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?

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.

  • $\begingroup$ I only vaguely understand this and I'm getting out of my lane, but to be clear you are initially in the ground state of $H(0)=-X$, which is $\vert +\rangle$, and you wish to evolve adiabatically to the ground state of $H(\pi/2)=-Z$, which is $\vert 0\rangle$. The gap throughout the evolution is $\Delta=1-(-1)=2$. I thought $\epsilon$ controlled how to guarantee that all instantaneous gaps are large. What does $\epsilon$ mean in this case? The gap is constant throughout the evolution; you don't need to make $\epsilon$ small enough? Isn't there some degeneracy somewhere? $\endgroup$ Apr 19, 2021 at 23:40
  • $\begingroup$ @MarkS Yes about the initial & final state. The $\epsilon$ is nothing to do with the gap directly. In adiabatic QC, the statement is that if you change the Hamiltonian slowly enough you stay in the ground state. The $\epsilon$ controls the speed at which you change the Hamiltonian, so $\epsilon\rightarrow 0$ is the correct limit to (theoretically) end up in the target state. $\endgroup$
    – DaftWullie
    Apr 20, 2021 at 6:23
  • $\begingroup$ Is your Hamiltonian analogous to letting a photon pass through a bunch of polarizing filters, with the first one initially diagonally, but each subsequent filter rotated by an angle of $\epsilon$ until eventually your last polarizing filter is horizontal? In the limit you’d need an infinite number of filters to guarantee that your photon passes through the last filter unscathed. $\endgroup$ Apr 21, 2021 at 4:18
  • $\begingroup$ @MarkS If it is, it would take some work to prove it. Conceptually, this is sort of what's supposed to happen, but there is never any active projection step, so we're trying to show that this is what happens. $\endgroup$
    – DaftWullie
    Apr 21, 2021 at 7:00

1 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).


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