I am trying to derive the non-abelian Aharonov-Bohm effect by generalising Michael Berry's derivation to the case of non-abelian gauge field $A$.

My derivation so far

We require a degenerate eigenspace in order to achieve a non-abelian Berry phase therefore I take my Hilbert space to be $\mathcal{H} = \mathcal{H}_\text{spatial} \otimes \mathcal{H}_\text{internal}$, where $\mathrm{dim}(\mathcal{H}_\text{internal})=N$. The wavefunctions will take the form

$$\Psi(x,t) = \psi(x,t) \mathbf{v} ,$$

where $\psi(x,t) $ is the spatial wavefunction and $\mathbf{v} $ is the internal state vector of the system. I now take my Hamiltonian to be

$$ H(X) = - \frac{1}{2m } (\nabla \mathbb{I} - ie A)^2 + V(X-x)\mathbb{I}$$

where $V(X-x)$ is the confining potential which traps our particle inside a small box centred at position $X$, $A$ is our gauge field and $\mathbb{I}$ is the identiy on $\mathcal{H}_\text{internal}$. This Hamiltonian is almost identical to the Hamiltonian used in Berry's derivation, except now I have upgraded this to an operator on $\mathcal{H}$ by allowing $H$ to have internal indices too and allowing $A$ to be a non-abelian gauge field.

Generalising the result of Berry's paper, the $N$ eigenstates of the Hamiltonian with energy $E$ in a region where the curvature of $A$ vanishes is given by

$$ \Psi_j(X;x,t) =P \exp \left( - i \int_X^x A \cdot \mathrm{d} l \right) \psi_E(X;x,t) e_j $$ where $P$ stands for path ordering, $\psi_E$ is the spatial wavefunction with energy $E$ and $e_j$ are the basis vectors of $\mathcal{H}_\text{internal}$. This is easy to show as the differential operator $\nabla$ only acts on the spatial degrees of freedom, so we have one eigenstate for every basis vector $\mathbf{e}_j$ and hence our desired degeneracy required for a non-abelian Berry connection. The correseponding Berry connection is given by

$$ [\mathcal{A}_\mu]_{ij}(X) = i\langle \Psi_i(X) | \frac{\partial}{\partial X^\mu} | \Psi_j(X) \rangle \\ = i\int \mathrm{d}^n x e_i^\dagger \bar{P} \exp \left( i \int_X^x A \cdot \mathrm{d} l \right) (iA_\mu) P \exp \left( - i \int_X^x A \cdot \mathrm{d} l \right) e_j \psi_E^*(X;x,t) \psi_E(X;x,t)$$

where $\bar{P}$ is the anti-path ordering operator, which is due to taking the Hermitian conjugate. For the case of an abelian gauge field $A$, the exponentials would commute past everything and the Berry connection would reduce to $\mathcal{A} \propto A$, however I do not know how to evaluate this for the case of non-abelian connections.

My problem

Multiple sources suggest the non-abelian Aharonov-Bohm effect would yield a Wilson line of the gauge field,

$$ U = P \exp \left( -i \oint_C A \cdot \mathrm{d} l \right) $$ e.g. this and this, which suggests to me that the Berry connection is proportional to the gauge field, i.e. $\mathcal{A} \propto A$, however from my derivation I get stuck at the last line above where I am required to evaluate

$$ \bar{P} \exp \left( i \int_X^x A \cdot \mathrm{d} l \right) A_\mu P \exp \left( - i \int_X^x A \cdot \mathrm{d} l \right)=? $$

Is there some sort of generalised Baker-Campbell-Hausdorff formula for path ordered exponentials, i.e. something like $e^X Y e^{-X} = Y + [X,Y] + \frac{1}{2} [X,[X,Y]] + \ldots $?

  • $\begingroup$ Do you literally mean the "curl" of $A$ to vanish or do you mean the curvature $F_{\mu\nu}$ vanishes. If you mean the latter then the gauge field has no effect at all. If the former, then the solution cannot be written in your factored form. $\endgroup$
    – mike stone
    Commented Aug 17, 2020 at 13:17
  • $\begingroup$ I mean that the curvature vanishes everywhere except one region where a non-trivial flux is present, just like the case for the usual AB effect with EM fields. I assume that because the curvature is zero outside of the flux, one can write the solution how I have as the integral defines a single-valued function? $\endgroup$ Commented Aug 17, 2020 at 13:26
  • $\begingroup$ It's not single valued if you go round a loop enclosing the flux. $\endgroup$
    – mike stone
    Commented Aug 17, 2020 at 13:35
  • $\begingroup$ Yes which is the Berry phase. I am generalising equation 33 of Berry's paper for the abelian case - it is also derived this way in Griffith's QM. This is because the particle is enclosed in the box outside of the solenoid by the potential $V(X-x)$, so the wavefunction only has support here. In this small region, the gauge field has no curvature so the integral is single valued here. $\endgroup$ Commented Aug 17, 2020 at 13:38

1 Answer 1


The wavefunction is not single valued if you go round a loop enclosing the flux. I don't think that solution for abelian BA effect on a particle of momentum $k$ scattering off a solenoid
$$ \psi(r,\theta)= \sum_{l=-\infty}^{\infty} e^{il \theta -(\pi/2)(l-\alpha)}J_{|l-\alpha|}(kr) $$ can be factored in your form, but I mght be wrong.

