Why do I get this extra factor when working out the dynamics of an adiabatic quantum computation?

Question

I was trying to revise my understanding of adiabatic quantum computation via a simple example. I'm familiar with the overall concept -- that you have an overall Hamiltonian $$ H(s)=(1-s)H_0+s H_f $$ where $s$ is a function of time, starting from $s=0$ and finishing at $s=1$. You prepare your system in the ground state of $H_0$ (known) and, provided the Hamiltonian changes slowly enough, your final state will be close to the ground state of $H_f$. The "slowly enough" condition is typically phrased in terms of the energy gap between the ground and first excited state, $\Delta$. A typical assumption is $$ \frac{ds}{dt}=\epsilon\Delta^2 $$ for small $\epsilon$ (I haven't been back to pick through more detailed statements).

So, my plan was to test $$ H(\theta)=-\cos\theta X-\sin\theta Z, $$ starting from $\theta=0$ and the system in $|+\rangle$. For all values of $\theta$, the gap is a constant (2) so, in essence, I have the conversion $\theta=4\epsilon t$ and the evolution time will be $\pi/(8\epsilon)$ and should leave the system in the state $|0\rangle$. However, when I tried to do this calculation, I didn't get this answer. Am I screwing up, or is there some condition that tells me I shouldn't expect the adiabatic theorem to hold in this case?

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Details of how I tried to do the calculation: Let $$ |y_{\pm}\rangle=(|0\rangle\pm i|1\rangle)/\sqrt{2}. $$ We have that $$ H|y_{\pm}\rangle=\pm ie^{\mp i\theta}|y_{\mp}\rangle. $$ We can decompose any state in this basis, $$ |\psi\rangle=a|y_+\rangle+b|y_-\rangle, $$ so we can talk about the time evolution of the coefficients $$ \frac{d}{dt}\begin{bmatrix} a \\ b \end{bmatrix}=4\epsilon\frac{d}{d\theta}\begin{bmatrix} a \\ b \end{bmatrix}=\begin{bmatrix} 0 & e^{-i\theta} \\ -e^{i\theta} & 0 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix} $$ I can perform a variable transformation $$ \tilde a=e^{i\theta/2}a,\quad \tilde b=e^{-i\theta/2}b $$ which then satisfy $$ 4\epsilon\frac{d}{d\theta}\begin{bmatrix} \tilde a \\ \tilde b \end{bmatrix}=i\begin{bmatrix} 2\epsilon & -i \\ i & -2\epsilon \end{bmatrix}\begin{bmatrix} \tilde a \\ \tilde b \end{bmatrix} $$ This is (finally!) something I can do something useful with! Taking the small $\epsilon$ limit, I essentially have $$ \frac{d}{d\theta}\begin{bmatrix} \tilde a \\ \tilde b \end{bmatrix}=i\frac{1}{4\epsilon}Y\begin{bmatrix} \tilde a \\ \tilde b \end{bmatrix} $$ I can put in my initial conditions and compare what I expected at the end. The problem is that I have an extra phase of the form $e^{i\pi/(8\epsilon)}$ floating around, and the taking of the small $\epsilon$ limit seems problematic as this varies rapidly between all possible values.

Answer 1

The resolution is essentially down to just global phases. Let us note that the initial state is $$ \begin{bmatrix} \tilde a \\ \tilde b \end{bmatrix}=\frac{1}{2}\begin{bmatrix} 1+i \\ 1-i \end{bmatrix}\equiv \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -i \end{bmatrix}. $$ In other words, this is just an eigenstate of $Y$. So, we we evolve under the Pauli $Y$ matrix, we only get a global phase (so who cares if it's rapidly oscillating?!)

Thus, at any later time, $$ \begin{bmatrix} a \\ b \end{bmatrix}=\begin{bmatrix} e^{-i2\epsilon t} \\ -i e^{i2\epsilon t} \end{bmatrix}. $$ For the target time of $t=\pi/8\epsilon$, we have $$ \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 1 \end{bmatrix}, $$ which is exactly what we expected (this is the representation of $|0\rangle$ in the $y$ basis).