非常简略的笔记：Chern-Simons Field Theory for IQHE
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Quantization of the Chern-Simons level
Large Gauge Transformation. With gauge transformation: \(A_\mu \to A_\mu + \partial_\mu \omega\). The electron wave function or field transforms as \(e^{ie\omega/\hbar}\).
- Wick rotation: \(\tau\) periodic in \(\mathbf{S}^1\)
- Only physical fields are single valued, \(\omega(t,x)\) is not need to be single valued. Hence, the real requirement is that \(e^{ie\omega/\hbar}\) is single valued, not \(\omega\).
Around the circle, setting \(\omega = \frac{2\pi\tau}{\beta} \frac{\hbar}{e}\), leaves the exponential \(e^{ie\omega/\hbar}\) single valued.
- Large gauge transformation, the name signify that they cannot be continuously connected to the identity.
Then the gauge transformation becomes \[ A_0 \to A_0 + \frac{2\pi}{\beta} \frac{\hbar}{e} \] Gauge tranformation does not change the physics, we can think of \(A_0\) as being a periodic variable, with periodicity \(2\pi\hbar/e\beta\).
Dirac quantization condition | Monopole \(\mathbf{R}^2 \to \mathbf{S}^2\).
We think about the spatial directions as forming a sphere \(\mathbf{S}^2\), rather than a plane \(\mathbf{R}^2\). By threading a background magnetic flux through the spatial \(\mathbf{S}^2\): \[\frac{1}{2 \pi} \int_{\mathbf{S}^{2}} F_{12}=\frac{\hbar}{e}\] where \(F_{12}=B_z\). This is tantamount to placing a Dirac magnetic monopole inside the \(\mathbf{S}^2\).
- It's hard to experiment by building a quantum Hall state on a sphere.
Chern-Simons Term.
\[S_{C S}=\frac{k}{4 \pi} \int d^{3} x A_{0} F_{12}+A_{1} F_{20}+A_{2} F_{01}\] We can safely set all terms with \(\partial_0\) to zero, but integrating by parts on the spatial derivatives we get an extra factor of 2, \[S_{C S}=\frac{k}{2 \pi} \int d^{3} x A_{0} F_{12}\] With the gauge potential \(A_0=a\) and Dirac magnetic monopole, we have \[S_{CS}=\beta a \frac{\hbar k}{e}\]
- Why integrate by parts before evaluating?
- think about the gauge field as being defined locally on different patches and glued together
Integral Hall Conductivity. Considering the periodic gauge transformation of \(A_0\), we have \[ \begin{aligned} S_{c s} &{}\to S_{c s} + \frac{k}{4\pi} \int d^{3} x \cdot \partial_{\mu}\left(\omega \varepsilon^{\mu \nu \rho} \partial_{\nu} A_{\rho}\right) \\ &= S_{c s} + \frac{k}{4\pi} \int d\tau d^{3} x \ \nabla\omega \cdot (\nabla \times A) \\ &= S_{c s} + \frac{2\pi k \hbar}{e} \cdot \frac{1}{2\pi} \int_{\mathbf{S}^2} F_{12} \\ &= S_{c s} + \frac{2\pi k \hbar^2}{e^2} \end{aligned} \] It’s ok if the Chern-Simons term itself is not gauge invariant, as long as the partition function \(Z=e^{iS_{CS}/\hbar}\) is, then we should have \[ \frac{k \hbar^2}{e^2} \in \mathbb{Z} \] By writting \(k=e^2v/\hbar\) with \(v \in \mathbb{Z}\), the Hall conductivity becomes \[\sigma_{xy} = \frac{k}{2\pi} = \frac{e^2}{2\pi\hbar}v\] There was no mention of Landau levels, no mention of whether the charge carriers were fermions or bosons, or whether they were free or strongly interacting.
Remarks.
APPENDIX: Dirac quantization condition
Without magnetic monopoles, \(q_{m}=0,\), \(\nabla \cdot \mathrm{B}=0\).