Ah -- I see what you are doing. You are not solving the non-abelian scatterig problem that Peter Horvathy does. You are interested only a particle in a small box that get's carried around the flux as Michal Berry does. So you can't get the full scattering solutions. As Berry says, his solution is single valued in ${\bf r}$ but only locally in ${\bf R}$.

In a simply connected region we can write $A_\mu(x) = U^\dagger(x)\partial_{x^\mu} U(x)$ and as $(\partial_\mu+A)U^{-1} \psi= U^{-1} \partial_\mu\psi$ we see that we can write $\psi(x)= U^{-1}(x)\psi_0(x-X)$ for the particle box centered at $X$ and where $\psi_0$ is the zero gauge field wavefunction. With this choice of wavefunction the Berry connection is zero as the wavefunctions is always is what it wants to be at that point. It does not need adiabatic Berry transport. To get a non-zero connection we can redefine our wavefunction so that in each box the wavefunction looks exactly the same. To do this we replace $\psi(x)$ with $U^{-1}(x) U(X)\psi_0$ so that in the center $x=X$ of each box the new wavefunction $\psi(X)=\psi_0(X)$ is the same independently of the position $X$ of the box. Now your computation directly gives ${\mathcal A}_\mu(X) = U^{-1}(X)\partial_{X^\mu} U(X)$.

Here are the details. Let the wavefunction in the box be $$ U^{-1}(x) U(X)\psi_0(x-X)\stackrel{\rm def}{=} \langle x |0,X\rangle $$ where $\psi_0$ is normalized. Then the Berry connection is $$ \langle 0,X|\partial_{X^\mu}|0,X\rangle = \int dx \psi_0^\dagger(x-X) U^{\dagger}(X) U(x) \partial_{X^\mu}\Big( U^{-1}(x)U(X) \psi_0(x-X)\Big)\\ =\int dx \psi_0^\dagger(x-X) U^{\dagger}(X) \partial_{X^\mu}\Big(U(X) \psi_0(x-X)\Big) $$ There are two terms to evaluate: one where the derivative hits $U(X)$ and one where it hits $\psi_0(x-X)$. The first is $$ \int dx \psi_0^\dagger(x-X) \partial_{X^\mu} \psi_0(x-X)= - \int dx \psi_0^\dagger(x-X) \partial_{x^\mu} \psi_0(x-X)\\ = \frac 12 \int dx \partial_{x^\mu}|\psi|^2\\ =0 $$ because you have set $\psi_{0,i} = v_i \psi_0$ where $v_i$ is the complex-vector amplitude that $U$ acts on and $\psi$, being a bound state, is real and vanishes on the boundary of the box. The second is $$ U^{-1}(X)\partial_{X_\mu} U(X) \int dx |\psi_0|^2\\ = U^{-1}(X)\partial_{X_\mu} U(X)=A_\mu(X). $$ Hence the Berry connection is just the gauge field evaluated at the centre of the box.

  • $\begingroup$ So the Berry phase is given by $ i \langle \Psi_i | \partial_\mu |\Psi_j \rangle = i \int d^n x e^\dagger_i \psi^*_E U^\dagger \partial_\mu( U \psi_E e_j) $. Using the fact that $\partial_\mu U = iA_\mu U$, we have the Berry connection $ i \int d^n x e^\dagger_i \psi^*_E U^\dagger i A_\mu U \psi_E e_j+ie^\dagger_i U^\dagger Ue_j\psi^*_E \partial_\mu \psi_E$. The second term in the integral vanishes because $U^\dagger U =1$ it's evaluating the expectation of momentum of the particle inside the box, so we're left with what I have above. I don't know how to evaluate this integral to extract $A$ $\endgroup$ Commented Aug 17, 2020 at 17:45
  • $\begingroup$ I modified my answer to give the derivation. $\endgroup$
    – mike stone
    Commented Aug 18, 2020 at 11:57
  • $\begingroup$ Thank you. What do you mean by "the wavefunctions is always is what it wants to be at that point"? Are you exploiting the gauge freedom to redefine our wavefunction? And what do you mean by $U(x)$ vs $U(X)$? $\endgroup$ Commented Aug 18, 2020 at 20:45
  • $\begingroup$ I mean that you choice of eigenstates has zero Berry connection. $X$ is the location of the box $x$ is where you evalute the wavefunction. Very different things! Remember that it is $\partial_X$ that defines the Berry connection--- not $\partial_x$. All this is in your $H(X)$, the hamiltonian of the box at centered $X$. This is all quite simple. What Horvathy does to solve the scattering problem is not trivial, and is not accessible by your route. $\endgroup$
    – mike stone
    Commented Aug 18, 2020 at 23:23
  • $\begingroup$ Yes I understand what the coordinates mean but the notation $U(x)$ vs $U(X)$ confuses me. If $U(x) = P \exp \left( - i \int_X^x A \cdot dl \right)$ as defined above then $U(X)$ would be the identity, what is your definition of $U$?. I also do not see why the original wavefunction $\psi(x) = U^{-1}(x) \psi_0(x-X)$ yields a zero Berry connection, because that calculation is nothing but the calculation I have tried in my original question above $\endgroup$ Commented Aug 19, 2020 at 10:30

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