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Almost every text on topological insulators have the Bloch sphere example of a two level system showing the non triviality of the bundle of an eigenvector over the sphere: we can't define an eigenstate over the whole Bloch sphere, instead of this, we must construct two local trivializations, namely

$$ \psi_{-}^{S}(\vec{n}) = \left(\begin{matrix} -\sin\frac{\theta}{2} \\ e^{i\varphi} \cos\frac{\theta}{2} \end{matrix}\right) \quad \psi_{-}^{N}(\vec{n}) = \left(\begin{matrix} - e^{-i\varphi}\sin\frac{\theta}{2} \\ \cos\frac{\theta}{2} \end{matrix}\right) $$

defined in the south and north hemispheres respectively to avoid the obstruction (to Stokes theorem). We can easily compute the Chern number via Berry curvature on the sphere on any of the states. But we are getting the Chern number integrating over the SPHERE, and not over the Brillouin TORUS. Does this result reflect directly the non-triviality of the fiber bundle over the Brillouin TORUS? (I think getting the pullback of the map from the sphere to the torus and using the chain rule may suffice to prove this, but I am not sure it's enough for a proof). Will we get the same result?

Aside: when in the Bloch sphere we choose the trivializations, we are defining a transition function like

$$ t_{NS} = e^{i\varphi} $$

that applies to the fiber space, with that definition, when going from the north to the south hemisphere. Is that $\varphi$ the berry phase? Quantitatively or qualitatively?

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I'll give you in the following a rather detailed answer, but let me first, shortly state the answers to your questions:

  1. The Chern number of the eigenvector bundle over the torus can be evaluated by integration over the sphere, the integrand will indeed be the monopole Berry curvature. However the integration region will not in general be a single sweep of the surface of the sphere, because the map from $T^2$ (the Brillouin zone) to $S^2$ may wind the sphere several times.

  2. The transition functions are rather related to the Chern number. Their winding number, i.e., the number of times they wind the equator is equal to the Chern number.

  3. Concerning the question paused in the title: The topology of all principal $U(1)$ bundles over the$2$-torus $T^2$ is determined solely by the Chern number. (This result is not general ; for example it is not true for the torus $T^3$ because it is of a higher dimension)

Details:

The Berry phase of a nondegenerate state is a holonomy of a principal $U(1)$ bundle over the parameter space $M$. There can be many topologically inequivalent $U(1)$ bundles over a given parameter space corresponding to inequivalent quantum systems. Thus the Berry phase allows a classification of parametrized quantum systems based upon the classification of principal bundles. This point of view was noticed by: Bohm, Boya, Mostafazadeh and Rudolph.

The classification theorem of principal bundles asserts the existence of a universal principal bundle $U(1)\rightarrow\eta\rightarrow B$, such that any principal $U(1)$ bundle $\lambda$ over the parameter space is the pullback of which under some map $f$. (Please see Nash and Sen for a more detailed explanation of the classification theory of principal bundles.)

$$\begin{array}{ccc} \lambda& \overset{f^*}\leftarrow & \eta\\ \downarrow & & \downarrow \\ \mathcal{M} & \overset{f}\rightarrow & B \end{array}$$

The classification theory of principal bundles specifies a base space (the classifying space) and a total space of the universal bundle depending only on the structure group (in our case U(1)). In the case of a nondegenerate state of a Hamiltonian unconstrained by any anti-unitary symmetry, the classifying space is known to be the infinite dimensional complex projective space $B = \mathbb{C}P^{\infty}$. (For a good account of complex projective spaces , please see chapter 4 in Bengtsson and Życzkowski . Also, the following exposition of John Baez of classifying spaces is very useful and contains some more explanation of the infinite dimensional case $\mathbb{C}P^{\infty}$ .

$\mathbb{C}P(\infty)$ is the space of all one dimensional projectors, and the map $f$ from the parameter space to the universal base space is:

$$P: \mathcal{M}\overset{P}\rightarrow \mathbb{C}P^{\infty}$$ $$ \mathcal{M} \ni m \mapsto P(m) = |u(m)\rangle\langle u(m)|\in \mathbb{C}P^{\infty}$$

Where $|u(m)\rangle $ is the state of the system. (The bundle $\lambda$ is sometimes called the Berry bundle, and when the state is an eigenstate of a Hamiltonian, Synonyms: the eigenbundle or the spectral bundle).

The construction of the Berry phase stems from the existence of a universal $U(1)$ connection over the infinite dimensional projective space whose curvature is the Fubini-Study form:

$$ dA = \frac{1}{2i}\mathrm{Tr} (P dP \wedge dP)$$

The pulled back Berry connection $P^*(A) = A(m)$ is the Berry connection whose holonomy is the Berry phase on $\mathcal{M}$ and the integral of its curvature on $M$ is the first Chen class of $\lambda$.

