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I read in this paper "Gauge fields and quantum entanglement" by J. Mielczarek and T. Trześniewski that there are two ways of deriving the transformation of a holonomy of a gauge field under a gauge transformation: one way is from the definition formula of the holonomy as the integral of the associated connection 1-form over a path$-$which treats holonomies as parallel-transported elements of the gauge group; the other way is from treating holonomies as isomorphisms between Hilbert spaces.

I think the latter way is very easy as shown in the following.

By treating a holonomy $h$ as a map between the Hilbert space at $s$, $\mathcal{H}_s$, and the Hilbert space at $t$, $\mathcal{H}_t$, of a spatial hypersurface $\Sigma$ of spacetime, we can write $h$ as $$h=h_{IJ}|I \rangle_{st} \langle J|\in \mathcal{H}_s\otimes\mathcal{H}^*_t.$$ Then under a unitary transformation at both $s$, $U_s$, and $t$, $U_t$—which aren't necessarily the same—$|I\rangle^{'}_s=U_{sKI}|K \rangle_{s}, |J\rangle^{'}_t=U_{tLJ}|L \rangle_{t}$, together with $h^{'}_{IJ}|I \rangle^{'}_{st} \langle J|^{'}=h_{KL}|K \rangle_{st} \langle L|$, one can find $U_{sKI}h^{'}_{IJ}U^\dagger_{tJL}=h_{KL}$, which, written in terms of the operator notation, is $U_sh^{'}U^\dagger_t=h$, which is equivalent to $$h'=U^\dagger_shU_t.$$

But I feel it's difficult to see how to derive it from the former way.

From the perspective of parallel transport of the gauge group, the holonomy of the gauge field $A^i_a$ associated with the connection $1$-form $A=A^i_a\tau_idx^a$ is defined as $$h_e[A]:=\mathcal{P}\exp \int_e A,$$ where $\tau_i$'s are the generators of the Lie algebra of the gauge group, $e$ is a path $e: [0,1]\rightarrow \Sigma$ intervened between $e(0)=s$ and $e(1)=t$ on $\Sigma$, and $\mathcal{P}$ denotes the path ordering. Then under a unitary transformation, the connection $1$-form transforms as $A^{'}\rightarrow U^\dagger AU+U^\dagger dU$, which leads the holonomy to transform as $$h^{'}_e[A]=U^\dagger(e(0))h_e[A]U(e(1))=U^\dagger_sh_e[A]U_t.$$ I cannot see how the transformation is reached:$$h^{'}_e[A]=\mathcal{P}\exp \int_e (U^\dagger AU+U^\dagger dU)\overset{?}{=}U^\dagger_s\left (\mathcal{P}\exp \int_e A\right)U_t.$$

Bohemian relativist
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1 Answers1

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I'd be interested if someone could provide a reference in the literature that takes the time to prove it? But anyway, here we go.

Background. Let's recap some of the stuff stated in the question, just so the answer is relatively self-contained. Let $\{A_{\mu}\}_{\mu=1}^{k}$ be the components of the Berry connection $\mathbf{A}$ on a rank $m$ vector bundle whose

  • Base space is $k$-manifold $\mathcal{M}$ (the parameter space, in which anticlockwise loop $C$ is adiabatically spanned; let $C$ be parametrised by smooth map $\lambda:[0,T]\to\mathcal{M}$).
  • Fibres are the ($m$-dimensional) ground state of the Hilbert space. Let $\mathcal{H}(\lambda)$ denote the fibres over $C$.

Recall that $A_{\mu}\in\mathfrak{u}(m)$ is a skew-Hermitian linear operator of rank $m$. Let $g:C\to U(\mathcal{H}(\lambda))$ be a smooth map that acts as a change of orthonormal basis on each fibre $\mathcal{H}(\lambda)$, i.e. $g$ is a gauge transformation. The connection transforms as $A_{\mu}\mapsto A_{\mu}'= g^{\dagger}A_{\mu}g - g^{\dagger}\partial_{\mu}g$.

OP's question (paraphrased). Let $\gamma:[0,T]\to\mathcal{M}$ be a smooth path from $a\in\mathcal{M}$ to $b\in\mathcal{M}$ with no self-intersections in the interval $(0,T)$. Let $\Gamma_{\mathbf{A}}(\gamma)$ denote the parallel transport along $\gamma$ with respect to connection $\mathbf{A}$. That is, $\Gamma_{\mathbf{A}}(\gamma)=\mathcal{P}\left(\exp\int_{a}^{b} A_{\mu}d\lambda^{\mu}\right)$ where we use Einstein summation. How does the parallel transport transform?

Claim. Gauge transformations only act on parallel transport along a path $\gamma$ at the endpoints. That is, $\Gamma_{\mathbf{A}'}(\gamma)=g^{\dagger}(b)\circ\Gamma_{\mathbf{A}}(\gamma)\circ g(a)$.

Proof. It suffices to consider parallel transport along a segment $[\lambda_0,\lambda_0 + \delta\lambda]$. Taking $\delta\lambda\to 0$ and starting with $g(\lambda_0 + \delta\lambda)\circ\Gamma_{\mathbf{A}'}(\gamma)\circ g^{\dagger}(\lambda_0)$,

$g(\lambda_0 + \delta\lambda)\circ\mathcal{P}\left(\exp\int_{\lambda_0}^{\lambda_0+\delta\lambda} A'_{\mu}d\lambda^{\mu}\right)\circ g^{\dagger}(\lambda)=g_{\lambda_0 + \delta\lambda}\circ\left(\mathbf{1}_{\mathcal{H}(\lambda_0)}+\mathbf{A}'(\lambda_0)\cdot\delta\lambda+O(\delta\lambda^2)\right)\circ g^{\dagger}_{\lambda_0}=\ldots$

Using $g_{\lambda_0+\delta\lambda}=g_{\lambda_0}+\partial_{\mu}g|_{\lambda_0}\delta\lambda^{\mu}+O(\delta\lambda^{2})$ and continuing,

$\ldots=\mathbf{1}_{\mathcal{H}(\lambda_0)}+\left(\partial_{\mu}g|_{\lambda_0}\circ g_{\lambda_0}^{\dagger}\right)\delta\lambda^{\mu} + \left(g_{\lambda_0}\circ A_{\mu}'(\lambda_0)\circ g_{\lambda_0}^{\dagger}\right)\cdot\delta\lambda^{\mu}+O(\delta\lambda^2)$
$\phantom{\ldots}=\mathcal{P}\left( \exp \int_{\lambda_0}^{\lambda_0+\delta\lambda}\left[ (\partial_{\mu}g)g^{\dagger}+gA_{\mu}'g^{\dagger} \right] d\lambda^{\mu} \right)=\Gamma_{\mathbf{A}}(\gamma) \quad \blacksquare$

Sachin Valera
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