I'm interested in how to capture the chiral anomaly effects in quasi-classical approach. Precisely, I want to derive Boltzmann equation for partition function for massless fermions with definite chirality, which contains the effect of chiral anomaly.

Here is an article which contains the method of performing the desired thing. An idea of their article is to obtain chiral anomaly effect through the Berry phase.

To get the Berry phase effect in Hamiltonian, the authors start from the path integral $$ \langle f \left| Te^{iH(t_{f} - t_{i})}\right|i\rangle \equiv \int Dp~\mathrm dx ~e^{i\int~\mathrm dt (\mathbf p \cdot \dot{\mathbf x} - H)}, $$ where $H = \sigma \cdot \mathbf p$ is the Weyl particle Hamiltonian. It can be expanded in a basis set of $\sigma \cdot \mathbf p$ matrix: for each given $\mathbf p$ $$ H = V_{\mathbf p}^{\dagger}\text{diag}\left(\mathbf{p}, -|\mathbf{p}|\right)V_{\mathbf p} $$ After few manipulations, discretising the continuous path by small steps $\delta t$ and inserting the identity $1 = V_{\mathbf p}^{\dagger}V_{\mathbf p}$ between two ''neighbours'' $e^{-iH(\mathbf p_{2})\Delta t}e^{-iH(\mathbf p_{1})\Delta t}$ with $\mathbf p_{2} = \mathbf p_{1} + \Delta \mathbf p$, the authors obtain $$\begin{align} \langle f \left| Te^{iH(t_{f} - t_{i})}\right|i\rangle &\equiv \langle f\left|e^{iH(t_{f}-t_{i})}\right|i\rangle\\ & \equiv \left(V_{\mathbf p_{f}}\int~ Dx Dp ~\text{exp}\left[i\int~\mathrm dt~(\mathbf p \cdot \mathbf x - |\mathbf p|\sigma_{3}-\hat{\mathbf{a}}\cdot \dot{\mathbf p})\right]V_{\mathbf p_{i}}^{\dagger}\right)_{fi},\end{align} $$ where $$ \hat{a}_{\mathbf p} = V_{p}\nabla_{\mathbf p}V^{\dagger}_{p} $$ is the desired part of Berry curvature (the integral $\int~\mathrm d^{3}~\mathbf p \hat{a}_{\mathbf p}$ of which gives the Berry phase matrix), which gives the effect of anomaly.

My question is: why these manipulations (i.e, performing the Hamiltonian diagonalization in the amplitude) generate the Berry phase, and - moreover - its effect in Hamiltonian?


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.