Assume that there is a magnetic monopole with charge \(q_{m},\) what will happen? The \(B\) fields is: \[ \vec{B}=q_{m} \frac{\vec{e}_{r}}{r^{2}}=q_{m} \frac{\vec{r}}{r^{3}}=q_{m} \frac{(x, y, z)}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \] Chosen a vector potential \[ \vec{A}=q_{m} \frac{(y,-x, 0)}{r(r-z)} \] The magnetic field \[ \begin{aligned} \nabla \times \vec{A} &=q_{m} \nabla \times \frac{(y,-x, 0)}{r(r-z)} = q_{m}\left(\partial_{x}, \partial_{y}, \partial_{z}\right) \times \frac{(y,-x, 0)}{r(r-z)} \\ &=q_{m}\left(\partial_{y} \frac{0}{r(r-z)}-\partial_{z} \frac{-x}{r(r-z)}, \partial_{z} \frac{y}{r(r-z)}-\partial_{x} \frac{0}{r(r-z)}, \partial_{x} \frac{-x}{r(r-z)}-\partial_{y} \frac{y}{r(r-z)}\right) \\ &=q_{m}\left(-\partial_{z} \frac{-x}{r(r-z)}, \partial_{z} \frac{y}{r(r-z)}, \partial_{x} \frac{-x}{r(r-z)}-\partial_{y} \frac{y}{r(r-z)}\right)\\ &=q_{m} \frac{(x, y, z)}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \end{aligned} \] It is singular at the north pole \(z=r\). In fact, one can prove that no matter which gauge one uses, there will always be a singularity point. * this singularity is not a physical singularity. All physical observables are smooth and non-singular functions at this point. * Only A (which is not a measurable quantity) shows singular behavior. * In addition, the location of this singularity point is gauge dependent.
So, we can use the first one to describe the south hemisphere and the second one to describe the north hemisp \[ \begin{aligned} \overrightarrow{A}_{N} &= q_{m} \frac{(-y, x, 0)}{r(r+z)} \\ \overrightarrow{A}_{S} &= q_{m} \frac{(y,-x, 0)}{r(r-z)} \end{aligned} \] At the equator, the vector potential is multivalued (depending on whether we use \(A_{N}\) or \(A_{S}\) ). The gauge transformation between \(A_{N}\) and \(A_{S}\) is: \[ \overrightarrow{A}_{N} = \overrightarrow{A}_{S} + 2 q_{m} \frac{(-y, x, 0)}{(r-z)(r+z)} \] At equator \(\mathrm{z}=0\) : \[ \begin{aligned} \overrightarrow{A}_{N} &=\overrightarrow{A_{S}}+2 q_{m} \frac{(-y, x, 0)}{r^{2}} = \overrightarrow{A}_{N}+2 q_{m} \nabla \varphi \\ \vec{A} &{} \rightarrow \vec{A}^{\prime}=\vec{A}+\nabla \Lambda(r, t) \\ \Psi(r, t) &{} \rightarrow \Psi^{\prime}(r, t)=\Psi(r, t) \exp \left(i \frac{q_{e}}{c \hbar} \Lambda\right) \\ \Phi &{} \rightarrow \Phi^{\prime} =\Phi-\frac{\partial \Lambda(r, t)}{c \partial t} \\ \end{aligned} \] Here \[ \begin{aligned} \Lambda(r, t) & =2 q_{m} \varphi \\ \Psi_{N}(r, t) &=\Psi_{S}(r, t) \exp \left(i \frac{q_{e}}{c \hbar} \Lambda\right)=\Psi_{S}(r, t) \exp \left[i \frac{2 q_{m} q_{e}}{c \hbar} \varphi\right] \end{aligned} \] Therefore, \(2 q_{m} q_{e}/c \hbar\) must be a integral. This tells us that the magnetic charge is quantized: \[ q_{m}=\frac{c \hbar}{2 q_{e}} n \] This is Dirac quantization condition.
- For a closed surface enclosing a magnetic monopole, no matter what gauge one uses, the vector potential must have some singularities.
- If \(A\) is a non-singular function on a closed manifold, the magnetic flux through this manifold must be zero.