The infinite dimensional projective space $\mathbb{C}P^{\infty}$ is the direct limit of a series of inclusions:

$$ \mathbb{C}P^{1} \subset \mathbb{C}P^{2} . . . \subset \mathbb{C}P^{\infty}$$

In the case of when the parameter space is two dimensional, it is sufficient to approximate the classifying space by its first component namely $ \mathbb{C}P^{1} = S^2$ and consider maps to the space of one dimensional projectors in two dimensions namely $S^2$, and consider the bundle map:

$$\begin{array}{ccc} \lambda& \overset{P^*}\leftarrow & \eta\\ \downarrow & & \downarrow \\ \mathcal{M} & \overset{P}\rightarrow & S^2 \end{array}$$

(Please see for example, Viennot, for a more detailed expposition of the finite dimensional approximations of the classifying spaces) Here, it is very easy to write the formula for the one dimensional projector map:

$$P(m) = \frac{1}{2} \begin{bmatrix} 1-\cos \theta(m) & \sin \theta(m) e^{i\phi(m)}\\ \sin \theta(m) e^{-i\phi(m)} & 1+\cos \theta(m) \end{bmatrix}$$ This projector gives rise to the pull back to $\mathcal{M}$ of well-known magnetic monopole Berry connection:

$$A_{\pm} = \frac{1\mp\cos \theta(m) }{\sin \theta(m) } d\phi(m) $$

whose Berry curvature is proportional to the area element of the sphere

$$F = \frac{1}{2} \sin \theta(m) d\theta(m) d\phi(m) $$

The integration is performed along a path in the manifold $\mathcal{M}$, thus the Berry phase corresponding to the path $\Gamma_M$ is given by:

$$\phi_B = \int_{\Gamma_M} A(m)$$

It can be pulled back to the two sphere (by change of the integration variable), but in this case we need to integrate on the image of the path $\Gamma = P(\Gamma_M)$:

$$\phi_B = \int_{\Gamma} A$$

The same is true for the Chern class:

$$ c = \int_{\mathcal{M}} F(m) = \int_{P(\mathcal{M})} F$$

The integration region may wind the two sphere several times and the Chern number will be equal to a multiple of the charge of the monopole.

For the second question, we observe that:

$$\begin{align*} \int_{P(\mathcal{M})} F & = \int_{P(\mathcal{M}) \cap S^1} \left (A_{+}-A_{-}\right ) &=\int_{P(\mathcal{M}) \cap S^1} d \phi &= \frac{1}{i}\int_{P(\mathcal{M}) \cap S^1} g^{-1}d g \end{align*} $$ Where $S^1$ is the equator and $g= e^{i \phi}$. The last term is the winding number of the mapping $g$. It is a one dimensional Wess-Zumino-Witten term.

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  • $\begingroup$ Thanks for such a complete answer. I think I need to get deeper in classification theory to fully understand it, however, let me see if I get the general idea. We have a non-trivial topology if the map from the torus covers the whole sphere at least one time. If we integrate over the sphere, the result is the same as if that map covered only one time the sphere completely (because this is the integration surface). So it shows if there is non-trivial topology, but the Chern number computed does not match in general. I see it now. $\endgroup$ – vbarcelo Dec 13 '17 at 12:16
  • $\begingroup$ ... So, can it, the Chern number, be viewed as the times we need to apply the transition function to cover the whole manifold? And it is the times we add the corresponding Berry phase to the wave-function? And, for the last eq. you wrote, I think there is a missing i in the last term. $\endgroup$ – vbarcelo Dec 13 '17 at 12:22
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    $\begingroup$ I think that your statements are correct, but let me emphasize: In order to find the Chern number for a system parametrized by the torus, we need to perform the integration on the torus. We can make a transformation to the sphere, because due to the classification theorem, the integrand will always have the form of the monopole curvature on the sphere. But in this case we must take into account the winding of the map over the sphere; once we do so, we will obtain the same Chern number that we would have obtained by integration on the torus. $\endgroup$ – David Bar Moshe Dec 14 '17 at 8:32
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    $\begingroup$ cont. All we did is a variable transformation that does not affect the result. We can compute the Chern number also, by computing the integral of the transition functions over the equator; here we must be also careful to the number of windings. There is no need to add the Berry phase to the wave functions (and it will not change the result, since the Chern number is gauge invariant). Thank you for the correction. $\endgroup$ – David Bar Moshe Dec 14 '17 at 8:33

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