1
$\begingroup$

I'm having some trouble with the exercise on the Stoner transition in the book by Altland & Simons (page 351, here).

The problem

Starting from the Hubbard Hamiltonian

$$ H=\sum_{\mathbf{k},\sigma}\xi_{\mathbf{k}}c_{\mathbf{k},\sigma}^\dagger c_{\mathbf{k},\sigma}+U\sum_i n_{i,\uparrow}n_{i,\downarrow} $$

with $\xi_{\mathbf{k}}=\epsilon_{\mathbf{k}}-\mu$, I need to show that the partition function in the functional formalism can be written as ($\hbar=1$)

$$ Z=Z_0 \int\mathcal{D}[m]\;\mathrm{exp}\left\{-\frac{U}{4}\int\mathrm{d}\tau\sum_i m_i^2(\tau)+\mathrm{tr}\left[\mathrm{ln}\left(1+\frac{U}{2}\sigma^z\hat{m}G_0\right)\right]\right\} $$

with $Z_0$ and $G_0$ resp. the partition and Green's function in the noninteracting theory, $\sigma^i$ Pauli matrices and $\hat{m}=m_i(\tau)\delta_{ij}\delta(\tau-\tau')$.

Own attempt

Writing the partition function as a field integral and performing a Hubbard-Stratonovich transformation, I obtained the following result (which is correct according to the answer in Altland & Simons):

$$ Z=\int\mathcal{D}[m]\int\mathcal{D}[\psi^*,\psi]\;\mathrm{exp}\left\{-\int\mathrm{d}\tau\left[\sum_{\mathbf{k},\sigma}\psi^*_{\mathbf{k},\sigma}(\partial_\tau+\xi_{\mathbf{k}})\psi_{\mathbf{k},\sigma} - \frac{U}{2}\sum_{i,\sigma}\psi^*_{i,\sigma}\sigma^zm_i\psi_{i,\sigma}+\frac{U}{4}\sum_i m_i^2(\tau)\right]\right\} $$

The next step is to perform the integration over the fields $\psi^*$ and $\psi$, but this is where I am stuck. My first guess is to Fourier transform using

$$ \psi_{\mathbf{k},\sigma}=\frac{1}{\sqrt{N}}\sum_j\mathrm{e}^{-i\mathbf{k}\cdot\mathbf{r}_j}\psi_{j,\sigma} $$

However, if I do this for the first term in the exponential, I am still left with $\xi$ which depends on $\mathbf{k}$, while doing it for the second term leaves me with $m$, which depends on i. Any help on this part would be appreciated.

$\endgroup$
  • $\begingroup$ You want to go the other way and express the fields $\psi_{j,\sigma}$ in the interaction term as Fourier modes. $\endgroup$ – Nephente Jan 28 '17 at 9:28

Your Answer

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

Browse other questions tagged or ask your own question.