0
$\begingroup$

The Berry connection $\mathcal{A}_{k}(\lambda)$ of a quantum system is given by

$$\mathcal{A}_{k}(\lambda) \equiv -i \langle n|\frac{\partial}{\partial \lambda^{k}}|n\rangle,$$

where the ket $|n(\lambda)\rangle$ depends on the parameters $\lambda^{k}, k=1,2,\dots$ in the system.

The field strength $\mathcal{F}_{kl}$ of the Berry connection $\mathcal{A}_{k}(\lambda)$ is defined by

$$\mathcal{F}_{kl} = \frac{\partial\mathcal{A}_k}{\partial\lambda_{l}}-\frac{\partial\mathcal{A}_l}{\partial\lambda_{k}}.$$

Therefore, we can define an analog of Maxwell's theory with the Berry connection $\mathcal{A}_{k}(\lambda)$. As such, we expect the Berry connection $\mathcal{A}_{k}(\lambda)$ to be gauge invariant. In other words, there must be a gauge redundancy in the definition of the Berry connection $\mathcal{A}_{k}(\lambda)$


What is the physical meaning of this gauge redundancy for some state $|n(\lambda)\rangle$?

What is the source of the gauge transformation for the Berry connection$\mathcal{A}_{k}(\lambda)$?

$\endgroup$
  • 1
    $\begingroup$ Here, the gauge redundancy just comes from the fact that quantum states are defined up to a phase factor, so you could replace $|n(\lambda) \rangle$ with $e^{i f(\lambda)} |n(\lambda) \rangle$. There's no physical difference. $\endgroup$ – knzhou Dec 1 '16 at 0:58
1
$\begingroup$

As was said in comments section, the reason for this gauge symmetry is arbitrariness in choosing of the phase factor. Precisely, the definition of the Berry phase $\phi$ reads $$ |n(\lambda(t + T))\rangle = e^{-i\int \limits_{t}^{t+T}E(t')dt' + i\phi}|n(\lambda(t))\rangle, $$ with $$ \phi \equiv \oint d \lambda \cdot A(\lambda), $$ where $A$ is the Berry connection.

You can redefine the state $n(\lambda(t))\rangle$ as $$ |n(\lambda(t))\rangle \to e^{i\alpha (\lambda (t))}|n(\lambda(t))\rangle , $$ so that the Berry connection is changed as $$ A(\lambda) \to A(\lambda) + \nabla_{\lambda}\alpha (\lambda) $$ However, the Berry phase remains unchanged, since $\lambda (t + T) = \lambda (t)$: $$ \phi \to \phi + \alpha(\lambda(t+T)) - \alpha(\lambda(t)) = \phi $$

The Berry phase as the result of impossibility of fixing the phase

Don't be confused with the above statement that the the state $|n(\lambda)\rangle$ phase definition is completely arbitrary. What about the Berry phase itself and why it can't be removed, You can ask?

The answer is that in general we can't fix the phases of eigenstates of the hamiltonian $H(\lambda)$ along the adiabatic transport. Let's sketch the general theory for understanding of this statement. Suppose as the general example the complex $n\times n$ non-degenerate Hamiltonian (degeneracy is possible only as the result of level crossing, i.e., $E_{i}(\lambda) = E_{j}(\lambda)$ only on the surface $S(\lambda)$). The set of eigenstates form manifold $C^{n}$. The adiabatic transport in $\lambda$ space generates the mapping $$ \tag 1 M_{\lambda} \to CP^{n-1}, $$ where $CP$ means complex and defined up to the phase. This mapping is responsible for the Berry phase. This phase can be made trivial only if the mapping $(1)$ can be lifted to the mapping $$ \tag 2 M_{\lambda}\to S^{2n-1} $$ of $M_{\lambda}$ on the space $S^{2n-1}$ of unit complex vectors with definite phases. This is nothing but the possibility of the global definition of phases.

The impossibility of the lifting is caused by topological obstructions. The formal reason of such obstructions is impossibility to define the phases of eigenstates at some points. Let me illustrate this statement on the case of $2\times 2$ Hamiltonian, which can be given in general form $$ H = \mathbf n \cdot \sigma , \quad \sigma = \sigma_{x},\sigma_{y},\sigma_{z} $$ Here $\mathbf n = (n_{x},n_{y},n_{z})$ plays the role of $\lambda$, with $M_{\mathbf n} \simeq S^{2}$.

The eigenvalues are $E_{\pm} = \pm |\mathbf n|$. You see that $E_{+} = E_{-}$ at $\mathbf n = 0$. Because of the degeneracy we have to surround this point by the sphere.

while corresponding eigenstates read ($\theta , \varphi$ set the direction of $\mathbf n$) $$ |\mathbf n,+\rangle = \begin{pmatrix}\cos(\theta/2)e^{-i\varphi} \\ \sin(\theta/2) \end{pmatrix}, \quad |\mathbf n,+\rangle = \begin{pmatrix} -\sin(\theta/2)e^{-i\varphi} \\ \cos(\theta/2) \end{pmatrix}, $$
The phase of $|+\rangle$ is bad defined on a North pole, while the phase of $|-\rangle$ is bad defined on a South pole. Therefore the mapping $$ S^{2} \to C^{2} $$ (where $C^{2}$ is the space of 2 complex vectors) is bad defined on North and South poles respectively, which creates the topological obstruction preventing the lifting after surrounding the degeneracy point $\mathbf n = 0$ by unit sphere. Therefore we can conclude that we have non-trivial Berry phase.

$\endgroup$
  • $\begingroup$ The mapping $S^2 \to CP^1$ is perfectly well-defined because $CP^1$ doesn't care about the phase. What you want to say to be consistent with what you said before is that this doesn't define a mapping to the sphere of complex unit vectors $S^3\subset\mathbb{C}^4$ because the phase is ill-defined on the poles of $S^2$. It is also not clear how this is supposed to be a topological obstruction (since $CP^1\cong S^2$ and maps $S^2\to S^3$ certainly exist (the "equator" of an $S^3$ is an $S^2$)), it's not clear what's topological about the failure of $S^2\to CP^1$ to lift. $\endgroup$ – ACuriousMind Dec 1 '16 at 16:58

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.