrm{tr} \ln \left(-i \hat\omega + \varepsilon + \frac{U}{\beta}\hat\Phi\right)\right]\,. $$

This expression looks similar to what one gets after using the Hubbard-Stratonovich transformation to get rid of a quartic field. The difference is that here, the $\phi$ field is fermionic, not bosonic.

I am not sure how to proceed from here to obtain the exact solution for the partition function. Generally, one would try to look for a saddle point and Taylor expand. However, in this case, there is a known solution.

Any suggestions are greatly appreciated.

$\endgroup$
1
$\begingroup$

This is one of the cases where the operator formalism is more powerful than the path integral.

As mentioned by the OP, the calculation of the partition function is trivial for the one-site (d=0) Hubbard model in the operator formalism, whereas the path-integral involves a quartic interaction which means that we cannot perform the path integral exactly. This is also true for the Bose-Hubbard model (with only one site).

What people usually do for larger dimensions is to do a mix of path integral for the hopping part, and a operator (exact) treatment of the one-site Hamiltonian, which allows to take advantage of the strength of both path-integral and operator formalism. This kind of calculation is equivalent to a strong-coupling RPA approximation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.