2.5 Charged particle on the Aharonov-Bohm ring
What is the use of the projective phase? It provides a way to find a qubit : if you can cook up a system with the \(\mathbb{Z}_2\times \mathbb{Z}_2\) with the nontrivial projective phase — or quantum anomaly —, you are guaranteed to have (at least) 2-fold degeneracy.
Let us see how this happens in an example with a infinite dimensional Hilbert space. Here we consider a charged free particle confined in a ring \(S^1\) with radius \(R\). We put the magnetic flux \(\Phi\) through the ring. The Hamiltonian is \[\begin{equation} \tag{2.30} H = \frac{1}{2m}(p_x + \frac{e}{c}A_x)^2, \end{equation}\] where \(x\) is the coordinate on \(S^1\) with the identification \(x \sim x + 2\pi R\), \(p_x = -\mathrm{i}\hbar \frac{\mathrm{d}}{\mathrm{d}x}\), and \(A_x = \Phi/(2\pi R)\) is the vector potential along the ring. We rewrite this as \[\begin{equation} \tag{2.31} H = \frac{\hbar^2}{2mR^2}\bigl(\hat{n} + \frac{\theta}{2\pi} \bigr)^2, \end{equation}\] were \(\hat{n} = \frac{R p_x}{\hbar}\) and \(\theta = \frac{e \Phi}{\hbar c}\).
A wave function \(\psi(x)\) can be expanded by the periodic exponential function \(\psi_n(x) \propto e^{\mathrm{i}n\,x/R}\), \(n \in \mathbb{Z}\). The corresponding state \(\ket{n}\) is the eigenstate of the operators \(\hat{n}\) and \(H\):4 \[\begin{equation} \tag{2.32} \hat{n} \ket{n} = n \ket{n}, \quad H \ket{n} = E_n(\theta) \ket{n} = \frac{\hbar^2}{2mR^2}(n+\frac{\theta}{2\pi})^2 \ket{n}. \end{equation}\]
This system has the symmetry shifting \(x\) as \(x \mapsto x+ \alpha R\) with a \(2\pi\)-periodic parameter \(\alpha\). The unitary corresponding to this symmetry can be directly read off from the wave function as5 \[\begin{equation} \tag{2.33} U_\lambda \ket{n} = e^{\mathrm{i}\lambda n}\ket{n}. \end{equation}\]
There are other symmetries, depending on \(\theta\). For example, when \(\theta = 0\), there is a \(\mathbb{Z}_2\) flipping \(x\): \(x\mapsto -x\), \(\ket{n} \mapsto \ket{-n}\). A more interesting case is \(\theta = \pi\), then the \(\ket{n} \mapsto \ket{-n}\) action is not a symmetry of the Hamiltonian, but the following is preserved: \[\begin{equation} \tag{2.34} U_{\mathbb{Z}_2}\ket{n} = \ket{-n-1}. \end{equation}\] Note that this is also the \(\mathbb{Z}_2\) flipping symmetry, in the sense of the following: \[\begin{equation} \tag{2.35} U_{\mathbb{Z}_2}^\dagger \, f(x) \, U_{\mathbb{Z}_2} = f(-x), \end{equation}\] for any periodic function \(f\) of \(x\).6 Now the multiplication between \(U_\lambda\) and \(\mathbb{Z}_2\) is \[\begin{equation} \tag{2.36} U_\lambda U_{\mathbb{Z}_2} = e^{-\mathrm{i}\lambda}U_{\mathbb{Z}_2}U_{-\lambda}. \end{equation}\] Note that in the right hand side the parameter \(\lambda\) of \(U_{\lambda}\) is flipped, so the two symmetry forms the semidirect product \(U(1) \rtimes \mathbb{Z}_2\). On top to that, we oberserve the projective phase \(e^{-\mathrm{i}\lambda}\). If we focus on the subgroup \(\mathbb{Z}_2^\text{shift}\) (\(\lambda = \pi\)) subgroup of the \(U(1)\), we have the subgroup \(\mathbb{Z}_2^\text{shift}\times \mathbb{Z}_2\) of \(U(1) \rtimes \mathbb{Z}_2\) and the projective phase \(e^{-\mathrm{i}\pi}=-1\) is the same phase we observed in Section 2.4!
Recall that, in Section 2.4, we have proven that there is no one-dimensional representation of \(\mathbb{Z}_2\times \mathbb{Z}_2\) with the nontrivial projective phase. Therefore every eigenspace of \(H\) at \(\theta = \pi\) has dimension at least two. This can be easily verified because \(E_n(\pi) = E_{-n-1}(\pi)\). In particular, the ground state has exact two-fold degeneracy that you might use as a qubit.7
Okay, but we could find the degeneracy without discussing the projective phase. The real advantage of the quantum anomaly is that it shows the degeneracy is robust; this is the result of the symmetry and its quantum anomaly, and as long as they are preserved, the degeneracy cannot be lifted. Also, the quantum anomaly is a discrete thing and therefore it cannot be changed by continuous deformation. (We will discuss about his a bit more in the next section.) Thus, even if we deform the Hamiltonian as \[\begin{equation} \tag{2.37} H' = \frac{\hbar^2}{2mR^2}\bigl(\hat{n} + \frac{1}{2} \bigr)^2 + V(x), \end{equation}\] with an arbitrary (periodic) potential \(V(x)\) that preserves \(\mathbb{Z}_2\times \mathbb{Z}_2\) : \(V(-x) = V(x)\) and \(V(x+\pi) = V(x)\), we know that there is a degeneracy, although in this system cannot be easily solved. This is the power of quantum anomaly: it tells us something nontrivial without solving the system!
References
\(E_n(\theta+2\pi) = E_{n+1}(\theta)\) and thus the spectrum is unchanged by the \(2\pi\)-shift of \(\theta\). This is closely related to the Dirac quantization of magnetic charge; \(\theta =2\pi\) is the flux created by the Dirac monopole.↩︎
If one does the \(2\pi\)-shift of theta adiabatically, the spectrum is unchanged as noticed in the above footnote, but the \(U(1)\) eigenvalue \(n\) carried by the state is shifted by 1. This (or something similar) is called Thouless pump. Indeed this is yet another quantum anomaly, “anomaly involving parameter space”. See [9] for more detail.↩︎
\(\braket{m|f(x)|n} = \int \mathrm{d}x f(x) e^{\mathrm{i}(n-m)x/R}\), and thus \(\braket{m|U^\dagger_{\mathbb{Z}_2}f(x)U_{\mathbb{Z}_2}|n} = \int \mathrm{d}x f(x) e^{\mathrm{i}(-(n-1-(m-1))x/R}= \braket{m|f(-x)|n}\).↩︎
I believe this is called the flux qubit, but do not trust me, I’m not an expert on quantum computing.↩︎