I was going through arXiv:quant-ph/0001106v1, the first paper by Farhi on adiabatic quantum computation. Equation 2.24 says, $$\tilde{H}(s) = (1-s)H_B + sH_P$$ which means the adiabatic evolution starts from the ground state of $H_B$ and slowly evolves until it arrives at the ground state of $H_P$. In section 3.1, the one qubit example has the adiabatic Hamiltonian as $$\tilde{H}(s) = \begin{pmatrix} s & \epsilon (1-s) \\ \epsilon (1-s) & 1-s \end{pmatrix}$$ I don't see how the plot of Figure 1 is drawn. In figure 1, Farhi plotted eigenvalues of the Hamiltonian for s while the range for s was 0 to 1. The Hamiltonian is supposed to evolve according to the Schrodinger equation (eq 2.1), $$i \frac{d}{dt} |\Psi (t) \rangle = H(t) |\Psi (t) \rangle$$ Was this evolution solved to draw the plot? Or did Farhi derive the formula for eigenvalues in terms of s using just matrix math and plotted accordingly?


2 Answers 2


If the parameter $s$ is varied adiabatically there should be no need to solve the time-dependent Schrodinger equation. The adiabatic theorem implies the instantaneous eigenvalues are always those of the parameterized Hamiltonian, provided its spectrum is separated.

Thus your final statement and hadsed's conclusion are correct: The figure just shows the eigenvalues of the matrix as a function of $s$.

  • $\begingroup$ Yes, I was able to draw the plot. $\endgroup$ Commented Mar 26, 2013 at 16:53

The latter. The reason is because it doesn't make a lot of sense to do the eigendecomposition of an approximated Hamiltonian, which is what you're doing when solving for the dynamics (your operator probably won't be Hermitian so your energies won't even be real). Doing the evolution gives you the final state probabilities, but it's not very good for analyzing the energy gap precisely.

So you just plug in the right $s$ values and find the eigenvalues of that $H(s)$, and plotting that gives you the eigenspectrum in Fig. 1.


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