# Holonomy of $\Lambda$-Configuration Hamiltonian

I'm currently reading through "Introduction to Topological Quantum Computing" By Jiannis Pachos in which the following is given as an exercise.

"Consider a quantum system with three states in $\Lambda$-configuration subject to the Hamiltonian

H = \begin{bmatrix} \Delta & \Omega_1 & \Omega_2 \\ \Omega_1^* & 0 & 0 \\ \Omega_2^* & 0 & 0 \end{bmatrix}

Here $\Omega_1$ and $\Omega_2$ are complex numbers that correspond to the coupling between the states and $\Delta$ is a real energy shift. Evaluate all possible Holonomies that can be produced by manipulating $\Omega_1$ and $\Omega_2$. [Hint: Determine first the degeneracy condition]"

It seems to me that the degeneracy condition is that $\Omega_1 = \Omega_2 = 0$ in which we have two ground states with energy $E =0$ and one excited state with $E = \Delta$. Now I do not see how to implement a unitary operator that will carry the degenerate Hamiltonian to one with nonzero $\Omega_1, \Omega_2$. That is, $U(\lambda(t))$ such that the full Hamiltonian $H = U^\dagger(\lambda(t)) H_0 U(\lambda(t))$ where $H_0$ is the degenerate Hamiltonian. Can anyone help me understand the adiabatic evolution I need to consider here? Thanks!

• According to my understanding, you need: (1) Define $\Omega_1,\Omega_2$ as a function of some parameters $\theta(t)$ so that you get a $H(\theta(t))$ (a loop); (2) Compute the 2 eigen states of $H(\theta)$ with 0 eigen value; (3) Compute the connection form $A(\theta)$ ; (4) Construct the unitary operation by computing the holonomy of the loop $H(\theta(t))$. Please refer to the paper "Geometric phases in Quantum Mechanics" at arxiv.org/ftp/arxiv/papers/0909/0909.3958.pdf, where an example on this problem is explained. – XXDD Aug 17 '16 at 9:04
• This paper is very useful, the calculation and the reference given here go through careful calculations. In replicating them I noticed that the smallest Hamiltonian they calculate the holonomies for is $4\times4$. For the Hamiltonian above there is a strict condition on when there is a degenerate subspace, as I described above. This does not seem to be the case for the $4 \times 4$ Hamiltoninan, as it has a degenerate subspace regardless of what values the coupling factors take. Hence my confusion on the problem as stated remains. – unclev Aug 17 '16 at 19:19
• I think you are working with a 1-qubit gate (2 x 2), the example I showed is a 2-qubit gate (4x4). Also I found an example in your book 'Introduction to Topological Quantum Computing' (P. 51), which seems to be exactly your case. Can that example solve your problem? – XXDD Aug 18 '16 at 4:20
• Yes absolutely, I just saw that section myself it clears it all up for me. Thanks for looking into this with me! – unclev Aug 18 '16 at 13:34
• I'm missing something about the connection between the $\Lambda$ Hamiltonian and the one on p. 51, $H(z)=U(z)\text{diag}(0,0,1)U^\dagger(z)$, $U(z)$ a three-dimensional unitary. It's true that their spectra are the same for $\Omega_1=\Omega_2=0$, but there's only a single point in the parameter space of the $\Lambda$ Hamiltonian for which this is the case, and as unclev pointed out, degeneracy of the ground states is lost away from this point. This seems to run counter to the idea of adiabatic isospectral evolution introduced at the beginning of 2.2.2.2 in the same source. What am I missing? – Jeffrey Nov 26 '16 at 3:16